Latex Code of Exercises in Hungerford's Algebra

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	\begin{document}
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	\chapter*{Prerequisites and Preliminaries}
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	\section{Logic}
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	\section{Sets and Classes}
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	\section{Functions}
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	\section{Relations and Partitions}
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	\section{Products}
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	\section{The Integers}
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	\section{The Axiom of Choice, Order and Zorn's Lemma}
	\begin{enumerate}
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	\item
	Let $(A, \leq)$ be a partially ordered set and $B$ a nonempty subset.
	A lower bound of $B$ is an element $d\in A$ such that $d\leq b$ for {\it every} $b\in B$.
	A greatest lower bound (g.l.b.) of $B$ is a lower bound $d_0$ of $B$ such that $d\leq d_0$ for every other lower bound $d$ of $B$.
	A least upper bound (l.u.b.) of $B$ is an upper bound $t_0$ of $B$ such that $t_0\leq t$ for every other upper bound $t$ of $B$.
	$(A, \leq)$ is a lattice if for all $a, b\in A$ the set $\{a, b\}$ has both a greatest lower bound and a least upper bound.
	\begin{enumerate}
	\item If $S\neq \emptyset$,
	then the power set $P(S)$ ordered by set-theoretic inclusion is a lattice,
	which has a unique maximal element.
	\item Give an example of a partially ordered set which is {\it not} a lattice.
	\item Give an example of a lattice with no maximal element and an example of a partially ordered set with two maximal elements.
	\end{enumerate}
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	\item
	A lattice $(A, \leq)$ (see Exercise {O.7.1}) is said to be complete if every nonempty subset of $A$ has both a least upper bound and a greatest lower bound.
	A map of partially ordered sets $f:A\to B$ is said to preserve order if $a\leq a'$ in $A$ implies $f(a)\leq f(a')$ in $B$.
	Prove that an order-preserving map $f$ of a complete lattice $A$ into itself has at least one fixed element (that is, an $a\in A$ such that $f(a)=a$).
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	\item
	Exhibit a well ordering of the set $\mathbb{Q}$ of rational numbers.
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	\item
	Let $S$ be a set.
	A choice function for $S$ is a function $f$ from the set of all nonempty subsets of $S$ to $S$ such that $f(A)\in A$ for all $A\neq \emptyset, A\subseteq S$.
	Show that the Axiom of Choice is equivalent to the statement that every set $S$ has a choice function.
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	\item
	Let $S$ be the set of all points $(x, y)$ in the plane with $y\leq 0$.
	Define an ordering by $(x_1, y_1)\leq (x_2, y_2)\Leftrightarrow x_1=x_2\text{ and }y_1\leq y_2$.
	Show that this is a partial ordering of $S$,
	and that $S$ has infinitely many maximal elements.
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	\item
	Prove that if all the sets in the family $\{A_i\mid i\in I\neq \emptyset\}$ are nonempty,
	then each of the projections $\pi_k:\prod_{i\in I}A_i\to A_k$ is surjective.
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	\item
	Let $(A, \leq)$ be a linearly ordered set.
	The immediate successor of $a\in A$ (if it exists) is the least element in the set $\{x\in A\mid a<x\}$.
	Prove that if $A$ is well ordered by $\leq $,
	then at most one element of $A$ has no immediate successor.
	Give an example of a linearly ordered set in which precisely two elements have no immediate successor.
	\end{enumerate}
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	\section{Cardinal Numbers}
	\begin{enumerate}
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	\item
	Let $I_0=\emptyset$ and for each $n\in \mathbb{N}^*$ let $I_n=\{1, 2, 3, ..., n\}$.
	\begin{enumerate}
	\item $I_n$ is not equipollent to any of its proper subsets [{\it Hint:} induction].
	\item $I_m$ are $I_n$ are equipollent if and only if $m=n$.
	\item $I_m$ is equipollent to a subset of $I_n$ but $I_n$ is not equipollent to any subset of $I_m$ if and only if $m<n$.
	\end{enumerate}
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	\item
	\begin{enumerate}
	\item Every infinite set is equipollent to one of its proper subsets.
	\item A set is finite if and only if it is not equipollent to one of its proper subsets [see Exercise {O.8.1}].
	\end{enumerate}
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	\item
	\begin{enumerate}
	\item $\mathbb{Z}$ is denumerable set.
	\item The set $\mathbb{Q}$ of rational numbers is denumerable.
	[{\it Hint:} show that $|\mathbb{Z}|\leq |\mathbb{Q}|\leq |\mathbb{Z}\times \mathbb{Z}|=|\mathbb{Z}|$.]
	\end{enumerate}
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	\item
	If $A, A', B, B'$ are sets such that $|A|=|A'|$ and $|B|=|B'|$,
	then $|A\times B|=|A'\times B'|$.
	If in addition $A\cap B=\emptyset =A'\cap B'$,
	then $|A\cup B|=|A'\cup B'|$.
	Therefore multiplication and addition of cardinals is well defined.
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	\item
	For all cardinal numbers $\alpha, \beta, \gamma$:
	\begin{enumerate}
	\item $\alpha+\beta=\beta+\alpha$ and $\alpha\beta=\beta\alpha$
	\item $(\alpha+\beta)+\gamma=\alpha+(\beta+\gamma)$ and $(\alpha\beta)\gamma=\alpha(\beta\gamma)$ (associative laws).
	\item $\alpha(\beta+\gamma)=\alpha\beta+\alpha\gamma$ and $(\alpha+\beta)\gamma=\alpha\gamma+\beta\gamma$ (distributive laws).
	\item $\alpha+0=\alpha$ and $\alpha 1=\alpha$.
	\item If $\alpha\neq 0$,
	then there is no $\beta$ such that $\alpha+\beta=0$ and if $\alpha\neq 1$,
	then there is no $\beta$ such that $\alpha\beta=1$.
	Therefore substraction and division of cardinal numbers cannot be defined.
	\end{enumerate}
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	\item
	Let $I_n$ be as in Exercise {O.8.1}.
	If $A\sim I_m$ and $B\sim I_n$ and $A\cap B=\emptyset$,
	then $(A\cup B)\sim I_{m+n}$ and $A\times B\sim I_{mn}$.
	Thus if we identify $|A|$ with $m$ and $|B|$ with $n$,
	then $|A|+|B|=m+n$ and $|A||B|=mn$.
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	\item
	If $A\sim A', B\sim B'$ and $f:A\to B$ is injective,
	then there is an injective map $A'\to B'$.
	Therefore the relation $\leq $ on cardinal numbers is well defined.
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	\item
	An infinite subset of a denumerable set is denumerable.
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	\item
	The infinite set of real numbers $\mathbb{R}$ is not denumerable (that is, $\aleph_0<|\mathbb{R}|$).
	[{\it Hint:} it suffices to show that the open interval $(0, 1)$ is not denumerable by Exercise {O.8.8}.
	You may assume each real number can be written as an infinite decimal.
	If $(0, 1)$ is denumerable there is a bijection $f:\mathbb{N}^*\to (0, 1)$.
	Construct an infinite decimal (real number) $.a_1 a_2\cdots$ in $(0, 1)$ such that $a_n$ is not the $n$th digit in the decimal expansion of $f(n)$.
	This number cannot be in $\text{Im }f$.]
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	\item
	If $\alpha, \beta$ are cardinals,
	define $\alpha^{\beta}$ to be the cardinal number of the set of all functions $B\to A$,
	where $A, B$ are sets such that $|A|=\alpha, |B|=\beta$.
	\begin{enumerate}
	\item $\alpha^{\beta}$ is independent of the choice of $A, B$.
	\item $\alpha^{\beta+\gamma}=(\alpha^{\beta})(\alpha^{\gamma})$; $(\alpha\beta)^{\gamma}=(\alpha^{\gamma})(\beta^{\gamma})$; $\alpha^{\beta\gamma}=(\alpha^{\beta})^{\gamma}$.
	\item If $\alpha\leq \beta$, then $\alpha^{\gamma}\leq \beta^{\gamma}$.
	\item If $\alpha, \beta$ are finite with $\alpha>1, \beta>1$ and $\gamma$ is infinite, then $\alpha^{\gamma}=\beta^{\gamma}$.
	\item For every finite cardinal $n$,
	$\alpha^n=\alpha\alpha\cdots \alpha$ ($n$ factors).
	Hence $\alpha^n=\alpha$ if $\alpha$ is infinite.
	\item If $P(A)$ is the power set of a set $A$,
	then $|P(A)|=2^{|A|}$.
	\end{enumerate}
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	\item
	If $I$ is an infinite set,
	and for each $i\in I$ $A_i$ is a finite set,
	then $|\bigcup_{i\in I}A_i|\leq |I|$.
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	\item
	Let $\alpha$ be a fixed cardinal number and suppose that for every $i\in I$,
	$A_i$ is a set with $|A_i|=\alpha$.
	Then $|\bigcup_{i\in I}A_i|\leq |I|\alpha$.
	\end{enumerate}
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	\chapter{Groups}
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	\section{Semigroups, Monoids and Groups}
	\begin{enumerate}
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	\item
	Give examples other than those in the text of semigroups and monoids that are not groups.
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	\item
	Let $G$ be a group (written additively),
	$S$ a nonempty set,
	and $M(S, G)$ the set of all functions $f:S\to G$.
	Define addition in $M(S, G)$ as follows:
	$(f+g):S\to G$ is given by $s\mapsto f(s)+g(s)\in G$.
	Prove that $M(S, G)$ is a group,
	which is abelian if $G$ is.
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	\item
	Is it true that a semigroup which has a left identity element and in which every element has a right inverse (see Propsition {I.1.3}) is a group?
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	\item
	Write out a multiplication table for the group $D_4^*$.
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	\item
	Prove that the symmetric group on $n$ letters,
	$S_n$,
	has order $n!$.
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	\item
	Write out an addition table for $\mathbb{Z}_2\oplus \mathbb{Z}_2$.
	$\mathbb{Z}_2\oplus \mathbb{Z}_2$ is called the Klein four group.
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	\item
	If $p$ is prime,
	then the nonzero elements of $\mathbb{Z}_p$ form a group of order $p-1$ under multiplication.
	[{\it Hint:} $\overline{a}\neq \overline{0}\Rightarrow \gcd{(a, p)}=1$;
	use Introduction, Theorem {O.6.5}.]
	Show that this statement is false if $p$ is not prime.
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	\item
	\begin{enumerate}
	\item The relation given by $a\sim b\Leftrightarrow a-b\in \mathbb{Z}$ is a cougruence relation on the additive group $\mathbb{Q}$ [see Theorem {I.1.5}].
	\item
	The set $\mathbb{Q}/\mathbb{Z}$ of equivalence classes is an infinite abelian group.
	\end{enumerate}
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	\item
	Let $p$ be a fixed prime.
	Let $R_p$ be the set of all those rational numbers whose denominator is relatively prime to $p$.
	Let $R^p$ be the set of rationals whose denominator is a power of $p$ ($p^i$, $i\geq 0$).
	Prove that both $R_p$ and $R^p$ are abelian groups under ordinary addition of rationals.
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	\item
	Let $p$ be a prime and let $Z(p^{\infty})$ be the following subset of the group $\mathbb{Q}/\mathbb{Z}$ (see pg. 27):
	$$Z(p^{\infty})=\{\overline{a/b}\in \mathbb{Q}/\mathbb{Z}\mid a, b\in \mathbb{Z}\text{ and }b=p^i\text{ for some }i\geq 0\}.$$
	Show that $Z(p^{\infty})$ is an infinite group under the addition operation of $\mathbb{Q}/\mathbb{Z}$.
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	\item
	The following conditions on a group $G$ are equivalent:
	\begin{enumerate}[(i)]
	\item $G$ is abelian;
	\item $(ab)^2=a^2 b^2$ for all $a, b\in G$;
	\item $(ab)^{-1}=a^{-1} b^{-1}$ for all $a, b\in G$;
	\item $(ab)^n=a^n b^n$ for all $n\in \mathbb{Z}$ and all $a, b\in G$;
	\item $(ab)^n=a^n b^n$ for three consecutive integers $n$ and all $a, b\in G$.
	\end{enumerate}
	Show that (v)$\Rightarrow$(i) is false if ``three'' is replaced by ``two.''
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	\item
	If $G$ is a group,
	$a, b \in G$ and $bab^{-1}=a^r$ for some $r\in \mathbb{N}$,
	then $b^j a b^{-j}=a^{r^j}$ for all $j\in \mathbb{N}$.
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	\item
	If $a^2=e$ for all elements $a$ of a group $G$,
	then $G$ is abelian.
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	\item
	If $G$ is a finite group of even order,
	then $G$ contains an element $a\neq e$ such that $a^2=e$.
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	\item
	Let $G$ be a nonempty finite set with an associative binary operation such that for all $a, b, c\in G$
	$ab=ac\Rightarrow b=c$
	and $ba=ca\Rightarrow b=c$.
	Then $G$ is a group.
	Show that this conclusion may be false if $G$ is infinite.
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	\item
	Let $a_1, a_2, ...$ be a sequence of elements in a semigroup $G$.
	Then there exists a unique function $\psi:\mathbb{N}^*\to G$ such that $\psi(1)=a_1, \psi(2)=a_1 a_2, \psi(3)=(a_1 a_2)a_3$
	and for $n\geq 1$,
	$\psi(n+1)=(\psi(n))a_{n+1}$.
	Note that $\psi(n)$ is precisely the standard $n$ product $\prod_{i=1}^{n}a_i$.
	[{\it Hint:} Applying the Recursion Theorem {O.6.2} of the Introduction with $a=a_1$,
	$S=G$ and $f_n:G\to G$ given by $x\mapsto xa_{n+2}$ yields a function $\varphi:\mathbb{N}\to G$.
	Let $\psi=\varphi \theta$,
	where $\theta:\mathbb{N}^*\to \mathbb{N}$ is given by $k\mapsto k-1$.]
	\end{enumerate}
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	\section{Homomorphisms and Subgroups}
	\begin{enumerate}
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	\item
	If $f:G\to H$ is a homomorphism of groups,
	then $f(e_G)=e_H$ and $f(a^{-1})=f(a)^{-1}$ for all $a\in G$.
	Show by example that the first conclusion may be false if $G, H$ are monoids that are not groups.
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	\item
	A group $G$ is abelian if and only if the map $G\to G$ given by $x\mapsto x^{-1}$ is an automorphism.
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	\item
	Let $Q_8$ be the group (under ordinary matrix multiplication) generated by the complex matrices $A=\begin{pmatrix}0&1\\-1&0\end{pmatrix}$ and
	$B=\begin{pmatrix}0&i\\i&0\end{pmatrix}$,
	where $i^2=-1$.
	Show that $Q_8$ is a nonabelian group of order $8$.
	$Q_8$ is called the quaternion group.
	[{\it Hint:}
	Observe that $BA=A^3B$,
	whence every element of $Q_8$ is of the form $A^i B^j$.
	Note also that $A^4=B^4=I$, where $I=\begin{pmatrix}1&0\\0&1\end{pmatrix}$ is the identity element of $Q_8$.]
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	\item
	Let $H$ be the group (under matrix multiplication) of real matrices generated by $C=\begin{pmatrix}0&1\\-1&0\\\end{pmatrix}$ and $D=\begin{pmatrix}0&1\\1&0\\\end{pmatrix}$.
	Show that $H$ is a nonabelian group of order $8$ which is not isomorphic to the quaternion group of Exercise {I.2.3},
	but is isomorphic to the group $D_4^*$.
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	\item
	Let $S$ be a nonempty subset of a group $G$ and define a relation on $G$ by $a\sim b$ if and only if $ab^{-1}\in S$.
	Show that $\sim$ is an equivalence relation if and only if $S$ is a subgroup of $G$.
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	\item
	A nonempty finite subset of a group is a subgroup if and only if it is closed under the product in $G$.
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	\item
	If $n$ is a fixed integer,
	then $\{kn\mid k\in \mathbb{Z}\}\subseteq \mathbb{Z}$ is an additive subgroup of $\mathbb{Z}$,
	which is isomorphic to $\mathbb{Z}$.
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	\item
	The set $\{\sigma\in S_n\mid \sigma(n)=n\}$ is a subgroup of $S_n$ which is isomorphic to $S_{n-1}$.
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	\item
	Let $f:G\to H$ be a homomorphism of groups,
	$A$ a subgroup of $G$,
	and $B$ a subgroup of $H$.
	\begin{enumerate}
	\item $\ker{f}$ and $f^{-1}(B)$ are subgroups of $G$.
	\item $f(A)$ is a subgroup of $H$.
	\end{enumerate}
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	\item
	List all subgroups of $\mathbb{Z}_2\oplus \mathbb{Z}_2$.
	Is $\mathbb{Z}_2\oplus \mathbb{Z}_2$ isomorphic to $\mathbb{Z}_4$?
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	\item
	If $G$ is a group,
	then $C=\{a\in G\mid ax=xa\text{ for all }x\in G\}$ is an abelian subgroup of $G$.
	$C$ is called the center of $G$.
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	\item
	The group $D_4^*$ is not cyclic,
	but can be generated by two elements.
	The same is true of $S_n$ (nontrivial).
	What is the minimal number of generators of the additive group $\mathbb{Z}\oplus \mathbb{Z}$?
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	\item
	If $G=\langle a\rangle$ is a cyclic group and $H$ is any group,
	then every homomorphism $f:G\to H$ is completely determined by the element $f(a)\in H$.
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	\item
	The following cyclic subgroups are all isomorphic:
	the multiplicative group $\langle i\rangle$ in $\mathbb{C}$,
	the additive group $\mathbb{Z}_4$ and the subgroup $\left\langle\begin{pmatrix}1&2&3&4\\2&3&4&1\\\end{pmatrix} \right\rangle$ of $S_4$.
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	\item
	Let $G$ be a group and $\text{Aut }G$ the set of all automorphisms of $G$.
	\begin{enumerate}
	\item $\text{Aut }G$ is a group with composition of functions s binary operation.
	[{\it Hint:} $1_G\in \text{Aut }G$ is an identity;
	inverses exists by Theorem {I.2.3}.]
	\item $\text{Aut }\mathbb{Z}\cong \mathbb{Z}_2$ and $\text{Aut }\mathbb{Z}_6\cong \mathbb{Z}_2$;
	$\text{Aut }\mathbb{Z}_8\cong \mathbb{Z}_2\oplus \mathbb{Z}_2$;
	$\text{Aut }\mathbb{Z}_p\cong \mathbb{Z}_{p-1}$ ($p$ prime).
	\item
	What is $\text{Aut }\mathbb{Z}_n$ for arbitrary $n\in \mathbb{N}^*$?
	\end{enumerate}
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	\item
	For each prime $p$ the additive subgroup $Z(p^{\infty})$ of $\mathbb{Q}/\mathbb{Z}$ (Exercise {I.1.10}) is generated by the set $\{\overline{1/p^n}\mid n\in \mathbb{N}^*\}$.
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	\item
	Let $G$ be an abelian group and let $H, K$ be subgroups of $G$.
	Show that the join $H\vee K$ is the set $\{ab\mid a\in H, b\in K\}$.
	Extend this result to any finite number of subgroups of $G$.
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	\item
	\begin{enumerate}
	\item Let $G$ be a group and $\{H_i\mid i\in I\}$ a family of subgroups.
	State and prove a condition that will imply that $\bigcup_{i\in I}H_i$ is a subgroup,
	that is,
	that $\bigcup_{i\in I}H_i=\langle \bigcup_{i\in I}H_i\rangle$.
	\item
	Give an example of a group $G$ and a family of subgroups $\{H_i\mid i\in I\}$ such that $\bigcup_{i\in I}H_i\neq \langle \bigcup_{i\in I}H_i\rangle$.
	\end{enumerate}
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	\item
	\begin{enumerate}
	\item The set of all subgroups of a group $G$,
	partially ordered by set theoretic inclusion,
	forms a complete lattice (Introduction,
	Exercise {O.7.1} and Exercise {O.7.2})
	in which the g.l.b. of $\{H_i\mid i\in I\}$ is $\bigcap_{i\in I}H_i$ and the l.u.b. is $\langle \bigcup_{i\in I}H_i\rangle$.
	\item
	Exhibit the lattice of subgroups of the groups $S_3, D_4^*, \mathbb{Z}_6, \mathbb{Z}_{27}$, and $\mathbb{Z}_{35}$.
	\end{enumerate}
	\end{enumerate}
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	\section{Cyclic Groups}
	\begin{enumerate}
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	Let $a, b$ be elements of group $G$.
	Show that $|a|=|a^{-1}|$; $|ab|=|ba|$, and $|a|=|cac^{-1}|$
	for all $c\in G$.
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	Let $G$ be an abelian group containing elements $a$ and $b$ of orders $m$ and $n$ respectively.
	Show that $G$ contains an element whose order is the least common multiple of $m$ and $n$.
	[{\it Hint:} first try the case when $\gcd{(m,n)}=1$.]
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	Let $G$ be an abelian group of order $pq$,
	with $\gcd{(p, q)}=1$.
	Assume there exist $a, b\in G$ such that $|a|=p, |b|=q$ and show that $G$ is cyclic.
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	If $f:G\to H$ is a homomorphism,
	$a\in G$,
	and $f(a)$ has finite order in $H$,
	then $|a|$ is infinite or $|f(a)|$ divides $|a|$.
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	Let $G$ be the multiplicative group of all nonsingular $2\times 2$ matrices with rational entries.
	Show that $a=\begin{pmatrix}0&-1\\1&0\\\end{pmatrix}$ has order $4$ and $b=\begin{pmatrix}0&1\\-1&-1\\\end{pmatrix}$ has order $3$,
	but $ab$ has infinite order.
	Conversely,
	show that the additive group $\mathbb{Z}_2\oplus \mathbb{Z}$ contains nonzero elements $a, b$ of infinite order such that $a+b$ has finite order.
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	If $G$ is a cyclic group of order $n$ and $k\mid n$,
	then $G$ has exactly one subgroup of order $k$.
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	Let $p$ be prime and $H$ a subgroup of $Z(p^{\infty})$ (Exercise {I.1.10}).
	\begin{enumerate}
	\item Every element of $Z(p^{\infty})$ has finite order $p^n$ for some $n\geq 0$.
	\item If at least one element of $H$ has order $p^k$ and no element of $H$ has order greater than $p^k$,
	then $H$ is the cyclic subgroup generated by $\overline{1/p^k}$,
	whence $H\cong \mathbb{Z}_{p^k}$.
	\item If there is no upper bound on the orders of elements of $H$,
	then $H=Z(p^{\infty})$;
	[see Exercise {I.2.16}].
	\item The only proper subgroups of $Z(p^{\infty})$ are the finite cyclic groups $C_n=\langle \overline{1/p^n}\rangle$ ($n=1, 2, ...$).
	Furthermore,
	$\langle 0\rangle=C_0\leq C_1\leq C_2\leq C_3\leq \cdots$.
	\item Let $x_1, x_2, ...$ be elements of an abelian group $G$ such that $|x_1|=p$,
	$px_2=x_1, px_3=x_2, ..., px_{n+1}=x_n, ...$.
	The subgroup generated by the $x_i$ ($i\geq 1$) is isomorphic to $Z(p^{\infty})$.
	[{\it Hint:} Verify that the map induced by $x_i\mapsto \overline{1/p^i}$ is a well-defined isomorphism.]
	\end{enumerate}
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	A group that has only a finite number of subgroups must be finite.
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	If $G$ is an abelian group,
	then the set $T$ of all elements of $G$ with finite order is a subgroup of $G$.
	[Compare Exercise {I.3.5}.]
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	An infinite group is cyclic if and only if it is isomorphic to each of its proper subgroups.
	\end{enumerate}
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\section{Cosets and Counting}
	\begin{enumerate}
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	Let $G$ be a group and $\{H_i\mid i\in I\}$ a family of subgroups.
	Then for any $a\in G$,
	$(\bigcap_i H_i)a=\bigcap_i H_i a$.
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	\begin{enumerate}
	%%%%%
	\item
	Let $H$ be the cyclic subgroup (of order $2$) of $S_3$ generated by $\begin{pmatrix}
	1 & 2 & 3\\
	2& 1 & 3\\
	\end{pmatrix}$.
	Then no left coset of $H$ (except $H$ itself) is also a right coset.
	There exists $a\in S_3$ such that $aH\cap Ha=\{a\}$.
	%%%%%
	\item
	If $K$ is the cyclic subgroup (of order $3$) of $S_3$ generated by $\begin{pmatrix}
	1 & 2 & 3 \\
	2 & 3 & 1\\
	\end{pmatrix}$,
	then every left coset of $K$ is also a right coset of $K$.
	\end{enumerate}
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	The following conditions on a finite group $G$ are equivalent.
	\begin{enumerate}[(i)]
	\item $|G|$ is prime.
	\item $G\neq \langle e\rangle$ and $G$ has no proper subgroups.
	\item $G\cong \mathbb{Z}_p$ for some prime $p$.
	\end{enumerate}
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	(Euler-Fermat) Let $a$ be an integer and $p$ a prime such that $p\nmid a$.
	Then $a^{p-1}\equiv 1\pmod{p}$).
	[{\it Hint:} Consider $\overline{a}\in \mathbb{Z}_p$ and the multiplicative group of nonzero elements of $\mathbb{Z}_p$;
	see Exercise {I.1.7}.]
	It follows that $a^{p}\equiv a\pmod{p}$ for any integer $a$.
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	Prove that there are only two distinct groups of order $4$ (up to isomorphism),
	namely $\mathbb{Z}_4$ and $\mathbb{Z}_2\oplus \mathbb{Z}_2$.
	[{\it Hint:} By Lagrange's Theorem (Corollary {I.4.6}) a group of order $4$ that is not cyclic must consist of an identity and three elements of order $2$.]
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	Let $H, K$ be subgroups of a group $G$.
	Then $HK$ is a subgroup of $G$ if and only if $HK=HK$.
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	Let $G$ be a group of order $p^k m$,
	with $p$ prime and $\gcd{(p,m)}=1$.
	Let $H$ be a subgroup of order $p^k$ and $K$ a subgroup of order $p^d$,
	with $0<d\leq k$ and $K\not\subseteq H$.
	Show that $HK$ is not a subgroup of $G$.
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	If $H$ and $K$ are subgroups of finite index of a group $G$
	such that $[G:H]$ and $[G:K]$ are relatively prime,
	then $G=HK$.
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	If $H, K$ and $N$ are subgroups of a group $G$ such that $H\leq N$,
	then $HK\cap N=H(K\cap N)$.
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	Let $H, K, N$ be subgroups of a group $G$ such that $H\leq K$,
	$H\cap N=K\cap N$, and $HN=KN$.
	Show that $H=K$.
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	Let $G$ be a goup of order $2n$;
	then $G$ contains an element of order $2$.
	If $n$ is odd and $G$ abelian,
	there is only one element of order $2$.
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	If $H$ and $K$ are subgroups of a group $G$,
	then $[H\vee K:H]\geq [K:H\cap K]$.
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	If $p>q$ are primes,
	a group of order $pq$ has at most one subgroup of order $p$.
	[{\it Hint:} Suppose $H, K$ are distinct subgroups of order $p$.
	Show that $H\cap K=\langle e\rangle$;
	use Exercise 12 to get a contradiction.]
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	Let $G$ be a group and $a, b\in G$ such that
	\begin{enumerate}[(i)]
	\item $|a|=4=|b|$;
	\item $a^2=b^2$;
	\item $ba=a^3 b=a^{-1}b$;
	\item $a\neq b$;
	\item $G=\langle a, b\rangle$.
	\end{enumerate}
	Show that $|G|=8$ and $G\cong Q_8$.
	(See Exercise 2.3;
	observe that the generators $A, B$ of $Q_8$ also satisfy (i)-(v).)
	\end{enumerate}
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\section{Normality, Quotient Groups, and Homomorphisms}
	\begin{enumerate}
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	If $N$ is a subgroup of index $2$ in a group $G$,
	then $N$ is normal in $G$.
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	If $\{N_i\mid i\in I\}$ is a family of normal subgroups of a group $G$,
	then $\bigcap_{i\in I}N_i$ is a normal subgroup of $G$.
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	Let $N$ be a subgroup of a group $G$.
	$N$ is normal in $G$ if and only if (right) congruence modulo $N$ is a congruence relation on $G$.
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	Let $\sim$ be an equivalence relation on a group $G$ and let $N=\{a\in G\mid a\sim e\}$.
	Then $\sim$ is a congruence relation on $G$ if and only if $N$ is a normal subgroup of $G$ and $\sim$ is congruence modulo $N$.
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	Let $N\leq S_4$ consist of all those permutations $\sigma$ such that $\sigma(4)=4$.
	Is $N$ normal in $S_4$?
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	Let $H\leq G$;
	then the set $aHa^{-1}$ is a subgroup for each $a\in G$,
	and $H\cong aHa^{-1}$.
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	Let $G$ be a finite group and $H$ a subgroup of $G$ of order $n$.
	If $H$ is the only subgroup of $G$ of order $n$,
	then $H$ is normal in $G$.
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	All subgroups of te quaternion group are normal (see Exercises {I.2.3} and {I.1.4}).
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	\begin{enumerate}
	\item
	If $G$ is a group,
	then the center of $G$ is a normal subgroup of $G$ (see Exercise {I.2.11});
	\item
	the center of $S_n$ is the identity subgroup for all $n>2$.
	\end{enumerate}
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	Find subgroups $H$ and $K$ of $D_4$ such that $H\lhd K$ and $K\lhd D_4$,
	but $H$ is not normal in $D_4$.
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	If $H$ is a cyclic subgroup of a group $G$ and $H$ is normal in $G$,
	then every subgroup of $H$ is normal in $G$. [Compare Exercise {I.5.10}.]
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	If $H$ is a normal subgroup of a group $G$ such that $H$ and $G/H$ are finitely generated,
	then so is $G$.
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	\begin{enumerate}
	\item Let $H\lhd G, K\lhd G$.
	Show that $H\vee K$ is normal in $G$.
	\item
	Prove that the set of all normal subgroups of $G$ form a complete lattice under inclusion (Introduction, Exercise 7.2).
	\end{enumerate}
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	If $N_1\lhd G_1, N_2\lhd G_2$ then $(N_1\times N_2)\lhd (G_1\times G_2)$ and $(G_1\times G_2)/(N_1\times N_2)\cong (G_1/N_1)\times (G_2/N_2)$.
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	Let $N\lhd G$ and $K\lhd G$.
	If $N\cap K=\langle e\rangle$ and $N\vee K=G$,
	then $G/N\cong K$.
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	If $f:G\to H$ is a homomorphism,
	$H$ is abelian and $N$ is a subgroup of $G$ containing $\ker{f}$,
	then $N$ is normal in $G$.
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	\begin{enumerate}
	\item Consider the subgroups $\langle 6\rangle$ and $\langle 30\rangle$ of $\mathbb{Z}$ and show that $\langle 6\rangle/\langle 30\rangle\cong \mathbb{Z}_5$.
	\item For any $k, m>0$,
	$\langle k\rangle/\langle km\rangle\cong \mathbb{Z}_m$;
	in particular,
	$\mathbb{Z}/\langle m\rangle=\langle 1\rangle/\langle m\rangle\cong \mathbb{Z}_m$.
	\end{enumerate}
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	If $f:G\to H$ is a homomorphism with kernel $N$ and $K\leq G$,
	then prove that $f^{-1}(f(K))=KN$.
	Hence $f^{-1}(f(K))=K$ if and only if $N\leq K$.
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	If $N\lhd G$,
	$[G:N]$ finite,
	$H\leq G$,
	$|H|$ finite,
	and $[G:N]$ and $|H|$ are relatively prime,
	then $H\leq N$.
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	If $N\lhd G$,
	$|N|$ finite,
	$H\leq G$,
	$[G:H]$ finite,
	and $[G:H]$ and $|N|$ are relatively prime,
	then $N\lhd H$.
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	If $H$ is a subgroup of $Z(p^{\infty})$ and $H\neq Z(p^{\infty})/H\cong Z(p^{\infty})$.
	[{\it Hint:} if $H=\langle \overline{1/p^n}\rangle$, let $x_i=\overline{1/p^{n+i}}+H$ and apply Exercise I.3.7(e).]
	\end{enumerate}
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\section{Symmetric, Alternating, and Dihedral Groups}
	\begin{enumerate}
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	Find four different subgroups of $S_4$ that are isomorphic to $S_3$ and nine isomorphic to $S_2$.
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	\begin{enumerate}
	\item $S_n$ is generated by the $n-1$ transpositions $(12), (13), (14), ..., (1n)$.
	[{\it Hint:} $(1i)(1j)(1i)=(ij)$.]
	\item $S_n$ is generated by the $n-1$ transpositions $(12), (23), (34), ..., (n-1~n)$.
	[{\it Hint:} $(1j)=(1~j-1)(j-1~j)(1~j-1)$; use (a).]
	\end{enumerate}
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	If $\sigma=(i_1 i_2\cdots i_r)\in S_n$ and $\tau\in S_n$,
	then $\tau\sigma\tau^{-1}$ is the $r$-cycle $(\tau(i_1)\tau(i_2)\cdots \tau(i_r))$.
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	\begin{enumerate}
	\item $S_n$ is generated by $\sigma_1=(12)$ and $\tau=(123\cdots n)$.
	[{\it Hint:} Apply Exercise 3 to $\sigma_1, \sigma_2=\tau\sigma_1 \tau^{-1}, \sigma_3=\tau\sigma_2\tau^{-1}, ..., \sigma_{n-1}=\tau\sigma_{n-2}\tau^{-1}$ and use Exercise {I.6.2}(b).]
	\item $S_n$ is generated by $(12)$ and $(23\cdots n)$.
	\end{enumerate}
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	Let $\sigma, \tau\in S_n$.
	If $\sigma$ is even (odd),
	then so is $\tau\sigma\tau^{-1}$.
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	$A_n$ is the only subgroup of $S_n$ of index $2$.
	[{\it Hint:} Show that a subgroup of index $2$ must contain all $3$-cycles of $S_n$ and apply Lemma {I.6.11}.]
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	Show that $N=\{(1), (12)(34), (13)(24), (14)(23)\}$ is a normal subgroup of $S_4$ contained in $A_4$ such that $S_4/N\cong S_3$ and $A_4/N\cong \mathbb{Z}_3$.
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	The group $A_4$ has no subgroup of order $6$.
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	For $n\geq 3$ let $G_n$ be the multiplicative group of complex matrices generated by $x=\begin{pmatrix}0&1\\1&0\\\end{pmatrix}$ and $y=\begin{pmatrix}e^{2\pi i/n}&0\\0&e^{-2\pi i/n}\\\end{pmatrix}$,
	where $i^2=-1$.
	Show that $G_n\cong D_n$.
	({\it Hint:} recall that $e^{2\pi i}=1$ and $e^{k2\pi i}\neq 1$,
	where $k$ is real,
	unless $k\in \mathbb{Z}$.)
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	Let $a$ be the generator of order $n$ of $D_n$.
	Show that $\langle a\rangle\lhd D_n$ and $D_n/\langle a\rangle\cong \mathbb{Z}_2$.
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	Find all normal subgroups of $D_n$.
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	The center (Exercise {I.2.11}) of the group $D_n$ is $\langle e\rangle$ if $n$ is odd and isomorphic to $\mathbb{Z}_2$ if $n$ is even.
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	For each $n\geq 3$ let $P_n$ be a regular polygon of $n$ sides (for $n=3$, $P_n$ is an equilateral triangle;
	for $n=4$, a square).
	A symmetry of $P_n$ is a bijection $P_n\to P_n$ that preserves distances and maps adjacent vertices onto adjacent vertices.
	\begin{enumerate}
	\item The set $D_n^*$ of all symmetries of $P_n$ is a group under the binary operation of compositiion of functions.
	\item Every $f\in D_n^*$ is completely determined by its action on the vertices of $P_n$.
	Number the vertices consecutively $1, 2, ..., n$;
	then each $f\in D_n^*$ determines a unique permutation $\sigma_f$ of $\{1, 2, ..., n\}$.
	The assignment $f\mapsto \sigma_f$ defines a monomorphism of groups $\varphi:D_n^*\to S_n$.
	\item $D_n^*$ is generated by $f$ and $g$,
	where $f$ is a rotation of $2\pi/n$ degrees about the center of $P_n$ and $g$ is a reflection about the ``diameter'' through the center and vertex $1$.
	\item $\sigma_f=(123\cdots n)$ and $\sigma_g=\begin{pmatrix}1&2&3&\cdots&n-1&n\\1&n&n-1&\cdots&3&2\\\end{pmatrix}$,
	whence $\text{Im }\varphi=D_n$ and $D_n^*\cong D_n$.
	\end{enumerate}
	\end{enumerate}
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\section{Categories: Products, Coproducts, and Free Objects}
	\begin{enumerate}
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	A pointed set is a pair $(S, x)$ with $S$ a set and $x\in S$.
	A morphism of pointed sets $(S, x)\to (S', x')$ is a triple $(f, x, x')$,
	where $f:S\to S'$ is a function such that $f(x)=x'$.
	Show that pointed sets form a category.
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	If $f:A\to B$ is an equivalence in a category $\mathcal{C}$ and $g:B\to A$ is the morphism such that $g\circ f=1_A$, $f\circ g=1_B$,
	show that $g$ is unique.
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	In the category $\mathcal{G}$ of groups,
	show that the group $G_1\times G_2$ together with the homomorphisms $\pi_1:G_1\times G_2\to G_1 $ and $\pi_2:G_1\times G_2\to G$ (as in the Example preceding Definition 2.2) is a product for $\{G_1, G_2\}$.
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	In the category $\mathfrak{a}$ of abelian groups,
	show that the group $A_1\times A_2$, together with the homomorphisms $\iota_1:A_1\times A_2$ and $\iota_2:A_2\to A_1\times A_2$ (as in the Example preceding Definition 2.2) is a coproduct for $\{A_1, A_2\}$.
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	Every family $\{A_i\mid i\in I\}$ in the category of sets has a coproduct.
	[{\it Hint:} consider $\bigcup A_i=\{(a, i)\in (\bigcup A_i)\times I\mid a\in A_i\}$ with $A_i\to \dot{\bigcup}A_i$ given by $a\mapsto (a, i)$.
	$\dot{\bigcup}A_i$ is called the disjoint union of the sets $A_i$.]
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	\begin{enumerate}
	\item Show that in the category $S_*$ of pointed sets (see Exercise 1) products always exist;
	describe them.
	\item Show that in $S_*$ every family of objects has a coproduct (often called a ``wedge product'');
	describe this coproduct.
	\end{enumerate}
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	Let $F$ be a free object on a set $X$ ($i:X\to F$) in a concrete category $\mathcal{C}$.
	If $\mathcal{C}$ contains an object whose underlying set has at least two elements in it,
	then $i$ is an injective map of sets.
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	Suppose $X$ is a set and $F$ is a free object on $X$ (with $i:X\to F$) in the category of groups (the existence of $F$ is proved in Section 9).
	Prove that $i(X)$ is a set of generators for the group $F$.
	[{\it Hint:} If $G$ is the subgroup of $F$ generated by $i(X)$,
	then there is a homomorphism $\varphi:F\to G$ such that $\varphi i=i$.
	Show that $F\stackrel{\varphi}{\to}G\stackrel{\subseteq}{\to}F$ is the identity map.]
	\end{enumerate}
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\section{Direct Products and Direct Sums}
	\begin{enumerate}
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	$S_3$ is not the dirct product of any family of its proper subgroups.
	The same is true of $\mathbb{Z}_{p^n}$ ($p$ prime, $n\geq 1$) and $\mathbb{Z}$.
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	Give an example of groups $H_i, K_j$ such that $H_1\times H_2\cong K_1\times K_2$ and no $H_i$ is isomorphic o any $K_j$.
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	Let $G$ be an (additive) abelian group with subgroups $H$ and $K$.
	Show that $G\cong H\oplus K$ if and only if there are homomorphisms
	$H\substack{\stackrel{\pi_1}{\longleftarrow}\\ \underset{\iota_1}{\longrightarrow}} G\substack{\stackrel{\pi_2}{\longrightarrow}\\ \underset{\iota_2}{\longleftarrow}}K$ such that
	$\pi_1\iota_1=1_H, \pi_2\iota_2=1_K, \pi_1 \iota_2=0$ and $\pi_2\iota_1=0$,
	where $0$ is the map sending every element onto the zero (identity) element,
	and $\iota_1 \pi_1(x)+\iota_2 \pi_2(x)=x$ for all $x\in G$.
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	Give an example to show that the weak direct product is not a coproduct in the category of all groups.
	({\it Hint:} it suffices to consider the case of two factors $G\times H$.)
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	Let $G, H$ be finite cyclic groups.
	Then $G\times H$ is cyclic if and only if $\gcd{(|G|, |H|)}=1$.
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	Every finitely generated abelian group $G\neq \langle e\rangle$ in which every element (except $e$)
	has order $p$ ($p$ prime) is isomorphic to $\mathbb{Z}_p\oplus \mathbb{Z}_p\oplus \cdots \oplus \mathbb{Z}_p$ ($n$ summands)
	for some $n\geq 1$.
	[{\it Hint:} Let $A=\{a_1, ..., a_n\}$ be a set of generators such that no proper subset of $A$ generates $G$.
	Show that $\langle a_i\rangle\cong \mathbb{Z}_p$
	and $G=\langle a_1\rangle\times \langle a_2\rangle\times \cdots\times \langle a_n\rangle$.]
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	Let $H, K, N$ be nontrivial normal subgroups of a group $G$ and suppose $G=H\times K$.
	Prove that $N$ is in the center of $G$ or $N$ intersects one of $H, K$ nontrivially.
	Give examples to show that both possibilities can actually occur when $G$ is nonabelian.
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	Corollary I.8.7 is false if one of the $N_i$ is not normal.
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	If a group $G$ is the (internal) direct product of its subgroups $H, K$,
	then $H\cong G/K$ and $G/H\cong K$.
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	If $\{G_i\mid i\in I\}$ is a family of groups,
	then $\prod^{w}G_i$ is the internal weak direct product its subgroups $\{\iota_i(G_i)\mid i\in I\}$.
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	Let $\{N_i\mid i\in I\}$ be a family of subgroups of a group $G$.
	Then $G$ is the internal weak direct product of $\{N_i\mid i\in I\}$ if and only if:
	\begin{enumerate}[(i)]
	\item
	$a_i a_j=a_j a_i$ for all $i\neq j$ and $a_i\in N_i, a_j\in N_j$;
	\item
	every nonidentity element of $G$ is uniquely a product $a_{i_1}\cdots a_{i_n}$,
	where $i_1, ..., i_n$ are distinct elements of $I$ and $e\neq a_{i_k}\in N_{i_k}$ for each $k$.
	[Compare Theorem I.8.9.]
	\end{enumerate}
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	A normal subgroup $H$ of a group $G$ is said to be a direct factor (direct summand if $G$ is additive abelian) if there exists a (normal) subgroup $K$ of $G$ such that $G=H\times K$.
	\begin{enumerate}
	\item If $H$ is a direct factor of $K$ and $K$ is a direct factor of $G$,
	then $H$ is normal in $G$.
	[Compare Exerecise {I.5.10}.]
	\item If $H$ is a direct factor of $G$,
	then every homomorphism $H\to G$ may be extended to an endomorphism $G\to G$.
	However,
	a monomorphism $H\to G$ need not be extendible to an automorphism $G\to G$.
	\end{enumerate}
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	Let $\{G_i\mid i\in I\}$ be a family of groups and $J\subseteq I$.
	The map $\alpha:\prod_{j\in J}G_j\to \prod_{i\in I}G_i$ given by $\{a_j\}\mapsto \{b_i\}$,
	where $b_j=a_j$ for $j\in J$ and $b_i=e_i$ (identity of $G_i$) for $i\notin J$,
	is a monomorphism of groups and $\prod_{i\in I}G_i/\alpha(\prod_{j\in J}G_j)\cong \prod_{i\in I-J}G_i$.
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item For $i=1, 2$ let $H_i\lhd G_i$ and give examples to show that each of the following statements may be false:
	\begin{enumerate}
	\item $G_1\cong G_2$ and $H_1\cong H_2\Rightarrow G_1/H_1\cong G_2/H_2$.
	\item $G_1\cong G_2$ and $G_1/H_1\cong G_2/H_2\Rightarrow H_1\cong H_2$.
	\item $H_1\cong H_2$ and $G_1/H_1\cong G_2/H_2\Rightarrow G_1\cong G_2$.
	\end{enumerate}
	\end{enumerate}
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\section{Free Groups, Free Products, Generators \& Relations}
	\begin{enumerate}
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	Every nonidentity element in a free group $F$ has infinite order.
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	Show that the free group on the set $\{a\}$ is an infinite cyclic group,
	and hence isomorphic to $\mathbb{Z}$.
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	Let $F$ be a free group and let $N$ be the subgroup generated by the set $\{x^n\mid x\in F, n\text{ a fixed integer}\}$.
	Show that $N\lhd F$.
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	Let $F$ be the free group on the set $X$,
	and let $Y\subseteq X$.
	If $H$ is the smallest normal subgroup of $F$ containing $Y$,
	then $F/H$ is a free group.
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	The group defined by generators $a, b$ and relations $a^8=b^2 a^4=ab^{-1}ab=e$ has order at most $16$.
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	The cyclic group of order $6$ is the group defined by generators $a, b$ and relations $a^2=b^3=a^{-1}b^{-1}ab=e$.
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	Show that the group defined by generators $a, b$ and relations $a^2=e, b^3=e$ is infinite and nonabelian.
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	The group defined by generators $a, b$ and relations $a^n=e$ ($3\leq n\in \mathbb{N}^*$),
	$b^2=e$ and $abab=e$ is the dihedral group $D_n$.
	[See Theorem {I.6.13}.]
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	The group defined by the generator $b$ and the relation $b^m=e$ ($m\in \mathbb{N}^*$) is the cyclic group $\mathbb{Z}_m$.
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	The operation of free product is commutative and associative:
	for any groups $A, B, C$,
	$A*B\cong B*A$ and $A*(B*C)\cong (A*B)*C$.
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	If $N$ is the normal subgroup of $A*B$ generated by $A$,
	then $(A*B)/N\cong B$.
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	If $G$ and $H$ each have more than one element,
	then $G*H$ is an infinite group with center $\langle e\rangle$.
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	A free group is a free product of infinite cyclic groups.
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	If $G$ is the group defined by generators $a, b$ and relations $a^2=e, b^3=e$,
	then $G\cong\mathbb{Z}_2*\mathbb{Z}_3$.
	[See Exercise {I.9.12} and compare Exercise {I.9.6}.]
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	If $f:G_1\to G_2$ and $g:H_1\to H_2$ are homomorphisms of groups,
	then there is a unique homomorphism $h:G_1*H_1\to G_2*H_2$ such that $h|_{G_1}=f$ and $h|_{H_1}=g$.
	\end{enumerate}
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\chapter{The Structure of Groups}
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\section{Free Abelian Groups}
	\begin{enumerate}
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	\begin{enumerate}
	\item If $G$ is an abelian group and $m\in \mathbb{Z}$,
	then $mG=\{mu\mid u\in G\}$ is a subgroup of $G$.
	\item
	If $G\cong \sum_{i\in I}G_i$,
	then $mG\cong \sum_{i\in I}mG_i$ and $G/mG\cong \sum_{i\in I}G_i/mG_i$.
	\end{enumerate}
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	A subset $X$ of an abelian group $F$ is said to be linearly independent if $n_1 x_1+\cdots+n_k x_k=0$ always implies $n_i=0$ for all $i$
	(where $n_i\in \mathbb{Z}$ and $x_1, ..., x_k$ are distinct elements of $X$).
	\begin{enumerate}
	\item $X$ is linearly independent if and only if every nonzero element of the subgroup $\langle X\rangle$ may be written uniquely in the form $n_1 x_1+\cdots+n_k x_k$ ($n_i\in \mathbb{Z}$, $n_i\neq 0$, $x_1, ..., x_k$ distinct elements of $X$).
	\item If $F$ is free abelian of finite rank $n$,
	it is {\it not} true that every linearly independent subset of $n$ elements is a basis
	[{\it Hint:} consider $F=\mathbb{Z}$].
	\item If $F$ is free abelian,
	it is {\it not} true that every linearly independent subset of $F$ may be extended to a basis of $F$.
	\item If $F$ is free abelian,
	it is {\it not} true that every generating set of $F$ contains a basis of $F$.
	However, if $F$ is also finitely generated by $n$ elements,
	$F$ has rank $m\leq n$.
	\end{enumerate}
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	Let $X=\{a_i\mid i\in I\}$ be a set.
	Then the free abelian group on $X$ is (isomorphic to) the group defined by the generators $X$ and the relations (in multiplicative notation) $\{a_i a_j a_i^{-1} a_j^{-1}=e\mid i, j\in I\}$.
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	A free abelian group is a free group (Section I.9) if and only if it is cyclic.
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	The direct sum of a family of free abelian groups is a free abelian group.
	(A direct product of free abelian groups need not be free abelian;
	see L. Fuchs [13, p. 168].)
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	If $F=\sum_{x\in X}\mathbb{Z}x$ is a free abelian group,
	and $G$ is the subgroup with basis $X'=X-\{x_0\}$ for some $x_0\in X$,
	then $F/G\cong \mathbb{Z}x_0$.
	Generalize this result to arbitrary subsets $X'$ of $X$.
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	A nonzero free abelian group has a subgroup of index $n$ for every positive integer $n$.
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	Let $G$ be the multiplicative group generated by the real matrices $a=\begin{pmatrix}2&0\\0&1\\\end{pmatrix}$ and $b=\begin{pmatrix}1&1\\0&1\\\end{pmatrix}$.
	If $H$ is the set of all matrices in $G$ whose (main) diagonal entries are $1$,
	then $H$ is a subgroup that is {\it not} finitely generated.
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	Let $G$ be a finitely generated abelian group in which no element (except $0$) has finite order.
	Then $G$ is a free abelian group.
	[{\it Hint:} Theorem II.1.6.]
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	\begin{enumerate}
	\item Show that the additive group of rationals $\mathbb{Q}$ is not finitely generated.
	\item Show that $\mathbb{Q}$ is not free.
	\item Conclude that Exercise II.1.9 is false if the hypothesis ``finitely generated'' is omitted.
	\end{enumerate}
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	\begin{enumerate}
	\item Let $G$ be the additive group of all polynomials in $x$ with integer coefficients.
	Show that $G$ is isomorphic to the group $\mathbb{Q}^*$ of all positive rationals (under multiplication).
	[{\it Hint:} Use the Fundamental Theorem of Arithmetic to construct an isomorphism.]
	\item The group $\mathbb{Q}^*$ is free abelian with basis $\{p\mid p\text{ is prime in }\mathbb{Z}\}$.
	\end{enumerate}
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	Let $F$ be the free (not necessarily abelian) group on a set $X$ (as in Section I.9) and $G$ the free group on a set $Y$.
	Let $F'$ be the subgroup of $F$ {\it generated} by $\{aba^{-1}b^{-1}\mid a, b\in F\}$ and similarly for $G'$.
	\begin{enumerate}
	\item $F'\lhd F, G'\lhd G$ and $F/F', G/G'$ are abelian [see Theorem II.7.8 below].
	\item $F/F'$ [resp. $G/G'$] is a free abelian group of rank $|X|$ [resp. $|Y|$].
	[{\it Hint:} $\{xF'\mid x\in X\}$ is a basis of $F/F'$.]
	\item $F\cong G$ if and only if $|X|=|Y|$.
	[{\it Hint:} if $\varphi:F\cong G$,
	then $\varphi$ induces an isomorphism $F/F'\cong G/G'$.
	Apply Proposition II.1.3 and (b).
	The converse is Theorem I.7.8.]
	\end{enumerate}
	\end{enumerate}
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\section{Finitely Generated Abelian Groups}
	\begin{enumerate}
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	Show that a finite abelian group that is not cyclic contains a subgroup which is isomorphic to $\mathbb{Z}_p\oplus \mathbb{Z}_p$ for some prime $p$.
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	Let $G$ be a finite abelian group and $x$ an element of maximal order.
	Show that $\langle x\rangle$ is a direct summand of $G$.
	Use this to obtain another proof of Theorem II.2.1.
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	Suppose $G$ is a finite abelian $p$-group (Exercise II.2.7) and $x\in G$ has maximal order.
	If $\overline{y}\in G/\langle x\rangle$ has order $p^r$,
	then there is a representative $y\in G$ of the coset $\overline{y}$ such that $|y|=p^r$.
	[Note that if $|x|=p^t$,
	then $p^t G=0$.]
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	Use Exercises II.2.3 and II.2.7 to obtain a proof of Theorem II.2.2 which is independent of Theorem II.2.1.
	[{\it Hint:} If $G$ is a $p$-group,
	let $x\in G$ have maximal order;
	$G/\langle x\rangle$ is a direct sum of cyclics by induction,
	$G/\langle x\rangle=\langle \overline{x}_1\rangle\oplus \cdots \oplus \langle \overline{x}_n\rangle$,
	with $|\overline{x}_i|=p^{r_i}$ and $1\leq r_1\leq r_2\leq \cdots\leq r_n$.
	Choose representatives $x_i$ of $\overline{x}_i$ such that $|x_i|=|\overline{x}_i|$.
	Show that $G=\langle x_1\rangle\oplus \cdots\oplus \langle x_n\rangle\oplus \langle x\rangle$ is the desired decomposition.]
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	If $G$ is a finitely generated abelian group such that $G/G_t$ has rank $n$,
	and $H$ is a subgroup of $G$ such that $H/H_t$ has rank $m$,
	then $m\leq n$ and $(G/H)/(G/H)_t$ has rank $n-m$.
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	Let $k, m\in \mathbb{N}^*$.
	If $\gcd{(k, m)}=1$,
	then $k\mathbb{Z}_m=\mathbb{Z}_m$ and $\mathbb{Z}_m[k]=0$.
	If $k\mid m$,
	say $m=kd$,
	then $k\mathbb{Z}_m\cong \mathbb{Z}_d$ and $\mathbb{Z}_m[k]\cong \mathbb{Z}_k$.
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	A (sub)group in which every element has order a power of a fixed prime $p$ is called a $p$-(sub)group ({\it note:} $|0|=1=p^0$).
	Let $G$ be an abelian torsion group.
	\begin{enumerate}
	%%%%%
	\item $G(p)$ is the unique maximum $p$-subgroup of $G$ (that is, every $p$-subgroup of $G$ is contained in $G(p)$).
	%%%%%
	\item $G=\sum G(p)$,
	where the sum is over all primes $p$ such that $G(p)\neq 0$.
	[{\it Hint:} If $|u|=p_1^{n_1}\cdots p_t^{n_t}$,
	let $m_i=|u|/p_i^{n_i}$.
	There exist $c_i\in \mathbb{Z}$ such that $c_1 m_1+\cdots+c_t m_t=1$,
	whence $u=c_1 m_1 u+\cdots+c_t m_t u$;
	but $c_i m_i u\in G(p_i)$.]
	%%%%%
	\item
	If $H$ is another abelian torsion group,
	then $G\cong H$ if and only if $G(p)\cong H(p)$ for all primes $p$.
	\end{enumerate}
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	A finite abelian $p$-group (Exercise {II.2.7}) is generated by its elements of maximal order.
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	How many subgroups of order $p^2$ does the abelian group $\mathbb{Z}_{p^3}\oplus \mathbb{Z}_{p^2}$ have?
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	\begin{enumerate}
	%%%%%
	\item Let $G$ be a finite abelian $p$-group (Exercise {II.2.7}).
	Show that for each $n\geq 0$,
	$p^{n+1}G\cap G[p]$ is a subgroup of $p^n G\cap G[p]$. %%%%%
	\item
	Show that $(p^n G\cap G[p])/(p^{n+1}G\cap G[p]$) is a direct sum of copies of $\mathbb{Z}_p$;
	let $k$ be the number of copies.
	%%%%%
	\item Write $G$ as a direct sum of cyclics;
	show that the number $k$ of part (b) is the number of summands of order $p^{n+1}$.
	\end{enumerate}
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	Let $G, H$, and $K$ be finitely generated abelian groups.
	\begin{enumerate}
	%%%%%
	\item If $G\oplus G\cong H\oplus H$,
	then $G\cong H$.
	%%%%%
	\item If $G\oplus H\cong G\oplus K$, then $H\cong K$.
	%%%%%
	\item If $G_1$ is a free abelian group of rank $\aleph_0$,
	then $G_1\oplus \mathbb{Z}\oplus \mathbb{Z}\cong G_1\oplus \mathbb{Z}$,
	but $\mathbb{Z}\oplus \mathbb{Z}\not\cong \mathbb{Z}$.
	\end{enumerate}
	{\it Note:} there exists an infinitely generated denumerable torsion-free abelian group $G$ such that $G\cong G\oplus G\oplus G$,
	but $G\not\cong G\oplus G$,
	whence (a) fails to hold with $H=G\oplus G$.
	See A.L.S. Corner [60].
	Also see Exercise {II.3.11}, Exercise {II.3.12}, and Exercise {IV.3.12}.
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	\begin{enumerate}
	%%%%%
	\item
	What are the elementary divisors of the group $\mathbb{Z}_2\oplus \mathbb{Z}_9\oplus \mathbb{Z}_{35}$;
	what are its invariant factors?
	Do the same for $\mathbb{Z}_{26}\oplus \mathbb{Z}_{42}\oplus \mathbb{Z}_{49}\oplus \mathbb{Z}_{200}\oplus \mathbb{Z}_{1000}$.
	%%%%%
	\item
	Determine up to isomorphism all abelian groups of order $64$;
	do the same for order $96$.
	%%%%%
	\item
	Determine all abelian groups of order $n$ for $n\leq 20$.
	\end{enumerate}
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	Show that the invariant factors of $\mathbb{Z}_m\oplus \mathbb{Z}_n$ are $\gcd{(m, n)}$ and $\text{lcm}(m,n)$ (the greatest common divisor and the least common multiple)
	if $\gcd{(m,n)}>1$ and $mn$ if $\gcd{(m, n)}=1$.
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	If $H$ is a subgroup of a finite abelian group $G$,
	then $G$ has a subgroup that is isomorphic to $G/H$.
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	Every finite subgroup of $\mathbb{Q}/\mathbb{Z}$ is cyclic [see Exercise {I.3.7} and Exercise {II.2.7}.]
	\end{enumerate}
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\section{The Krull-Schmidt Theorem}
	\begin{enumerate}
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	A group $G$ is indecomposable if and only if $G\neq \langle e\rangle$ and $G\cong H\times K$ implies $H=\langle e\rangle$ or $K=\langle e\rangle$.
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	$S_n$ is indecomposable for all $n\geq 2$.
	[{\it Hint:} If $n\geq 5$ Theorem {I.6.8} and Theorem {I.6.10} and Exercise {I.8.7} amy be helpful.]
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	The additive group $\mathbb{Q}$ is indecomposable.
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	A nontrivial homomorphic image of an indecomposable group need not be indecomposable.
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	\begin{enumerate}
	\item $\mathbb{Z}$ satisfies the ACC but not the DCC on subgroups.
	\item
	Every finitely generated abelian group satisfies the ACC on subgroups.
	\end{enumerate}
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	Let $H, K$ be normal subgroups of a group $G$ such that $G=H\times K$.
	\begin{enumerate}
	\item
	If $N$ is a normal subgroup of $H$,
	then $N$ is normal in $G$ (compare Exercise {I.5.10}).
	\item
	If $G$ satisfies the ACC or DCC on normal subgroups,
	then so do $H$ and $K$.
	\end{enumerate}
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	If $f$ and $g$ are endomorphisms of a group $G$,
	then $f+g$ need not be an endomorphism.
	[{\it Hint:} Let $a=(123), b=(132)\in S_3$ and define $f(x)=axa^{-1}, g(x)=bxb^{-1}$.]
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	Let $f$ and $g$ be normal endomorphisms of a group $G$.
	\begin{enumerate}
	\item $fg$ is a normal endomorphism.
	\item $H\lhd G$ implies $f(H)\lhd G$.
	\item If $f+g$ is an endomorphism,
	then it is normal.
	\end{enumerate}
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	Let $G=G_1\times \cdots \times G_n$.
	For each $i$ let $\lambda_i:G_i\to G$ be the inclusion map and $\pi_i:G\to G_i$ the canonical projection (see page 59).
	Let $\varphi_i=\lambda_i \pi_i$.
	Then the ``sum'' $\varphi_{i_1}+\cdots+\varphi_{i_k}$ of any $k$ ($1\leq k\leq n$) {\it distinct} $\varphi_i$ is a normal endomorphism of $G$.
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	Use the Krull-Schmidt Theorem to prove Theorem {II.2.2} and Theorem {II.2.6} (iii) for {\it finite} abelian groups.
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	If $G$ and $H$ are groups such that $G\times G\cong H\times H$ and $G$ satisfies both the ACC and DCC on normal subgroups,
	then $G\cong H$ [see Exercise {II.2.11}].
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	If $G, H, K$ and $J$ are groups such that $G\cong H\times K$ and $G\cong H\times J$ and $G$ satisfies both the ACC and DCC on normal subgroups,
	then $K\cong J$ [see Exercise {II.2.11}]
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	For each prime $p$ the group $Z(p^{\infty})$ satisfies the descending but not the ascending chain condition on subgroups [see Exercise {I.3.7}].
	\end{enumerate}
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\section{The Action of a Group on a Set}
	\begin{enumerate}
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	Let $G$ be a group and $A$ a normal abelian subgroup.
	Show that $G/A$ operates on $A$ by conjugation and obtain a homomorphism $G/A\to \text{Aut }{A}$.
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	If $H, K$ are subgroups of $G$ such that $H\lhd K$,
	show that $K\leq N_G(H)$.
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	If a group $G$ contains an element $a$ having exactly two conjugates,
	then $G$ has a proper normal subgroup $N\neq \langle e\rangle$.
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	Let $H$ be a subgroup of $G$.
	The centralizer of $H$ is the set $C_G(H)=\{g\in G\mid hg=gh\text{ for all }h\in H\}$.
	Show that $C_G(H)$ is a subgroup of $N_G(H)$.
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	If $H$ is a subgroup of $G$,
	the factor group $N_G(H)/C_G(H)$ (see Exercise {II.4.4}) is isomorphic to a subgroup of $\text{Aut }H$.
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	Let $G$ be a group acting on a set $S$ containing at least two elements.
	Assume that $G$ is transitive;
	that is,
	given any $x, y\in S$,
	there exists $g\in G$ such that $gx=y$.
	Prove that
	\begin{enumerate}
	\item for $x\in S$, the orbit $\overline{x}$ of $x$ is $S$;
	\item all the stabilizers $G_x$ (for $x\in S$) are conjugate;
	\item if $G$ has the property: $\{g\in G\mid gx=x\text{ for all }x\in S\}=\langle e\rangle$
	(which is the case if $G\leq S_n$ for some $n$ and $S=\{1, 2, ..., n\}$) and if $N\lhd G$ and $N\leq G_x$ for some $x\in S$,
	then $N=\langle e\rangle$;
	\item for $x\in S$, $|S|=[G:G_x]$;
	hence $|S|$ divides $|G|$.
	\end{enumerate}
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	Let $G$ be a group and let $\text{Inn }G$ be the set of all inner automorphism of $G$.
	Show that $\text{Inn }G$ is a normal subgroup of $\text{Aut }G$.
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	Exhibit an automorphism of $\mathbb{Z}_6$ that is {\it not} an inner automorphism.
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	If $G/C(G)$ is cyclic,
	then $G$ is abelian.
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	Show that the center of $S_4$ is $\langle e\rangle$;
	conclude that $S_4$ is isomorphic to the group of all inner automorphisms of $S_4$.
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	Let $G$ be a group containing an element $a$ not of order $1$ or $2$.
	Show that $G$ has a nonidentity automorphism.
	[{\it Hint:} Exercise {I.2.2} and Corollary {II.4.7}.]
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	Any finite group is isomorphic to a subgroup of $A_n$ for some $n$.
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	If a group $G$ contains a subgroup ($\neq G$) of finite index,
	it contains a normal sbgroup ($\neq G$) of finite index.
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	If $|G|=pn$, with $p>n$, $p$ prime,
	and $H$ is a subgroup of order $p$,
	then $H$ is normal in $G$.
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	If a normal subgroup $N$ of order $p$ ($p$ prime) is contained in a group $G$ of order $p^n$,
	then $N$ is in the center of $G$.
	\end{enumerate}
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\section{The Sylow Theorems}
	\begin{enumerate}
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	If $N\lhd G$ and $N, G/N$ are both $p$-groups,
	then $G$ is a $p$-group.
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	If $G$ is a finite $p$-group,
	$H\lhd G$ and $H\neq \langle e\rangle$,
	then $H\cap C(G)\neq \langle e\rangle$.
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	Let $|G|=p^n$.
	For each $k$,
	$0\leq k\leq n$,
	$G$ has a {\it normal} subgroup of order $p^k$.
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	If $G$ is an infinite $p$-group ($p$ prime),
	then either $G$ has a subgroup of order $p^n$ for each $n\geq 1$ or there exists $m\in \mathbb{N}^*$ such that every finite subgroup of $G$ has order $\leq p^m$.
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	If $P$ is a normal Sylow $p$-subgroup of a finite group $G$ and $f:G\to G$ is an endomorphism,
	then $f(P)\leq P$.
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	If $H$ is a normal subgroup of order $p^k$ of a finite group $G$,
	then $H$ is contained in every Sylow $p$-subgroup of $G$.
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	Find the Sylow $2$-subgroups and Sylow $3$-subgroups of $S_3, S_4, S_5$.
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	If every Sylow $p$-subgroup of a finite group $G$ is normal for every prime $p$,
	then $G$ is the direct product of its Sylow subgroups.
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	If $|G|=p^n q$,
	with $p>q$ primes,
	then $G$ contains a unique normal subgroup of index $q$.
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	Every group of order $12, 28, 56$, and $200$ must contain a normal Sylow subgroup,
	and hence is not simple.
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	How many elements of order $7$ are there in a simple group of order $168$?
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	Show that every automorphism of $S_4$ is an inner automorphism,
	and hence $S_4\cong\text{Aut }S_4$.
	[{\it Hint:} see Exercise {II.4.10}.
	Every automorphism of $S_4$ induces a permutation of the set $\{P_1, P_2, P_3, P_4\}$ of Sylow $3$-subgroups of $S_4$.
	If $f\in \text{Aut }S_4$ has $f(P_i)=P_i$ for all $i$,
	then $f=1_{S_4}$.]
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	Every group $G$ of order $p^2$ ($p$ prime) is abelian
	[{\it Hint:} Exercise {II.4.9} and Corollary {II.5.4}].
	\end{enumerate}
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\section{Classification of Finite Groups}
	\begin{enumerate}
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	Let $G$ and $H$ be groups and $\theta:H\to \text{Aut }G$ a homomorphism.
	Let $G\times_{\theta} H$ be the set $G\times H$ with the following binary operation:
	$(g, h)(g', h')=(g[\theta(h)(g')], hh')$.
	Show that $G\times_{\theta}H$ is a group with identity element $(e, e)$ and $(g, h)^{-1}=(\theta(h^{-1})(g^{-1}), h^{-1})$.
	$G\times_{\theta} H$ is called the semidirect product of $G$ and $H$.
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	Let $C_p=\langle a\rangle$ and $C_q=\langle b\rangle$ be (multiplicative) cyclic groups of prime orders $p$ and $q$ respectively such that $p>q$ and $q\mid p-1$.
	Let $s$ be an integer such that $s\not\equiv 0 \pmod{p}$.
	Elementary number theory shows that such an $s$ exists (see J. E. Shockley [51; Corollary 6.1, p.67]).
	\begin{enumerate}
	\item The map $\alpha:C_p\to C_p$ given by $a^i\mapsto a^{si}$ is an automorphism.
	\item The map $\theta:C_q\to \text{Aut }C_p$ given by $\theta(b^i)=\alpha^i$ ($\alpha$ as in part (a)) is a homomorphism $(\alpha^0=1_{C_p}$).
	\item If we write $a$ for $(a, e)$ and $b$ for $(e, b)$,
	then the group $C_p\times_{\theta} C_q$ (see Exercise {II.6.1}) is a group of order $pq$,
	generated by $a$ and $b$ subject to the relations:
	$|a|=p, |b|=q, ba=a^s b$,
	where $s\not\equiv 1\pmod{p}$,
	and $s^q\equiv 1\pmod{p}$.
	The group $C_p\times_{\theta}C_q$ is called the metacyclic group.
	\end{enumerate}
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	Consider the set $G=\{\pm 1, \pm i, \pm j, \pm k\}$ with multiplication given by $i^2=j^2=k^2=-1$;
	$ij=k$;
	$jk=i, ki=j$;
	$ji=-k, kj=-i, ik=-j$,
	and the usual rules for multiplying by $\pm 1$.
	Show that $G$ is a group isomorphic to the quaternion group $Q_8$.
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	What is the center of the quaternion group $Q_8$?
	Show that $Q_8/C(Q_8)$ is abelian.
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	\begin{enumerate}
	\item Show that there is a nonabelian subgroup $T$ of $S_3\times \mathbb{Z}_4$ of order $12$ generated by elements $a, b$ such that $|a|=6, a^3=b^2, ba=a^{-1}b$.
	\item Any group of order $12$ with generators $a, b$ such that $|a|=6, a^3=b^2, ba=a^{-1}b$ is isomorphic to $T$.
	\end{enumerate}
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	No two of $D_6, A_4$, and $T$ are isomorphic,
	where $T$ is the group of order $12$ described in Proposition {II.6.4} and Exercise {II.6.5}.
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	If $G$ is a nonabelian group of order $p^3$ ($p$ prime),
	then the center of $G$ is the subgroup generated by all elements of the form $aba^{-1}b^{-1}$ ($a, b\in G$).
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	Let $p$ be an {\it odd} prime.
	Prove that there, at most, two nonabelian groups of order $p^3$.
	[One has generators $a, b$ satisfying $|a|=p^2$;
	$|b|=p$;
	$b^{-1}ab=a^{1+p}$;
	the other has generators $a, b, c$ satisfying $|a|=|b|=c|=p$;
	$c=a^{-1}b^{-1}ab$;
	$ca=ac$;
	$cb=bc$.]
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	Classify up to isomorphism all groups of order $18$.
	Do the same for orders $20$ and $30$.
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	Show that $D_4$ is not isomorphic to $Q_8$.
	[{\it Hint:} Count elements of order $2$.]
	\end{enumerate}
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\section{Nilpotent and Solvable Groups}
	\begin{enumerate}
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	\begin{enumerate}
	\item $A_4$ is not the direct product of its Sylow subgroups,
	but $A_4$ does have the property:
	$mn=12$ and $\gcd{(m, n)}=1$ imply there is a subgroup of order $m$.
	\item $S_3$ has subgroups of orders $1, 2, 3$, and $6$ but is not the direct product of its Sylow subgroups.
	\end{enumerate}
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	Let $G$ be a group and $a, b\in G$.
	Denote the commutator $aba^{-1}b^{-1}\in G$ by $[a,b]$.
	Show that for any $a, b, c, \in G$,
	$[ab,c]=a[b,c]a^{-1}[a,c]$.
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	If $H$ and $K$ are subgroups of a group $G$,
	let $(H, K)$ be the subgroup of $G$ generated by the elements $\{hkh^{-1}k^{-1}\mid h\in H, k\in K\}$.
	Show that
	\begin{enumerate}
	\item $(H, K)$ is normal in $H\vee K$.
	\item If $(H, G')=\langle e\rangle$, then $(H', G)=\langle e\rangle$.
	\item $H\lhd G$ if and only if $(H, G)\leq H$.
	\item Let $K\lhd G$ and $K\leq H$;
	then $H/K\leq C(G/K)$ if and only if $(H, G)\leq K$.
	\end{enumerate}
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	Define a chain of subgroups $\gamma_i(G)$ of a group $G$ as follows:
	$\gamma_1(G)=G, \gamma_2(G)=(G, G), \gamma_i(G)=(\gamma_{i-1}(G), G)$
	(see Exercise {II.7.3}).
	Show that $G$ is nilpotent if and only if $\gamma_m(G)=\langle e\rangle$ for some $m$.
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	Every subgroup and every quotient group of a nilpotent group is nilpotent.
	[{\it Hint:} Theorem {II.7.5} or Exercise {II.7.4}.].
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	(Wielandt)
	Prove that a finite group $G$ is nilpotent if and only if every maximal proper subgroup of $G$ is normal.
	Conclude that every maximal proper subgroup has prime index.
	[{\it Hint:} if $P$ is a Sylow $p$-subgroup of $G$,
	show that any subgroup containing $N_G(P)$ is its own normalizer;
	see Theorem {II.5.11}.]
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	If $N$ is a nontrivial normal subgroup of a nilpotent group $G$,
	then $N\cap C(G)\neq \langle e\rangle$.
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	If $D_n$ is the dihedral group with generators $a$ of order $n$ and $b$ of order $2$,
	then
	\begin{enumerate}
	\item $a^2\in D_n'$.
	\item If $n$ is odd, $D_n'\cong \mathbb{Z}_n$.
	\item If $n$ is even, $D_n'\cong \mathbb{Z}_m$, where $2m=n$.
	\item $D_n$ is nilpotent if and only if $n$ is a power of $2$.
	\end{enumerate}
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	Show that the commutator subgroup of $S_4$ is $A_4$.
	What is the commutator group of $A_4$?
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	$S_n$ is solvable for $n\leq 4$,
	but $S_3$ and $S_4$ are not nilpotent.
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	A nontrivial finite solvable group $G$ contains a normal abelian subgroup $H\neq \langle e\rangle$.
	If $G$ is not solvable then $G$ contains a normal subgroup $H$ such that $H'=H$.
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	There is no group $G$ such that $G'=S_4$.
	[{\it Hint:} Exercise {II.7.9} and Exercise {II.5.12} may be helpful.]
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	If $G$ is a group,
	then the $i$th derived subgroup $G^{(i)}$ is a fully invariant subgroup,
	whence $G^{(i)}$ is normal.
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	If $N\lhd G$ and $N\cap G'=\langle e\rangle$,
	then $N\leq C(G)$.
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	If $H$ is a maximal proper subgroup of a finite solvable group $G$,
	then $[G:H]$ is a prime power.
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	For any group $G$,
	$C(G)$ is characteristic,
	but not necessarily fully invariant.
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	If $G$ is an abelian $p$-group,
	then the subgroup $G[p]$ (see Lemma {II.2.5}) is fully invariant in $G$.
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	If $G$ is a finite nilpotent group,
	then every minimal normal subgroup of $G$ is contained in $C(G)$ and has prime order.
	\end{enumerate}
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\section{Normal and Subnormal Series}
	\begin{enumerate}
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	\begin{enumerate}
	\item Find a normal series of $D_4$ consisting of $4$ subgroups.
	\item Find all composition series of the group $D_4$.
	\item Do part (b) f ro the group $A_4$.
	\item Do part (b) for the group $S_3\times \mathbb{Z}_2$.
	\item Find all composition factors of $S_4$ and $D_6$.
	\end{enumerate}
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	If $G=G_0\geq G_1\geq \cdots \geq G_n$ is a subnormal series of a finite group $G$,
	then $|G|=\left(\prod_{i=0}^{n-1}|G_i/G_{i+1}\right)|G_n|$.
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	If $N$ is a simiple normal subgroup of a group $G$ and $G/N$ has a composition series,
	then $G$ has a composition series.
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	A composition series of a group is a subnormal series of maximal (finite) length.
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	An abelian group has a composition series if and only if it is finite.
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	If $H\lhd G$,
	where $G$ has a composition series,
	then $G$ has a composition series on of whose terms is $H$.
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	A solvable group with a composition series is finite.
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	If $H$ and $K$ are solvable subgroups of $G$ with $H\lhd G$,
	then $HK$ is a solvable subgroup of $G$.
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	Any group of order $p^2 q$ ($p, q$ primes) is solvable.
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	A group $G$ is nilpotent if and only if there is a normal series $G=G_0\geq G_1\geq \cdots \geq G_n=\langle e\rangle$ such that $G_i/G_{i+1}\leq C(G/G_{i+1})$ for every $i$.
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	\begin{enumerate}
	\item Show that the analogue of Theorem {II.7.11} is false for nilpotent groups [Consider $S_3$].
	\item If $H\leq C(G)$ and $G/H$ is nilpotent,
	then $G$ is nilpotent.
	\end{enumerate}
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	Prove the Fundamental Theorem of Arithmetic,
	Introduction,
	Theorem {O.6.7},
	by applying the Jordan-H{\" o}lder Theorem to the group $\mathbb{Z}_n$.
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	Any simple group $G$ of order $60$ is isomorphic to $A_5$.
	[{\it Hint:} use Corollary {II.4.9};
	if $H\leq G$,
	then $[G:H]\geq 5$
	(since $|S_n|<60$ for $n\leq 4$);
	if $[G:H]=5$ then $G\cong A_5$ by Theorem {I.6.8}.
	The assumption that there is no subgroup of index $5$ leads to a contradiction.]
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	There are no nonabelian simple groups of order $<60$.
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	Let $G$ be the subgroup of $S_7$ generated by $(1234567)$ and $(26)(34)$.
	Show that $|G|=168$.
	\end{enumerate}
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	Exercises 16-20 outline a proof of the fact that the group $G$ of Exercise {II.8.15} is simple.
	We consider $G$ as acting on the set $S=\{1, 2, 3, 4, 5, 6, 7\}$ as in the first example after Definition {II.4.1} and make use of Exercise {II.4.6}.
	\begin{enumerate}[resume]
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item The group $G$ is transitive (see Exercise {II.4.6}).
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item For each $x\in S$,
	$G_x$ is a maximal (proper) subgroup of $G$.
	The proof of this fact proceeds in several steps:
	\begin{enumerate}
	\item A {\it block} of $G$ is a subset $T$ of $S$ such that for each $g\in G$ either $gT\cap T=\emptyset$ or $gT=T$,
	where $gT=\{gx\mid x\in T\}$.
	Show that if $T$ is a block,
	then $|T|$ divides $7$.
	[{\it Hint:} let $H=\{g\in G\mid gT=T\}$ and show that for $x\in T$,
	$G_x\leq H$ and $[H:G_x]=|T|$.
	Hence $|T|$ divides $[G:G_x]=[G:H][H:G_x]$.
	But $[G:G_x]=7$ by Exercise {II.4.6}(a) and Theorem {II.4.3}.]
	\item
	If $G_x$ is not maximal,
	then there is a block $T$ of $G$ such that $|T|\nmid 7$,
	contradicting part (a).
	[{\it Hint:} If $G_x\lneq H\leq G$,
	show that $H$ is not transitive on $S$
	(since $1\leq [H:G_x]<|S|$,
	which contradicts Exercise {II.4.6}.(d)).
	Let $T=\{hx\mid h\in H\}$.
	Since $H$ is not transitive,
	$|T|<|S|=7$ and since $H\neq G_x$,
	$|T|>1$.
	Show that $T$ is a block.]
	\end{enumerate}
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	If $\langle 1\rangle\neq N\lhd G$,
	then $7$ divides $|N|$.
	[{\it Hint:}
	Exercise {II.4.6} (c) $\Rightarrow$
	$G_x\lneq NG_x$ for all $x\in S$
	$\Rightarrow$
	$NG_x=G$ for all $x\in S$ by Exercise {II.8.17}
	$\Rightarrow$
	$N$ is transitive on $S$
	$\Rightarrow$
	$7$ divides $|N|$ by Exercise {II.4.6} (d).]
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	The group $G$ contains a subgroup $P$ of order $7$ such that the smallest normal subgroup of $G$ containing $P$ is $G$ itself.
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	If $\langle 1\rangle\neq N\lhd G$,
	then $N=G$;
	hence $G$ is simple.
	[Use Exercise {I.5.19} and Exercise {II.8.18} to show $P\leq N$;
	apply Exercise {II.8.19}.]
	\end{enumerate}
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\chapter{Rings}
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\section{Rings and Homomorphisms}
	\begin{enumerate}
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	\begin{enumerate}
	\item Let $G$ be an (additive) abelian group.
	Define an operation of multiplication in $G$ by $ab=0$ (for all $a ,b\in G$).
	Then $G$ is a ring.
	\item
	Let $S$ be the set of all subsets of some fixed set $U$.
	For $A, B\in S$,
	define $A+B=(A-B)\cup (B-A)$ and $AB=A\cap B$.
	Then $S$ is a ring.
	Is $S$ commutative?
	Does it have an identity?
	\end{enumerate}
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	Let $\{R_i\mid i\in I\}$ be a family of rings with identity.
	Make the direct sum of abelian groups $\sum_{i\in I}R_i$ into a ring by defining multiplication coordinatewise.
	Does $\sum_{i\in I}R_i$ have an identity?
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	A ring $R$ such that $a^2=a$ for all $a\in R$ is called a Boolean ring.
	Prove that every Boolean ring $R$ is commutative and $a+a=0$ for all $a\in R$.
	[For an example of a Boolean ring,
	see Exercise {III.1.1}(b).]
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	Let $R$ be a ring and $S$ a nonempty set.
	Then the group $M(S, R)$ (Exercise {I.1.2}) is a ring with multiplication defined as follows:
	the product of $f, g\in M(S, R)$ is the function $S\to R$ given by $s\mapsto f(s)g(s)$.
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	If $A$ is the abelian group $\mathbb{Z}\oplus \mathbb{Z}$,
	then $\text{End }A$ is a noncommutative ring (see page 116).
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	A finite ring with more than one element and no zero divisors is a division ring.
	(Special case:
	a finite integral domain is a field.)
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	Let $R$ be a ring with more than one element such that for each nonzero $a\in R$
	there is a unique $b\in R$ such that $aba=a$.
	Prove:
	\begin{enumerate}
	\item $R$ has no zero divisors.
	\item $bab=b$.
	\item $R$ has an identity.
	\item $R$ is a division ring.
	\end{enumerate}
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	Let $R$ be the set of all $2\times 2$ matrices over the complex field $\mathbb{C}$ of the form
	$$\begin{pmatrix}
	z & w \\
	-\overline{w} & \overline{z}\\
	\end{pmatrix},$$
	where $\overline{z}, \overline{w}$ are the complex conjugates of $z$ and $w$ respectively
	(that is,
	$c=a+b\sqrt{-1}\Leftrightarrow \overline{c}=a-b\sqrt{-1}$).
	Then $R$ is a division ring that is isomorphic to the division ring $K$ of real quaternions.
	[{\it Hint:} Define an isomorphism $K\to R$ by letting the images of $1, i, j, k\in K$ be respectively the matrices
	$$
	\begin{pmatrix}
	1 & 0 \\
	0 & 1 \\
	\end{pmatrix},
	\begin{pmatrix}
	\sqrt{-1} & 0 \\
	0 & -\sqrt{-1} \\
	\end{pmatrix},
	\begin{pmatrix}
	0 & 1 \\
	-1 & 0 \\
	\end{pmatrix},
	\begin{pmatrix}
	0 & \sqrt{-1} \\
	\sqrt{-1} & 0 \\
	\end{pmatrix}.$$
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	\begin{enumerate}
	\item The subset $G=\{1, -1, i, -i, j, -j, k, -k\}$ of the division ring $K$ of real quaternions forms a group under multiplication.
	\item $G$ is isomorphic to the quaternion group (Exercise {I.4.14} and Exercise {I.2.3}).
	\item What is the difference between the ring $K$ and the group ring $\mathbb{R}(G)$
	($\mathbb{R}$ the field of real numbers)?
	\end{enumerate}
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	Let $k, n$ be integers such that $0\leq k\leq n$ and ${n\choose k}$ the binomial coefficient $n!/(n-k)!k!$,
	where $0!=1$ and for $n>0$,
	$n!=n(n-1)(n-2)\cdots 2\cdot 1$.
	\begin{enumerate}
	\item
	${n \choose k}={n \choose n-k}$
	\item
	${n \choose k}<{n \choose k+1}$ for $k+1\leq n/2$.
	\item
	${n \choose k}+{n \choose k+1}={n+1 \choose k+1}$ for $k<n$.
	\item
	${n \choose k}$ is an integer.
	\item
	if $p$ is prime and $1\leq k\leq p^n-1$,
	then ${p^n \choose k}$ is divisible by $p$.
	[{\it Hints:} (b) observe that ${n \choose k+1}={n \choose k}\frac{n-k}{k+1}$;
	(d) note that ${m \choose 0}={m \choose m}=1$ and use induction on $n$ in part (c).]
	\end{enumerate}
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	(The Freshman's Dream).
	Let $R$ be a commutative ring with identity of prime characteristic $p$.
	If $a, b\in R$,
	then $(a\pm b)^{p^n}=a^{p^n}+b^{p^n}$ for all integers $n\geq 0$
	[see Theorem {III.1.6} and Exercise {III.1.10};
	note that $b=-b$ if $p=2$].
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	An element of a ring is nilpotent if $a^n=0$ for some $n$.
	Prove that in a commutative ring $a+b$ is nilpotent if $a$ and $b$ are.
	Show that this result may be false if $R$ is not commutative.
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	In a ring $R$ the following conditions are equivalent.
	\begin{enumerate}
	\item $R$ has no nonzero nilpotent elements (see Exercise {III.1.12}).
	\item If $a\in R$ and $a^2=0$,
	then $a=0$.
	\end{enumerate}
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	Let $R$ be a commutative ring with identity and prime characteristic $p$.
	The map $R\to R$ given by $r\mapsto r^p$ is a homomorphism of rings called the Frobenius homomorphism [see Exercise {III.1.11}].
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	\begin{enumerate}
	\item Give an example of a nonzero homomorphism $f:R\to S$ of rings with identity such that $f(1_R)\neq 1_S$.
	\item If $f:R\to S$ is an epimorphism of rings with identity,
	then $f(1_R)=1_S$.
	\item If $f:R\to S$ is a homomorphism of rings with identity and $u$ is a unit in $R$ such that $f(u)$ is a unit in $S$,
	then $f(1_R)=1_S$ and $f(u^{-1})=f(u)^{-1}$.
	[Note: there are easy examples which show that $f(u)$ need not be a unit in $S$ even though $u$ is a unit in $R$.]
	\end{enumerate}
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	Let $f:R\to S$ be a homomorphism of rings such that $f(r)\neq 0$ for some nonzero $r\in R$.
	If $R$ has an identity and $S$ has no zero divisors,
	then $S$ is a ring with identity $f(1_R)$.
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	\begin{enumerate}
	\item If $R$ is a ring,
	then so is $R^{op}$,
	where $R^{op}$ is defined as follows.
	The underlying set of $R^{op}$ is precisely $R$ and addition in $R^{op}$ coincides with addition in $R$.
	Multiplication in $R^{op}$,
	denoted $\circ$,
	is defined by $a\circ b=ba$,
	where $ba$ is the product in $R$.
	$R^{op}$ is called the opposite ring of $R$.
	\item $R$ has an identity if and only if $R^{op}$ does.
	\item $R$ is a division ring if and only if $R^{op}$ is.
	\item $(R^{op})^{op}=R$.
	\item If $S$ is a ring,
	then $R\cong S$ if and only if $R^{op}\cong S^{op}$.
	\end{enumerate}
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	Let $\mathbb{Q}$ be the field of rational numbers and $R$ any ring.
	If $f, g:\mathbb{Q}\to R$ are homomorphisms of rings such that $f|_{\mathbb{Z}}=g|_{\mathbb{Z}}$,
	then $f=g$.
	[{\it Hint:} show that for $n\in \mathbb{Z}$ ($n\neq 0$),
	$f(1/n)g(n)=g(1)$,
	whence $f(1/n)=g(1/n)$.]
	\end{enumerate}
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\section{Ideals}
	\begin{enumerate}
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	The set of all nilpotent elements in a commutative ring forms an ideal [see Exercise {III.1.12}].
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	Let $I$ be an ideal in a commutative ring $R$ and let $\text{Rad }I=\{r\in R\mid r^n\in I\text{ for some }n\}$.
	Show that $\text{Rad }I$ is an ideal.
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	If $R$ is a ring and $a\in R$,
	then $J=\{r\in R\mid ra=0\}$ is a left ideal and $K=\{r\in R\mid ar=0\}$ is a right ideal in $R$.
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	If $I$ is a left ideal of $R$,
	then $A(I)=\{r\in R\mid rx=0\text{ for every }x\in I\}$ is an ideal in $R$.
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	if $I$ is an ideal in a ring $R$,
	let $[R:I]=\{r\in R\mid xr\in I\text{ for every }x\in R\}$.
	Prove that $[R:I]$ is an ideal of $R$ which contains $I$.
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	\begin{enumerate}
	\item The center of the ring $S$ of all $2\times 2$ matrices over a field $F$ consists of all matrices of the form $\begin{pmatrix}
	a&0\\
	0&a\\
	\end{pmatrix}$.
	\item The center of $S$ is not an ideal in $S$.
	\item
	What is the center of the ring of all $n\times n$ matrices over a division ring?
	\end{enumerate}
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	\begin{enumerate}
	\item A ring $R$ with identity is a division ring if and only if $R$ has no proper left ideals.
	[Proposition {I.1.3} may be helpful.]
	\item
	If $S$ is a ring (possibly without identity) with no proper left ideals,
	then either $S^2=0$ or $S$ is a division ring.
	[{\it Hint:} show that $\{a\in S\mid Sa=0\}$ is an ideal.
	If $cd\neq 0$,
	show that $\{r\in S\mid rd=0\}=0$.
	Find $e\in S$ such that $ed=d$ and show that $e$ is a (two-sided) identity.]
	\end{enumerate}
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	Let $R$ be a ring with identity and $S$ the ring of all $n\times n$ matrices over $R$.
	$J$ is an ideal of $S$ if and only if $J$ is the ring of all $n\times n$ matrices over $I$ for some ideal $I$ in $R$.
	[{\it Hint:} Given $J$,
	let $I$ be the set of all those elements of $R$ that appear as the row $1$-column $1$ entry of some matrix in $J$.
	Use the matrices $E_{r, s}$,
	where $1\leq r\leq n$,
	$1\leq s\leq n$,
	and $E_{r, s}$ has $1_R$ as the row $r$-column $s$ entry and $0$ elsewhere.
	Observe that for a matrix $A=(a_{ij})$,
	$E_{p, r}AE_{s, q}$ is the matrix with $a_{rs}$ in the row $p$-column $q$ entry and $0$ elsewhere.]
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	Let $S$ be the ring of all $n\times n$ matrices over a division ring $D$.
	\begin{enumerate}
	\item $S$ has no proper ideals (that is, $0$ is a maximal ideal).
	[{\it Hint:} apply Exercise {III.2.8} or argue directly,
	using the matrices $E_{r, s}$ mentioned there.]
	\item $S$ has zero divisors.
	Consequently,
	\begin{enumerate}[(i)]
	\item $S\cong S/0$ is not a division ring and
	\item $0$ is a prime ideal which does not satisfy condition (1) of Theorem {III.2.15}.
	\end{enumerate}
	\end{enumerate}
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	\begin{enumerate}
	\item Show that $\mathbb{Z}$ is a principal ideal ring [see Theorem {I.3.1}].
	\item Every homomorphic image of a principal ideal ring is also a principal ideal ring.
	\item $\mathbb{Z}_m$ is a principal ideal ring for every $m>0$.
	\end{enumerate}
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	If $N$ is the ideal of all nilpotent elements in a commutative ring $R$ (see Exercise {III.2.1}),
	then $R/N$ is a ring with no nonzero nilpotent elements.
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	Let $R$ be a ring without identity and with no zero divisors.
	Let $S$ be the ring whose additive group is $R\times \mathbb{Z}$ as in the proof of Theorem {III.1.10}.
	Let $A=\{(r, n)\in S\mid rx+nx=0\text{ for every }x\in R\}$.
	\begin{enumerate}
	\item $A$ is an ideal in $S$.
	\item $S/A$ has an identity and contains a subring isomorphic to $R$.
	\item $S/A$ has no zero divisors.
	\end{enumerate}
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	Let $f:R\to S$ be a homomorphism of rings,
	$I$ an ideal in $R$,
	and $J$ an ideal in $S$.
	\begin{enumerate}
	\item
	$f^{-1}(J)$ is an ideal in $R$ that contains $\ker{f}$.
	\item If $f$ is an epimorphism,
	then $f(I)$ is an ideal in $S$.
	If $f$ is not surjective,
	$f(I)$ need not be an ideal in $S$.
	\end{enumerate}
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	If $P$ is an ideal in a not necessarily commutative ring $R$,
	then the following conditions are equivalent.
	\begin{enumerate}
	\item $P$ is a prime ideal.
	\item If $r, s\in R$ are such that $rRs\subseteq P$,
	then $r\in P$ or $s\in P$.
	[{\it Hint:} If (a) holds and $rRs\subseteq P$,
	then $(RrR)(RsR)\subseteq P$,
	whence $RrR\subseteq P$ or $RsR\subseteq P$,
	say $RrR\subseteq P$.
	If $A=\langle r\rangle$,
	then $A^3\subseteq RrR\subseteq P$,
	whence $r\in A\subseteq P$.]
	\item If $\langle r\rangle$ and $\langle s\rangle$ are principal ideals of $R$ such that $\langle r\rangle \langle s\rangle\subseteq P$,
	then $r\in P$ or $s\in P$.
	\item If $U$ and $V$ are right ideals in $R$ such that $UV\subseteq P$,
	then $U\subseteq P$ or $V\subseteq P$.
	\item If $U$ and $V$ are left ideals in $R$ such that $UV\subseteq P$,
	then $U\subseteq P$ or $V\subseteq P$.
	\end{enumerate}
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	The set consisting of zero and all zero divisors in a commutative ring with identity contains at least one prime ideal.
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	Let $R$ be a commutative ring with identity and suppose that the ideal $A$ of $R$ is contained in a finite union of prime ideals $P_1\cup \cdots \cup P_n$.
	Show that $A\subseteq P_i$ for some $i$.
	[{\it Hint:} otherwise one may assume that $A\cap P_j\not\subseteq \bigcup_{i\neq j}P_i$ for all $j$.
	Let $a_j\in (A\cap P_j)-(\bigcup_{i\neq j}P_i)$.
	Then $a_1+a_2 a_2\cdots a_n$ is in $A$ but not in $P_1\cup \cdots \cup P_n$.]
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	Let $f:R\to S$ be an epimorphism of rings with kernel $K$.
	\begin{enumerate}
	\item If $P$ is a prime idal in $R$ that contains $K$,
	then $f(P)$ is a prime ideal in $S$ [see Exercise {III.2.13}].
	\item If $Q$ is a prime ideal in $S$,
	then $f^{-1}(Q)$ is a prime ideal in $R$ that contains $K$.
	\item There is a one-to-one correspondence between the set of all prime ideals in $R$
	that contains $K$ and the set of al prime ideals in $S$,
	given by $P\mapsto f(P)$.
	\item If $I$ is an ideal in a ring $R$,
	then every prime ideal in $R/I$ is of the form $P/I$,
	where $P$ is a prime ideal in $R$ that contains $I$.
	\end{enumerate}
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	An ideal $M\neq R$ in a commutative ring $R$ with identity is maximal if and only if for every $r\in R-M$,
	there exists $x\in R$ such that $1_R-rx\in M$.
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	The ring $E$ of even integers contains a maximal ideal $M$ such that $E/M$ is {\it not} a field.
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	In the ring $\mathbb{Z}$ the following conditions on a nonzero ideal $I$ are equivalent:
	\begin{enumerate}[(i)]
	\item $I$ is prime;
	\item $I$ is maximal;
	\item $I=\langle p\rangle$ with $p$ prime.
	\end{enumerate}
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	Determine all prime and maximal ideals in the ring $\mathbb{Z}_m$.
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	\begin{enumerate}
	\item If $R_1, ..., R_n$ are rings with identity and $I$ is an ideal in $R_1\times \cdots \times R_n$,
	then $I=A_1\times \cdots \times A_m$,
	where each $A_i$ is an ideal in $R_i$.
	[{\it Hint:} Given $I$ let $A_k=\pi_k(I)$,
	where $\pi_k:R_1\times \cdots R_n\to R_k$ is the canonical epimorphism.]
	\item Show that the conclusion of (a) need not hold if the rings $R_i$ do not have identities.
	\end{enumerate}
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	An element $e$ in a ring $R$ is said to be idempotent if $e^2=e$.
	An element of the center of the ring $R$ is said to be central.
	If $e$ is a central idempotent in a ring $R$ with identity,
	then
	\begin{enumerate}
	\item $1_R-e$ is a central idempotent;
	\item $eR$ and $(1_R-e)R$ are ideals in $R$ such that $R=eR\times (1_R-e)R$.
	\end{enumerate}
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	Idempotent elements $e_1, ..., e_n$ in a ring $R$ [see Exercise {III.2.23}] are said to be orthogonal if $e_i e_j=0$ for $i\neq j$.
	If $R, R_1, ..., R_n$ are rings with identity,
	then the following conditions are equivalent:\
	\begin{enumerate}
	\item $R\cong R_1\times \cdots R_n$.
	\item $R$ contains a set of orthogoanl central idempotents [Exercise {III.2.23}] $\{e_1, ..., e_n\}$ such that $e_1+e_2+\cdots +e_n=1_R$ and $e_i R\cong R_i$ for each $i$.
	\item $R$ is the internal direct product $R=A_1\times \cdots \times A_n$ where each $A_i$ is an ideal of $R$ such that $A_i\cong R_i$.
	[{\it Hint:} (a)$\$ightarrow$(b) The elements $\overline{e}_1=(1_{R_1}, 0, ..., 0), \overline{e}_2=(0, 1_{R_2}, 0, ..., 0), ..., \overline{e}_n=(0, ..., 0, 1_{R_n})$ are orthogonal central idempotents in $S=R_1\times \cdots \times R_n$ such that $\overline{e}_1+\cdots \overline{e}_n=1_S$ and $\overline{e}_i S\cong R_i$.
	(b)$\Rightarrow$(c) Note that $A_k=e_k R$ is the principal ideal $\langle e_k\rangle$ in $R$ and that $e_k R$ is itself a ring with identity $e_k$.]
	\end{enumerate}
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	If $m\in \mathbb{Z}$ has a prime decomposition $m=p_1^{k_1}\cdots p_t^{k_t}$ ($k_i>0$; $p_i$ distinct primes),
	then there is an isomorphism of rings $\mathbb{Z}_m\cong \mathbb{Z}_{p_1^{k_1}}\times \cdots \times \mathbb{Z}_{p_t^{k_t}}$.
	[{\it Hint:} Corollary {III.2.27}.]
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	If $R=\mathbb{Z}, A_1=\langle 6\rangle$ and $A_2=\langle 4\rangle$,
	then the map $\theta:R/A_1\cap A_2\mapsto R/A_1\times R/A_2$ of Corollary {III.2.27} is not surjective.
	\end{enumerate}
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\section{Factorization in Commutative Rings}
	\begin{enumerate}
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	A nonzero ideal in a principal ideal domain is maximal if and only if it is prime.
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	An integral domain $R$ is a unique factorization domain if and only if every nonzero prime ideal in $R$ contains a nonzero principal ideal that is prime.
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	Let $R$ be the subring $\{a+b\sqrt{10}\mid a, b\in \mathbb{Z}\}$ of the field of real numbers.
	\begin{enumerate}
	\item The map $N:R\to \mathbb{Z}$ given by $a+b\sqrt{10}\mapsto (a+b\sqrt{10})(a-b\sqrt{10})=a^2-10b^2$ is such that that $N(uv)=N(u)N(v)$ for all $u, v\in R$ and $N(u)=0$ if and only if $u=0$.
	\item $u$ is a unit in $R$ if and only if $N(u)=\pm 1$.
	\item $2, 3, 4+\sqrt{10}$ and $4-\sqrt{10}$ are irreducible elements of $R$.
	\item $2, 3, 4+\sqrt{10}$ and $4-\sqrt{10}$ are not prime elements of $R$.
	[{\it Hint:} $3\cdot 2=6=(4+\sqrt{10})(4-\sqrt{10})$.]
	\end{enumerate}
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	Show that in the integral domain of Exercise III.3.3 every element can be factored into a product of irreducibles,
	but this factorization need not be unique (in the sense of Definition III.3.5 (ii)).
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	Let $R$ be a principal ideal domain.
	\begin{enumerate}
	\item Every proper ideal is product $P_1 P_2\cdots P_n$ of maximal ideals,
	which are uniquely determined up to order.
	\item An ideal $P$ in $R$ is said to be primary if $ab\in P$ and $a\notin P$ imply $b^n\in P$ for some $n$.
	Show that $P$ is primary if and only if for some $n$,
	$P=\langle p^n\rangle$,
	where $p\in R$ is prime ($=$ irreducible) or $p=0$.
	\item If $P_1, P_2, ..., P_n$ are primary ideals such that $P_i=\langle p_i^{n_i}\rangle$ and the $p_i$ are distinct primes,
	then $P_1 P_2\cdots P_n=P_1\cap P_2\cap \cdots \cap P_n$.
	\item Every proper ideal in $R$ can be expressed (uniquely up to order) as the intersection of a finite number of primary ideals.
	\end{enumerate}
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	\begin{enumerate}
	\item If $a$ and $n$ are integers,
	$n>0$,
	then there exist integers $q$ and $r$ such that $a=qn+r$,
	where $|r|\leq n/2$.
	\item The Gaussian integers $\mathbb{Z}[i]$ form a Euclidean domain with $\varphi(a+bi)=a^2+b^2$.
	[{\it Hint:} to show that Definition III.3.8(ii) holds,
	first let $y=a+bi$ and assume $x$ is a positive integer.
	By part (a) there are integers such that $a=q_1 x+r_1$ and $b=q_2 x+r_2$ with $|r_1|\leq x/2, |r_2|\leq x/2$.
	Let $q=q_1+q_2 i$ and $r=r_1+r_2 i$;
	then $y=qx+r$,
	with $r=0$ or $\varphi(r)<\varphi(x)$.
	In the general case,
	observe that for $x=c+di\neq 0$ and $\overline{x}=c-di$,
	$x\overline{x}>0$.
	There are $q, r_0\in \mathbb{Z}[i]$ such that $y\overline{x}=q(x\overline{x})+r_0$,
	with $r_0=0$ or $\varphi(r_0)<\varphi(x\overline{x})$.
	Let $r=y-qx$;
	then $y=qx+r$ and $r=0$ or $\varphi(r)<\varphi(x)$.]
	\end{enumerate}
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	What are the units in the ring of Gaussian integers $\mathbb{Z}[i]$?
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	Let $R$ be the following subring of the complex numbers:
	$R=\{a+b(1+\sqrt{19}i)/2\mid a, b\in \mathbb{Z}\}$.
	Then $R$ is a principal ideal domain that is not a Euclidean domain.
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	Let $R$ be a unique factorization domain and $d$ a nonzero element of $R$.
	There are only a finite number of distinct principal ideals that contain the ideal $\langle d\rangle$.
	[{\it Hint:} $\langle d\rangle\subseteq \langle k\rangle\Rightarrow k\mid d$.]
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	If $R$ is a unique factorization domain and $a, b\in R$ are relatively prime and $a\mid bc$,
	then $a\mid c$.
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	Let $R$ be a Euclidean ring and $a\in R$.
	Then $a$ is a unit in $R$ if and only if $\varphi(a)=\varphi(1_R)$.
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	Every nonempty set of elements (possibly infinite) in a commutative principal ideal ring with identity has a greatest common divisor.
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	(Euclidean algorithm).
	Let $R$ be a Euclidean domain with associated function $\varphi:R-\{0\}\to \mathbb{N}$.
	If $a, b\in R$ and $b\neq 0$,
	here is a method for finding the greatest common divisor of $a$ and $b$.
	By repeated use of Definition III.3.8(ii) we have
	$$
	\begin{array}{lllll}
	a=q_0 b+r_1, & \text{ with } & r_1=0 & \text{ or } & \varphi(r_1)<\varphi(b); \\
	b=q_1 r_1+r_2, & \text{ with } & r_2=0 & \text{ or } & \varphi(r_2)<\varphi(r_1); \\
	r_1=q_2 r_2+r_3, & \text{ with } & r_3=0 & \text{ or } & \varphi(r_3)<\varphi(r_2); \\
	\vdots \\
	r_k=q_{k+1} r_{k+1}+r_{k+2}, & \text{ with } & r_{k+2}=0 & \text{ or } & \varphi(r_{k+2})<\varphi(r_{k+1}); \\
	\vdots
	\end{array}
	$$
	Let $r_0=b$ and let $n$ be the least integer such that $r_{n+1}=0$ (such an $n$ exists since the $\varphi(r_k)$ form a strictly decreasing sequence of nonnegative integers).
	Show that $r_n$ is the greatest common divisor $a$ and $b$.
	\end{enumerate}
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\section{Rings of Quotients and Localization}
	\begin{enumerate}
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	Determine the complete ring of quotients of the ring $\mathbb{Z}_n$ for each $n\geq 2$.
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	Let $S$ be a multiplicative subset of a commutative ring $R$ with identity and let $T$ be a multiplicative subset of the ring $S^{-1}R$.
	Let $S_*=\{r\in R\mid r/s\in T\text{ for some }s\in S\}$.
	Then $S_*$ is a multiplicative subset of $R$ and there is a ring isomorphism ${S_*}^{-1}R\cong T^{-1}(S^{-1}R)$.
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	\begin{enumerate}
	\item The set $E$ of positive even integers is a multiplicative subset of $\mathbb{Z}$ such that $E^{-1}(\mathbb{Z})$ is the field of rational numbers.
	\item State and prove condition(s) on a multiplicative subset $S$ of $\mathbb{Z}$ which insure that $S^{-1}\mathbb{Z}$ is the field of rationals.
	\end{enumerate}
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	If $S=\{2, 4\}$ and $R=\mathbb{Z}_6$,
	then $S^{-1}R$ is isomorphic to the field $\mathbb{Z}_3$.
	Consequently,
	the converse of Theorem {III.4.3}(ii) is false.
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	Let $R$ be an integral domain with quotient field $F$.
	If $T$ is an integral domain such that $R\subseteq T\subseteq F$,
	then $F$ is (isomorphic to) the quotient field of $T$.
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	Let $S$ be a multiplicative subset of an integral domain $R$ such that $0\notin S$.
	If $R$ is a principal ideal domain [resp. unique factorization domain],
	then so is $S^{-1}R$.
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	Let $R_1$ and $R_2$ be integral domains with quotient fields $F_1$ and $F_2$ respectively.
	If $f:R_1\to R_2$ is an isomorphism,
	then $f$ extends to an isomorphism $F_1\cong F_2$.
	[{\it Hint:} Corollary {III.4.6}.]
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	Let $R$ be a commutative ring with identity,
	$I$ an ideal of $R$ an $\pi:R\to R/I$ the canonical projection.
	\begin{enumerate}
	\item If $S$ is a multiplicative subset of $R$,
	then $\pi S=\pi(S)$ is a multiplicative subset of $R/I$.
	\item The mapping $\theta:S^{-1}R\to (\pi S)^{-1}(R/I)$ given by $r/s\mapsto \pi(r)/\pi(s)$ is a well-defined function.
	\item $\theta$ is a ring epimorphism with kernel $S^{-1}I$ and hence induces a ring isomorphism $S^{-1}R/S^{-1}I\cong (\pi S)^{-1}(R/I)$.
	\end{enumerate}
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	Let $S$ be a multiplicative subset of a commutative ring $R$ with identity.
	If $I$ is an ideal in $R$,
	then $S^{-1}(\text{Rad }I)=\text{Rad }(S^{-1}I)$.
	[See Exercise {III.2.2}.]
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	Let $R$ be an integral domain and for each maximal ideal $M$ (which is also prime,
	of course),
	consider $R_M$ as a subring of the quotient field of $R$.
	Show that $\cap R_M=R$,
	where the intersection is taken over all maximal ideals $M$ of $R$.
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	Let $p$ be a prime in $\mathbb{Z}$;
	then $\langle p\rangle$ is a prime ideal.
	What can be said about the relationship of $\mathbb{Z}_p$ and the localization $\mathbb{Z}_{\langle p\rangle}$?
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	A commutative ring with identity is local if and only if for all $r, s\in R$,
	$r+s=1_R$ implies $r$ or $s$ is a unit.
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	The ring $R$ consisting of all rational numbers with denominators not divisible by some (fixed) prime $p$ is a local ring.
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	If $M$ is a maximal ideal in a commutative ring $R$ with identity and $n$ is a positive integer,
	then the ring $R/M^n$ has a unique prime ideal and therefore is local.
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	In a commutative ring $R$ with identity the following conditions are equivalent:
	\begin{enumerate}[(i)]
	\item $R$ has a unique prime ideal;
	\item every nonunit is nilpotent (see Exercise {III.1.12});
	\item $R$ has a minimal prime ideal which contains all zero divisors,
	and all nonunits of $R$ are zero divisors.
	\end{enumerate}
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	Every nonzero homomorphic image of a local ring is local.
	\end{enumerate}
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\section{Rings of Polynomials and Formal Power Series}
	\begin{enumerate}
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	\begin{enumerate}
	\item If $\varphi:R\to S$ is a homomorphism of rings,
	then the map $\overline{\varphi}:R[[x]]\to S[[x]]$ given by $\overline{\varphi}(\sum a_i x^i)=\sum \varphi(a_i)x^i$ is a homomorphism of rings such that $\overline{\varphi}(R[x])\subseteq S[x]$.
	\item $\overline{\varphi}$ is a monomorphism [epimorphism] if and only if $\varphi$ is.
	In this case $\overline{\varphi}:R[x]\to S[x]$ is also a monomorphism [epimorphism].
	\item Extend the results of (a) and (b) to the polynomial rings $R[x_1, ..., x_n], S[x_1, ..., x_n]$.
	\end{enumerate}
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	Let $\text{Mat}_n R$ be the ring of $n\times n$ matrices over a ring $R$.
	Then for each $n\geq 1$:
	\begin{enumerate}
	\item $(\text{Mat}_n R)[x]\cong \text{Mat}_n R[x]$.
	\item $(\text{Mat}_n R)[[x]]\cong \text{Mat}_n R[[x]]$.
	\end{enumerate}
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	Let $R$ be a ring and $G$ an infinite multiplicative cyclic group with generator denoted $x$.
	Is the group ring $R(G)$ (see page 117) isomorphic to the polynomial ring in one indeterminate over $R$?
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	\begin{enumerate}
	\item Let $S$ be a nonempty set and let $\mathbb{N}^S$ be the set of all functions $\varphi:S\to \mathbb{N}$ such that $\varphi(s)\neq 0$ for at most a finite number of elements $s\in S$.
	Then $\mathbb{N}^S$ is a multiplicative abelian monoid with product defined by
	$$(\varphi \psi)(s)=\varphi(s)+\psi(s)~(\varphi, \psi\in \mathbb{N}^S; s\in S).$$
	The identity element in $\mathbb{N}^S$ is the zero function.
	\item
	For each $x\in S$ and $i\in \mathbb{N}$ let $x^i\in \mathbb{N}^S$ be defined by $x^i(x)=i$ and $x^i(s)=0$ for $s\neq x$.
	If $\varphi\in \mathbb{N}^S$ and $x_1, ..., x_n$ are the only elements of $S$ such that $\varphi(x_i)\neq 0$,
	then in $\mathbb{N}^S$,
	$\varphi=x_1^{i_1}x_2^{i_2}\cdots x_n^{i_n}$,
	where $i_j=\varphi(x_j)$.
	\item If $R$ is a ring with identity let $R[S]$ be the set of all functions $f:\mathbb{N}^S\to R$ such that $f(\varphi)\neq 0$ for at most a finite number of $\varphi\in \mathbb{N}^S$.
	Then $R[S]$ is a ring with identity,
	where addition and multiplication are defined as follows:
	\begin{eqnarray*}
	(f+g)(\varphi) &=& f(\varphi)+g(\varphi)~(f, g\in R[S]; \varphi\in \mathbb{N}^S);\\
	(fg)(\varphi) &=& \sum f(\theta)g(\zeta)~(f, g\in R[S]; \theta, \zeta, \varphi\in \mathbb{N}^S),
	\end{eqnarray*}
	where the sum is over all pairs $(\theta, \zeta)$ such that $\theta \zeta=\varphi$.
	$R[S]$ is called the ring of polynomials in $S$ over $R$.
	\item
	For each $\varphi=x_1^{i_1}\cdots x_n^{i_n}\in \mathbb{N}^S$ and each $r\in R$ we denote by $rx_1^{i_1}\cdots x_n^{i_n}$ the function $\mathbb{N}^S\to R$ which is $r$ at $\varphi$ and $0$ elsewhere.
	Then every nonzero element $f$ of $R[S]$ can be written in the form $f=\sum_{i=0}^{m}r_i x_1^{k_{i_1}}x_2^{k_{i_2}}\cdots x_n^{k_{i_n}}$ with the $r_i\in R, x_i\in S$ and $k_{i_j}\in \mathbb{N}$ all uniquely determined.
	\item If $S$ is finite of cardinality $n$,
	then $R[S]\cong R[x_1, ..., x_n]$.
	[{\it Hint:} if $\mathbb{N}^n$ is considered as an additive abelian monoid as in the text,
	then there is an isomorphism of monoids $\mathbb{N}^S\cong \mathbb{N}^n$ given by $\varphi\mapsto (\varphi(s_1), ..., \varphi(s_n))$,
	where $S=\{s_1, ..., s_n\}$.]
	\item State and prove n analogue of Theorem {III.5.5} for $R[S]$.
	\end{enumerate}
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	Let $R$ and $S$ be rings with identity,
	$\varphi:R\to S$ a homomorphism of rings with $\varphi(1_R)=1_S$,
	and $s_1, s_2, ..., s_n\in S$ such that $s_i s_j=s_j s_i$ for all $i, j$ and $\varphi(r)s_i=s_i \varphi(r)$ for all $r\in R$ and all $i$.
	Then there is a unique homomorphism $\overline{\varphi}:R[x_1, ..., x_n]\to S$ such that $\overline{\varphi}|_{R}=\varphi$ and $\overline{\varphi}(x_i)=s_i$.
	This property completely determines $R[x_1, ..., x_n]$ up to isomorphism.
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	\begin{enumerate}
	\item If $R$ is the ring of all $2\times 2$ matrices over $\mathbb{Z}$,
	then for any $A\in R$,
	$$(x+A)(x-A)=x^2-A^2\in R[x].$$
	\item There exist $C, A\in R$ such that $(C+A)(C-A)\neq C^2-A^2$.
	Therefore, Corollary {III.5.6} is alse if the rings involved are not commutative.
	\end{enumerate}
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	If $R$ is a commutative ring with identity and $f=a_n x^n+\cdots +a_0$ is a zero divisor in $R[x]$,
	then there exists a nonzero $b\in R$ such that $ba_n=ba_{n-1}=\cdots =ba_0=0$.
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	\begin{enumerate}
	\item The polynomial $x+1$ is a unit in the power series ring $\mathbb{Z}[[x]]$,
	but is not a unit in $\mathbb{Z}[x]$.
	\item $x^2+3x+2$ is irreducible in $\mathbb{Z}[[x]]$,
	but not in $\mathbb{Z}[[x]]$.
	\end{enumerate}
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	If $F$ is a field,
	then $\langle x\rangle$ is a maximal ideal in $F[x]$,
	but it is not the only maximal ideal (compare Corollary {III.5.10}).
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	\begin{enumerate}
	\item If $F$ is a field then every nonzero element of $F[[x]]$ is of the form $x^k u$ with $u\in F[[x]]$ a unit.
	\item $F[[x]]$ is a principal ideal domain whose only ideals are $0, F[[x]]=\langle 1_F\rangle=\langle x^0\rangle$ and $\langle x^k\rangle$ for each $k\geq 1$.
	\end{enumerate}
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	Let $\mathcal{C}$ be the category with objects all commutative rings with identity and morphisms all ring homomorphisms $f:R\to S$ such that $f(1_R)=1_S$.
	Then the poynomial ring $\mathbb{Z}[x_1, ..., x_n]$ is a free object on the set $\{x_1, ..., x_n\}$ in the category $\mathcal{C}$.
	[{\it Hint:} for any $R$ in $\mathcal{C}$ the map $\mathbb{Z}\to R$ given by $n\mapsto n1_R$ is a ring homomorphism;
	use Theorem {III.5.5}.]
	\end{enumerate}
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\section{Factorization in Polynomial Rings}
	\begin{enumerate}
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	\begin{enumerate}
	\item If $D$ is an integral domain and $c$ is an irreducible element in $D$,
	then $D[x]$ is not a principal ideal domain.
	[{\it Hint:} consider the ideal $\langle x, c\rangle$ generated by $x$ and $c$.]
	\item $\mathbb{Z}[x]$ is not a principal ideal domain.
	\item If $F$ is a field and $n\geq 2$,
	then $F[x_1, ..., x_n]$ is not a principal ideal domain.
	[{\it Hint:} show that $x_1$ is irreducible in $F[x_1, ..., x_{n-1}]$.]
	\end{enumerate}
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	If $F$ is a field and $f, g\in F[x]$ with $\deg{g}\geq 1$,
	then there exist unique polynomials $f_0, f_1, ..., f_r\in F[x]$ such that $\deg{f_i}<\deg{g}$ for all $i$ and
	$$f=f_0+f_1 g+f_2 g^2+\cdots +f_r g^r.$$
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	Let $f$ be a polynomial of positive degree over an integral $D$.
	\begin{enumerate}
	\item If $\text{char }D=0$,
	then $f'\neq 0$.
	\item If $\text{char }D=p\neq 0$,
	then $f'=0$ if and only if $f$ is a polynomial in $x^p$ (that is,
	$f=a_0+a_p x^p+a_{2p}x^{2p}+\cdots +a_{jp}x^{jp}$).
	\end{enumerate}
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	If $D$ is a unique factorization domain,
	$a\in D$ and $f\in D[x]$,
	then $C(af)$ and $aC(f)$ are associates in $D$.
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	Let $R$ be a commutative ring with identity and $f=\sum_{i=0}^n a_i x^i\in R[x]$.
	Then $f$ is a unit in $R[x]$ if and only if $a_0$ is a unit in $R$ and $a_1, ..., a_n$ are nilpotent elements of $R$ (Exercise {III.1.12}).
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item{}
	[Probably impossible with the tools at hand.]
	Let $p\in \mathbb{Z}$ be a prime;
	let $F$ be a field and let $c\in F$.
	Then $x^p-c$ is irreducible in $F[x]$ if and only if $x^p-c$ has no root in $F$.
	[{\it Hint:} consider two cases: $\text{char }F=p$ and $\text{char }F\neq p$.]
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	If $f=\sum a_i x^i\in \mathbb{Z}[x]$ and $p$ is prime,
	let $\overline{f}=\sum \overline{a}_i x^i\in \mathbb{Z}_p[x]$,
	where $\overline{a}$ is the image of $a$ under the canonical epimorphism $\mathbb{Z}\to \mathbb{Z}_p$.
	\begin{enumerate}
	\item If $f$ is monic and $\overline{f}$ is irreducible in $\mathbb{Z}_p[x]$ for some prime $p$,
	then $f$ is irreducible in $\mathbb{Z}[x]$.
	\item Give an example to show that (a) may be false if $f$ is not minic.
	\item Extend (a) to polynomials over a unique factorization domain.
	\end{enumerate}
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item{}
	[Probably impossible with the tools at hand.]
	\begin{enumerate}
	\item Let $c\in F$,
	where $F$ is a field of characteristic $p$ ($p$ prime).
	Then $x^p-x-c$ is irreducible in $F[x]$ if and only if $x^p-x-c$ has no root in $F$.
	\item If $\text{char }F=0$,
	part (a) is false.
	\end{enumerate}
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	Let $f=\sum_{i=0}a_i x^o\in \mathbb{Z}[x]$ have degree $n$.
	Suppose that for some $k$ ($0<k<n$) and some prime $p$:
	$p\nmid a_n$;
	$p\nmid a_k$;
	$p\mid a_i$ for all $0\leq i\leq k-1$;
	and $p^2\nmid a_0$.
	Show that $f$ has a factor $g$ of degree at least $k$ that is irreducible in $\mathbb{Z}[x]$.
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	\begin{enumerate}
	\item Let $D$ be an integral domain and $c\in D$.
	Let $f(x)=\sum_{i=0}^n a_i x^i\in D[x]$ and $f(x-c)=\sum_{i=0}^n a_i(x-c)^i\in D[x]$.
	Then $f(x)$ is irreducible in $D[x]$ if and only if $f(x-c)$ is irreducible.
	\item For each prime $p$,
	the cyclotomic polynomial $f=x^{p-1}+x^{p-2}+\cdots +x+1$ is irreducible in $\mathbb{Z}[x]$.
	[{\it Hint:} observe that $f=(x^p-1)/(x-1)$,
	whence $f(x+1)=((x+1)^p-1)/x$.
	Use the Binomial Theorem {III.1.6} and Eisenstein's Criterion to show that $f(x+1)$ is irreducible in $\mathbb{Z}[x]$.]
	\end{enumerate}
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	If $c_0, c_1, ..., c_n$ are distinct elements of an integral domain $D$ and $d_0, ..., d_n$ are any elements of $D$,
	then there is at most one polynomial $f$ of degree $\leq n$ in $D[x]$ such that $f(c_i)=d_i$ for $i=0, 1, ..., n$.
	[For the existence of $f$,
	see Exercise {III.6.12}].
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	{\it Lagrange's Interpolation Formula}.
	If $F$ is a field,
	$a_0, a_1, ..., a_n$ are distinct elements of $F$ and $c_0, c_1, ..., c_n$ are any elements of $F$,
	then $$f(x)=\sum_{i=0}^n \frac{(x-a_0)\cdots (x-a_{i-1})(x-a_{i+1})\cdots (x-a_n)}{(a_i-a_0)\cdots (a_i-a_{i-1})(a_i-a_{i+1})\cdots (a_i-a_n)}c_i$$
	is the unique polynomial of degree $\leq n$ in $F[x]$ such that $f(a_i)=c_i$ for all $i$ [see Exercise {III.6.11}].
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	Let $D$ be a unique factorization domain with a finite number of units and quotient field $F$.
	If $f\in D[x]$ has degree $n$ and $c_0, c_1, ..., c_n$ are $n+1$ distinct elements of $D$,
	then $f$ is completely determined by $f(c_0), f(c_1), ..., f(c_n)$ according to Exercise {III.6.11}.
	Here is Kronecker's Method for finding all the irreducible factors of $f$ in $D[x]$.
	\begin{enumerate}
	\item It suffices to find only those factors $g$ of degree at most $n/2$.
	\item If $g$ is a factor of $f$,
	then $g(c)$ is a factor of $f(c)$ for all $c\in D$.
	\item Let $m$ be the largest integer $\leq n/2$ and choose distinct elements $c_0, c_1, ..., c_m\in D$.
	Choose $d_0, d_1, ..., d_m\in D$ such that $d_i$ is a factor of $f(c_i)$ in $D$ for all $i$.
	Use Exercise {III.6.12} to construct a polynomial $g\in F[x]$ such that $g(c_i)=d_i$ for all $i$;
	it is unique by Exercise {III.6.11}.
	\item Check to see if the polynomial $g$ of part (c) is a factor of $f$ in $F[x]$.
	If not, make a new choice of $d_0, ..., d_m$ and repeat part (c).
	(Since $D$ is a unique factorization domain with only finitely many units there are only a finite number of possible choices for $d_0, ..., d_m$.)
	If $g$ is a factor of $f$,
	say $f=gh$,
	then repeat the entire process on $g$ and $h$.
	\item After a finite number of steps,
	all the (irreducible) factors of $f$ in $F[x]$ will have been found.
	If $g\in F[x]$ is such a factor (of positive degree) then choose $r\in D$ such that $rg\in D[x]$ (for example, let $r$ be the product of the denominators of the coefficients of $g$).
	Then $r^{-1}(rg)$ and hence $rg$ is a factor of $f$.
	Then $rg=C(rg)g_1$ with $g_1\in D[x]$ primitive and irreducible in $F[x]$.
	By Lemma {III.6.13},
	$g_1$ is an irreducible factor of $f$ in $D[x]$.
	Proceed in this manner to obtain all the nonconstant irreducible factors of $f$;
	the constants are then easily found.
	\end{enumerate}
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	Let $R$ be a commutative ring with identity and $c, b\in R$ with $c$ a unit.
	\begin{enumerate}
	\item Show that the assignment $x\mapsto cx+b$ induces a unique automorphism of $R[x]$ that is the identity of $R$.
	What is its inverse?
	\item If $D$ is an integral domain,
	then show that every automorphism of $D[x]$ that is the identity on $D$ is of the type described in (a).
	\end{enumerate}
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	If $F$ is a field,
	then $x$ and $y$ are relatively prime in the polynomial domain $F[x, y]$,
	but $F[x, y]=\langle 1_F\rangle \supsetneq \langle x\rangle +\langle y\rangle$
	[compare Theorem {III.3.11} (i)].
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	Let $f=a_n x^n+\cdots +a_0$ be a polynomial over the field $\mathbb{R}$ of real numbers and let $\varphi=|a_n|x^n+\cdots +|a_0|\in \mathbb{R}[x]$.
	\begin{enumerate}
	\item If $|u|\leq d$,
	then $|f(u)|\leq \varphi(d)$.
	[Recall that $|a+b|\leq |a|+|b|$ and that $|a|\leq a', |b|\leq b'\Rightarrow |ab|\leq a'b'$.]
	\item Given $a, c\in \mathbb{R}$ with $c>0$ there exists $M\in \mathbb{R}$ such that $|f(a+h)-f(a)|\leq M |h|$ for all $h\in \mathbb{R}$ with $|h|\leq c$.
	[{\it Hint:} use part (a).]
	\item (Intermediate Value Theorem)
	If $a<b$ and $f(a)<d<f(b)$,
	then there exists $c\in \mathbb{R}$ such that $a<c<b$ and $f(c)=d$.
	[{\it Hint:} Let $c$ be the least upper bound of $S=\{x\mid a<x<b\text{ and }f(x)\leq d\}$.
	Use part (b).]
	\item Every polynomial $g$ of odd degree in $\mathbb{R}[x]$ has a real root.
	[{\it Hint:} for suitable $a, b\in \mathbb{R}$, $g(a)<0$ and $g(b)>0$;
	use part (c).]
	\end{enumerate}
	\end{enumerate}
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\chapter{Modules}
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\section{Modules, Homomorphisms and Exact Sequences}
	{\it Note:} $R$ is a ring.
	\begin{enumerate}
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	If $A$ is an abelian group and $n>0$ an integer such that $na=0$ for all $a\in A$,
	then $A$ is a unitary $\mathbb{Z}_n$-module,
	with the action of $\mathbb{Z}_n$ on $A$ given by $\overline{k}a=ka$,
	where $k\in \mathbb{Z}$ and $k\mapsto \overline{k}\in \mathbb{Z}_n$ under the canonical projection $\mathbb{Z}\to \mathbb{Z}_n$.
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	Let $f:A\to B$ be an $R$-module homomorphism.
	\begin{enumerate}
	%%%%%
	\item $f$ is a monomorphism if and only if for every pair of $R$-module homomorphisms $g, h:D\to A$ such that $fg=fh$,
	we have $g=h$.
	[{\it Hint:} to prove ($\Leftarrow$),
	let $D=\ker{f}$,
	with $g$ the inclusion map and $h$ the zero map.]
	%%%%%
	\item $f$ is an epimorphism if and only if for every pair of $R$-module homomorphisms $k, t:B\to C$ such that $kf=tf$,
	we have $k=t$.
	[{\it Hint:} to prove ($\Leftarrow$),
	let $k$ be the canonical epimorphism $B\to B/\text{Im }f$ and $t$ the zero map.]
	\end{enumerate}
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	Let $I$ be a left ideal of a ring $R$ and $A$ an $R$-module.
	\begin{enumerate}
	\item If $S$ is a nonempty subset of $A$,
	then $IS=\{\sum_{i=1}^{n} r_i a_i \mid n\in \mathbb{N}^*; r_i\in I; a_i\in S\}$ is a submodule of $A$.
	Note that if $S=\{a\}$,
	then $IS=Ia=\{ra\mid r\in I\}$.
	\item
	If $I$ is a two-sided ideal,
	then $A/IA$ is an $R/I$-module with the action of $R/I$ given by $(r+I)(a+IA)=ra+IA$.
	\end{enumerate}
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	If $R$ has an identity,
	then every unitary cyclic $R$-module is isomorphic to an $R$-module of the form $R/J$,
	where $J$ is a left ideal of $R$.
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	If $R$ has an identity,
	then a nonzero unitary $R$-module $A$ is simple if its only submodules are $0$ and $A$.
	\begin{enumerate}
	\item Every simple $R$-module is cyclic.
	\item If $A$ is simple every $R$-module endomorphism is either the zero map or an isomorphism.
	\end{enumerate}
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	A finitely generated $R$-module need not be finitely genreated as an abelian group.
	[{\it Hint:} Exercise {II.1.10}.]
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	\begin{enumerate}
	\item If $A$ and $B$ are $R$-modules,
	then the set $\text{Hom}_{R}(A, B)$ of all $R$-module homomorphisms $A\to B$ is an abelian group with $f+g$ given on $a\in A$ by $(f+g)(a)=f(a)+g(a)\in B$.
	The identity element is the zero map.
	\item $\text{Hom}_R(A, A)$ is a ring with identity,
	where multiplication is composition of functions.
	$\text{Hom}_R(A, A)$ is called the endomorphism ring of $A$.
	\item $A$ is a left $\text{Hom}_R(A, A)$-module with $fa$ defined to be
	$$f(a)~(a\in A, f\in \text{Hom}_R(A, A)).$$
	\end{enumerate}
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	Prove that the obvious analogues of Theorem {I.8.10} and Corollary {I.8.11} are valid for $R$-modules.
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	If $f:A\to A$ is an $R$-module homomorphism such that $ff=f$,
	then $$A=\ker{f}\oplus \text{Im }f.$$
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	Let $A, A_1, ..., A_n$ be $R$-modules.
	Then $A\cong A_1\oplus \cdots \oplus A_n$ if and only if for each $i=1, 2, ..., n$ there is an $R$-module homomorphism $\varphi_i:A\to A$ such that $\text{Im }\varphi_i\cong A_i$;
	$\varphi_i \varphi_j=0$ for $i\neq j$;
	and $\varphi_1+\varphi_2+\cdots +\varphi_n=1_A$.
	[{\it Hint:} If $A\cong A_1\oplus \cdots \oplus A_n$ let $\pi_i, \iota_i$ be as in Theorem {IV.1.14} and define $\varphi_i=\iota_i\pi_i$.
	Conversely,
	given $\{\varphi_i\}$,
	show that $\varphi_i\varphi_i=\varphi_i$.
	Let $\psi_i=\varphi_i|_{\text{Im }\varphi_i}:\text{Im }\varphi_i\to A$ and apply Theorem {IV.1.14} with
	$A, \text{Im }\varphi_i, \varphi_i$, and $\psi_i$ in place of $A, A_i, \pi_i$, and $\iota_i$.
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	\begin{enumerate}
	\item If $A$ is a module over a commutative ring $R$ and $a\in A$,
	then $\mathcal{O}_a=\{r\in R\mid ra=0\}$ is an ideal of $R$.
	If $\mathcal{O}_a\neq 0$,
	$a$ is said to be a torsion element of $A$.
	\item If $R$ is an integral domain,
	then the set $T(A)$ of all torsion elements of $A$ is a submodule of $A$.
	($T(A)$ is called the torsion submodule.)
	\item Show that (b) may be false for a commutative ring $R$,
	which is not an integral domain.
	\end{enumerate}
	In (d) - (f) $R$ is an integral domain.
	\begin{enumerate}[resume]
	\item If $f:A\to B$ is an $R$-module homomorphism,
	then $f(T(A))\subseteq T(B)$;
	hence the restriction $f_T$ of $f$ to $T(A)$ is an $R$-module homomorphism $T(A)\to T(B)$.
	\item If $0\to A\stackrel{f}{\to}B\stackrel{g}{\to}C$ is an exact sequence of $R$-modules,
	then so is $0\to T(A)\stackrel{f_T}{\to}T(B)\stackrel{g_T}{\to}T(C)$.
	\item If $g:B\to C$ is an $R$-module epimorphism,
	then $g_T:T(B)\to T(C)$ need not be an epimorphism.
	[{\it Hint:} consider abelian groups.]
	\end{enumerate}
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	(The Five Lemma).
	Let
	$$\xymatrix{
	A_1 \ar[r]\ar[d]^{\alpha_1} & A_2 \ar[r]\ar[d]^{\alpha_2} & A_3 \ar[r]\ar[d]^{\alpha_3} & A_4 \ar[r]\ar[d]^{\alpha_4} & A_5 \ar[d]^{\alpha_5} \\
	B_1 \ar[r] & B_2 \ar[r] & B_3 \ar[r] & B_4 \ar[r] & B_5
	}$$
	be a commutative diagram of $R$-modules and $R$-module homomorphisms,
	with exact rows.
	Prove that:
	\begin{enumerate}
	\item $\alpha_1$ an epimorphism and $\alpha_2, \alpha_4$ monomorphisms $\Rightarrow$ $\alpha_3$ is a monomorphism;
	\item $\alpha_5$ a monomorphism and $\alpha_2, \alpha_4$ epimorphisms $\Rightarrow$ $\alpha_3$ is an epimorphism.
	\end{enumerate}
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	\begin{enumerate}
	\item If $0\to A\to B\stackrel{f}{\to}C\to 0$ and $0\to C\stackrel{g}{\to}D\to E\to 0$ are short exact sequences of modules,
	then the sequence $0\to A\to B\stackrel{gf}{\to}D\to E\to 0$ is exact.
	\item Show that every exact sequence may be obtained by splicing together suitable short exact sequences as in (a).
	\end{enumerate}
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	Show that isomorphism of short exact sequences is an equivalence relation.
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	If $f:A\to B$ and $g:B\to A$ are $R$-module homomorphisms such that $gf=1_A$,
	then $B=\text{Im }f\oplus \ker{g}$.
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	Let $R$ be a ring and $R^{op}$ its opposite ring (Exercise {III.1.17}).
	If $A$ is a left [resp. right] $R$-module,
	then $A$ is a right [resp. left] $R^{op}$-module such that $ra=ar$ for all $a\in A, r\in R, r\in R^{op}$.
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	\begin{enumerate}
	\item If $R$ has an identity and $A$ is an $R$-module,
	then there are submodules $B$ and $C$ of $A$ such that $B$ is unitary,
	$RC=0$ and $A=B\oplus C$.
	[{\it Hint:} let $B=\{1_R a\mid a\in A\}$ and $C=\{a\in A\mid 1_R a=0\}$ and observe that for {\it all} $a\in A$,
	$a-1_R a\in C$.]
	\item Let $A_1$ be another $R$-module,
	with $A_1=B_1\oplus C_1$ ($B_1$ unitary,
	$RC_1=0$).
	If $f:A\to A_1$ is an $R$-module homomorphism then $f(B)\subseteq B_1$ and $f(C)\subseteq C_1$.
	\item If the map $f$ of part (b) is an epimorphism [resp. isomorphism],
	then so are $f|_B:B\to B_1$ and $f|_C:C\to C_1$.
	\end{enumerate}
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	Let $R$ be a ring without identity.
	Embed $R$ in a ring $S$ with identity and characteristic zero as in the proof of Theorem {III.1.10}.
	Identity $R$ with its image in $S$.
	\begin{enumerate}
	\item Show that every element of $S$ may be uniquely expressed in the form $r1_S+n1_S$ ($r\in R, n\in \mathbb{Z}$).
	\item If $A$ is an $R$-module and $a\in A$,
	show that there is a unique $R$-module homomorphism $f:S\to A$ such that $f(1_S)=a$.
	[{\it Hint:} Let $f(r1_S+n1_S)=ra+na$.]
	\end{enumerate}
	\end{enumerate}
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\section{Free Modules and Vector Spaces}
	\begin{enumerate}
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	\begin{enumerate}
	\item A set of vectors $\{x_1, ..., x_n\}$ in a vector space $V$ over a division ring $R$ is linearly dependent if and only if some $x_k$ is a linear combination of the preceding $x_i$.
	\item If $\{x_1, x_2, x_3\}$ is a linearly independent subset of $V$,
	then the set $\{x_1+x_2, x_2+x_3, x_3+x_1\}$ is linearly independent if and only if $\text{Char }R\neq 2$.
	[See Definition {III.1.8}].
	\end{enumerate}
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	Let $R$ be any ring (possibly without identity) and $X$ a nonempty set.
	In this exercise an $R$-module $F$ is called a free module on $X$ if $F$ is a free object on $X$ in the category of {\it all} left $R$-module.
	Thus by Definition {I.7.7},
	$F$ is the free module on $X$ if there is a function $\iota:X\to F$ such that for any left $R$-module $A$ and function $f:X\to A$ there is a unique $R$-module homomorphism $\overline{f}:F\to A$ with $\overline{f}\iota=f$.
	\begin{enumerate}
	\item
	Let $\{X_i\mid i\in I\}$ be a collection of mutually disjoint sets and for each $i\in I$,
	suppose $F_i$ is a free module on $X_i$,
	with $\iota_i:X_i\to F_i$.
	Let $X=\bigcup_{i\in I}X_i$ and $F=\sum_{i\in I}F_i$,
	with $\phi_i:F_i\to F$ the canonical injection.
	Define $\iota:X\to F$ by $\iota(x)=\phi_i \iota_i(x)$ for $x\in X_i$;
	($\iota$ is well defined since the $X_i$ are disjoint).
	Prove that $F$ is a free module on $X$.
	[{\it Hint:} Theorem {IV.1.13} may be useful.]
	\item
	Assume $R$ has an identity.
	Let the abelian group $\mathbb{Z}$ be given the trivial $R$-module structure ($rm=0$ for all $r\in R, m\in \mathbb{Z}$),
	so that $R\oplus \mathbb{Z}$ is an $R$-module with $r(r', m)=(rr', 0)$ for all $r, r'\in R, m\in \mathbb{Z}$.
	If $X$ is any one element set,
	$X=\{t\}$, let $\iota:X\to R\oplus \mathbb{Z}$ be given by $\iota(t)=(1_R, 1)$.
	Prove that $R\oplus \mathbb{Z}$ is a free module on $X$.
	[{\it Hint:} given $f:X\to A$,
	let $A=B\oplus C$ as in Exercise {IV.1.17},
	so that $f(t)=b+c$ ($b\in B, c\in C$).
	Define $f(r, m)=rb+mc$.]
	\item
	If $R$ is an arbitrary ring and $X$ is any set,
	then there exists a free module on $X$.
	[{\it Hint.} Since $X$ is the disjoint union of the sets $\{t\}$ with $t\in X$,
	it suffices by (a) to assume $X$ has only one element.
	If $R$ has an identity,
	use (b).
	If $R$ has no identity,
	embed $R$ in a ring $S$ with identity and characteristic $0$ as in the proof of Theorem {III.1.10}.
	Use Exercise {IV.1.18} to show that $S$ is a free $R$-module on $X$.]
	\end{enumerate}
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	Let $R$ be any ring (possibly without identity and $F$ a free $R$-module on the set $X$,
	with $\iota:X\to F$,
	as in Exercise {IV.2.2}.
	Show that $\iota(X)$ is a set of generators of the $R$-module $F$.
	[{\it Hint:} let $G$ be the submodule of $F$ generated by $\iota(X)$ and use the definition of ``free module'' to show that there is a module homomorphism $\varphi$ such that
	$$\xymatrix{
	& F \ar[d]^{\varphi} \\
	X \ar[ru]^{\iota} \ar[r]^{\iota} \ar[rd]^{\iota} & G \ar[d]^{\subseteq} \\
	& F
	}$$
	is commutative.
	Conclude that $\varphi=1_F$.]
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	Let $R$ be a principal ideal domain,
	$A$ a unitary left $R$-module,
	and $p\in R$ a prime ($=$ irreducible).
	Let $pA=\{pa\mid a\in A\}$ and $A[p]=\{a\in A\mid pa=0\}$.
	\begin{enumerate}
	\item $R/\langle p\rangle$ is a field (Theorem {III.2.20} and Theorem {III.3.4}).
	\item $pA$ and $A[p]$ are submodules of $A$.
	\item $A/pA$ is a vector space over $R/\langle p\rangle$, with $(r+\langle p\rangle)(a+pA)=ra+pA$.
	\item $A[p]$ is a vector space over $R/\langle p\rangle$, with $(r+\langle p\rangle)a=ra$.
	\end{enumerate}
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	Let $V$ be a vector space over a division ring $D$ and $S$ the set of all subspaces of $V$,
	partially ordered by set theoretic inclusion.
	\begin{enumerate}
	\item $S$ is a complete lattice (see Introduction, Exercise {O.7.2}; the l.u.b. of $V_1, V_2$ is $V_1+V_2$ and the g.l.b. $V_1\cap V_2$).
	\item $S$ is a complemented lattice; that is, for each $V_1\in S$ there exists $V_2\in S$ such that $V=V_1+V_2$ and $V_1\cap V_2=0$, so that $V=V_1\oplus V_2$.
	\item $S$ is a modular lattice; that is, if $V_1, V_2, V_3\in S$ and $V_3\subseteq V_1$, then $$V_1\cap (V_2+V_3)=(V_1\cap V_2)+V_3.$$
	\end{enumerate}
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	Let $\mathbb{R}$ and $\mathbb{C}$ be the fields of real and complex numbers respectively.
	\begin{enumerate}
	\item $\dim_{\mathbb{R}}{\mathbb{C}}=2$ and $\dim_{\mathbb{R}}{\mathbb{R}}=1$.
	\item There is no field $K$ such that $\mathbb{R}\subseteq K\subseteq \mathbb{C}$.
	\end{enumerate}
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	If $G$ is a nontrivial group that is not cyclic of order $2$,
	then $G$ has a nonidentity automorphism.
	[{\it Hint:} Exercise {II.4.11} and Exercise {IV.2.4}(d) above.]
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	If $V$ is a finite dimensional vector space and $V^m$ is the vector space
	$$V\oplus V\oplus \cdots \oplus V ~(m\text{ summands}),$$
	then for each $m\geq 1$, $V^m$ is finite dimensional and $\dim{V^m}=m(\dim{V})$.
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	If $F_1$ and $F_2$ are free modules over a ring with the invariant dimension property,
	then $\text{rank }(F_1\oplus F_2)=\text{rank }F_1+\text{rank }F_2$.
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	Let $R$ be a ring with no zero divisors such that for all $r, s\in R$ there exist $a, b\in R$,
	not both zero,
	with $ar+bs=0$.
	\begin{enumerate}
	\item If $R=K\oplus L$ (module direct sum),
	then $K=0$ or $L=0$.
	\item If $R$ has an identity,
	then $R$ has the invariant dimension property.
	\end{enumerate}
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	Let $F$ be a free module of infinite rank $\alpha$ over a ring $R$ that has the invariant dimension property.
	For each cardinal $\beta$ such that $0\leq \beta\leq \alpha$,
	$F$ has infinitely many proper free submodules of rank $\beta$.
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	If $F$ is a free module over a ring with identity such that $F$ has a basis of finite cardinality $n\geq 1$ and another basis of cardinality $n+1$,
	then $F$ has a basis of cardinality $m$ for every $m\geq n$ ($m\in \mathbb{N}^*$).
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	Let $K$ be a ring with identity and $F$ a free $K$-module with an infinite denumerable basis $\{e_1, e_2, ...\}$.
	Then $R=\text{Hom}_{K}(F, F)$ is a ring by Exercise {IV.1.7}(b).
	If $n$ is any positive integer,
	then the free left $R$-module $R$ has a basis of $n$ elements;
	that is, as an $R$-module,
	$R\cong R\oplus \cdots \oplus R$ for any finite number of summands.
	[{\it Hint:} $\{1_R\}$ is a basis of one element;
	$\{f_1, f_2\}$ is a basis of two elements,
	where $f_1(e_{2n})=e_n, f_1(e_{2n-1})=0, f_2(e_{2n})=0$
	and $f_2(e_{2n-1})=e_n$.
	Note that for any $g\in R$,
	$g=g_1 f_1+g_2 f_2$,
	where $g_1(e_n)=g(e_{2n})$ and $g_2(e_n)=g(e_{2n-1})$.]
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	Let $f:V\to V'$ be a linear transformation of finite dimensional vector space $V$ and $V'$ such that $\dim{V}=\dim{V'}$.
	Then the following conditions are equivalent:
	\begin{enumerate}[(i)]
	\item $f$ is an isomorphism;
	\item $f$ is an epimorphism;
	\item $f$ is a monomorphism.
	\end{enumerate}
	[{\it Hint:} Corollary {IV.2.14}.]
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	Let $R$ be a ring with identity.
	Show that $R$ is {\it not} a free module on any set in the category of {\it all} $R$-modules (as defined in Exercise {IV.2.2}).
	[{\it Hint:} Consider a nonzero abelian group $A$ with the trivial $R$-module structure ($ra=0$ for all $r\in R, a\in A$).
	Observe that the only module homomorphism $R\to A$ is the zero map.]
	\end{enumerate}
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\section{Projective and Injective Modules}
	{\it Note:} $R$ is a ring.
	If $R$ has an identity,
	all $R$-modules are assumed to be unitary.
	\begin{enumerate}
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	The following conditions on a ring $R$ [with identity] are equivalent:
	\begin{enumerate}
	\item Every [unitary] $R$-module is projective.
	\item Every short exact sequence of [unitary] $R$-modules is split exact.
	\item Every [unitary] $R$-module is injective.
	\end{enumerate}
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	Let $R$ be a ring with identity.
	An $R$-module $A$ is injective if and only if for every left ideal $L$ of $R$ and $R$-module homomorphism $g:L\to A$,
	there exists $a\in A$ such that $g(r)=ra$ for every $r\in L$.
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	Every vector space over a division ring $D$ is both a projective and an injective $D$-module.
	[See Exercise {IV.3.1}.]
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	\begin{enumerate}
	\item For each prime $p$, $\mathbb{Z}(p^{\infty})$ (see Exercise {I.1.10}) is a divisible group.
	\item No nonzero finite abelian group is divisible.
	\item No nonzero free abelian group is divisible.
	\item $\mathbb{Q}$ is a divisible abelian group.
	\end{enumerate}
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	$\mathbb{Q}$ is not a projective $\mathbb{Z}$-module.
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	If $G$ is an abelian group,
	then $G=D\oplus N$,
	with $D$ divisible and $N$ reduced (meaning that $N$ has no nontrivial divisible subgroups).
	[{\it Hint:} Let $D$ be the subgroup generated by the set theoretic union of all divisible subgroups of $G$.]
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	Without using Lemma {IV.3.9} prove that:
	\begin{enumerate}
	\item Every homomorphic image of a divisible abelian group is divisible.
	\item Every direct summand (Exercise {I.8.12}) of a divisible abelian group is divisible.
	\item A direct sum of divisible abelian groups is divisible.
	\end{enumerate}
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	Every torsion-free divisible abelian group $D$ is a direct sum of copies of the rationals $\mathbb{Q}$.
	[{\it Hint:} if $0\neq n\in \mathbb{Z}$ and $a\in D$,
	then there exists a unique $b\in D$ such that $nb=a$.
	Denote $b$ by $(1/n)a$.
	For $m, n\in \mathbb{Z}$ ($n\neq 0$),
	define $(m/n)a=m(1/n)a$.
	Then $D$ is a vector space over $\mathbb{Q}$.
	Use Theorem {IV.2.4}.]
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	\begin{enumerate}
	\item If $D$ is an abelian group with torsion subgroup $D_t$,
	then $D/D_t$ is torsion free.
	\item If $D$ is divisible,
	then so is $D_t$,
	whence $D=D_t\oplus E$,
	with $E$ torsion free.
	\end{enumerate}
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	Let $p$ be a prime and $D$ a divisible abelian $p$-group.
	Then $D$ is a direct sum of copies of $\mathbb{Z}(p^{\infty})$.
	[{\it Hint:} let $X$ be a basis of the vector space $D[p]$ over $\mathbb{Z}_p$ (see Exercise {IV.2.4}).
	If $x\in X$,
	then there exists $x_1, x_2, x_3, ...\in D$ such that $x_1=x$, $|x_1|=p$,
	$px_2=x_1, px_3=x_2, ..., px_{n+1}=x_n, ...$.
	If $H_x$ is the subgroup generated by the $x_i$,
	then $H_x\cong \mathbb{Z}(p^{\infty})$ by Exercise {I.3.7}.
	Show that $D\cong \sum_{x\in X}H_x$.]
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	Every divisible abelian group is a direct sum of copies of the rationals $\mathbb{Q}$ and copies of $\mathbb{Z}(p^{\infty})$ for various primes $p$.
	[{\it Hint:} apply Exercise {IV.3.9} to $D$ and Exercise {IV.3.7} and Exercise {IV.3.8} to the torsion-free summand so obtained.
	The other summand $D_t$ is a direct sum of copies of various $\mathbb{Z}(p^{\infty})$ by Exercise {IV.3.7}, Exercise {IV.3.10} and Exercise {II.2.7}.]
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	Let $G, H, K$ be divisible abelian groups.
	\begin{enumerate}
	\item If $G\oplus G\cong H\oplus H$, then $G\cong H$ [see Exercise {IV.3.11}].
	\item If $G\oplus H\cong G\oplus K$, then $H\cong K$ [see Exercise {IV.3.11} and Exercise {II.2.11}.].
	\end{enumerate}
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	If one attempted to dualize the notion of free module over a ring $R$ (and called the object so defined ``co-free'') the definition would read:
	An $R$-module $F$ is co-free on a set $X$ if there exists a function $\iota:F\to X$ such that for any $R$-module $A$ and function $f:A\to X$,
	there exists a unique module homomorphism $\overline{f}:A\to F$ such that $\iota\overline{f}=f$ (see Theorem {IV.2.1}(iv)).
	Show that for any set $X$ with $|X|\geq 2$ no such $R$-module $F$ exists.
	If $|X|=1$,
	then $0$ is the only co-free module.
	[{\it Hint:} If $F$ exists and $|X|\geq 2$,
	arrive at a contradiction by considering possible images of $0$ and constructing $f:R\to X$ such that $\iota\overline{f}\neq f$ for every homomorphism $\overline{f}:R\to F$.]
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	If $D$ is a ring with identity such that every unitary $D$-module is free,
	then $D$ is a division ring.
	[{\it Hint:} it suffices by Exercise {III.2.7} and Theorem {III.2.18} to show that $D$ has no nonzero maximal left ideals.
	Note that every left ideal of $D$ is a free $D$-module and hence a (module) direct summand of $D$ by Theorem {IV.3.2}, Exercise {IV.3.1}, and Proposition {IV.3.13}.]
	\end{enumerate}
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\section{Hom and Duality}
	{\it Note:} $R$ is a ring.
	\begin{enumerate}
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	\begin{enumerate}
	\item For any abelian group $A$ and positive integer $m$,
	$\text{Hom}(\mathbb{Z}_m, A)\cong A[m]=\{a\in A\mid ma=0\}$.
	\item $\text{Hom}(\mathbb{Z}_m, \mathbb{Z}_n)\cong \mathbb{Z}_{\gcd{(m, n)}}$.
	\item The $\mathbb{Z}$-module $\mathbb{Z}_m$ has ${\mathbb{Z}_m}^*=0$.
	\item For each $k\geq 1$, $\mathbb{Z}_m$ is a $\mathbb{Z}_{mk}$-module (Exercise {IV.1.1}); as a $\mathbb{Z}_{mk}$-module, ${\mathbb{Z}_m}^*\cong \mathbb{Z}_m$.
	\end{enumerate}
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	If $A, B$ are abelian groups and $m, n$ integers such that $mA=0=nB$,
	then every element of $\text{Hom}(A, B)$ has order dividing $\gcd{(m, n)}$.
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	Let $\pi:\mathbb{Z}\to \mathbb{Z}_2$ be the canonical epimorphism.
	The induced map $\overline{\pi}:\text{Hom}(\mathbb{Z}_2, \mathbb{Z})\to \text{Hom}(\mathbb{Z}_2, \mathbb{Z}_2)$ is the zero map.
	Since $\text{Hom}(\mathbb{Z}_2, \mathbb{Z}_2)\neq 0$ (Exercise {IV.4.1}(b)),
	$\overline{\pi}$ is not an epimorphism.
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	Let $R, S$ be rings and $A_R, {}_S B_R, {}_S C_R, D_R$ (bi)modules as indicated.
	Let $\text{Hom}_R$ denote all {\it right} $R$-module homomorphisms.
	\begin{enumerate}
	\item $\text{Hom}_R(A, B)$ is a left $S$-module, with the action of $S$ given by $(sf)(a)=s(f(a))$.
	\item If $\varphi:A\to A'$ is an homomorphism of right $R$-modules, then the induced map $\overline{\varphi}:\text{Hom}_R(A', B)\to \text{Hom}_R(A, B)$ is an homomorphism at left $S$-modules.
	\item $\text{Hom}_R(C, D)$ is a irght $S$-module, with the action of $S$ given by $(gs)(c)=g(sc)$.
	\item If $\psi:D\to D'$ is an homomorphism of right $R$-modules, then $\overline{\psi}:\text{Hom}_R(C, D)\to \text{Hom}_R(C, D')$ is an homomorphism of right $S$-modules.
	\end{enumerate}
	Let $\text{Hom}_R$ denote all {\it right} $R$-module homomorphisms.
	$$\begin{array}{llll}
	\text{Hom}_R(A_R, {}_S B_R)\text{ is a} & \text{left}& S\text{-module}, & (sf)(a)=s(f(a)); \\
	\text{Hom}_R({}_S C_R, D_R)\text{ is a} & \text{right}& S\text{-module}, & (gs)(c)=g(sc).
	\end{array}$$
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	Let $R$ be a ring with identity;
	then there is a {\it ring} isomorphism $\text{Hom}_R(R, R)\cong R^{op}$ where $\text{Hom}_R$ denotes left $R$-module homomorphisms (see Exercise {III.1.17} and Exercise {IV.1.7}).
	In particular, if $R$ is commutative,
	then there is a ring isomorphism $\text{Hom}_R(R, R)\cong R$.
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	Let $S$ be a nonempty subset of a vector space $V$ over a division ring.
	The annihilator of $S$ is the subset $S^0$ of $V^*$ given by $S^0=\{f\in V^*\mid \langle s, f\rangle=0\text{ for all }s\in S\}$.
	\begin{enumerate}
	\item $0^0=V^*$; $V^0=0$; $S\neq \{0\}\Rightarrow S^0\neq V^*$.
	\item If $W$ is a subspace of $V$, then $W^0$ is a subspace of $V^*$.
	\item If $W$ is a subspace of $V$ and $\dim{V}$ is finite, then $\dim{W^0}=\dim{V}-\dim{W}$.
	\item Let $W, V$ be as in (c). There is an isomorphism $W^*\cong V^*/W^0$.
	\item Let $W, V$ be as in (c) and identify $V$ with $V^{**}$ under the isomorphism $\theta$ of Theorem {IV.4.12}. Then $(W^0)^0=W\subseteq V^{**}$.
	\end{enumerate}
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	If $V$ is a vector space over a division ring and $f\in V^*$,
	let $W=\{a\in V\mid \langle a, f\rangle=0\}$,
	then $W$ is a subspace of $V$.
	If $\dim{V}$ is finite,
	what is $\dim{W}$?
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	If $R$ has an identity and we denote the left $R$-module $R$ by ${}_R R$ and the right $R$-module $R$ by $R_R$,
	then $({}_R R)^*\cong R_R$ and $(R_R)^*\cong {}_R R$.
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	For any homomorphism $f:A\to B$ of left $R$-modules the diagram
	$$\xymatrix{
	A \ar[r]^{\theta_A} \ar[d]_{f} & A^{**} \ar[d]^{f^*} \\
	B \ar[r]^{\theta_B} & B^{**}
	}$$
	is commutative,
	where $\theta_A, \theta_B$ are as in Theorem {IV.4.12} and $f^*$ is the map induced on $A^{**}=\text{Hom}_R(\text{Hom}_R(A, R), R)$ by the map $\overline{f}:\text{Hom}_R(B, R)\to \text{Hom}_R(A, R)$.
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	Let $F=\sum_{x\in X}\mathbb{Z}x$ be a free $\mathbb{Z}$-module with an infinite basis $X$.
	Then $\{f_x\mid x\in X\}$ (Theorem {IV.4.11}) does not form a basis of $F^*$.
	[{\it Hint:} by Theorem {IV.4.7} and Theorem {IV.4.9},
	$F^*\cong \prod_{x\in X}\mathbb{Z}x$;
	but under this isomorphism $f_y\mapsto \{\delta_{xy}x\}\in \prod_{x\in X}\mathbb{Z}x$.]
	{\it Note:} $F^*=\prod \mathbb{Z}x$ is not a free $\mathbb{Z}$-module; see L. Fuchs [13; p. 168].
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	If $R$ has an identity and $P$ is a finitely generated projective unitary left $R$-module,
	then
	\begin{enumerate}
	\item $P^*$ is a finitely generated projective right $R$-module.
	\item $P$ is reflexive.
	\end{enumerate}
	This proposition may be false if the words ``finitely generated'' are omitted;
	see Exercise {IV.4.10}.
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	Let $F$ be a field,
	$X$ an infinite set,
	and $V$ the free left $F$-module (vector space) on the set $X$.
	Let $F^X$ be the set of all functions. $f:X\to F$.
	\begin{enumerate}
	\item $F^X$ is a (right) vector space over $F$ (with $(f+g)(x)=f(x)+g(x)$ and $(fr)(x)=rf(x)$).
	\item There is a vector-space isomorphism $V^*\cong F^X$.
	\item $\dim_{F}F^X=|F|^{|X|}$ (see Introduction, Exercise {O.8.10}).
	\item $\dim_F V^*>\dim_F V$ [{\it Hint:} by Introduction, Exercise {O.8.10} and Introduction, Theorem {O.8.5} $\dim_F V^*=\dim_F F^X=|F|^{|X|}\geq 2^{|X|}=|P(X)|>|X|=\dim_F V$.]
	\end{enumerate}
	\end{enumerate}
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\section{Tensor Products}
	{\it Note:}
	$R$ is a ring and $\otimes=\otimes_{\mathbb{Z}}$.
	\begin{enumerate}
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	If $R=\mathbb{Z}$,
	then condition (iii) of Definition {IV.5.1} is superfluous (that is, (i) and (ii) imply (iii)).
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	Let $A$ and $B$ be abelian groups.
	\begin{enumerate}
	\item For each $m>0$, $A\otimes \mathbb{Z}_m\cong A/mA$.
	\item $\mathbb{Z}_m\otimes \mathbb{Z}_n\cong \mathbb{Z}_c$, where $c=\gcd{(m, n)}$.
	\item Describe $A\otimes B$, when $A$ and $B$ are finitely generated.
	\end{enumerate}
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	If $A$ is a torsion abelian group and $\mathbb{Q}$ the (additive) group of rationals,
	then
	\begin{enumerate}
	\item $A\otimes \mathbb{Q}=0$.
	\item $\mathbb{Q}\otimes \mathbb{Q}\cong \mathbb{Q}$.
	\end{enumerate}
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	Give examples to show that each of the following may actually occur for suitable rings $R$ and modules $A_R, {}_R B$.
	\begin{enumerate}
	\item $A\otimes_R B\neq A\otimes_{\mathbb{Z}} B$.
	\item $u\in A\otimes_R B$, but $u\neq a\otimes b$ for any $a\in A, b\in B$.
	\item $a\otimes b=a_1\otimes b_1$ but $a\neq a_1$ and $b\neq b_1$.
	\end{enumerate}
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	If $A'$ is a submodule of the right $R$-module $A$ and $B'$ is a submodule of the left $R$-module $B$,
	then $A/A'\otimes_R B/B'\cong (A\otimes_R B)/C$,
	where $C$ is the subgroup of $A\otimes_R B$ generated by all elements $a'\otimes b$ and $a\otimes b'$ with $a\in A, a'\in A', b\in B, b'\in B'$.
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	Let $f:A_R\to {A_R}'$ and $g:{}_R B\to {}_R B'$ be $R$-module homomorphisms.
	What is the difference between the homomorphism $f\otimes g$ (as given by Corollary {IV.5.3}) and the element $f\otimes g$ of the tensor product of abelian groups
	$$\text{Hom}_R(A, A')\otimes \text{Hom}_R (B, B')?$$
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	The usual injection $\alpha:\mathbb{Z}_2\to \mathbb{Z}_4$ is a monomorphism of abelian groups.
	Show that $1\otimes \alpha:\mathbb{Z}_2\otimes \mathbb{Z}_2\to \mathbb{Z}_2\otimes \mathbb{Z}_4$ s the zero map (but $\mathbb{Z}_2\otimes \mathbb{Z}_2\neq 0$ and $\mathbb{Z}_2\otimes \mathbb{Z}_4\neq 0$;
	see Exercise {IV.5.2}).
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	Let $0\to A\stackrel{f}{\to} B\stackrel{g}{\to}C\to 0$ be a short exact sequence of left $R$-modules and $D$ a right $R$-module.
	Then $0\to D\otimes_R A\stackrel{1_D\otimes f}{\longrightarrow}D\otimes_R B\stackrel{1_D\otimes g}{\longrightarrow}D\otimes_R C\to 0$ is a short exact sequence of abelian groups under any one of the following hypothesis:
	\begin{enumerate}
	\item $0\to A\stackrel{f}{\to} B\stackrel{g}{\to}C\to 0$ is split exact.
	\item $R$ has an identity and $D$ is a free right $R$-module.
	\item $R$ has an identity and $D$ is a projective unitary right $R$-module.
	\end{enumerate}
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	\begin{enumerate}
	\item If $I$ is a right ideal of a ring $R$ with identity and $B$ a left $R$-module,
	then there is a group isomorphism $R/I\otimes_R B\cong R/IB$,
	where $IB$ is the subgroup of $B$ generated by all elements $rb$ with $r\in I, b\in B$.
	\item If $R$ is commutative and $I, J$ are ideals of $R$,
	then there is an $R$-module isomorphism $R/I\otimes_R R/J\cong R/(I+J)$.
	\end{enumerate}
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	If $R, S$ are rings,
	$A_R, {}_R B_S, {}_S C$ are (bi)modules and $D$ an abelian group,
	define a {\it middle linear map} to be a function $f:A\times B\times C\to D$ such that
	\begin{enumerate}[(i)]
	\item $f(a+a', b, c)=f(a, b, c)+f(a', b, c)$;
	\item $f(a, b+b', c)=f(a, b, c)+f(a, b', c)$;
	\item $f(a, b, c+c')=f(a, b, c)+f(a, b, c')$;
	\item $f(ar, b, c)=f(a, rb, c)$ for $r\in R$;
	\item $f(a, bs, c)=f(a, b, sc)$ for $s\in S$.
	\end{enumerate}
	\begin{enumerate}
	\item The map $i:A\times B\times C\to (A\otimes_R B)\otimes_S C$ given by $(a, b, c)\mapsto (a\otimes b)\otimes c$ is middle linear.
	\item The middle linear map $i$ is {\it universal};
	that is,
	given a middle linear map $g:A\times B\times C\to D$,
	there exists a unique group homomorphism $\overline{g}:(A\otimes_R B)\otimes_S C\to D$ such that $\overline{g}i=g$.
	\item The map $j:A\times B\times C\to A\otimes_R (B\otimes_S C)$ given by $(a, b, c)\mapsto a\otimes (b\otimes c)$ is also a universal middle linear map.
	\item $(A\otimes_R B)\otimes_S C\cong A\otimes_R (B\otimes_S C)$ by (b), (c), and Theorem {I.7.10}.
	\item Define a middle linear function on $n$ (bi)modules ($n\geq 4$) in the obvious way and sketch a proof of the extension of the above results to the case of $n$ (bi)modules (over $n-1$ rings).
	\item If $R=S$, $R$ is commutative and $A, B, C, D$ are $R$-modules,
	define a trilinear map $A\times B\times C\to D$ and extend the results of (a), (b), (c) to such maps.
	\end{enumerate}
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	Let $A, B, C$ be modules over a commutative ring $R$.
	\begin{enumerate}
	\item The set $\mathcal{L}(A, B; C)$ of all $R$-bilinear maps $A\times B\to C$ is an $R$-module with $(f+g)(a, b)=f(a, b)+g(a, b)$ and $(rf)(a, b)=rf(a, b)$.
	\item Each one of the following $R$-modules is isomorphic to $\mathcal{L}(A, B; C)$:
	\begin{enumerate}[(i)]
	\item $\text{Hom}_R(A\otimes_R B, C)$;
	\item $\text{Hom}_R(A, \text{Hom}_R(B, C))$;
	\item $\text{Hom}_R(B, \text{Hom}_R(A, C))$.
	\end{enumerate}
	\end{enumerate}
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	Assume $R$ has an identity.
	Let $\mathcal{C}$ be the category of all unitary $R$-$R$ bimodules and bimodule homomorphisms (that is, group homomorphisms $f:A\to B$ such that $f(ras)=rf(a)s$ for all $r, s\in R$).
	Let $X=\{1_R\}$ and let $\iota:X\to R$ be the inclusion map.
	\begin{enumerate}
	\item If $R$ is noncommutative,
	then $R$ (equipped with $\iota:X\to R$) is not a free object on the set $X$ in the category $\mathcal{C}$.
	\item $R\otimes_{\mathbb{Z}} R$ is an $R$-$R$ bimodule (Theorem {IV.5.5}).
	If $\iota:X\to R\otimes_{\mathbb{Z}} R$ is given by $1_R\mapsto 1_R\otimes 1_R$,
	then $R\otimes_{\mathbb{Z}}R$ is a free object on the set $X$ in the category $\mathcal{C}$.
	\end{enumerate}
	\end{enumerate}
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\section{Modules over a Principal Ideal Domain}
	{\it Note:} Unless stated otherwise,
	$R$ is a principal ideal domain and all modules are unitary.
	\begin{enumerate}
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	If $R$ is a nonzero commutative ring with identity and every submodule of every free $R$-module is free,
	then $R$ is a principal ideal domain.
	[{\it Hint:} Every ideal $I$ of $R$ is a free $R$-module.
	If $u, v\in I$ ($u\neq 0, v\neq 0$),
	then $uv+(-v)u=0$,
	which implies that $I$ has a basis of one element;
	that is,
	$I$ is principal.]
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	Every free module over an arbitrary integral domain with identity is torsion-free.
	The converse is false (Exercise {II.1.10}).
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	Let $A$ be a cyclic $R$-module of order $r\in R$.
	\begin{enumerate}
	\item If $s\in R$ is relatively prime to $r$,
	then $sA=A$ and $A[s]=0$.
	\item If $s$ divides $r$,
	say $sk=r$,
	then $sA\cong R/\langle k\rangle$ and $A[s]\cong R/\langle s\rangle$.
	\end{enumerate}
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	If $A$ is a cyclic $R$-module of order $r$,
	then
	\begin{enumerate}[(i)]
	\item every submodule of $A$ is cyclic,
	with order dividing $r$;
	\item for every ideal $\langle s\rangle$ containing $\langle r\rangle$,
	$A$ has exactly one submodule,
	which is cyclic of order $s$.
	\end{enumerate}
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	If $A$ is a finitely generated torsion module,
	then $\{r\in R\mid rA=0\}$ is a nonzero ideal in $R$,
	say $\langle r_1\rangle$.
	$r_1$ is called the minimal annihilator of $A$.
	Let $A$ be a finite abelian group with minimal annihilator $m\in \mathbb{Z}$.
	Show that a cyclic subgroup of $A$ of order properly dividing $m$ need not be a direct summand of $A$.
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	If $A$ and $B$ are cyclic modules over $R$ of nonzero orders $r$ and $s$ respectively,
	and $r$ is {\it not} relatively prime to $s$,
	then the invariant factors of $A\oplus B$ are the greatest common divisor of $r, s$ and the least common multiple of $r, s$.
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	Let $A$ and $a\in A$ satisfy the hypotheses of Lemma {IV.6.8}.
	\begin{enumerate}
	\item Every $R$-module of $A$ is an $R/\langle p^n\rangle$-module with $(r+\langle p^n\rangle)a=ra$.
	Conversely,
	every $R/\langle p^n\rangle$-submodule of $A$ is an $R$-submodule by pullback along $R\to R/\langle p^n\rangle$.
	\item The submodule $Ra$ is isomorphic to $R/\langle p^n\rangle$.
	\item The only proper ideals of the ring $R/\langle p^n\rangle$ are the ideals generated by $p^i+\langle p^n\rangle$ ($i=1, 2, ..., n-1$).
	\item $R/\langle p^n\rangle$ (and hence $Ra$) is an injective $R/\langle p^n\rangle$-module.
	[{\it Hint:} use (c) and Lemma {IV.3.8}.]
	\item There exists an $R$-submodule $C$ of $A$ such that $A=Ra\oplus C$.
	[{\it Hint:} Proposition {IV.3.13}.]
	\end{enumerate}
	\end{enumerate}
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\section{Algebras}
	{\it Note:} $K$ is always a commutative right with identity.
	\begin{enumerate}
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	Let $\mathcal{C}$ be the category whose objects are all commutative $K$-algebras with identity and whose morphisms are all $K$-algebra homomorphisms $f:A\to B$ such that $f(1_A)=1_B$.
	Then any two $K$-algebras $A, B$ of $\mathcal{C}$ have a coproduct.
	[{\it Hint:} consider $A\to A\otimes_K B\leftarrow B$,
	where $a\mapsto a\otimes 1_B$ and $b\mapsto 1_A\otimes b$.]
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	If $A$ and $B$ are unitary $K$-modules [resp. $K$-algebras],
	then there is an isomorphism of $K$-modules [resp. $K$-algebras] $\alpha:A\otimes_K B\to B\otimes_K A$ such that $\alpha(a\otimes b)=b\otimes a$ for all $a\in A, b\in B$.
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	Let $A$ be a ring with identity.
	Then $A$ is a $K$-algebra with identity if and only if there is a ring homomorphism of $K$ into the center of $A$ such that $1_K\mapsto 1_A$.
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	Let $A$ be a one-dimensional vector space over the rational field $\mathbb{Q}$.
	If we define $ab=0$ for all $a, b\in A$,
	then $A$ is a $\mathbb{Q}$-algebra.
	Every proper additive subgroup of $A$ is an ideal of the ring $A$,
	but not an algebra ideal.
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	Let $\mathcal{C}$ be the category of Exercise {IV.7.1}.
	If $X$ is the set $\{x_1, ..., x_n\}$,
	then the polynomial algebra $K[x_1, ..., x_n]$ is a free object on the set $X$ in the category $\mathcal{C}$.
	[{\it Hint:} Given an algebra $A$ in $\mathcal{C}$ and a map $g:\{x_1, ..., x_n\}\to A$,
	apply Theorem {III.5.5} to the unit map $I:K\to A$ and the elements $g(x_1), ..., g(x_n)\in A$.]
	\end{enumerate}
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\chapter{Fields and Galois Theory}
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\section{Field Extensions}
	{\it Note:} Unless specified otherwise $F$ is always an extension field of the field $K$ and $\mathbb{Q}, \mathbb{R}, \mathbb{C}$ denote the fields of rational, real, and complex numbers respectively.
	\begin{enumerate}
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	\begin{enumerate}
	\item $[F:K]=1$ if and only if $F=K$.
	\item If $[F:K]$ is prime, then there are no intermediate fields between $F$ and $K$.
	\item If $u\in F$ has degree $n$ over $K$, then $n$ divides $[F:K]$.
	\end{enumerate}
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	Give an example of a finitely generated field extension, which is not finite dimensional.
	[{\it Hint:} think transcendental.]
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	If $u_1, ..., u_n\in F$ then the field $K(u_1, ..., u_n)$ is (isomorphic to) the quotient field of the ring $K[u_1, ..., u_n]$.
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	\begin{enumerate}
	\item For any $u_1, ..., u_n\in F$ and any permutation $\sigma\in S_n$, $K(u_1, ..., u_n)=K(u_{\sigma(1)}, ..., u_{\sigma(n)})$.
	\item $K(u_1, ..., u_{n-1})(u_n)=K(u_1, ..., u_n)$.
	\item State and prove the analogues of (a) and (b) for $K[u_1, ..., u_n]$.
	\item If each $u_i$ is algebraic over $K$, then $K(u_1, ..., u_n)=K[u_1, ..., u_n]$.
	\end{enumerate}
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	Let $L$ and $M$ be subfields of $F$ and $LM$ their composite.
	\begin{enumerate}
	\item If $K\subseteq L\cap M$ and $M=K(S)$ for some $S\subseteq M$, then $LM=L(S)$.
	\item When is it true that $LM$ is the set theoretic union $L\cup M$?
	\item If $E_1, ..., E_n$ are subfields of $F$, show that $$E_1 E_2\cdots E_n=E_1(E_2(E_3(\cdots (E_{n-1}(E_n)))\cdots ).$$
	\end{enumerate}
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	Every element of $K(x_1, ..., x_n)$ which is not in $K$ is transcendental over $K$.
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	If $v$ is algebraic over $K(u)$ for some $u\in F$ and $v$ is transcendental over $K$,
	then $u$ is algebraic over $K(v)$.
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	If $u\in F$ is algebraic of odd degree over $K$,
	then so is $u^2$ and $K(u)=K(u^2)$.
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	If $x^n-a\in K[x]$ is irreducible and $u\in F$ is a root of $x^n-a$ and $m$ divides $n$,
	then prove that the degree of $u^m$ over $K$ is $n/m$.
	What is the irreducible polynomial for $u^m$ over $K$?
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	If $F$ is algebraic over $K$ and $D$ is an integral domain such that $K\subseteq D\subseteq F$,
	then $D$ is a field.
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	\begin{enumerate}
	\item Give an example of a field extension $K\subseteq F$ such that $u, v\in F$ are transcendental over $K$,
	but $K(u, v)\not\cong K(x_1, x_2)$.
	[{\it Hint:} consider $v$ over the field $K(u)$.]
	\item State and prove a generalization of Theorem {V.1.5} to the case of $n$ transcendental elements $u_1, ..., u_n$.
	\end{enumerate}
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	If $d\geq 0$ is an integer that is not a square describe the field $\mathbb{Q}(\sqrt{d})$ and find a set of elements that generate the whole field.
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	\begin{enumerate}
	\item Consider the extension $\mathbb{Q}(u)$ of $\mathbb{Q}$ generated by a real root $u$ of $x^3-6x^2+9x+3$.
	(Why is this irreducible?)
	Express each of the following elements in terms of the basis $\{1, u, u^2\}$: $u^4$; $u^5$; $3u^5-u^4+2$; $(u+1)^{-1}$; $(u^2-6u+8)^{-1}$.
	\item Do the same with respect to the basis $\{1, u, u^2, u^3, u^4\}$ of $\mathbb{Q}(u)$ where $u$ is a real root of $x^5+2x+2$ and the elements in question are:
	$(u^2+2)(u^3+3u)$; $u^{-1}$; $u^4(u^4+3u^2+7u+5)$; $(u+2)(u^2+3)^{-1}$.
	\end{enumerate}
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	\begin{enumerate}
	\item If $F=\mathbb{Q}(\sqrt{2}, \sqrt{3})$, find $[F:\mathbb{Q}]$ and a basis of $F$ over $\mathbb{Q}$.
	\item Do the same for $F=\mathbb{Q}(i, \sqrt{3}, \omega)$, where $i\in \mathbb{C}$, $i^2=-1$, and $\omega$ is a complex (nonreal) cube root of $1$.
	\end{enumerate}
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	In the field $K(x)$,
	let $u=x^3/(x+1)$.
	Show that $K(x)$ is a simple extension of the field $K(u)$.
	What is $[K(x):K(u)]$?
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	In the field $\mathbb{C}$, $\mathbb{Q}(i)$ and $\mathbb{Q}(\sqrt{2})$ are isomorphic as vector spaces,
	but not as fields.
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	Find an irreducible polynomial $f$ of degree $2$ over the field $\mathbb{Z}_2$.
	Adjoin a root $u$ of $f$ to $\mathbb{Z}_2$ to obtain a field $\mathbb{Z}_2(u)$ of order $4$.
	Use the same method to construct a field of order $8$.
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	A complex number is said to be an algebraic number if it is algebraic over $\mathbb{Q}$ and an algebraic integer if it is the root of a monic polynomial in $\mathbb{Z}[x]$.
	\begin{enumerate}
	\item If $u$ is an algebraic number, there exists an integer $n$ such that $nu$ is an algebraic integer.
	\item If $r\in \mathbb{Q}$ is an algebraic integer, then $r\in \mathbb{Z}$.
	\item If $u$ is an algebraic number and $n\in \mathbb{Z}$, then $u+n$ and $nu$ are algebraic integers.
	\item The sum and product of two algebraic integers are algebraic integers.
	\end{enumerate}
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	If $u, v\in F$ are algebraic over $K$ of degrees $m$ and $n$ respectively,
	then $[K(u, v):K]\leq mn$.
	If $\gcd{(m, n)}=1$,
	then $[K(u, v):K]=mn$.
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	Let $L$ and $M$ be intermediate fields in the extension $K\subseteq F$.
	\begin{enumerate}
	\item $[LM:K]$ is finite if and only if $[L:K]$ and $[M:K]$ are finite.
	\item If $[LM:K]$ is finite, then $[L:K]$ and $[M:K]$ divide $[LM:K]$ and $$[LM:K]\leq [L:K][M:K].$$
	\item If $[L:K]$ and $[M:K]$ are finite and relatively prime, then $$[LM:K]=[L:K][M:K].$$
	\item If $L$ and $M$ are algebraic over $K$, then so is $LM$.
	\end{enumerate}
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	\begin{enumerate}
	\item Let $L$ and $M$ be intermediate fields of the extension $K\subseteq F$, of finite dimension over $K$.
	Assume that $[LM:K]=[L:K][M:K]$ and prove that $L\cap M=K$.
	\item The converse of (a) holds if $[L:K]$ or $[M:K]$ is $2$.
	\item Using a real and a nonreal cube root of $2$ give an example where $L\cap M=K$,
	$[L:K]=[M:K]=3$, but $[LM:K]<9$.
	\end{enumerate}
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	$F$ is an algebraic extension of $K$ if and only if for every intermediate field $E$ every monomorphism $\sigma:E\to E$ which is the identity on $K$ is in fact an automorphism of $E$.
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	If $u\in F$ is algebraic over $K(X)$ for some $X\subseteq F$ then there exists a finite subset $X'\subseteq X$ such that $u$ is algebraic over $K(X')$.
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	Let $F$ be a subfield of $\mathbb{R}$ and $P, Q$ points in the Eulidean plane whose coordinates lie in $F$.
	\begin{enumerate}
	\item The straight line through $P$ and $Q$ has an equation of the form $ax+by+c=0$, with $a, b, c\in F$.
	\item The circle with center $P$ and radius the line segment $PQ$ has an equation of the form $x^2+y^2+ax+by+c=0$ with $a, b, c\in F$.
	\end{enumerate}
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	Let $c, d$ be contructible real numbers.
	\begin{enumerate}
	\item $c+d$ and $c-d$ are constructible.
	\item If $d\neq 0$, then $c/d$ is constructible.
	[{\it Hint:} If $(x, 0)$ is the intersection of the $x$ axis and the straight line through $(0, 1)$ that is parallel the line through $(0, d)$ and $(c, 0)$, then $x=c/d$.]
	\item $cd$ is constructible [{\it Hint:} use (b)].
	\item The constructible real numbers form a subfield containing $\mathbb{Q}$.
	\item If $c\geq 0$, then $\sqrt{c}$ is constructible.
	[{\it Hint:} If $y$ is the length of the straight line segment perpendicular to the $x$ axis that joins $(1, 0)$ with the (upper half of the) circle with center $((c+1)/2, 0)$ and radius $(c+1)/2$ then $y=\sqrt{c}$.]
	\end{enumerate}
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	Let $E_1$ and $E_2$ be subfields of $F$ and $X$ a subset of $F$.
	If every element of $E_1$ is algebraic over $E_2$,
	then every element of $E_1(X)$ is algebraic over $E_2(X)$.
	[{\it Hint:} $E_1(X)\subseteq (E_2(X))(E_1)$;
	use Theorem {V.1.12}.]
	\end{enumerate}
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\section{The Fundamental Theorem}
	{\it Note:}
	Unless stated otherwise $F$ is always an extension field of the field $K$ and $E$ is an intermediate field of the extension.
	\begin{enumerate}
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	\begin{enumerate}
	\item If $F$ is a field and $\sigma:F\to F$ a (ring) homomorphism,
	then $\sigma=0$ or $\sigma$ is a monomorphism.
	If $\sigma \neq 0$,
	then $\sigma(1_F)=1_F$.
	\item The set $\text{Aut }F$ of all field automorphisms $F\to F$ forms a group under the operation of composition of functions.
	\item $\text{Aut}_K F$, the set of all $K$-automorphisms of $F$ is a subgroup of $\text{Aut }F$.
	\end{enumerate}
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	$\text{Aut}_{\mathbb{Q}}\mathbb{R}$ is the identity group.
	[{\it Hint:} Since every positive element of $\mathbb{R}$ is a square,
	it follows that an automorphism of $\mathbb{R}$ sends positives to positives and hence that it preserves the order in $\mathbb{R}$.
	Trap a given real number between suitable rational numbers.]
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	If $0\leq d\in \mathbb{Q}$,
	then $\text{Aut}_{\mathbb{Q}}\mathbb{Q}(\sqrt{d})$ is the identity or is isomorphic to $\mathbb{Z}_2$.
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	What is the Galois group of $\mathbb{Q}(\sqrt{2}, \sqrt{3}, \sqrt{5})$ over $\mathbb{Q}$?
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	\begin{enumerate}
	\item If $0\leq d\in \mathbb{Q}$, then $\mathbb{Q}(\sqrt{d})$ is Galois over $\mathbb{Q}$.
	\item $\mathbb{C}$ is Galois over $\mathbb{R}$.
	\end{enumerate}
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	Let $f/g\in K(x)$ with $f/g\notin K$ and $f, g$ relatively prime in $K[x]$ and consider the extension of $K$ by $K(x)$.
	\begin{enumerate}
	\item $x$ is algebraic over $K(f/g)$ and $[K(x):K(f/g)]=\max{(\deg{f}, \deg{g})}$.
	[{\it Hint:} $x$ is a root of the nonzero polynomial $\varphi(y)=(f/g)g(y)=f(y)\in K(f/g)[y]$;
	show that $\varphi$ has degree $\max{(\deg{f}, \deg{g})}$.
	Show that $\varphi$ is irreducible as follows.
	Since $f/g$ is transcendental over $K$ (why?)
	we may for convenience replace $K(f/g)$ by $K(z)$ ($z$ an indeterminate) and consider $\varphi=zg(y)-f(y)\in K(z)[y]$.
	By Lemma {III.6.13} $\varphi$ is irreducible in $K(z)[y]$ provided it is irreducible in $K[z][y]$.
	The truth of this latter condition follows from the fact that $\varphi$ is linear in $z$ and $f, g$ are relatively prime.]
	\item If $E\neq K$ is an intermediate field, then $[K(x):E]$ is finite.
	\item The assignment $x\mapsto f/g$ induces a homomorphism $\sigma:K(x)\to K(x)$ such that $\varphi(x)/\psi(x)\mapsto \varphi(f/g)/\psi(f/g)$.
	$\sigma$ is a $K$ automorphism of $K(x)$ if and only if $\max{(\deg{f}, \deg{g})}=1$.
	\item $\text{Aut}_K K(x)$ consists of all those automorphisms induced (as in (c)) by the assignment
	$$x\mapsto (ax+b)/(cx+d),$$
	where $a, b, c, d\in K$ and $ad-bc\neq 0$.
	\end{enumerate}
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	Let $G$ be the subset of $\text{Aut}_K K(x)$ consisting of the three automorphisms induced (as in Exercise {V.2.6} (c)) by $x\mapsto x$, $x\mapsto 1_K/(1_K-x)$, $x\mapsto (x-1_K)/x$.
	Then $G$ is a subgroup of $\text{Aut}_K K(x)$.
	Determine the fixed field of $G$.
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	Assume $\text{char }K=0$ and let $G$ be the subgroup of $\text{Aut}_K K(x)$ that is generated by the automorphism induced by $x\mapsto x+1_K$.
	Then $G$ is an infinite cyclic group.
	Determine the fixed field $E$ of $G$.
	What is $[K(x):E]$?
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	\begin{enumerate}
	\item If $K$ is an infinite field,
	then $K(x)$ is Galois over $K$.
	[{\it Hint:} If $K(x)$ is not Galois over $K$,
	then $K(x)$ is finite dimensional over the fixed field $E$ of $\text{Aut}_K K(x)$ by Exercise {V.2.6}(b).
	But $\text{Aut}_E K(x)=\text{Aut}_K K(x)$ is infinite by Exercise {V.2.6}(d),
	which contradicts Lemma {V.2.8}.]
	\item If $K$ is finite,
	then $K(x)$ is {\it not} Galois over $K$.
	[{\it Hint:} If $K(x)$ were Galois over $K$,
	then $\text{Aut}_K K(x)$ would be infinite by Lemma {V.2.9}.
	But $\text{Aut}_K K(x)$ is finite by Exercise {V.2.6}(d).]
	\end{enumerate}
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	If $K$ is an infinite field,
	then the only closed subgroups of $\text{Aut}_K K(x)$ are itself and its finite subgroups.
	[{\it Hint:} see Exercise {V.2.6}(b) and Exercise {V.2.9}.]
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	In the extension of $\mathbb{Q}$ by $\mathbb{Q}(x)$,
	the intermediate field $\mathbb{Q}(x^2)$ is closed,
	but $\mathbb{Q}(x^3)$ is not.
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	If $E$ is an intermediate field of the extension such that $E$ is Galois over $K$,
	$F$ is Galois over $E$,
	and every $\sigma\in \text{Aut}_K E$ is extendible to $F$,
	then $F$ is Galois over $K$.
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	In the extension of an infinite field $K$ by $K(x, y)$,
	the intermediate field $K(x)$ is Galois over $K$,
	but not stable (relative to $K(x, y)$ and $K$).
	[See Exercise {V.2.9}; compare this result with Lemma {V.2.13}.]
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	Let $F$ be a finite dimensional Galois extension of $K$ and let $L$ and $M$ be two intermediate fields.
	\begin{enumerate}
	\item $\text{Aut}_{LM} F=\text{Aut}_L F\cap \text{Aut}_M F$;
	\item $\text{Aut}_{L\cap M}F=\text{Aut}_L F\vee\text{Aut}_M F$;
	\item What conclusion can be drawn if $\text{Aut}_L F\cap \text{Aut}_M F=1$?
	\end{enumerate}
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	If $F$ is a finite dimensional Galois extension of $K$ and $E$ is an intermediate field,
	then there is a unique smallest field $L$ such that $E\subseteq L\subseteq F$ and $L$ is Galois over $K$;
	furthermore
	$$\text{Aut}_L F=\bigcap_{\sigma}\sigma(\text{Aut}_E F)\sigma^{-1},$$
	where $\sigma$ runs over $\text{Aut}_K F$.
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	If $\sigma\in S_n$,
	then the map $K(x_1, ..., x_n)\mapsto K(x_1, ..., x_n)$ given by
	$$\frac{f(x_1, ..., x_n)}{g(x_1, ..., x_n)}\mapsto \frac{f(x_{\sigma(1)}, ..., x_{\sigma(n)})}{g(x_{\sigma(1)}, ..., x_{\sigma(n)})}$$
	is a $K$-automorphism of $K(x_1, ..., x_n)$.
	\end{enumerate}
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\section{Splitting Fields, Algebraic Closure and Normality}
	{\it Note:}
	Unless stated otherwise $F$ is always an extension field of the field $K$ and $S$ is a set of polynomials (of positive degree) in $K[x]$.
	\begin{enumerate}
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	$F$ is a splitting field over $K$of a finite set $\{f_1, ..., f_n\}$ of polynomials in $K[x]$ if and only if $F$ is a splitting field over $K$ of the single polynomial $f=f_1 f_2\cdots f_n$.
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	If $F$ is a splitting field of $S$ over $K$ and $E$ is an intermediate field,
	then $F$ is a splitting field of $S$ over $E$.
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	\begin{enumerate}
	\item Let $E$ be an intermediate field of the extension $K\subseteq F$ and assume that $E=K(u_1, ..., u_r)$ where the $u_i$ are (some of the) roots of $f\in K[x]$.
	Then $F$ is a splitting field of $f$ over $K$ if and only if $F$ is a splitting field of $f$ over $E$.
	\item Extend part (a) to splitting fields of arbitrary sets of polynomials.
	\end{enumerate}
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	If $F$ is a splitting field over $K$ of $S$,
	then $F$ is also a splitting field over $K$ of the set $T$ of all irreducible factors of polynomials in $S$.
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	If $f\in K[x]$ has degree $n$ and $F$ is a splitting field of $f$ over $K$,
	then $[F:K]$ divides $n!$.
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	Let $K$ be a field such that for every extension field $F$ the maximal algebraic extension of $K$ contained in $F$ (see Theorem {V.1.14}) is $K$ itself.
	Then $K$ is algebraically closed.
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	If $F$ is algebraically closed and $E$ consists of all elements in $F$ that are algebraic over $K$,
	then $E$ is an algebraic closure of $K$ [see Theorem {V.1.14}].
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	No finite field $K$ is algebraically closed.
	[{\it Hint:} If $K=\{a_0, ..., a_n\}$ consider $a_1+(x-a_0)(x-a_1)\cdots (x-a_n)\in K[x]$,
	where $a_1\neq 0$.]
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	$F$ is an algebraic closure of $K$ if and only if $F$ is algebraic over $K$ and for every algebraic extension $E$ of $K$ there exists a $K$-monomorphism $E\to F$.
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	$F$ is an algebraic closure of $K$ if and only if $F$ is algebraic over $K$ and for every algebraic field extension $E$ of another field $K_1$ and isomorphism of fields $\sigma:K_1\to K$,
	$\sigma$ etends to a monomorphism $E\to F$.
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	\begin{enumerate}
	\item If $u_1, ..., u_n\in F$ are separable over $K$,
	then $K(u_1, ..., u_n)$ is a separable extension of $K$.
	\item If $F$ is generated by a (posibly infinite) set of separable elements over $K$,
	then $F$ is a separable extension of $K$.
	\end{enumerate}
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	Let $E$ be an intermediate field.
	\begin{enumerate}
	\item If $u\in F$ is separable over $K$, then $u$ is separable over $E$.
	\item If $F$ is separable over $K$, then $F$ is separable over $E$ and $E$ is separable over $K$.
	\end{enumerate}
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	Suppose $[F:K]$ is finite. Then the following conditions are equivalent:
	\begin{enumerate}[(i)]
	\item $F$ is Galois over $K$;
	\item $F$ is separable over $K$ and a splitting field of a polynomial $f\in K[x]$;
	\item $F$ is a splitting field over $K$ of a polynomial $f\in K[x]$ whose irreducible factors are separable.
	\end{enumerate}
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	(Lagrange's Theorem on Natural Irrationalities).
	If $L$ and $M$ are intermediate fields such that $L$ is a finite dimensional Galois extension of $K$,
	then $LM$ is finite dimensional and Galois over $M$ and $\text{Aut}_M LM\cong \text{Aut}_{L\cap M}L$.
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	Let $E$ be an intermediate field.
	\begin{enumerate}
	\item If $F$ is algebraic Galois over $K$,
	then $F$ is algebraic Galois over $E$.
	[Exercise {V.2.9} and Exercise {V.2.11} show that the ``algebraic'' hypothesis is necessary.]
	\item If $F$ is Galois over $E$,
	$E$ is Galois over $K$ and $F$ is a splitting field over $E$ of a family of polynomials in $K[x]$,
	then $F$ is Galois over $K$ [see Exercise {V.2.12}].
	\end{enumerate}
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	Let $F$ be an algebraic closure of the field $\mathbb{Q}$ of rational numbers and let $E\subseteq F$ be a splitting field over $\mathbb{Q}$ of the set $S=\{x^2+a\mid a\in \mathbb{Q}\}$ so that $E$ is algebraic and Galois over $\mathbb{Q}$ (Theorem {V.3.11}).
	\begin{enumerate}
	\item $E=\mathbb{Q}(X)$ where $X=\{\sqrt{p}\mid p=-1\text{ or }p\text{ is a prime integer}\}$.
	\item If $\sigma\in \text{Aut}_{\mathbb{Q}}E$, then $\sigma^2=1_E$. Therefore, the group $\text{Aut}_{\mathbb{Q}}E$ is actually a vector space over $\mathbb{Z}_2$ [see Exerxise {I.1.13} and Exercise {IV.1.1}].
	\item $\text{Aut}_{\mathbb{Q}}E$ is infinite and not denumerable.
	[{\it Hint:} for each subset $Y$ of $X$ there exists $\sigma\in \text{Aut}_{\mathbb{Q}}E$ such that $\sigma(\sqrt{p})=-\sqrt{p}$ for $\sqrt{p}\in Y$ and $\sigma(\sqrt{p})=\sqrt{p}$ for $\sqrt{p}\in X-Y$.
	Therefore, $|\text{Aut}_{\mathbb{Q}}E|=|P(X)|>|X|$ by Introduction, Theorem {O.8.5}. But $|X|=\aleph_0$.]
	\item If $B$ is a basis of $\text{Aut}_{\mathbb{Q}}E$ over $\mathbb{Z}_2$, then $B$ is infinite and not denumerable.
	\item $\text{Aut}_{\mathbb{Q}}E$ has an infinite nondenumerable number of subgroups of index $2$.
	[{\it Hint:} If $b\in B$, then $B-\{b\}$ generates a subgroup of index $2$.]
	\item The set of extension fields of $\mathbb{Q}$ contained in $E$ of dimension $2$ over $\mathbb{Q}$ is denumerable.
	\item The set of closed subgroups of index $2$ in $\text{Aut}_{\mathbb{Q}}E$ is denumerable.
	\item $[E:\mathbb{Q}]\leq \aleph_0$, whence $[E:\mathbb{Q}]<|\text{Aut}_{\mathbb{Q}}E|$.
	\end{enumerate}
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	If an intermediate field $E$ is normal over $K$,
	then $E$ is stable (relative to $F$ and $K$).
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	Let $F$ be normal over $K$ and $E$ an intermediate field.
	Then $E$ is normal over $K$ if and only if $E$ is stable [see Exercise {V.3.17}].
	Furthermore $\text{Aut}_K F/E'\cong \text{Aut}_K E$.
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	Part (ii) or (ii)' of the Fundamental Theorem (Theorem {V.2.5} or Theorem {V.3.12}) is equivalent to:
	an intermediate field $E$ is normal over $K$ if and only if the corresponding subgroup $E'$ is normal in $G=\text{Aut}_K F$ in which case $G/E'\cong \text{Aut}_K E$.
	[See Exercise {V.3.18}.]
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	If $F$ is normal over an intermediate field $E$ and $E$ is normal over $K$,
	then $F$ need not be normal over $K$.
	[{\it Hint:} Let $\sqrt[4]{2}$ be a real fourth root of $2$ and consider $\mathbb{Q}(\sqrt[4]{2})\supseteq \mathbb{Q}(\sqrt{2})\supseteq\mathbb{Q}$; use Exercise {V.3.23}.]
	Compare Exercise {V.3.2}.
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	Let $F$ be algebraic over $K$.
	$F$ is normal over $K$ if and only if for every $K$-monomorphism of field $\sigma:F\to N$,
	where $N$ is any normal extension of $K$ containing $F$,
	$\sigma(F)=F$ so that $\sigma$ is a $K$-automorphism of $F$.
	[{\it Hint:} Adapt the proof of Theorem {V.3.14},
	using Theorem {V.3.16}.]
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	If $F$ is algebraic over $K$ and every element of $F$ belongs to an intermediate field that is normal over $K$,
	then $F$ is normal over $K$.
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	If $[F:K]=2$,
	then $F$ is normal over $K$.
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	An algebraic extension $F$ of $K$ is normal over $K$ if and only if for every irreducible $f\in K[x]$,
	$f$ factors in $F[x]$ as a product of irreducible factors all of which have the same degree.
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	Let $F$ be a splitting field of $f\in K[x]$.
	Without using Theorem {V.3.14} show that $F$ is normal over $K$.
	\end{enumerate}
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\section{The Galois Group of a Polynomial}
	{\it Note:} Unless stated otherwise $K$ is a field,
	$f\in K[x]$ and $F$ is a splitting field of $f$ over $K$.
	\begin{enumerate}
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	Suppose $f\in K[x]$ splits in $F$ as $f=(x-u_1)^{n_1}\cdots (x-u_k)^{n_k}$ ($u_i$ distinct; $n_i\geq 1$).
	Let $v_0, ..., v_k$ be the coefficients of the polynomial $g=(x-u_1)(x-u_2)\cdots (x-u_k)$ and let $E=K(v_0, ..., v_k)$.
	Then
	\begin{enumerate}
	\item $F$ is a splitting field of $g$ over $E$.
	\item $F$ is Galois over $E$.
	\item $\text{Aut}_E F=\text{Aut}_K F$.
	\end{enumerate}
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	Suppose $K$ is a subfield of $\mathbb{R}$ (so that $F$ may be taken to be a subfield of $\mathbb{C}$) and that $f$ is irreducible of degree $3$.
	Let $D$ be the discriminant of $f$.
	Then
	\begin{enumerate}
	\item $D>0$ if and only if $f$ has three real roots.
	\item $D<0$ if and only if $f$ has precisely one real root.
	\end{enumerate}
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	Let $f$ be a separable cubic with Galois group $S_3$ and roots $u_1, u_2, u_3\in F$.
	Then the distinct intermediate fields of the extension of $K$ by $F$ are $F, K(\Delta), K(u_1), K(u_2), K(u_3), K$.
	The corresponding subgroups of the Galois group are $1, A_3, T_1, T_2, T_3$ and $S_3$ where $T_i=\{(1), (jk)\mid j\neq i\neq k\}$.
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	If $\text{char }K\neq 2, 3$ then the discriminant of $x^3+bx^2+cx+d$ is $-4c^3-27d^2+b^2(c^2-4bd)+18bcd$.
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	If $\text{char }K\neq 2$ and $f\in K[x]$ is a cubic whose discriminant is a square in $K$,
	then $f$ is either irreducible or factors completely in $K$.
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	Over any base field $K$,
	$x^3-3x+1$ is either irreducible or splits over $K$.
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	$S_4$ has no transitive subgroup of order $6$.
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	Let $f$ be an (irreducible) separable quartic over $K$ and $u$ a root of $f$.
	There is no field properly between $K$ and $K(u)$ if and only if the Galois group of $f$ is either $A_4$ or $S_4$.
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	Let $x^4+ax^2+b\in K[x]$ (with $\text{char }K\neq 2$) be irreducible with Galois group $G$.
	\begin{enumerate}
	\item If $b$ is a square in $K$, then $G=V$.
	\item If $b$ is not a square in $K$ and $b(a^2-4b)$ is a square in $K$, then $G\cong \mathbb{Z}_4$.
	\item If neither $b$ nor $b(a^2-4b)$ is a square in $K$, then $G\cong D_4$.
	\end{enumerate}
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	Determine the Galois groups of the following polynomials over the fields indicated:
	\begin{enumerate}
	\item $x^4-5$ over $\mathbb{Q}$; over $\mathbb{Q}(\sqrt{5})$; over $\mathbb{Q}(\sqrt{5}i)$.
	\item $(x^3-2)(x^2-3)(x^2-5)(x^2-7)$ over $\mathbb{Q}$.
	\item $x^3-x-1$ over $\mathbb{Q}$; over $\mathbb{Q}(\sqrt{23}i)$.
	\item $x^3-10$ over $\mathbb{Q}$; over $\mathbb{Q}(\sqrt{2})$.
	\item $x^4+3x^3+3x-2$ over $\mathbb{Q}$.
	\item $x^5-6x+3$ over $\mathbb{Q}$.
	\item $x^3-2$ over $\mathbb{Q}$.
	\item $(x^3-2)(x^2-5)$ over $\mathbb{Q}$.
	\item $x^4-4x^2+5$ over $\mathbb{Q}$.
	\item $x^4+2x^2+x+3$ over $\mathbb{Q}$.
	\end{enumerate}
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	Determine all the subgroups of the Galois group and all of the intermediate fields of the splitting field (over $\mathbb{Q}$) of the polynomial $(x^3-2)(x^2-3)\in \mathbb{Q}[x]$.
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	Let $K$ be a subfield of the real numbers and $f\in K[x]$ an irreducible quartic.
	If $f$ has exactly two real roots,
	the Galois group of $f$ is $S_4$ or $D_4$.
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	Assume that $f(x)\in K[x]$ has distinct roots $u_1, u_2, ..., u_n$ in the splitting field $F$ and let $G=\text{Aut}_K F\leq S_n$ be the Galois group of $f$.
	Let $y_1, ..., y_n$ be indeterminates and define:
	$$g(x)=\prod_{\sigma\in S_n}(x-(u_{\sigma(1)}y_1+u_{\sigma(2)}y_2+\cdots +u_{\sigma(n)}y_n))$$
	\begin{enumerate}
	\item Show that $$g(x)=\prod_{\sigma\in S_n}(x-(u_1 y_{\sigma(1)}+u_2 y_{\sigma(2)}+\cdots +u_n y_{\sigma(n)})).$$
	\item Show that $g(x)\in K[y_1, ..., y_n, x]$.
	\item Suppose $g(x)$ factors as $g_1(x) g_2(x)\cdots g_r(x)$ with $g_i(x)\in K(y_1, ..., y_n)[x]$ monic irreducible.
	If $x-\sum_{i}u_{\sigma(i)}y_i$ is a factor of $g_1(x)$,
	then show that $$g_1(x)=\prod_{\tau\in G}(x-\sum_{i}u_{\tau\sigma(i)}y_i).$$
	Show that this implies that $\deg{g_i(x)}=|G|$.
	\item If $K=\mathbb{Q}$,
	$f\in \mathbb{Z}[x]$ is monic,
	and $p$ is a prime,
	let $\overline{f}\in \mathbb{Z}_p[x]$ be the polynomial obtained from $f$ by reducing the coefficients of $f\pmod{p}$.
	Assume $\overline{f}$ has distinct roots $\overline{u}_1, ..., \overline{u}_n$ in some splitting field $\overline{F}$ over $\mathbb{Z}_p$.
	Show that $$\overline{g}(x)=\prod_{\tau\in S_n}(x-\sum_{i}\overline{u}_i y_{\tau(i)})\in \overline{F}[x, y_1, ..., y_n].$$
	If the $\overline{u}_i$ are suitably ordered,
	then prove that the Galois group $\overline{G}$ of $\overline{f}$ is a subgroup of the Galois group $G$ of $f$.
	\item Show that $x^6+22x^5-9x^4+12x^3-37x^2-29x-15\in \mathbb{Q}[x]$ has Galois group $S_6$.
	[{\it Hint:} apply (d) with $p=2, 3, 5$.]
	\item The Galois group of $x^5-x-1\in \mathbb{Q}[x]$ is $S_5$.
	\end{enumerate}
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	Here is a method for constructing a polynomial $f\in \mathbb{Q}[x]$ with Galois group $S_n$ for a given $n>3$.
	It depends on the fact that there exist irreducible polynomials of every degree in $\mathbb{Z}_p[x]$ ($p$ prime; Corollary {V.5.9} below).
	First choose $f_1, f_2, f_3\in \mathbb{Z}[x]$ such that
	\begin{enumerate}[(i)]
	\item $\deg{f_1}=n$ and $\overline{f}_1\in \mathbb{Z}_2[x]$ is irreducible (notation as in Exercise {V.4.13}(d)).
	\item $\deg{f_2}=n$ and $\overline{f}_2\in \mathbb{Z}_3[x]$ factors in $\mathbb{Z}_3[x]$ as $gh$ with $g$ an irreducible of degree $n-1$ and $h$ linear;
	\item $\deg{f_3}=n$ and $\overline{f}_3\in \mathbb{Z}_5[x]$ factors as $gh$ or $gh_1 h_2$ with $g$ an irreducible quadratic in $\mathbb{Z}_5[x]$ and $h, h_1, h_2$ irreducible polynomials of odd degree in $\mathbb{Z}_5[x]$.
	\end{enumerate}
	\begin{enumerate}
	\item Let $f=-15f_1+10f_2+6f_3$. Then $f\equiv f_1\pmod{2}, f\equiv f_2\pmod{3}$, and $f\equiv f_3\pmod{5}$.
	\item The Galois group $G$ of $f$ is transitive (since $\overline{f}$ is irreducible in $\mathbb{Z}_2[x]$).
	\item $G$ contains a cycle of the type $\zeta=(i_1 i_2\cdots i_{n-1})$ and element $\sigma \lambda$ where $\sigma$ is a transposition and $\lambda$ a product of cycles of odd order.
	Therefore $\sigma\in G$, whence $(i_k i_n)\in G$ for some $k$ ($1\leq k\leq n-1$) by Exercise {I.6.3} and transitivity.
	\item $G=S_n$ (see part (c) and Exercise {I.6.4}(b)).
	\end{enumerate}
	\end{enumerate}
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\section{Finite Fields}
	{\it Note:} $F$ always denotes an extension field of a field $K$.
	\begin{enumerate}
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	If $F$ is a finite field of characteristic $p$,
	describe the structure of the additive group of $K$.
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	(Fermat) If $p\in \mathbb{Z}$ is prime,
	then $a^p=a$ for all $a\in \mathbb{Z}_p$ or equivalently,
	$c^p\equiv c\pmod{p}$ for all $c\in \mathbb{Z}$.
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	If $|K|=p^n$,
	then every element of $K$ has a unique $p$th root in $K$.
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	If the roots of a monic polynomial $f\in K[x]$ (in some splitting field of $f$ over $K$) are distinct and form a field,
	then $\text{char }K=p$ and $f=x^{p^n}-x$ for some $n\geq 1$.
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	\begin{enumerate}
	\item Construct a field with $9$ elements and give its addition and multiplication tables.
	\item Do the same for a field of $25$ elements.
	\end{enumerate}
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	If $|K|=q$ and $\gcd{(n, q)}=1$ and $F$ is a splitting field of $x^n-1_K$ over $K$,
	then $[F:K]$ is the least positive integer $k$ such that $n\mid (q^k-1)$.
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	If $|K|=q$ and $f\in K[x]$ is irreducible,
	then $f$ divides $x^{q^n}-x$ if and only if $\deg{f}$ divides $n$.
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	If $|K|=p^r$ and $|F|=p^n$,
	then $r\mid n$ and $\text{Aut}_K F$ is cyclic with generator $\varphi$ given by $u\mapsto u^{p^r}$.
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	If $n\geq 3$,
	then $x^{2^n}+x+1$ is reducible over $\mathbb{Z}_2$.
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	Every element in a finite field may be written as the sum of two squares.
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	Let $F$ be an algebraic closure of $\mathbb{Z}_p$ ($p$ prime).
	\begin{enumerate}
	\item $F$ is algebraic Galois over $\mathbb{Z}_p$.
	\item The map $\varphi:F\to F$ given by $u\mapsto u^p$ is a nonidentity $\mathbb{Z}_p$-automorphism of $F$.
	\item The subgroup $H=\langle \varphi\rangle$ is a proper subgroup of $\text{Aut}_{\mathbb{Z}_p}F$ whose fixed field is $\mathbb{Z}_p$,
	which is also the fixed field of $\text{Aut}_{\mathbb{Z}_p}F$ by (a).
	\end{enumerate}
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	If $K$ is finite and $F$ is an algebraic closure of $K$,
	then $\text{Aut}_K F$ is abelian.
	Every element of $\text{Aut}_K F$ (except $1_F$) has infinite order.
	\end{enumerate}
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\section{Separability}
	{\it Note:}
	Unless stated otherwise $F$ is always an extension field of a field $K$.
	\begin{enumerate}
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	Let $\text{char }K=p\neq 0$ and let $n\geq 1$ be an integer such that $\gcd{(p, n)}=1$.
	If $v\in F$ and $nv\in K$, then $v\in K$.
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	If $u\in F$ is purely inseparable over $K$,
	then $u$ is purely inseparable over any intermediate field $E$.
	Hence if $F$ is purely inseparable over $K$,
	then $F$ is purely inseparable over $E$.
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	If $F$ is purely inseparable over an intermediate field $E$ and $E$ is purely inseparable over $K$,
	then $F$ is purely inseparable over $K$.
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	If $u\in F$ is separable over $K$ and $v\in F$ is purely inseparable over $K$,
	then $K(u, v)=K(u+v)$.
	If $u\neq 0, v\neq 0$,
	then $K(u, v)=K(uv)$.
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	If $\text{char }K=p\neq 0$ and $a\in K$ but $a\notin K^p$,
	then $x^{p^n}-a\in K[x]$ is irreducible for every $n>1$.
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	If $f\in K[x]$ is monic irreducible,
	$\deg{f}\geq 2$, and $f$ has all its roots equal (in a splitting field),
	then $\text{char }K=p\neq 0$ and $f=x^{p^n}-a$ for some $n\geq 1$ and $a\in K$.
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	Let $F, K, S, P$ be as in Theorem {V.6.7} and suppose $E$ is an intermediate field.
	Then
	\begin{enumerate}
	\item $F$ is purely inseparable over $E$ if and only if $S\subseteq E$.
	\item If $F$ is separable over $E$, then $P\subseteq E$.
	\item If $E\cap S=K$, then $E\subseteq P$.
	\end{enumerate}
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	If $\text{char }K=p\neq 0$ and $[F:K]$ is finite and not divisible by $p$,
	then $F$ is separable over $K$.
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	Let $\text{char }K=p\neq 0$.
	Then an algebraic element $u\in F$ is separable over $K$ if and only if $K(u)=K(u^{p^n})$ for all $n\geq 1$.
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	Let $\text{char }K=p\neq 0$ and let $f\in K[x]$ be irreducible of degree $n$.
	Let $m$ be the largest nonnegative integer such that $f$ is a polynomial in $x^{p^m}$ but is not a polynomial in $x^{p^{m+1}}$.
	Then $n=n_0 p^m$.
	If $u$ is a root of $f$,
	then $[K(u):K]_s=n_0$ and $[K(u):K]_i=p^m$.
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	If $f\in K[x]$ is irreducible of degree $m>0$,
	and $\text{char }K$ does not divide $m$,
	then $f$ is separable.
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	$F$ is purely inseparable over $K$ if and only if $F$ is algebraic over $K$ and for any extension field $E$ of $F$,
	the only $K$-monomorphism $F\to E$ is the inclusion map.
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	\begin{enumerate}
	\item
	The following conditions on a field $K$ are equivalent:
	\begin{enumerate}[(i)]
	\item every irreducible polynomial in $K[x]$ is separable;
	\item every algebraic closure $\overline{K}$ of $K$ is Galois over $K$;
	\item every algebraic extension field of $K$ is separable over $K$;
	\item either $\text{char }K=0$ or $\text{char }K=p$ and $K=K^p$.
	\end{enumerate}
	A field $K$ that satisfies (i)-(iv) is said to be perfect.
	\item Every finite field is perfect.
	\end{enumerate}
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	If $F=K(u, v)$ with $u, v$ algebraic over $K$ and $u$ separable over $K$,
	then $F$ is a simple extension of $K$.
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	Let $\text{char }K=p\neq 0$ and assume $F=K(u, v)$ where $u^p\in K, v^p\in K$ and $[F:K]=p^2$.
	Then $F$ is not a simple extension of $K$.
	Exhibit an infinite number of intermediate fields.
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	Let $F$ be an algebraic extension of $K$ such that every polynomial in $K[x]$ has a root in $F$.
	Then $F$ is an algebraic closure of $K$.
	[{\it Hint:} Theorem {V.3.14} and Theorem {V.6.7} and Proposition {V.6.15} may be helpful.]
	\end{enumerate}
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\section{Cyclic Extensions}
	\begin{enumerate}
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	If $\overline{K}$ is replaced by any normal extension $N$ of $K$ containing $F$ in Definition {V.7.1},
	then this new definition of norm and trace is equivalent to the orginal one.
	In particular,
	the new definition does not depend on the choice of $N$.
	See Exercise {V.3.21}.
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	Let $F$ be a finite dimensional extension of a finite field $K$.
	The norm ${N_K}^F$ and the trace ${T_K}^F$ (considered as maps $F\to K$) are surjective.
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	Let $\overline{\mathbb{Q}}$ be a (fixed) algebraic closure of $\mathbb{Q}$ and $v\in \overline{\mathbb{Q}}, v\notin \mathbb{Q}$.
	Let $E$ be a subfield of $\overline{\mathbb{Q}}$ maximal with respect to the condition $v\notin E$.
	Prove that every finite dimensional extension of $E$ is cyclic.
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	Let $K$ be a field,
	$\overline{K}$ an algebraic closure of $K$ and $\sigma\in \text{Aut}_K\overline{K}$.
	Let $$F=\{u\in \overline{K}\mid \sigma(u)=u\}.$$
	Then $F$ is a field and every finite dimensional extension of $F$ is cyclic.
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	If $F$ is a cyclic extension of $K$ of degree $p^n$ ($p$ prime) and $L$ is an intermediate field such that $F=L(u)$ and $L$ is cyclic over $K$ of degree $p^{n-1}$,
	then $F=K(u)$.
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	If $\text{char }F=p\neq 0$,
	let $K_p=\{u^p-u\mid u\in K\}$.
	\begin{enumerate}
	\item A cyclic extension field $F$ of $K$ of degree $p$ exists if and only if $K\neq K_p$.
	\item If there exists a cyclic extension of degree $p$ of $K$,
	then there exists a cyclic extension of degree $p^n$ for every $n\geq 1$.
	[{\it Hint:} Use induction;
	if $E$ is cyclic over $K$ of degree $p^{n-1}$ with $\text{Aut}_K E$ generated by $\sigma$,
	show that there exist $u, v\in E$ such that ${T_K}^E(v)=1_K$ and $\sigma(u)-u=v^p-v$.
	Then $x^p-x-u\in E[x]$ is irreducible and if $w$ is a root,
	then $K(w)$ is cyclic of degree $p^n$ over $K$.]
	\end{enumerate}
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	If $n$ is an odd integer such that $K$ contains a primitive $n$th root of unity and $\text{char }K\neq 2$,
	then $K$ also contains a primitive $2n$th root of unity.
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	If $F$ is a finite dimensional extension of $\mathbb{Q}$,
	then $F$ contains only a finite number of roots of unity.
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	Which roots of unity are contained in the following fields:
	$\mathbb{Q}(i)$, $\mathbb{Q}(\sqrt{2})$, $\mathbb{Q}(\sqrt{3})$, $\mathbb{Q}(\sqrt{5})$, $\mathbb{Q}(\sqrt{-2})$, $\mathbb{Q}(\sqrt{-3})$?
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	\begin{enumerate}
	\item Let $p$ be a prime and assume either
	\begin{enumerate}[(i)]
	\item $\text{char }K=p$ or
	\item $\text{char }K\neq p$ and $K$ contains a primitive $p$th root of unity.
	\end{enumerate}
	Then $x^p-a\in K[x]$ is either irreducible or splits in $K[x]$.
	\item If $\text{char }K=p\neq 0$,
	then for any root $u$ of $x^p-a\in K[x]$,
	$K(u)\neq K(u^p)$ if and only if $[K(u):K]=p$.
	\end{enumerate}
	\end{enumerate}
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\section{Cyclotomic Extensions}
	\begin{enumerate}
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	If $i\in \mathbb{Z}$,
	let $\overline{i}$ denote the image of $i$ in $\mathbb{Z}_n$ under the canonical projection $\mathbb{Z}\to \mathbb{Z}_n$.
	Prove that $\overline{i}$ is a unit in the ring $\mathbb{Z}_n$ if and only if $\gcd{(i, n)}=1$.
	Therefore the multiplicative group of units in $\mathbb{Z}_n$ has order $\varphi(n)$.
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	Establish the following properties of the Euler function $\varphi$.
	\begin{enumerate}
	\item If $p$ is prime and $n>0$,
	then $\varphi(p^n)=p^n(1-1/p)=p^{n-1}(p-1)$.
	\item If $\gcd{(m, n)}=1$, then $\varphi(mn)=\varphi(m)\varphi(n)$.
	\item If $n=p_1^{k_1}\cdots p_r^{k_r}$ ($p_i$ distinct primes; $k_i>0$),
	then $\varphi(n)=n(1-1/p_1)(1-1/p_2)\cdots (1-1/p_r)$.
	\item $\sum_{d\mid n}\varphi(d)=n$.
	\item $\varphi(n)=\sum_{d\mid n}d\mu(n/d)$,
	where $\mu$ is the Moebius function defined by
	$$\mu(n)=
	\left\{\begin{array}{ll}
	1 & \text{if }n=1 \\
	(-1)^t & \text{if }n \text{ is a product of }t\text{ distinct primes}\\
	0 & \text{if }p^2\text{ divides }n\text{ for some prime }p.
	\end{array}\right.$$
	\end{enumerate}
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	Let $\varphi$ be the Euler function.
	\begin{enumerate}
	\item $\varphi(n)$ is even for $n>2$.
	\item Find all $n>0$ such that $\varphi(n)=2$.
	\item Find all pairs $(n, p)$ (where $n, p>0$, and $p$ is prime) such that $\varphi(n)=n/p$. [See Exercise {V.8.2}.]
	\end{enumerate}
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	\begin{enumerate}
	\item If $p$ is an odd prime and $n>0$,
	then the multiplicative group of units in the ring $\mathbb{Z}_{p^n}$ is cyclic of order $p^{n-1}(p-1)$.
	\item Part (a) is also true if $p=2$ and $1\leq n\leq 2$.
	\item If $n\geq 3$,
	then the multiplicative grou pof units in $\mathbb{Z}_{2^n}$ is isomorphic to $\mathbb{Z}_2\oplus \mathbb{Z}_{2^{n-2}}$.
	\end{enumerate}
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	If $f(x)=\sum_{i=0}^{t}a_i x^i$,
	let $f(x^s)$ be the polynomial $\sum_{i=0}^{t}a_i x^{is}$.
	Establish the following properties of the cyclotomic polynomials $g_n(x)$ over $\mathbb{Q}$.
	\begin{enumerate}
	\item If $p$ is prime and $k\geq 1$,
	then $g_{p^k}(x)=g_p(x^{p^k-1})$.
	\item If $n=p_1^{r_1}\cdots p_k^{r_k}$ ($p_i$ distinct primes; $r_i>0$),
	then $$g_n(x)=g_{p_1\cdots p_k}(x^{p_1^{r_1-1}\cdots p_k^{r_k-1}}).$$
	\item If $n$ is odd, then $g_{2n}(x)=g_n(-x)$.
	\item If $p$ is a prime and $p\nmid n$, then $g_{pn}(x)=g_n(x^p)/g_n(x)$.
	\item $g_n(x)=\prod_{d\mid n}(x^{n/d}-1)^{\mu(d)}$, where $\mu$ is the Moebius function of Exercise {V.8.2}(e).
	\item $g_n(1)=p$ if $n=p^k$ ($k>0$), $0$ if $n=1$, and $1$ otherwise.
	\end{enumerate}
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	Calculate the $n$th cyclotomic polynomials over $\mathbb{Q}$ for all positive $n$ with $n\leq 20$.
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	Let $F_n$ be a cyclotomic extension of $\mathbb{Q}$ of order $n$.
	Determine the structure of $\text{Aut}_{\mathbb{Q}}F_n$ for every $n$.
	[{\it Hint:} if ${U_n}^*$ denotes the multiplicative group of units in $\mathbb{Z}_n$,
	then show that ${U_n}^*=\prod_{i=1}^{r}{U_{p_i^{n_i}}}^*$ where $n$ has prime decomposition $n=p_1^{n_1}\cdots p_r^{n_r}$.
	Apply Exercise {V.8.4}.]
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item Let $F_n$ be a cyclotomic extension of $\mathbb{Q}$ of order $n$.
	\begin{enumerate}
	\item Determine $\text{Aut}_{\mathbb{Q}}F_5$ and all intermediate fields.
	\item Do the same for $F_8$.
	\item Do the same for $F_7$;
	if $\zeta$ is a primitive $7$th root of unity what is the irreducible polynomial over $\mathbb{Q}$ of $\zeta+\zeta^{-1}$?
	\end{enumerate}
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	If $n>2$ and $\zeta$ is a primitive $n$th root of unity over $\mathbb{Q}$,
	then $[\mathbb{Q}(\zeta+\zeta^{-1}):\mathbb{Q}]=\varphi(n)/2$.
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	(Wedderburn) A finite division ring $D$ is a field.
	Here is an outline of the proof (in which $E^*$ denotes the multiplicative group of nonzero elements of a division ring $E$).
	\begin{enumerate}
	\item The center $K$ of $D$ is a field and $D$ is a vector space over $K$,
	whence $|D|=q^n$ where $q=|K|\geq 2$.
	\item If $0\neq a\in D$, then $N(a)=\{d\in D\mid da=ad\}$ is a subdivision ring of $D$ containing $K$.
	Furthermore, $|N(a)|=q^r$ where $r\mid n$.
	\item If $0\neq a\in D-K$, then $N(a)^*$ is the centralizer of $a$ in the group $D^*$ and $[D^*:N(a)^*]=(q^n-1)/(q^r-1)$ for some $r$ such that $1\leq r<n$ and $r\mid n$.
	\item $q^n-1=q-1+\sum_{r}(q^n-1)/(q^r-1)$, where the last sum taken over a finite number of integers $r$ such that $1\leq r<n$ and $r\mid n$.
	[{\it Hint:} use the class equation of $D^*$;
	see pp. 90-91.]
	\item For each primitive $n$th root of unity $\zeta\in \mathbb{C}$,
	$|q-\zeta|>q-1$,
	where $|a+bi|=\sqrt{a^2+b^2}$ for $a+bi\in \mathbb{C}$.
	Consequently,
	$|g_n(q)|>q-1$,
	where $g_n$ is the $n$th cyclotomic polynomial over $\mathbb{Q}$.
	\item The equation in (d) is impossible unless $n=1$,
	whence $K=D$.
	[{\it Hint:} Use Proposition {V.8.2} to show that for each positive divisor $r$ of $n$ with $r\neq n$,
	$f_r(x)=(x^n-1)/(x^r-1)$ is in $\mathbb{Z}[x]$ and $f_r(x)=g_n(x)h_r(x)$ for some $h_r(x)\in \mathbb{Z}[x]$.
	Consequently, for each such $rg_n(q)$ divides $f_r(q)$ in $\mathbb{Z}$,
	whence $g_n(q)\mid (q-1)$ by (d).
	This contradicts (e).]
	\end{enumerate}
	\end{enumerate}
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\section{Rational Extensions}
	\begin{enumerate}
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	If $F$ is a radical extension field of $K$ and $E$ is an intermediate field,
	then $F$ is a radical extension of $E$.
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	Suppose that ``radical extension'' is defined as follows:
	$F$ is a radical extension of $K$ if there is a finite tower of fields $K=E_0\subseteq E_1\subseteq \cdots \subseteq E_n=F$ such that for each $1\leq i\leq n$,
	$E_i=E_{i-1}(u_i)$ and one of the following is true:
	\begin{enumerate}[(i)]
	\item $u_i^{m_i}\in E_{i-1}$ for some $m_i>0$;
	\item $\text{char }K=p$ and $u^p-u\in E_{i-1}$.
	State and prove the analogues of Theorem {V.9.4}. Proposition {V.9.6}, Corollary {V.9.7}, and Proposition {V.9.8}.
	\end{enumerate}
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	Let $K$ be a field,
	$f\in K[x]$ an irreducible polynomial of degree $n\geq 5$ and $F$ a splitting field of $f$ over $K$.
	Assume that $\text{Aut}_K F\cong S_n$.
	(See the example following Theorem {V.4.12}).
	Let $u$ be a root of $f$ in $F$.
	Then
	\begin{enumerate}
	\item $K(u)$ is not Galois over $K$;
	$[K(u):K]=n$ and $\text{Aut}_K K(u)=1$ (and hence is solvable).
	\item Every normal closure over $K$ that contains $u$ also contains an isomorphic copy of $F$.
	\item There is no radical extension field $E$ of $K$ such that $E\supseteq K(u)\supseteq K$.
	\end{enumerate}
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	If $F$ is a radical extension field of $E$ and $E$ is a radical extension field of $K$,
	then $F$ is a radical extension of $K$.
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	(Cardan) Let $K$ be a field with $\text{char }K\neq 2, 3$ and consider the cubic equation $x^3+a_1 x^2+a_2 x+a_3=0$ ($a_i\in K$).
	Let $p=a_2-\frac{a_1^2}{3}$ and $q=\frac{2a_1^3}{27}-\frac{a_1 a_2}{3}+a_3$.
	Let $P=\sqrt[3]{-q/2+\sqrt{p^3/27+q^2/4}}$ and $Q=\sqrt[3]{-q/2-\sqrt{p^3/27+q^2/4}}$ (with cube roots chosen properly).
	Then the solutions of the given equation are $P+Q-a_1/3$; $\omega P+\omega^2 Q-a_1/3$;
	and $\omega^2 P+\omega Q-a_1/3$ where $\omega$ is a primitive cube root of unity.
	\end{enumerate}
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\chapter{The Structure of Fields}
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\section{Transcendence Bases}
	{\it Note:}
	$F$ is always an extension field of a field $K$.
	\begin{enumerate}
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	(Exchange property) Let $S$ be a subset of $F$.
	If $u\in F$ is algebraic over $K(S)$ and $u$ is not algebraic over $K(S-\{v\})$,
	where $v\in S$,
	then $v$ is algebraic over $K((S-\{v\})\cup \{u\})$.
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	\begin{enumerate}
	\item Use Zorn's Lemma to show that every field extension possesses a transcendence base.
	\item Every algebraically independent subset of $F$ is contained in a transcendence base.
	\end{enumerate}
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	$\{x_1, ..., x_n\}$ is a transcendence base of $K(x_1, ..., x_n)$.
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	If $E_1, E_2$ are intermediate fields, then
	\begin{enumerate}
	\item $\text{tr.d.}E_1 E_2/K\geq \text{tr.d.}E_i/K$ for $i=1, 2$;
	\item $\text{tr.d.}E_1 E_2/K\leq (\text{tr.d.}E_1/K)+(\text{tr.d.}E_2/K)$.
	\end{enumerate}
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	If $F=K(u_1, ..., u_n)$ is a finitely generated extension of $K$ and $E$ is an intermediate field,
	then $E$ is a finitely generated extension of $K$.
	[{\it Note:} the algebraic case is trivial by Theorem {V.1.11} and Theorem {V.1.12}.]
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	\begin{enumerate}
	\item If $S$ is a transcendence base of the field $\mathbb{C}$ of complex numbers over the field $\mathbb{Q}$ of rationals,
	then $S$ is infinite.
	[{\it Hint:} Show that if $S$ is finite, then
	$$|\mathbb{Q}(S)|=|\mathbb{Q}(x_1, ..., x_n)|=|\mathbb{Q}[x_1, ..., x_n]|=|\mathbb{Q}|<|\mathbb{C}|$$
	see Exercise {O.8.3} and Exercise {O.8.9} of the Introduction and Theorem {VI.1.2}).
	But Lemma {V.3.5} implies $|\mathbb{Q}(S)|=|\mathbb{C}|$.
	\item There are infinitely many distinct automorphisms of the field $\mathbb{C}$.
	\item $\text{tr.d.}\mathbb{C}/\mathbb{Q}=|\mathbb{C}|$.
	\end{enumerate}
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	If $F$ is algebraically closed and $E$ an intermediate field such that $\text{tr.d.}E/K$ is finite,
	then any $K$-monomorphism $E\to F$ extends to a $K$-automorphism of $F$.
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	If $F$ is algebraically closed and $\text{tr.d.}F/K$ is finite,
	then every $K$-monomorphism $F\to F$ is in fact an automorphism.
	\end{enumerate}
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\section{Linear Disjointness and Separability}
	{\it Note.} $E$ and $F$ are always extension fields of a field $K$,
	and $C$ is an algebraically closed field containing $E$ and $F$.
	\begin{enumerate}
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	The subring $E[F]$ generated by $E$ and $F$ is a vector space over $K$ in the obvious way.
	The tensor product $E\otimes_K F$ is also a $K$-vector space (see Theorem {IV.5.5} and Corollary {IV.5.12}).
	$E$ and $F$ are linearly disjoint over $K$ if and only if the $K$-linear transformation $E\otimes_K F\to E[F]$ (given on generators of $E\otimes_K F$ by $a\otimes b\mapsto ab$) is an isomorphism.
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	Assume $E$ and $F$ are the quotient fields of integral domains $R$ and $S$ respectively.
	Then $C$ is an $R$-module and and $S$-module in the obviously way.
	\begin{enumerate}
	\item $E$ and $F$ are linearly disjoint over $K$ if and only if every subset of $R$ that is linearly independent over $K$ is also linearly independent over $S$.
	\item Assume further that $R$ is a vector space over $K$.
	Then $E$ and $F$ are linearly disjoint over $K$ if and only if every basis of $R$ over $K$ is linearly independent over $F$.
	\item Assume that both $R$ and $S$ are vector spaces over $K$.
	Then $E$ and $F$ are linearly disjoint over $K$ if and only if for every basis $X$ of $R$ over $K$ and basis $Y$ of $S$ over $K$,
	the set $\{uv\mid u\in X; v\in Y\}$ is linearly independent over $K$.
	\end{enumerate}
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	Use Exercise {VI.2.1} to prove Theorem {VI.2.2}.
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	Use Exercise {VI.2.1} and the associativity of the tensor product to prove Theorem {VI.2.4}.
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	If $\text{char }K=p\neq 0$,
	then
	\begin{enumerate}
	\item $K^{1/p^n}$ is a field for every $n\geq 0$. See Exercise {III.1.11}.
	\item $K^{1/p^{\infty}}$ is a field.
	\item $K^{1/p^n}$ is a splitting field over $K$ of $\{x^{p^n}-k\mid k\in K\}$.
	\end{enumerate}
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	If $\{u_1, ..., u_n\}$ is algebraically independent over $F$,
	then $F$ and $K(u_1, ..., u_n)$ are linearly disjoint over $K$.
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	If $E$ is a purely transcendental extension of $K$ and $F$ is algebraic over $K$,
	then $E$ and $F$ are linearly disjoint over $K$.
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	Let $K=\mathbb{Z}_p, F=\mathbb{Z}_p(x)$, and $E=\mathbb{Z}_p(x^p)$.
	\begin{enumerate}
	\item $F$ is separably generated and separable over $K$.
	\item $E\neq F$.
	\item $F$ is algebraic and purely inseparable over $E$.
	\item $\{x^p\}$ is a transcendence base of $F$ over $K$ which is not a separating transcendence base.
	\end{enumerate}
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	Let $\text{char }K=p\neq 0$ and let $u$ be transcendental over $K$.
	Suppose $F$ is generated over $K$ by $\{u, v_1, v_2, ...\}$,
	where $v_i$ is a root of $x^{p^i}-u\in K(u)[x]$ for $i=1, 2, ...$.
	Then $F$ is separable over $K$,
	but $F$ is not separably generated over $K$.
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	\begin{enumerate}
	\item $K$ is a perfect field if and only if every field extension of $K$ is separable (see Exercise {V.6.13}).
	\item (Mac Lane) Assume $K$ is a perfect field,
	$F$ is not perfect and $\text{tr.d.}F/K=1$.
	Then $F$ is separably generated over $K$.
	\end{enumerate}
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	$F$ is a purely inseparable over $K$ if and only if the only $K$-monomorphism $F\to C$ is the inclusion map.
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	$E$ and $F$ are free over $K$ if every subset $X$ of $E$ that is algebraically independent over $K$ is also algebraically independent over $F$.
	\begin{enumerate}
	\item The definition is symmetric (that is, $E$ and $F$ are free over $K$ if and only if $F$ and $E$ are free over $K$).
	\item If $E$ and $F$ are linearly disjoint over $K$,
	then $E$ and $F$ are free over $K$.
	Show by example that the converse if false.
	\item If $E$ is separable over $K$ and $E$ and $F$ are free over $K$,
	then $EF$ is separable over $F$.
	\item If $E$ and $F$ are free over $K$ and both separable over $K$,
	then $EF$ is separable over $K$.
	\end{enumerate}
	\end{enumerate}
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\chapter{Linear Algebra}
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\section{Matrices and Maps}
	{\it Note:} All matrices are assumed to have entries in a ring $R$ with identity.
	\begin{enumerate}
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	Let $R$ be commutative.
	\begin{enumerate}
	\item If the matrix product $AB$ is defined,
	then so is the product $B^t A^t$ and $(AB)^t=B^t A^t$.
	\item If $A$ is invertible, then so is $A^t$ and $(A^t)^{-1}=(A^{-1})^t$.
	\item If $R$ is not commutative, then (a) and (b) may be false.
	\end{enumerate}
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	A matrix $(a_{ij})\in \text{Mat}_n R$ is said to be
	$$\begin{array}{lllll}
	\text{(upper) triangular} & \Leftrightarrow & a_{ij}=0 & \text{for} & j<i;\\
	\text{strictly triangular} & \Leftrightarrow & a_{ij}=0 & \text{for} & j\leq i.
	\end{array}$$
	Prove that the set of all diagonal matrices is a subring of $\text{Mat}_n R$ which is (ring) isomorphic to $R\times \cdots \times R$ ($n$ factors).
	Show that the set $T$ of all triangular matrices is a subring of $\text{Mat}_n R$ and the set $I$ of all strictly triangular matrices is an ideal in $T$.
	Identify the quotient ring $T/I$.
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	\begin{enumerate}
	\item The center of the ring $\text{Mat}_n R$ consists of all matrices of the form $r I_n$,
	where $r$ is in the center of $R$.
	[{\it Hint:} every matrix in the center of $\text{Mat}_n R$ must commute with each of the matrices $B_{r, s}$,
	where $B_{r, s}$ has $1_R$ in position $(r, s)$ and $0$ elsewhere.]
	\item The center of $\text{Mat}_n R$ is isomorphic to the center of $R$.
	\end{enumerate}
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	The set of all $m\times n$ matrices over $R$ is a free $R$-module with a basis of $mn$ elements.
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	A matrix $A\in \text{Mat}_n R$ is symmetric if $A=A^t$ and skew-symmetric if $A=-A^t$.
	\begin{enumerate}
	\item If $A$ and $B$ are [skew] symmetric, then $A+B$ is [skew] symmetric.
	\item Let $R$ be commutative. If $A, B$ are symmetric, then $AB$ is symmetric if and only if $AB=BA$.
	Also show that for any matrix $B\in \text{Mat}_n R$, $BB^t$ and $B+B^t$ are symmetric and $B-B^t$ is skew-symmetric.
	\end{enumerate}
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	If $R$ is a division ring and $A, B\in \text{Mat}_n R$ are such that $BA=I_n$,
	then $AB=I_n$ and $B=A^{-1}$.
	[{\it Hint:} use linear transformations.]
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	Similarity of matrices is an equivalence relation on $\text{Mat}_n R$.
	Equivalence of matrices is an equivalence relation on the set of all $m\times n$ matrices over $R$.
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	Let $E, F$ be finite dimensional (left) vector spaces over a field and consider the dual spaces to be {\it left} vector spaces in the usual way.
	If $A$ is the matrix of a linear transformation $f:E\to F$,
	then $A^t$ is the matrix of the dual map $\overline{f}:F^* \to E^*$.
	\end{enumerate}
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\section{Rank and Equivalence}
	\begin{enumerate}
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	Let $f, g:E\to E, h:E\to F, k:F\to G$ be linear transformations of left vector spaces over a division ring $D$ with $\dim_D E=n, \dim_D F=m, \dim_D G=p$.
	\begin{enumerate}
	\item $\text{Rank }(f+g)\leq \text{rank }f+\text{rank }g$.
	\item $\text{Rank }(kh)\leq \min{\{\text{rank }h, \text{rank }k\}}$.
	\item $\text{Nullity }kh\leq \text{nullity }h+\text{nullity }k$.
	\item $\text{Rank }f+\text{rank }g-n\leq \text{rank }fg\leq \min{\{\text{rank }f, \text{rank }g\}}$.
	\item $\max{\{\text{nullity }g, \text{nullity }h\}}\leq \text{nullity }hg$.
	\item If $m\neq n$, then (e) is false for $h$ and $k$.
	\end{enumerate}
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	An $n\times m$ matrix $A$ over a division ring $D$ has an $m\times n$ left inverse $B$ (that is, $BA=I_m$) if and only if $\text{rank }A=m$.
	$A$ has an $m\times n$ right inverse $C$ (with $AC=I_n$) if and only if $\text{rank }A=n$.
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	If $(c_{i1}, c_{i2}, ..., c_{im})$ is a nonzero row of a matrix $(c_{ij})$,
	then its leading entry is $c_{it}$ where $t$ is the first integer such that $c_{it}\neq 0$.
	A matrix $C=(c_{ij})$ over a division ring $D$ is said to be in reduced row echelon form provided:
	\begin{enumerate}[(i)]
	\item for some $r\geq 0$ the first $r$ rows of $C$ are nonzero (row vectors) and all other rows are zero;
	\item the leading entry of each nonzero row is $1_D$;
	\item if $c_{ij}=1_D$ is the leading entry of row $i$, then $c_{kj}=0$ for all $k\neq i$;
	\item if $c_{1 j_1}, c_{2 j_2}, ..., c_{r j_r}$ are the leading entries of rows $1, 2, ..., r$,
	then $j_1<j_2<\cdots <j_r$.
	\end{enumerate}
	\begin{enumerate}
	\item If $C$ is in reduced row echelon form,
	then $\text{rank }C$ is the number of nonzero rows.
	\item If $A$ is any matrix over $D$,
	then $A$ may be changed to a matrix in reduced row echelon form by a finite sequence of elementary {\it row} operations.
	\end{enumerate}
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	\begin{enumerate}
	\item The system of $n$ linear equations in $m$ unknowns $x_i$ over a field $K$
	$$\begin{array}{ccccccccc}
	a_{11}x_1 & + & a_{12} x_2 & + & \cdots & + & a_{1m} x_m & = & b_1 \\
	\vdots & \vdots & \vdots & \vdots & \ddots & \vdots & \vdots & \vdots & \vdots \\
	a_{n1}x_1 & + & a_{n2} x_2 & + & \cdots & + & a_{nm} x_m & = & b_n \\
	\end{array}$$
	has a (simultaneous) solution if and only if the matrix equation $AX=B$ has a solution $X$,
	where $A$ is the $n\times m$ matrix $(a_{ij})$,
	$X$ is the $m\times 1$ column vector $(x_1 x_2 \cdots x_m)^t$ and $B$ is the $n\times 1$ column vector $(b_1 b_2 \cdots b_n)^t$.
	\item If $A_1, B_1$ are matrices obtained from $A, B$ respectively by performing the same sequence of elementary {\it row} operations on both $A_1$ and $B_1$ then $X$ is a solution of $AX=B$ if and only if $X$ is a solution of $A_1 X=B_1$.
	\item Let $C$ be the $n\times (m+1)$ matrix
	$$\begin{pmatrix}
	a_{11} & \cdots & a_{1m} & b_1 \\
	\vdots & \ddots & \vdots & \vdots \\
	a_{n1} & \cdots & a_{nm} & b_n
	\end{pmatrix}$$
	Then $AX=B$ has solution if and only if $\text{rank }A=\text{rank }C$.
	In this case the solution is unique if and only if $\text{rank }A=m$.
	[{\it Hint:} use (b) and Exercise {VII.2.3}.]
	\item The system $AX=B$ is homogeneous if $B$ is the zero column vector.
	A homogeneous system $AX=B$ has a nontrivial solution (that is,
	not all $x_i=0$)
	if and only if $\text{rank }A<m$ (in particular, if $n<m$).
	\end{enumerate}
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	Let $R$ be a principal ideal domain.
	For each positive integer $r$ and sequence of nonzero ideals $I_1\supseteq I_2\supseteq \cdots \supseteq I_r$ choose a sequence $d_1, ..., d_r\in R$ such that $\langle d_j\rangle=I_j$ and $d_1\mid d_2\mid \cdots \mid d_r$.
	For a given pair of positive integers $\gcd{(n, m)}$,
	let $S$ be the set of all $n\times m$ matrices of the form $\begin{pmatrix}
	L_r & 0 \\
	0 & 0 \\
	\end{pmatrix}$,
	where $r=1, 2, ..., \min{(n, m)}$ and $L_r$ is an $r\times r$ diagonal matrix with main diagonal one of the chosen sequences $d_1, ..., d_r$.
	Show that $S$ is a set of canonical forms under equivalence for the set of all $n\times m$ matrices over $R$.
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	\begin{enumerate}
	\item If $f:E\to F$ is a linear transformation of finite dimensional vector spaces over a division ring,
	then there exist bases $\{u_1, ..., u_n\}$ of $E$ and $\{v_1, ..., v_m\}$ of $F$ and an integer $r$ ($r\leq \min{(m, n)}$) such that $f(u_i)=v_i$ for $i=1, 2, ..., r$ and $f(u_i)=0$ for $i=r+1, ..., n$.
	\item State and prove a similar result for free modules of finite rank over a principal ideal domain [see Proposition {VII.2.11}].
	\end{enumerate}
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	Let $R$ be a Euclidean domain with ``degree function'' $\phi:R\to \{0\}\to \mathbb{N}$ (Definition {III.3.8}).
	(For example, let $R=\mathbb{Z}$).
	\begin{enumerate}
	\item If $A=\begin{pmatrix}
	a & b \\
	c & d \\
	\end{pmatrix}$ is a $2\times 2$ matrix over $R$ then $A$ can be changed to a diagonal matrix $D$ by a finite sequence of elementary row and column operations.
	[{\it Hint:} If $a\neq 0, b\neq 0$,
	then $b=aq+r$ with $r=0$, or $r\neq 0$ and $\phi(r)<\phi(a)$.
	Performing suitable elementary column operations yields:
	$$\begin{pmatrix}
	a&b\\c&d\\
	\end{pmatrix}
	\to
	\begin{pmatrix}
	a&b-aq\\c&d-cq\\
	\end{pmatrix}
	=
	\begin{pmatrix}
	a&r\\c&*\\
	\end{pmatrix}
	\to
	\begin{pmatrix}
	r&a\\*&c\\
	\end{pmatrix}.$$
	Since $\phi(r)<\phi(a)$,
	repetitions of this argument change $A$ to $B=\begin{pmatrix}
	s&0\\u&*\\
	\end{pmatrix}$
	with $\phi(s)<\phi(a)$ if $s\neq 0$.
	If $u\neq 0$,
	a similar argument with rows changes $B$ to $C=\begin{pmatrix}
	t&w\\0&*\\
	\end{pmatrix}$ with $\phi(t)<\phi(s)<\phi(a)$ if $t\neq 0$;
	(and possibly $w\neq 0$).
	Since the degrees of the $(1, 1)$ entries are strictly decreasing,
	a repetition of these arguments must yield a diagonal matrix $D=\begin{pmatrix}
	d_1&0\\0&d_2\\
	\end{pmatrix}$
	after a finite number of steps.]
	\item If $A$ is invertible,
	then $A$ is a product of elementary matrices.
	[{\it Hint:} By (a) and the proof of Corollary {VII.2.10} $D=PAQ$ with $P, Q$ invertible,
	whence $D$ is invertible and $d_1, d_2$ are units in $R$.
	Thus $A=P^{-1}\begin{pmatrix}
	d_1&0\\0&1_R\\
	\end{pmatrix}
	\begin{pmatrix}
	1_R&0\\
	0&d_2\\
	\end{pmatrix}
	Q^{-1}$;
	use Corollary {VII.2.9}.]
	\item Every $n\times m$ secondary matrix (see the proof of Proposition {VII.2.11}) over a Euclidean domain is a product of elementary matrices.
	\end{enumerate}
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	\begin{enumerate}
	\item An invertible matrix over a principal ideal domain is a product of elementary and secondary matrices.
	\item An invertible matrix over a Euclidean domain is a product of elementary matrices [see Exercise {VII.2.7}].
	\end{enumerate}
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	Let $n_1, n_2, ..., n_t, n$ be positive integers such that $n_1+n_2+\cdots +n_t=n$ and for each $i$ let $M_i$ be an $n_i\times n_i$ matrix.
	Let $M$ be the $n\times n$ matrix
	$$\begin{pmatrix}
	M_1 & & & & \\
	& M_2 & & O & \\
	& & \ddots & & \\
	& O & & \ddots & \\
	& & & & & M_t\\
	\end{pmatrix}$$
	where the main diagonal of each $M_i$ lies on the main diagonal of $M$.
	For each permutation $\sigma$ of $\{1, 2, ..., t\}$,
	$M$ is similar to the matrix
	$$\sigma M=\begin{pmatrix}
	M_{\sigma 1} & & & & \\
	& M_{\sigma 2} & & O & \\
	& & \ddots & & \\
	& O & & \ddots & \\
	& & & & & M_{\sigma t}\\
	\end{pmatrix}.$$
	[{\it Hint:} If $t=3, \sigma=(13)$, and $P=\begin{pmatrix}
	0 & & I_{n_3} \\
	& I_{n_2} & \\
	I_{n_1} & & 0 \\
	\end{pmatrix}$,
	then $P^{-1}=\begin{pmatrix}
	0 & & I_{n_1} \\
	& I_{n_2} & \\
	I_{n_3} & & 0 \\
	\end{pmatrix}$ and $PMP^{-1}=\sigma M$.
	In the general case adapt the proof of results 2.8-2.10 (Theorem {VII.2.8}, Corollary {VII.2.9}, Corollary {VII.2.10}.]
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	Given the set $\{a_1, ..., a_n\}$ and the words $w_1, w_2, ..., w_r$ (on the $a_i$),
	let $F^*$ be the free (nonabelian multiplicative) group on the set $\{a_1, ..., a_n\}$ and let $M$ be the normal subgroup generated by the words $w_1, w_2, ..., w_r$ (see Section {I.9}).
	Let $N$ be the normal subgroup generated by all words of the form $a_i a_j a_i^{-1} a_j^{-1}$.
	\begin{enumerate}
	\item $F^*/M$ is the group defined by generators $\{a_1, ..., a_n\}$ and relations $\{w_1=w_2=\cdots =w_r=e\}$ (Definition {I.9.4}).
	\item $F^*/N$ is the free abelian group on $\{a_1, ..., a_n\}$ (see Exercise {II.1.12}).
	\item $F^*/(M\vee N)$ is (in multiplicative notation) the abelian group defined by generators $\{a_1, ..., a_n\}$ and relations $\{w_1=w_2=\cdots =w_r=e\}$ (see p. 343).
	\item There are group epimorphisms $F^*\to F^*/N\to F^*/(M\vee N)$.
	\end{enumerate}
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	Let $F$ be a free abelian group with basis $\{a_1, ..., a_m\}$.
	Let $K$ be the subgroup of $F$ generated by $b_1=r_{11}a_1+\cdots +r_{1m}a_m, ..., b_n=r_{n1}a_1+\cdots +r_{nm}a_m$ ($r_{ij}\in \mathbb{Z}$).
	\begin{enumerate}
	\item For each $i$, both $\{b_1, ..., b_{i-1}, -b_i, b_{i+1}, ..., b_n\}$ and $\{b_1, ..., b_{i-1}, b_i+rb_j, b_{i+1}, ..., b_n\}$ ($r\in \mathbb{Z}$; $i\neq j$) generate $K$.
	[See Lemma {II.1.5}.]
	\item For each $i$ $\{a_1, ..., a_{i-1}, -a_i, a_{i+1}, ..., a_n\}$ is a basis of $F$ relative to which $b_j=r_{j1}a_1+\cdots +r_{j, i-1}a_{i-1}-r_{ji}(-a_i)+r_{j, i+1}a_{i+1}+\cdots +r_{jm}a_m$.
	\item For each $i$ and $j\neq i$ $\{a_1, ..., a_{j-1}, a_j-ra_i, a_{j+1}, ..., a_m\}$ ($r\in \mathbb{Z}$) is a basis of $F$ relative to which $b_k=r_{k1}a_1+\cdots +r_{k, i-1}a_{i-1}+(r_{ki}+rr_{kj})a_i+r_{k, i+1}a_{i+1}+\cdots +r_{k, j-1}a_{j-1}+r_{kj}(a_j-ra_i)+r_{k, j+1}a_{j+1}+\cdots +r_{km}a_m$.
	\end{enumerate}
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	Determine the structure of the abelian group $G$ defined by generators $\{a, b\}$ and relations $2a+4b=0$ and $3b=0$.
	Do the same for the group with generators $\{a, b, c, d\}$ and relations $2a+3b=4a=5c+11d=0$ and for the group with generators $\{a, b, c, d, e\}$ and relations
	$\{a-7b+14d-21c=0;
	5a-7b-2c+10d-15e=0;
	3a-3b-2c+6d-9e=0;
	a-b+2d-3e=0\}$.
	\end{enumerate}
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\section{Determinants}
	{\it Note:} Unless stated otherwise all matrices have entries in a commutative ring $R$ with identity.
	\begin{enumerate}
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	If $r+r\neq 0$ for all nonzero $r\in R$,
	then prove that an $n$-linear form $B^n\to R$ is alternating if and only if it is skew-symmetric.
	What if $\text{char }R=2$?
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	\begin{enumerate}
	\item If $m>n$, then every alternating $R$-multilinear form on $(R^n)^m$ is zero.
	\item If $m<n$, then there is a nonzero alternating $R$-multilinear form on $(R^n)^m$.
	\end{enumerate}
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	Use Exercise {VII.3.2} to prove directly that if there is an $R$-module isomorphism $R^m\cong R^n$, then $m=n$.
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	If $A\in \text{Mat}_n R$,
	then $|A^n|=|A|^{n-1}$ and $(A^a)^a=|A|^{n-2}A$.
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	If $R$ is a field and $A, B\in \text{Mat}_n R$ are invertible then the matrix $A+rB$ is invertible for all but a finite number of $r\in R$.
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	Let $A$ be an $n\times n$ matrix over a field.
	Without using Proposition {VII.3.7} prove that $A$ is invertible if and only if $|A|\neq 0$.
	[{\it Hint:} Theorem {VII.2.6} and Theorem {VII.3.5} (viii) and Proposition {VII.2.12}.]
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	Let $F$ be a free $R$-module with basis $U=\{u_1, ..., u_n\}$.
	If $\phi:F\to F$ is an $R$-module endomorphism with matrix $A$ relative to $U$,
	then the determinant of the endomorphism $\phi$ is defined to be $|A|\in R$ and is denoted $|\phi|$.
	\begin{enumerate}
	\item $|\phi|$ is independent of the choice of $U$.
	\item $|\phi|$ is the unique element of $R$ such that $f(\phi(b_1), \phi(b_2), ..., \phi(b_n))=|\phi|f(b_1, ..., b_n)$ for every alternating $R$-multilinear form on $F^n$ and all $b_i\in F$.
	\end{enumerate}
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	Suppose that $(b_1, ..., b_n)$ is a solution of the system of homogeneous linear equations
	$$\begin{array}{ccccccccc}
	a_{11}x_1 & + & \cdots & + & a_{1n} x_n & = & 0 \\
	\vdots & \vdots & \ddots & \vdots & \vdots & \vdots & \vdots \\
	a_{n1}x_1 & + & \cdots & + & a_{nn} x_n & = & 0 \\
	\end{array}$$
	and that $A=(a_{ij})$ is the $n\times n$ matrix of coefficients.
	Then $|A|b_i=0$ for every $i$.
	[{\it Hint:} If $B_i$ is the $n\times n$ diagonal matrix with diagonal entries $1_R, ..., 1_R, b_i, 1_R, ..., 1_R$,
	then $|AB_i|=|A|b_i$.
	To show that $|AB_i|=0$ add $b_j$ times column $j$ of $AB_i$ to column $i$ for every $j\neq i$.
	The resulting matrix has determinant $|AB_i|$ and $(k, i)$ entry $a_{k1}b_1+a_{k2}b_2+\cdots +a_{kn}b_n=0$ for $k=1, 2, ..., n$.]
	\end{enumerate}
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\section{Decomposition of a Single Linear Transformation and Similarity}
	{\it Note:} Unless stated otherwise,
	$E$ is an $n$-dimensional vector space over a field $K$.
	\begin{enumerate}
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	If $A$ and $B$ are $n\times n$ matrices over $K$ with minimum polynomials $q_1$ and $q_2$ respectivel,
	then the minimal polynomial of the direct sum of $A$ and $B$ (a $2n\times 2n$ matrix) is the least common multiple of $q_1$ and $q_2$.
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	The $0$ linear transformation $E\to E$ has invariant factors [resp. elementary divisors] $q_1=x, q_2=x, ..., q_n=x$.
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	\begin{enumerate}
	\item Let $a, b, c$ be distinct elements of $K$ and let $D\in \text{Mat}_6 K$ be the diagonal matrix with main diagonal $a, a, a, b, b, c$.
	Then the invariant factors of $D$ are $q_1=x-a, q_2=(x-a)(x-b)$ and $q_3=(x-a)(x-b)(x-c)$.
	\item Describe the invariant factors of any diagonal matrix in $\text{Mat}_n K$.
	\end{enumerate}
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	If $q$ is the minimal polynomial of a linear transformation $\phi:E\to E$,
	with $\dim_K E=n$,
	then $\deg{q}\leq n$.
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	The minimal polynomial of the companion matrix of a monic polynomial $f\in K[x]$ is precisely $f$.
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	Let $F$ be an extension field of $K$.
	The invarian factors in $K[x]$ of a matrix $A\in \text{Mat}_n K$ are the same as the invariant factors in $F[x]$ of $A$ considered as a matrix over $F$.
	[{\it Hint:} A $K$-basis of $K^n$ is an $F$-basis of $F^n$.
	Use linear transformations.]
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	Let $F$ be an extension field of $K$.
	$A, B\in \text{Mat}_n K\subseteq \text{Mat}_n F$ are similar over $F$ if and only if they are similar over $K$ [see Exercise {VII.4.6}].
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	$A\in \text{Mat}_n K$ is similar to a diagonal matrix if and only if the elementary divisors of $A$ are all linear.
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	If $A\in \text{Mat}_n K$ is nilpotent (that is, $A^r=0$ for some $r>0$),
	then $A$ is similar to a matrix all of whose entries are zero except for certain entries $1_K$ on the diagonal next {\it above} the main diagonal.
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	Find all possible [primary] rational canonical forms for a matrix $A\in \text{Mat}_n \mathbb{Q}$ such that
	\begin{enumerate}[(i)]
	\item $A$ is $6\times 6$ with minimal polynomial $(x-2)^2 (x+3)$;
	\item $A$ is $7\times 7$ with minimal polynomial $(x^2+1)(x-7)$.
	\end{enumerate}
	Find all possible Jordan canonical forms of $A$ considered as a matrix over $\mathbb{C}$.
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	If $A$ is the companion matrix of a monic polynomial $f\in K[x]$,
	with $\deg{f}=n$,
	show explicitly that $A-xI_n$ is similar to a diagonal matrix with main diagonal $1_K, 1_K, ..., 1_K, f$.
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	$A\in \text{Mat}_n K$ is {\it idempotent} provided $A^2=A$.
	Show that two idempotent matrices in $\text{Mat}_n K$ are similar if and only if they are equivalent.
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	An $n\times n$ matrix $A$ is similar to its transpose $A^t$.
	\end{enumerate}
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\section{The Characteristic Polynomial, Eigenvectors and Eigenvalues}
	{\it Note:} Unless stated otherwise $K$ is a commutative ring with identity.
	\begin{enumerate}
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	Prove directly that a matrix over $K$ and its transpose have the same characteristic polynomial.
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	(Cayley-Hamilton) If $\phi$ is an endomorphism of a free $K$-module $E$ of finite rank,
	then $p_{\phi}(\phi)=0$.
	[{\it Hint:} if $A$ is the matrix of $\phi$ and $B=xI_n-A$,
	then $B^a B=|B|I_n=p_{\phi}I_n$ in $\text{Mat}_n K[x]$.
	If $E$ is a $K[x]$-module with structure induced by $\phi$ and $\psi$ is the $K[x]$-module endomorphism $E\to E$ with matrix $B$,
	then $\psi(u)=xu-\phi(u)=\phi(u)-\phi(u)=0$ for all $u\in E$.]
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	If $A$ is an $n\times m$ matrix over $K$ and $B$ an $m\times n$ matrix over $K$,
	then $x^m p_{AB}=x^n p_{BA}$.
	Furthermore,
	if $m=n$,
	then $p_{AB}=p_{BA}$.
	[{\it Hint:} let $C, D$ be the $(m+n)\times (m+n)$ matrices over $K[x]$: $C=\begin{pmatrix}
	xI_n & A \\ B & I_m \\
	\end{pmatrix}$ and $D=\begin{pmatrix}
	I_n & 0 \\ -B & xI_m \\
	\end{pmatrix}$ and observe that $|CD|=|DC|$.]
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	\begin{enumerate}
	\item Exhibit three $3\times 3$ matrices over $\mathbb{Q}$ no two of which are similar such that $-2$ is the only eigenvalue of each of the matrices.
	\item Exhibit a $4\times 4$ matrix whose eigenvalues over $\mathbb{R}$ are $\pm 1$ and whose eigenvalues over $\mathbb{C}$ are $\pm 1$ and $\pm i$.
	\end{enumerate}
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	Let $K$ be a field and $A\in \text{Mat}_n K$.
	\begin{enumerate}
	\item $0$ is an eigenvalue of $A$ if and only if $A$ is not invertible.
	\item If $k_1, ..., k_r\in K$ are the (not necessarily distinct) eigenvalues of $A$ and $f\in K[x]$,
	then $f(A)\in \text{Mat}_n K$ has eignevalues $f(k_1), ..., f(k_r)$.
	\end{enumerate}
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	\begin{enumerate}
	\item If $\phi$ and $\psi$ are endomorphisms of a finite dimensional vector space $E$ such that $\phi \psi=\psi \phi$.
	If $E$ has a basis of eigenvectors of $\phi$ and a basis of eigenvectors of $\psi$,
	then $E$ has a basis consisting of vectors that are eigenvectors for both $\phi$ and $\psi$.
	\item Interpret (a) as a statement about matrices that are similar to a diagonal matrix.
	\end{enumerate}
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	Let $\phi:E\to E$ be a linear transformation of a vector space $E$ over a field $K$.
	If $U$ is a set of eigenvectors of $\phi$ whose corresponding eigenvalues are all distinct,
	then $U$ is linearly independent.
	[{\it Hint:} If $U$ were linearly dependent,
	there would be a relation $r_1 u_1+\cdots +r_t u_t=0$ ($u_i\in U$; $0\neq r_i\in K$) with $t$ minimal.
	Apply the transformation $k_1 1_E-\phi$,
	where $\varphi(u_1)=k_1 u_1$,
	and reach a contradiction.]
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	Let $F$ be an extension field of a field $K$ and $u\in F$.
	Let $\phi:F\to F$ be the endomorphism of the vector space $F$ given by $v\mapsto uv$.
	\begin{enumerate}
	\item Then $\text{Tr }\phi$ is the trace of $u$,
	${T_K}^F(u)$, as in Definition {V.7.1}.
	[{\it Hint:} first try the case when $F=K(u)$].
	\item The determinant of $\phi$ is the norm of $u$, ${N_K}^F(u)$.
	\end{enumerate}
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	Let $K$ be a field and $A\in \text{Mat}_n K$.
	\begin{enumerate}
	\item If $A$ is nilpotent (that is, $A^m=0$ for some $m$),
	then $\text{Tr }A^r=0$ for all $r\geq 1$.
	[{\it Hint:} the minimal polynomial of $A^r$ has the form $x^t$ an $A^r$ is similar to a matrix in rational or Jordan canonical form.]
	\item If $\text{char }K=0$ and $\text{Tr }A^r=0$ for all $r\geq 1$, then $A$ is nilpotent.
	\end{enumerate}
	\end{enumerate}
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\chapter{Commutative Rings and Modules}
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\section{Chain Conditions}
	\begin{enumerate}
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	\begin{enumerate}
	\item The ring of all $2\times 2$ matrices $\begin{pmatrix}a&b\\0&c\\\end{pmatrix}$
	such that $a$ is an integer and $b, c$ are rational is right Noetherian but not left Noetherian.
	\item The ring of all $2\times 2$ matrices $\begin{pmatrix}d&r\\0&s\\\end{pmatrix}$
	such that $d$ is rational and $r, s$ are real is right Artinian but not left Artinian.
	\end{enumerate}
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	If $I$ is a nonzero ideal in a principal ideal domain $R$,
	then the ring $R/I$ is both Noetherian and Aritnian.
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	Let $S$ be a multiplicative subset of a commutative Noetherian ring $R$ with identity.
	Then the ring $S^{-1}R$ is Noetherian.
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	Let $R$ be a commutative ring with identity.
	If an ideal $I$ of $R$ is not finitely generated,
	then there is an infinite properly ascending chain of ideals $J_1\subsetneq J_2\subsetneq \cdots $
	such that $J_k\subseteq I$ for all $k$.
	The union of the $J_k$ need not be $I$.
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	Every homomorphic image of a left Noetherian [resp. Artinian] ring is left Noetherian [resp. Artinian].
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	A ring $R$ is left Noetherian [resp. Artinian] if and only if $\text{Mat}_n R$ is left Noetherian [resp. Artinian] for every $n\geq 1$ [nontrivial].
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	An Artinian integral domain is a field.
	[{\it Hint:} to find an inverse for $a\neq 0$,
	consider $\langle a\rangle \supseteq \langle a^2\rangle\supseteq \langle a^3\rangle\cdots$.]
	\end{enumerate}
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\section{Prime and Primary Ideals}
	{\it Note:}
	$R$ is always a commutative ring.
	\begin{enumerate}
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	Let $R$ be a commutative Artinian ring with identity.
	\begin{enumerate}
	\item Every prime ideal of $R$ is maximal
	[{\it Hint:} Theorem {III.2.16} and Theorem {III.2.20} and Exercise {VIII.1.5} and Exercise {VIII.1.7}].
	\item
	$R$ has only a finite number of distinct prime ideals.
	\end{enumerate}
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	If $R$ has an identity and $\{P_i\mid i\in I\}$ is a nonempty family of prime ideals of $R$ which is linearly ordered by inclusion,
	then $\bigcup_{i\in I}P_i$ and $\bigcup_{i\in I}P_i$ are prime ideals.
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	If $P_1, P_2, ..., P_n$ are prime ideals in $R$ and $I$ is any ideal such that $I\not\subseteq P_i$ for all $i$,
	then there exists $r\in I$ such that $r\not\in P_i$ for all $i$.
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	If $R$ has an identity and $M_1, ..., M_r$ are distinct maximal ideals in $R$,
	then show that $M_1\cap M_2\cap \cdots \cap M_r=M_1 M_2\cdots M_r$.
	Is this true if ``maximal'' is replaced by ``prime''?
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	If $R$ has an identity,
	then the set of all zero divisors of $R$ is a union of prime ideals.
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	Let $R$ have an identity.
	A prime ideal $P$ in $R$ is called a minimal prime ideal of the ideal $I$ if $I\subseteq P$ and there is no prime ideal $P'$ such that $I\subseteq P'\subsetneq P$.
	\begin{enumerate}
	\item If an ideal $I$ of $R$ is contained in a prime ideal $P$ of $R$,
	then $P$ contains a minimal prime ideal of $I$.
	[{\it Hint:} Zornify the set of all prime ideals $P'$ such that $I\subseteq P'\subseteq P$.]
	\item Every proper ideal possesses at least one minimal prime ideal.
	\end{enumerate}
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	The radical of an ideal $I$ in a ring $R$ with identity is the intersection of all its minimal prime ideals [see Exercise {VIII.2.6}].
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	If $R$ has an identity,
	$I$ is an ideal and $J$ is a finitely generated ideal such that $J\subseteq \text{Rad }I$,
	then there exists a positive integer $n$ such that $J^n\subseteq I$.
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	What is the radical of the zero ideal in $\mathbb{Z}_n$?
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	If $S$ is a multiplicative subset of a commutative ring $R$ and $I$ is an ideal of $R$,
	then $S^{-1}(\text{Rad }I)=\text{Rad }(S^{-1}I)$ in the ring $S^{-1}R$.
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	Let $Q$ ($\neq R$) be an ideal in $R$.
	Then $Q$ is primary if and only if every zero divisor in $R/Q$ is nilpotent (see Exercise {III.1.12}).
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	If $F$ is a field,
	then:
	\begin{enumerate}
	\item the ideal $\langle x, y\rangle$ is maximal in $F[x, y]$;
	\item $\langle x, y\rangle^2=\langle x^2, xy, y^2\rangle\subsetneq \langle x^2, y\rangle\subsetneq \langle x, y\rangle$;
	\item the ideal $\langle x^2, y\rangle$ is primary and the only proper prime ideal containing it is $\langle x, y\rangle$.
	\end{enumerate}
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	In the ring $\mathbb{Z}[x, y]$ the ideals $\langle x^2, y\rangle, \langle x^2, y^2\rangle, \langle x^2, y^3\rangle, ..., \langle x^i, y^j\rangle, ...$ are all primary ideals belonging to the prime ideal $\langle x, y\rangle$.
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	The conclusion of Theorem {VIII.2.11} is false if infinite intersections are allowed [{\it Hint:} consider $\mathbb{Z}$.]
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	Let $f:R\to S$ be an epimorphism of commutative rings with identity.
	If $J$ is an ideal of $S$,
	let $I=f^{-1}(J)$.
	\begin{enumerate}
	\item Then $I$ is primary in $R$ if and only if $J$ is primary in $S$.
	\item If $J$ is primary for $P$,
	then $I$ is primary for the prime ideal $f^{-1}(P)$.
	\end{enumerate}
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	Find a reduced primary decomposition for the ideal $I=\langle x^2, xy, 2\rangle$ in $\mathbb{Z}[x, y]$ and determine the associated primes of the primary ideals appearing in this decomposition.
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	\begin{enumerate}
	\item If $p$ is prime and $n>1$,
	then $\langle p^n\rangle$ is a primary,
	but not a prime ideal of $\mathbb{Z}$.
	\item Obtain a reduced primary decomposition of the ideal $\langle 12600\rangle$ in $\mathbb{Z}$.
	\end{enumerate}
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	If $F$ is a field and $I$ is the ideal $\langle x^2, xy\rangle$ in $F[x, y]$,
	then there are at least three distinct reduced primary decompositions of $I$;
	three such are:
	\begin{enumerate}[(i)]
	\item $I=\langle x\rangle\cap \langle x^2, y\rangle$;
	\item $I=\langle x\rangle\cap \langle x^2, x+y\rangle$;
	\item $I=\langle x\rangle\cap \langle x^2, xy, y^2\rangle$.
	\end{enumerate}
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	\begin{enumerate}
	\item In the ring $\mathbb{Z}[x]$,
	the following are primary decompositions:
	\begin{eqnarray*}
	\langle 4, 2x, x^2\rangle &=& \langle 4, x\rangle\cap \langle 2, x^2\rangle;\\
	\langle 9, 3x+3\rangle &=& \langle 3\rangle \cap \langle 9, x+1\rangle.
	\end{eqnarray*}
	\item Are the primary decompositions of part (a) reduced?
	\end{enumerate}
	\end{enumerate}
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\section{Primary Decomposition}
	{\it Note:}
	Unless otherwise stated $R$ is always a commutative ring with identity.
	\begin{enumerate}
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	Consider the ring $R$ as an $R$-module.
	If $Q$ is a primary submodule of $R$,
	then $Q$ is a primary ideal.
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	\begin{enumerate}
	\item Let $f:B\to D$ be an $R$-module epimorphism and $C$ ($\neq D$) a submodule of $D$.
	Then $C$ is a primary submodule of $D$ if and only if $f^{-1}(C)$ is a primary submodule of $B$.
	\item If $C$ and $f^{-1}(C)$ are primary,
	then they both belong to the same prime ideal $P$.
	\end{enumerate}
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	If $A$ ($\neq B$) is a submodule of the $R$-module $B$ and $P$ is an ideal of $R$ such that
	\begin{enumerate}[(i)]
	\item $rx\in A$ and $x\notin A~(r\in R, x\in B) \Rightarrow r\in P$; and
	\item $r\in P\Rightarrow r^n B\subseteq A$ for some positive integer $n$,
	\end{enumerate}
	then $P$ is a prime ideal and $A$ is a $P$-primary submodule of $B$.
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	If $A$ is a $P$-primary submodule of an $R$-module $B$ and $rx\in A$ ($r\in R, x\in B$),
	then either $r\in P$ or $x\in A$.
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	If $A$ is a $P$-primary submodule of an $R$-module $B$ and $C$ is any submodule of $B$ such that $C\not\subseteq A$ then $\{r\in R\mid rC\subseteq A\}$ is a $P$-primary ideal.
	[{\it Hint:} Exercise {VIII.3.3} may be helpful.]
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	Let $A$ be a $P$-primary submodule of the $R$-module $B$ and let $C$ be any submodule of $B$ such that $C\not\subseteq A$.
	Then $A\cap C$ is a $P$-primary submodule of $C$.
	[{\it Hint:} Exercise {VIII.3.3} may be helpful.]
	{VIII.3.3} may be helpful.]
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	If $B$ is an $R$-module and $x\in B$,
	the annihilator of $x$,
	denoted $\text{ann }x$,
	is $\{r\in R\mid rx=0\}$.
	Show that $\text{ann }x$ is an ideal.
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	If $B\neq 0$ is an $R$-module and $P$ is maximal in the set of ideals $\{\text{ann }x\mid 0\neq x\in B\}$ (see Exercise {VIII.3.7}),
	then $P$ is prime.
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	Let $R$ be Noetherian and let $B$ be an $R$-module.
	If $P$ is a prime ideal such that $P=\text{ann }x$ for some nonzero $x\in B$ (see Exercise {VIII.3.7}),
	then $P$ is called an associated prime of $B$.
	\begin{enumerate}
	\item If $B\neq 0$,
	then there exists an associated prime of $B$.
	[{\it Hint:} use Exercise {VIII.3.8}.]
	\item If $B\neq 0$ and $B$ satisfies the ascending chain condition on submodules,
	then there exist prime ideals $P_1, ..., P_{r-1}$ and a sequence of submodules $B=B_1\supseteq B_2\supseteq \cdots \supseteq B_r=0$ such that $B_i/B_{i+1}\cong R/P_i$ for each $i<r$.
	\end{enumerate}
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	Let $R$ and $B$ be as in Exercise {VIII.3.9}(b).
	Then the following conditions on $r\in R$ are equivalent:
	\begin{enumerate}[(i)]
	\item for each $x\in B$ there exists a positive integer $n(x)$ such that $r^{n(x)}x=0$;
	\item $r$ lies in every associated prime of $B$ (see Exercise {VIII.3.9} and Exercise {VIII.3.15}).
	\end{enumerate}
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item Let $R$ be Noetherian,
	$r\in R$,
	and $B$ an $R$-module.
	Then $rx=0$ ($x\in B$) implies $x=0$ if and only if $r$ does not lie in any associated prime of $B$ (see Exercise {VIII.3.8} and Exercise {VIII.3.9}).
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	Let $R$ be Noetherian and let $B$ be an $R$-module satisfying the ascending chain condition on submodules.
	Then the following are equivalent:
	\begin{enumerate}[(i)]
	\item There exists exactly one associated prime of $B$ (see Exercise {VIII.3.9});
	\item $B\neq 0$ and for each $r\in R$ one of the following is true:
	either $rx=0$ implies $x=0$ for all $x\in B$ or for each $x\in B$ there exists a positive integer $n(x)$ such that $r^{n(x)}x=0$.
	[See Exercise {VIII.3.10} and Exercise {VIII.3.11}.]
	\end{enumerate}
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	Let $R$ and $B$ be as in Exercise {VIII.3.12}.
	Then a submodule $A$ of $B$ is primary if and only if $B/A$ has exactly one associated prime $P$ and in that case $A$ is $P$-primary;
	(see Exercise {VIII.3.9} and Exercise {VIII.3.12}).
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	Let $R$ and $B$ be as in Exercise {VIII.3.12}.
	If $A$ ($\neq B$) is a submodule of $B$,
	then every associated prime of $A$ is an associated prime of $B$.
	Every associated prime of $B$ is an associated prime of either $A$ or $B/A$; (see Exercise {VIII.3.9}).
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	Let $R$ and $B$ be as in Exercise {VIII.3.12}.
	Then the associated primes of $B$ are precisely the primes $P_1, ..., P_n$,
	where $0=A_1\cap \cdots \cap A_n$ is a reduced primary decomposition of $0$ with each $A_i$ $P_i$-primary.
	In particular,
	the set of associated primes of $B$ is finite.
	[{\it Hint:} see Exercise {VIII.3.9}, Exercise {VIII.3.13}, Exercise {VIII.3.14}.]
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	Let $S$ be a multiplicative subset of $R$ and let $A$ be a $P$-primary submodule of an $R$-module $B$.
	If $P\cap S=\emptyset$,
	then $S^{-1}A$ is an $S^{-1}P$-primary submodule of the $S^{-1}R$-module $S^{-1}B$.
	\end{enumerate}
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\section{Noetherian Rings and Modules}
	\begin{enumerate}
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	Let $R$ be a commutative ring with identity and $I$ a finitely generated ideal of $R$.
	Let $C$ be a submodule of an $R$-module $A$.
	Assume that for each $r\in I$ there exists a positive integer $m$ (depending on $r$) such that $r^m A\subseteq C$.
	Show that for some integer $n$,
	$I^n A\subseteq C$.
	[{\it Hint:} see Theorem {III.1.1}(v) and Theorem {III.2.5}(vi)].
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	Without using primary decomposition,
	prove this version of the Krull Intersection Theorem.
	If $R$ is a commutative Noetherian ring with identity,
	$I$ and ideal of $R$.
	$A$ a finitely generated $R$-module,
	and $B=\bigcap_{n=1}^{\infty}I^n A$,
	then $IB=B$.
	[{\it Hint:} Let $C$ be maximal in the set $\mathcal{S}$ of all submodules $S$ of $A$ such that $B\cap S=IB$.
	It suffices to show $I^m A\subseteq C$ for some $m$.
	By Exercise {VIII.4.1} it suffices to show that for each $r\in I$,
	$r^n A\subseteq C$ for some $n$ (depending on $r$).
	For each $k$,
	let $D_k=\{a\in A\mid r^k a\in C\}$.
	$D_0\subseteq D_1\subseteq D_2\subseteq \cdots$ is an ascending chain of $R$-submodules;
	hence for some $n$,
	$D_k=D_n$ for all $k\geq n$.
	Show that $(r^n A+C)\cap B=IB$.
	The maximality of $C$ implies $r^n A+C=C$,
	that is, $r^n A\subseteq C$.]
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	Let $R$ be a Noetherian local ring with maximal ideal $M$.
	If the ideal $M/M^2$ in $R/M^2$ is generated by $\{a_1+M^2, ..., a_n+M^2\}$,
	then the ideal $M$ is generated in $R$ by $\{a_1, ..., a_n\}$.
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	(Nakayama's Lemma, second version ) Let $R$ be a commutative ring with identity,
	$J$ an ideal that is contained in every maximal ideal of $R$,
	and $A$ a finitely generated $R$-module.
	If $R/J\otimes_R A=0$,
	then $A=0$.
	[{\it Hint:} use the exact sequence $0\to J\to R\to R/J\to 0$ and the natural isomorphism $R\otimes_R A\cong A$ to show $JA=A$.]
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	Let $R$ and $J$ be as in Exercise {VIII.4.4};
	let $A$ be a finitely generated $R$-module and $f:C\to A$ an $R$-module homomorphism.
	Then $f$ induces a homomorphism $\overline{f}:C/JC\to A/JA$ in the usual way (Corollary {IV.1.8}).
	Show that if $\overline{f}$ is an epimorphism,
	then $f$ is an epimorphism.
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	\begin{enumerate}
	\item Let $R$ be a commutative ring with identity.
	If every ideal of $R$ can be generated by a finite or denumerable subset,
	then the same is true of $R[x]$.
	\item State and prove an analogue of part (a) for $R[[x]]$;
	(the answer is not quite the same here).
	\end{enumerate}
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	Let $R$ be a commutative ring with identity and let $f, g\in R[[x]]$.
	Denote by $\text{In }f$,
	then initial degree of $f$ (that is, the smallest $n$ such that $a_n\neq 0$,
	where $f=\sum_{i=0}^{\infty}a_i x^i$).
	Show that
	\begin{enumerate}
	\item $\text{In }(f+g)\geq \min{(\text{In }f, \text{In }g)}$.
	\item $\text{In }(fg)\geq \text{In }f+\text{In }g$.
	\item If $R$ is an integral domain, $\text{In }(fg)=\text{In }f+\text{In }g$.
	\end{enumerate}
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	Let $R$ be a commutative Noetherian ring with identity and let $Q_1\cap \cdots \cap Q_n=0$ be a reduced primary decomposition of the ideal $0$ of $R$ with $Q_i$ belonging to the prime ideal $P_i$.
	Then $P_1\cup P_2\cup \cdots \cup P_n$ is the set of zero divisors in $R$.
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	Let $R$ be a commutative ring with identity.
	If every maximal ideal of $R$ is of the form $\langle c\rangle$,
	where $c^2=c$,
	for some $c\in R$,
	then $R$ is Noetherian.
	[{\it Hint:} show that every primary ideal is maximal;
	use Proposition {VIII.4.1}.]
	\end{enumerate}
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\section{Ring Extensions}
	{\it Note:} Unless otherwise specified,
	$S$ is always an extension ring of $R$.
	\begin{enumerate}
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	Let $S$ be an integral extension ring of $R$ and suppose $R$ and $S$ are integral domains.
	Then $S$ is a field if and only if $R$ is a field.
	[{\it Hint:} Corollary {III.2.21}.]
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	Let $R$ be an integral domain.
	If the quotient field $F$ of $R$ is integral over $R$,
	then $R$ is a field.
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	Let $R$ be an integral domain with quotient field $F$.
	If $0\neq a\in R$ and $1_R/a\in F$ is integral over $R$,
	then $a$ is a unit in $R$.
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	\begin{enumerate}
	%%%%%
	\item Let $R$ be an integral domain with quotient field $F$.
	If $0\neq a\in R$,
	then the following are equivalent:
	\begin{enumerate}[(i)]
	\item every nonzero prime ideal of $R$ contains $a$;
	\item every nonzero ideal of$R$ contains some power of $a$;
	\item $F=R[1_R/a]$ (ring extension).
	\end{enumerate}
	An integral domain $R$ that contains an element $a\neq 0$ satisfying (i)-(iii) is called a Goldmann ring.
	%%%%%
	\item
	A principal ideal domain is a Goldmann ring if and only if it has only finitely many distinct primes.
	%%%%%
	\item
	Is the homomorphic image of a Goldmann ring also a Goldmann ring?
	\end{enumerate}
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	If $S$ is an integral extension ring of $R$ and $f:S\to S$ is a ring homomorphism,
	such that $f(1_S)=1_S$,
	then $f(S)$ is an integral extension ring of $f(R)$.
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	If $S$ is an integral extension ring of $R$,
	then $S[x_1, ..., x_n]$ is an integral extension ring of $R[x_1, ..., x_n]$.
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	If $S$ is an integral extension ring of $R$ and $T$ is a multiplicative subset of $R$ ($0\notin T$),
	then $T^{-1}S$ is an integral extension of $T^{-1}R$.
	[{\it Hint:} If $s/t\in T^{-1}S$,
	then $s/t=\phi_T(s)(1_R/t)$,
	where $\phi_T:S\to T^{-1}S$ is the canonical map (Theorem {III.4.4}).
	Show that $\phi_T(s)$ and $1_R/t$ are integral over $T^{-1}R$,
	whence $s/t$ is integral over $T^{-1}R$ by Theorem {VIII.5.5}.]
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	Every unique factorization domain is integrally closed.
	[{\it Hint:} Proposition {III.6.8}.]
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	Let $T$ be a commutative ring with identity and $\{S_i\mid i\in I\}, \{R_i\mid i\in I\}$ families of subrings such that $T$ is an extension ring of $S_i$
	and $S_i$ is an extension ring of $R_i$ for every $i$.
	If each $R_i$ is integrally closed in $S_i$,
	then $\bigcap_{i}R_i$ is integrally closed in $\bigcap_{i}S_i$.
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	\begin{enumerate}
	\item If $I$ ($\neq S$) is an ideal of $S$,
	then $I\cap R\neq R$ and $I\cap R$ is an ideal of $R$.
	\item If $Q$ is a prime ideal of $S$,
	then $Q\cap R$ is a prime ideal of $R$.
	\end{enumerate}
	\end{enumerate}
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\section{Dedekind Domains}
	\begin{enumerate}
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	The ideal generated by $3$ and $1+\sqrt{5}i$ in the subdomain $\mathbb{Z}[\sqrt{5}i]$ of $\mathbb{C}$ is invertible.
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	An invertible ideal in an integral domain that is a local ring is principal.
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	If $I$ is an invertible ideal in an integral domain $R$ and $S$ is a multiplicative set in $R$ with $0\notin S$,
	then $S^{-1}I$ is invertible in $S^{-1}R$.
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	Let $R$ be any ring with identity and $P$ an $R$-module.
	Then $P$ is projective if and only if there exist sets $\{a_i\mid i\in I\}\subseteq P$ and $\{f_i\mid i\in I\}\subseteq \text{Hom}_R (P, R)$ such that for all $a\in P$,
	$a=\sum_{i\in I}f_i(a) a_i$.
	[See the proof of Theorem {VIII.6.8}.]
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	(Converse of Lemma {VIII.6.9})
	A discrete valuation ring $R$ is Noetherian and integrally closed.
	[{\it Hint:} Exercise {VIII.5.8}.]
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	\begin{enumerate}
	\item If every prime ideal in an integral domain $R$ is invertible,
	then $R$ is Dedekind.
	\item If $R$ is a Noetherian integral domain in which every maximal ideal is invertible,
	then $R$ is Dedekind.
	\end{enumerate}
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	If $S$ is a multiplicative subset of a Dedekind domain $R$ (with $1_R\in S$, $0\notin S$),
	then $S^{-1}R$ is a Dedekind domain.
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	If $R$ is an integral domain and $P$ a prime ideal in $R[x]$ such that $P\cap R=0$,
	then $R[x]_P$ is a discrete valuation ring.
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	If a Dedekind domain $R$ has only a finite number of nonzero prime ideals $P_1, ..., P_n$,
	then $R$ is a principal ideal domain.
	[{\it Hint:} There exists $a_i\in P_i-P_i^2$ and by the Chinese Remainder Theorem {III.2.25} there exists $b_i\in P_i$ such that $b_i\equiv a_i\pmod{P_i}$ and $b_i\equiv 1_R\pmod{P_i}$ for $j\neq i$.
	Show that $P_i=\langle b_i\rangle$,
	which implies that every ideal is principal.]
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	If $I$ is a nonzero ideal in a Dedekind domain may be generated by at most two elements.
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	An $R$-module $A$ is divisible if $rA=A$ for all nonzero $r\in R$.
	If $R$ is a Dedekind domain,
	every divisible $R$-module is injective.
	[N.B. the converse is also true,
	but harder.]
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	(Nontrivial) If $R$ is a Dedekind domain with quotient field $K$,
	$F$ is a finite dimensional extension field of $K$ and $S$ is the integral closure of $R$ in $F$
	(that is, the ring of all elements of $F$ that are integral over $R$),
	then $S$ is a Dedekind domain.
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	\begin{enumerate}
	\item Prove that the integral domain $\mathbb{Z}[\sqrt{10}]$ is an integral extension ring of $\mathbb{Z}$ with quotient field $\mathbb{Q}(\sqrt{10})$.
	\item Let $u\in \mathbb{Q}(\sqrt{10})$ be integral over $\mathbb{Z}[\sqrt{10}]$.
	Then $u$ is integral over $\mathbb{Z}$ (Theorem {VIII.5.6}).
	Furthermore if $u\in \mathbb{Q}$,
	then $u\in \mathbb{Z}$ (Exercise {VIII.5.8}).
	Prove the if $u\in \mathbb{Q}(\sqrt{10})$ and $u\notin \mathbb{Q}$,
	then $u$ is the root of an irreducible monic polynomial of degree $2$ in $\mathbb{Z}[x]$.
	[{\it Hint:} Corollary {III.6.13} and Theorem {V.1.6}.]
	\item Prove that if $u=r+s\sqrt{10}\in \mathbb{Q}(\sqrt{10})$ and $u$ is a root of $x^2+ax+b\in \mathbb{Z}[x]$,
	then $a=-2r$ and $b=r^2-10s^2$.
	[{\it Hint:} note that $u^2-2ru+(r^2-10s^2)=0$;
	if $u\notin \mathbb{Q}$ use Theorem {V.1.6}.]
	\item Prove that $\mathbb{Z}[\sqrt{10}]$ is integrally closed.
	[{\it Hint:} if $u=r+s\sqrt{10}\in \mathbb{Q}(\sqrt{10})$ is a root of $x^2+ax+b\in \mathbb{Z}[x]$ and $a$ is even,
	then $r\in \mathbb{Z}$ by (c);
	it follows that $s\in \mathbb{Z}$.
	The assumption that $a$ is odd leads to a contradiction.]
	\end{enumerate}
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	\begin{enumerate}
	\item If $P$ is a nonzero prime ideal of the ring $\mathbb{Z}[\sqrt{10}]$,
	then $P\cap \mathbb{Z}$ is a nonzero prime ideal of $\mathbb{Z}$.
	[{\it Hint:} if $0\neq u\in P$,
	then $u$ is a root of $x^2+ax+b\in \mathbb{Z}[x]$ by Exercise {VIII.6.14}.
	Show that one of $a, b$ is nonzero and lies in $P$.]
	\item Every nonzero prime ideal of $\mathbb{Z}[\sqrt{10}]$ is maximal.
	[Use (a),
	Theorem {III.3.4} and either an easy direct argument or Theorem {VIII.5.12}.]
	\end{enumerate}
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	A valuation domain is an integral domain $R$ such that for all $a ,b\in R$ either $a\mid b$ or $b\mid a$.
	(Clearly a discrete valuation ring is a valuation domain.)
	A Pr{\"u}fer domain is an integral domain in which every finitely generated ideal is invertible.
	\begin{enumerate}
	\item The following are equivalent:
	\begin{enumerate}[(i)]
	\item $R$ is a Pr{\"u}fer domain;
	\item for every prime ideal $P$ in $R$,
	$R_P$ is a valuation domain;
	\item for every maximal ideal $M$ in $R$, $R_M$ is a valuation domain.
	\end{enumerate}
	\item A Pr{\"u}fer domain is Dedekind if and only if it is Noehterian.
	\item If $R$ is a Pr{\"u}fer domain with quotient field $K$,
	then any domain $S$ such that $R\subseteq S\subseteq K$ is Pr{\"u}fer.
	\end{enumerate}
	\end{enumerate}
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\section{The Hilbert Nullstellensatz}
	{\it Note:}
	$F$ is always an algebraically closed extension field of a field $K$;
	$J, V$, and $F^n$ are as above.
	\begin{enumerate}
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	A subset $Y$ of $F^n$ is closed (that is, $V(J(Y))=Y$) if and only if $Y$ is an affine $K$-variety determined by some subset $S$ of $K[x_1, ..., x_n]$.
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	A subset $S$ of $K[x_1, ..., x_n]$ is closed (that is, $J(V(S))=S$) if and ony if $S$ is a radical ideal (that is, $S$ is an ideal and $S=\text{Rad }S$).
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	There is a one-to-one inclusion reversing correspondence between the set of affine $K$-varieties in $F^n$ and the set of radical ideals of $K[x_1, ..., x_n]$.
	[See Exercise {VIII.7.1}, Exercise {VIII.7.2}.]
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	Every affine $K$-variety in $F^n$ is of the form $V(S)$ where $S$ is a finite ubset of $K[x_1, ..., x_n]$.
	[{\it Hint:} Theorem {VIII.1.9} and Theorem {VIII.4.9} and Exercise {VIII.7.3}.]
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	If $V_1\supseteq V_2\supseteq \cdots$ is a descending chain of $K$-varieties in $F^n$,
	ten $V_m=V_{m+1}=\cdots$ for some $m$.
	[{\it Hint:} Theorem {VIII.4.9} and Exercise {VIII.7.3}.]
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	Show that the Nullstellensatz implies Lemma {VIII.7.3}.
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	If $I_1, ..., I_k$ are ideals of $K[x_1, ..., x_n]$,
	then $V(I_1\cap I_2\cap \cdots \cap I_k)=V(I_1)\cup V(I_2)\cup \cdots \cup V(I_k)$ and $V(I_1 I_2 \cdots I_k)=V(I_1)\cap V(I_2)\cap \cdots \cap V(I_k)$.
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	A $K$-variety $V$ in $F^n$ is irreducible provided that whenever $V=W_1\cup W_2$ with each $W_i$ a $K$-variety in $F^n$,
	either $V=W_1$ or $V=W_2$.
	\begin{enumerate}
	\item Prove that $V$ is irreducible if and only if $J(V)$ is a prime ideal in $K[x_, ..., x_n]$.
	\item Let $F=\mathbb{C}$ and $S=\{x_1^2-2x_2^2\}$.
	Then $V(S)$ is irreducible as a $\mathbb{Q}$-variety but not as an $\mathbb{R}$-variety.
	\end{enumerate}
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	Every nonempty $K$-variety in $F^n$ may be written uniquely as a finite union $V_1\cup C_2\cup \cdots \cup V_k$ of affine $K$-varieties in $F^n$ such that $V_j\not\subseteq V_i$ for $i\neq j$ and each $V_i$ is irreducible (Exercise {VIII.7.8}).
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	The coordinate ring of an affine $K$-variety $V(I)$is isomorphic to $K[x_1, ..., x_n]/J(V(I))$.
	\end{enumerate}
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\chapter{Structure of Rings}
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\section{Simple and Primitive Rings}
	\begin{enumerate}
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	Let $F$ be a field of characteristic $0$ and $R=F[x, y]$ the additive group of polynomials in two indeterminates.
	Define multiplication in $R$ by requiring that multiplication be distributive,
	that $ax=xa, ay=ya$ for all $a\in F$,
	that the product of $x$ and $y$ (in that order) be the polynomial $xy$ as usual,
	but that the product of $y$ and $x$ be the polynomial $xy+1$.
	\begin{enumerate}
	\item $R$ is a ring.
	\item $yx^k=x^k y+kx^{k-1}$ and $y^k x=xy^k+ky^{k-1}$
	\item $R$ is simple.
	({\it Hint:} Let $f$ be a nonzero element in an ideal $I$ of $R$;
	then either $f$ has no terms involving $y$ or $g=xf-fx$ is a nonzero element of $I$ that has lower degree in $y$ than does $f$.
	In the latter case,
	consider $xg-gx$.
	Eventually,
	find a nonzero $h\in I$,
	which is free of $y$.
	If $h$ is nonconstant,
	consider $hy-yh$.
	In a finite number of steps,
	obtain a nonzero constant element of $I$;
	hence $I=R$.)
	\item $R$ has no zero divisors.
	\item $R$ is not a division ring.
	\end{enumerate}
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	\begin{enumerate}
	\item If $A$ is an $R$-module,
	then $A$ is also a well-defined $R/\mathfrak{a}(A)$-module with $(r+\mathfrak{a}(A))a=ra$ ($a\in A$).
	\item If $A$ is a simple left $R$-module,
	then $R/\mathfrak{a}(A)$ is a primitive ring.
	\end{enumerate}
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	Let $V$ be an infinite dimensional vector space over a division ring $D$.
	\begin{enumerate}
	\item If $F$ is the set of all $\theta\in \text{Hom}_D(V, V)$ such that $\text{Im }\theta$ is finite dimensional,
	then $F$ is a proper ideal of $\text{Hom}_D(V, V)$.
	Therefore $\text{Hom}_D(V, V)$ is not simple.
	\item $F$ is itself a simple ring.
	\item $F$ is contained in every nonzero ideal of $\text{Hom}_D(V, V)$.
	\item $\text{Hom}_D(V, V)$ is not (left) Artinian.
	\end{enumerate}
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	Let $V$ be a vector space over a division ring $D$.
	A subring $R$ of $\text{Hom}_D(V, V)$ is said to be $n$-fold transitive if for every $k$ ($1\leq k\leq n$) and every linearly independent subset $\{u_1, ..., u_k\}$ of $V$ and every arbitrary subset $\{v_1, ..., v_k\}$ of $V$,
	there exists $\theta\in R$ such that $\theta(u_i)=v_i$ for $i=1, 2, ..., k$.
	\begin{enumerate}
	\item If $R$ is one-fold transitive,
	then $R$ is primitive.
	[{\it Hint:} examine the example after Definition {IX.1.5}.]
	\item If $R$ is two-fold transitive,
	then $R$ is dense in $\text{Hom}_D(V, V)$.
	[{\it Hints:} Use (a) to show that $R$ is a dense subring of $\text{Hom}_{\Delta}(V, V)$,
	where $\Delta=\text{Hom}_R(V, V)$.
	Use two-fold transitivity to show that $\Delta=\{\beta_d\mid d\in D\}$,
	where $\beta_d:V\to V$ is given by $x\mapsto dx$.
	Consequently $\text{Hom}_{\Delta}(V, V)=\text{Hom}_D(V, V)$.]
	\end{enumerate}
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	If $R$ is a primitive ring such that for all $a, b\in R$,
	$a(ab-ba=(ab-ba)a$,
	then $R$ is a division ring.
	[{\it Hint:} show that $R$ is isomorphic to a dense ring of endomorphisms of a vector space $V$ over a division ring $D$ with $\dim_D{V}=1$,
	whence $R\cong D$.]
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	If $R$ is a primitive ring with identity and $e\in R$ is such that $e^2=e\neq 0$,
	then
	\begin{enumerate}
	\item $eRe$ is a subring of $R$,
	with identity $e$.
	\item $eRe$ is primitive [{\it Hint:} if $R$ is isomorphic to a dense ring of endomorphisms of the vector space $V$ over a division ring $D$,
	then $Ve$ is a $D$-vector space and $eRe$ is isomorphic to a dense ring of endomorphisms of $Ve$.]
	\end{enumerate}
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	If $R$ is a dense ring of endomorphisms of a vector space $V$ and $K$ is a nonzero ideal of $R$,
	then $K$ is also a dense ring of endomorphisms of $V$.
	\end{enumerate}
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\section{The Jacobson Radical}
	{\it Note:} $R$ is always a ring.
	\begin{enumerate}
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	For each $a, b\in R$ let $a\circ b=a+b+ab$.
	\begin{enumerate}
	\item $\circ$ is an associative binary operation with identity element $0\in R$.
	\item The set $G$ of all elements of $R$ that are {\it both} left and right quasi-regular forms a group under $\circ$.
	\item If $R$ has an identity,
	then $a\in R$ is left [resp. right] quasi-regular if and only if $1_R+a$ is left [resp. right] invertible.
	[{\it Hint:} $(1_R+r)(1_R+a)=(1_R+r\circ a$ and $r(1_R+a)-1_R=(r-1_R)\circ a$.]
	\end{enumerate}
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	(Kaplansky) $R$ is a division ring if and only if every element of $R$ except one is left quasi-regular.
	[Note that the only element in a division ring $D$ that is {\it not} left quasi-regular is $-1_D$;
	also see Exercise {IX.2.1}.]
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	Let $I$ be a left ideal of $R$ and let $(I:R)=\{r\in R\mid rR\subseteq I\}$.
	\begin{enumerate}
	\item $(I:R)$ is an ideal of $R$.
	If $I$ is regular,
	then $(I:R)$ is the largest ideal of $R$ that is contained in $I$.
	\item If $I$ is a regular maximal left ideal of $R$ and $A\cong R/I$,
	then $\mathfrak{a}(A)=(I:R)$.
	Therefore $J(R)=\bigcap (I:R)$,
	where $I$ runs over all the regular maximal left ideals of $R$.
	\end{enumerate}
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	The radical $J(R)$ contains no nonzero idempotents.
	However,
	a nonzero idempotent may be left quasi-regular.
	[{\it Hint:} Exercise {IX.2.1} and Exercise {IX.2.2}].
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	If $R$ has an identity,
	then
	\begin{enumerate}
	\item $J(R)=\{r\in R\mid 1_R+sr\text{ is left invertible for all }s\in R\}$.
	\item $J(R)$ is the largest ideal $K$ such that for all $r\in K$,
	$1_R+r$ is aunit.
	\end{enumerate}
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	\begin{enumerate}
	\item The homomorphic image of a semisimple ring need not be semisimple.
	\item If $f:R\to S$ is a ring epimorphism,
	then $f(J(R))\subseteq J(S)$.
	\end{enumerate}
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	If $R$ is the ring of all rational numbers with odd denominators,
	then $J(R)$ consists of all rational numbers with odd denominator and even numerator.
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	Let $R$ be the ring of all upper triangular $n\times n$ matrices over a division ring $D$ (see Exercise {VII.1.2}).
	Find $J(R)$ and prove that $R/J(R)$ is isomorphic to the direct product $D\times D\times \cdots \times D$ ($n$ factors).
	[{\it Hint:} show that a strictly triangular matrix is nilpotent.]
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	A principal ideal domain $R$ is semisimple if and only if $R$ is a field or $R$ contains an infinite number of distinct nonassociate irreducible elements.
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	Let $D$ be a principal ideal domain and $d$ a nonzero nonunit element of $D$.
	Let $R$ be the quotient ring $D/\langle d\rangle$.
	\begin{enumerate}
	\item $R$ is semisimple if and only if $d$ is the product of distinct nonassociate irreducible elements of $D$.
	[{\it Hint:} Exercise {VIII.1.2}.]
	\item What is $J(R)$?
	\end{enumerate}
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	If $p$ is a prime,
	let $R$ be the subring $\sum_{n\geq 1}\mathbb{Z}_{p^n}$ of $\prod_{n\geq 1}\mathbb{Z}_{p^n}$.
	The ideal $I=\sum_{n\geq 1}I_n$,
	where $I_n$ is the ideal of $\mathbb{Z}_{p^n}$ generated by $p\in \mathbb{Z}_{p^n}$,
	is a nil ideal of $R$ that is not nilpotent.
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	Let $R$ be a ring without identity.
	Embed $R$ in a ring $S$ with identity which has characteristic zero,
	as in Theorem {III.1.10}.
	Prove that $J(R)=J(S)$.
	Consequently every semisimple ring may be embedded in a semisimple ring with identity.
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	$J(\text{Mat}_n R=\text{Mat}_n J(R)$.
	Here is an outline of a proof:
	\begin{enumerate}
	\item If $A$ is a left $R$-module,
	consider the elements of $A^n=A\oplus A\oplus \cdots\oplus A$ ($n$ summands) as column vectors;
	then $A^n$ is a left $(\text{Mat}_n R)$-module (under ordinary matrix multiplication).
	\item If $A$ is a simple $R$-module,
	$A^n$ is a simple $(\text{Mat}_n R)$-module.
	\item $J(\text{Mat}_n R)\subseteq \text{Mat}_n J(R)$.
	\item $\text{Mat}_n J(R)\subseteq J(\text{Mat}_n R)$.
	[{\it Hint:} prove that $\text{Mat}_n J(R)$ is a left quasi-regular ideal of $\text{Mat}_n R$ as follows.
	For each $k=1, 2, ..., n$ let $K_k$ consist of all matrices $(a_{ij})$ such that $a_{ij}\in J(R)$ and $a_{ij}=0$ if $j\neq k$.
	Show that $K_k$ is a left quasi-regular left ideal of $\text{Mat}_n R$ and observe that $K_1+K_2+\cdots+K_n=\text{Mat}_n J(R)$.]
	\end{enumerate}
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	\begin{enumerate}
	\item Let $I$ be a nonzero ideal of $R[x]$ and $p(x)$ a nonzero polynomial of least degree in $I$ with leading coefficient $a$.
	If $f(x)\in R[x]$ and $a^m f(x)=0$,
	then $a^{m-1}p(x)f(x)=0$.
	\item If a ring $R$ has no nonzero nil ideals (in particular,
	if $R$ is semisimple),
	then $R[x]$ is semisimple.
	[{\it Hint:} Let $M$ be the set of nonzero polynomials of least degree in $J(R[x])$.
	Let $N$ be the set consisting of $0$ and the leading coefficients of polynomials in $M$.
	Use (a) to show that $N$ is a nil ideal of $R$,
	whence $J(R[x])=0$.]
	\item There exist rings $R$ such that $R[x]$ is semisimple,
	but $R$ is not.
	[{\it Hint}, consider $R=F[[x]]$,
	with $F$ a field.]
	\end{enumerate}
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	Let $L$ be a left ideal and $K$ a right ideal of $R$.
	Let $M(R)$ be the ideal generated by {\it all} nilpotent ideals of $R$.
	\begin{enumerate}
	\item $L+LR$ is an ideal such that $(L+LR)^n\subseteq L^n+L^n R$ for all $n\geq 1$.
	\item $K+RK$ is an ideal such that $(K+RK)^n\subseteq K^n+RK^n$ for all $n\geq 1$.
	\item If $L$ [resp. $K$] is nilpotent,
	so is the ideal $L+LR$ [resp. $K+RK$],
	whence $L\subseteq M(R)$ [resp. $K\subseteq M(R)$].
	\item If $N$ is a maximal nilpotent ideal of $R$,
	then $R/N$ has no nonzero nilpotent left or right ideals.
	[{\it Hint:} first show that $R/N$ has no nonzero nilpotent ideals;
	then apply (c) to the ring $R/N$.]
	\item If $K$ [resp. $L$] is nil,
	but not nilpotent and $\pi:R\to R/N$ is the canonical epimorphism,
	then $\pi(K)$ [resp. $\pi(L)$] is a nil right [resp. left] ideal of $R/N$ which is not nilpotent.
	\end{enumerate}
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	(Levitsky) Every nil left or right ideal $I$ in a left Noetherian ring $R$ is nilpotent.
	[{\it Sketch of Proof.} It suffices by Exercise {IX.2.15} to assume that $R$ has no nonzero nilpotent left or right ideals.
	Suppose $I$ is a left or a right ideal which is not nilpotent and $0\neq a\in I$.
	Show that $aR$ is a nil right ideal (even though $I$ may be a left ideal),
	whence the left ideal $\mathfrak{a}(u)$ is nonzero for all $u\in aR$.
	There exists a nonzero $u_0\in aR$ with $\mathfrak{a}(u_0)$ maximal,
	whence $\mathfrak{a}(u_0)=\mathfrak{a}(u_0 x)$ for all $x\in R$ such that $u_0 x\neq 0$.
	Show that $(u_0 y)u_0=0$ for all $y\in R$,
	so that $(Ru_0)^2=0$.
	Therefore $Ru_0=0$,
	which implies that $\{r\in R\mid Rr=0\}$ is a nonzero nilpotent right ideal of $R$;
	contradiction.]
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	Show that Nakayama's Lemma {VIII.4.5} is valid for any ring $R$ with identity,
	provided condition (i) is replaced by the condition
	$$
	\text{(i') } J\text{ is contained in the Jacobson radical of }R.
	$$
	[{\it Hint:} Use Theorem {IX.2.3}(iv) and Exercise {IX.2.1} (c) to show (i') $\Rightarrow$ (ii).]
	\end{enumerate}
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\section{Semisimple Rings}
	\begin{enumerate}
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	A ring $R$ is isomorphic to a subdirect product of the family of rings $\{R_i\mid i\in I\}$ if nd only if there exists for each $i\in I$ an ideal $K_i$ of $R$ such that $R/K_i\cong R_i$ and $\bigcap_{i\in I}K_i=0$.
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	A ring $R$ is subdirectly irreducible if the intersection of all nonzero ideals of $R$ is nonzero.
	\begin{enumerate}
	\item $R$ is subdirectly irreducible if and only if whenever $R$ is isommorphic to a subdirect product of $\{R_i\mid i\in I\}$,
	$R\cong R_i$ for some $i\in I$ [see Exercise {IX.3.1}].
	\item (Birkhoff) Every ring is isomorphic to a subdirect product of a family of subdirectly irreducible rings.
	\item The zero divisors in a commutative subdirectly irreducible ring (together with $0$) form an ideal.
	\end{enumerate}
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	A commutative semisimple left Artinian ring is a direct product of fields.
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	Determine up to isomorphism all semisimple rings of order $1008$.
	How many of them are commutative?
	[{\it Hint:} Exercise {V.8.10}.]
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	An element $a$ of a ring $R$ is regular (in the sense of Von Neumann) if there exists $x\in R$ such that $axa=a$.
	If every element of $R$ is regular,
	then $R$ is said to be a regular ring.
	\begin{enumerate}
	\item Every division ring is regular.
	\item A finite direct product of regular rings is regular.
	\item Every regular ring is semisimple.
	[The converse is false (for example, $\mathbb{Z}$).]
	\item The ring of all linear transformations on a vector space (not necessarily finite dimensional) ove a division ring is regular.
	\item A semisimple left Artinian ring is regular.
	\item $R$ is regular if and only if every principal left [resp. right] ideal of $R$ is generated by an idempotent element.
	\item A nonzero regular ring $R$ with identity is a division ring if and only if its only idempotents are $0$ and $1_R$.
	\end{enumerate}
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	\begin{enumerate}
	\item Every nonzero homomorphic image and every nonzero submodule of a semisimple module is semisimple.
	\item The intersection of two semisimple submodules is $0$ or semisimple.
	\end{enumerate}
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	The following conditions on a semisimple module $A$ are equivalent:
	\begin{enumerate}
	\item $A$ is finitely generated.
	\item $A$ is a direct sum of a finite number of simple submodules.
	\item $A$ has a composition series (see p. 375).
	\item $A$ satisfies both the ascending and descending chain conditions on submodules (see Theorem {VIII.1.11}).
	\end{enumerate}
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	Let $A$ be a module over a left Artinian ring $R$ such that $Ra\neq 0$ for all nonzero $a\in A$ and let $J=J(R)$.
	Then $JA=0$ if and only if $A$ is semisimple.
	[{\it Hint:} if $JA=0$,
	then $A$ is an $R/J$-module,
	with $R/J$ semisimple left Artinian;
	see Exercise {IV.1.17}.]
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	Let $R$ be a ring that (as a left $R$-module) is the sum of its minimal left ideals.
	Assume that $\{r\in R\mid Rr=0\}=0$.
	If $A$ is an $R$-module such that $RA=A$,
	then $A$ is semisimple.
	[{\it Hint:} if $I$ is a minimal left ideal and $a\in A$,
	show that $Ia$ is either zero or a simple submodule of $A$.]
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	Let $R$ be the ring of $2\times 2$ matrices over an infinite field.
	\begin{enumerate}
	\item $R$ has an infinite number of distinct proper left ideals,
	and two of which are isomorphic as left $R$-modules.
	\item There are infinitely many distinct pairs $(B, C)$ such that $B$ and $C$ are minimal left ideals of $R$ and $R=B\oplus C$.
	\end{enumerate}
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	A left Artinian ring $R$ has the same number of nonisomorphic simple left $R$-modules as nonisomorphic simple right $R$-modules.
	[{\it Hint:} Show that $A$ is a simple $R$-module if and only if $A$ is a simple $R/J(R)$-module;
	use Theorem {IX.2.14} and Theorem {IX.3.10}.]
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	\begin{enumerate}
	\item (Hopkins) If $R$ is a left Artinian ring with identity,
	then $R$ is left Noetherian.
	[{\it Hint:} Let $n$ be the least positive integer such that $J^n=0$ (Proposition {IX.2.13}).
	Let $J^0=R$.
	Since $J(J^i/J^{i+1}=0$ and $R$ is left Artinian each $J^i/J^{i+1}$ ($0\leq i\leq n-1$) has a composition series by Exercise {IX.3.7} and Exercise {IX.3.8}.
	Use these and Theorem {IV.1.10} to construct a composition series for $R$;
	apply Theorem {VIII.1.11}.]
	{\it Remark.} Hopkins' Theorem is valid even if the hypothesis ``$R$ has an identity'' is replaced by the much weaker hypothesis that $\{r\in R\mid rR=0\text{ and }Rr=0\}=0$;
	see L. Fuchs [13; pp. 283-286].
	\item The converse of Hopkins' Theorem is false.
	\end{enumerate}
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	\end{enumerate}
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\section{The Prime Radical; Prime and Semiprime Rings}
	{\it Note:} $R$ is always a ring.
	\begin{enumerate}
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	A subset $T$ of $R$ is said to be an $m$-system (generalized multiplicative system) if
	$$
	c, d\in T\Rightarrow cxd\in T\text{ for some }x\in R.
	$$
	\begin{enumerate}
	\item $P$ is a prime ideal of $R$ if and only if $R-P$ is an $m$-system .
	[{\it Hint:} Exercise {III.2.14}.]
	\item Let $I$ be an ideal of $R$ that is disjoint from an $m$-system $T$.
	Show that $I$ is contained in an ideal $Q$ which is maximal respect to the property that $Q\cap T=\emptyset$.
	Then show that $Q$ is a prime ideal.
	[{\it Hint:} Adapt the proof of Theorem {VIII.2.2}.]
	\item An element $r$ of $R$ is said to have the {\it zero property} if every $m$-system that contains $r$ also contains $0$.
	Show that the prime radical $P(R)$ is the set $M$ of all elements of $R$ that have the zero property.
	[{\it Hint:} use (a) to show $M\subseteq P(R)$ and (b) to show $P(R)\subseteq M$.]
	\item Every element $c$ of $P(R)$ is nilpotent.
	[{\it Hint:} $\{c^i\mid i\geq 1\}$ is an $m$-system.]
	If $R$ is commutative,
	$P(R)$ consists of all nilpotent elements of $R$.
	\end{enumerate}
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	\begin{enumerate}
	\item If $I$ is an ideal of $R$,
	then $P(I)=I\cap P(R)$.
	In particular,
	$P(P(R))=P(R)$.
	[{\it Hint:} Exercise {IX.4.1}(c).]
	\item $P(R)$ is the smallest ideal $K$ of $R$ such that $P(R/K)=0$.
	In particular,
	$P(R/P(R))=0$,
	whence $R/P(R)$ is semiprime.
	[{\it Hint:} Exercise {III.2.17}(d).]
	\item An ideal $I$ is said to have the {\it zero property} if every element of $I$ has the zero property (Exercise {IX.4.1}(c)).
	Show that the zero property is a radical property (as defined in the introduction to Section {IX.2}),
	whose radical is precisely $P(R)$.
	\end{enumerate}
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	\begin{enumerate}
	\item Every semisimple ring is semiprime.
	\item $P(R)\subseteq J(R)$.
	[{\it Hint:} Exercise {IX.4.1}(d); or (a) and Exercise {IX.4.2}(b).]
	\item If $R$ is left Artinian,
	$P(R)=J(R)$.
	[{\it Hint:} Proposition {IX.2.13}.]
	\end{enumerate}
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	$R$ is semiprime if and only if for all ideals $A, B$
	$$
	AB=0\Rightarrow A\cap B=0.
	$$
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	\begin{enumerate}
	\item Let $R$ be a ring with identity.
	The matrix ring $\text{Mat}_n R$ is prime if and only if $R$ is prime.
	\item If $R$ is any ring,
	then $P(\text{Mat}_n R)=\text{Mat}_n P(R)$.
	[{\it Hint:} Use Exercise {IX.4.2} and part (a) if $R$ has an identity.
	In the general case,
	embed $R$ in a ring $S$ with identity via Theorem {III.1.10};
	then $P(R)=R\cap P(S)$ by Exercise {IX.4.2}.]
	\end{enumerate}
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	If $R$ is semisimple left Artinian,
	then $R$ is its own quotient ring.
	[{\it Hint:} Since $R$ has an identity by Theorem {IX.3.3},
	it suffices to show that every regular element of $R$ is actually a unit.
	By Theorem {IX.3.3} and a direct argument it suffices to assume $R=\text{Mat}_n D$ for some division ring $D$.
	Theorem {VII.2.6} and Proposition {VII.2.12} may be helpful.]
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	The following are equivalent:
	\begin{enumerate}
	\item $R$ is prime;
	\item $a, b\in R$ and $aRb=0$ imply $a=0$ or $b=0$.
	\item the right annihilator of every nonzero right ideal of $R$ is $0$;
	\item the left annihilator of every nonzero left ideal of $R$ is $0$.
	\end{enumerate}
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	Every primitive ring is prime [see Exercise {IX.4.7}].
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	The center of a prime ring with identity is an integral domain.
	[See Exercise {IX.4.7}; for the converse see Exercise {IX.4.10}.]
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	Let $J$ be an integral domain and let $F$ be the complete field of quotients of $J$.
	Let $R$ be the set of all infinite matrices (row, columns indexed by $\mathbb{N}^*$) of the form
	$$
	\begin{pmatrix}
	A_n & & & & \\
	& d & & & & & O & \\
	& & d & & \\
	& & & d & \\
	& & & & & \ddots \\
	& O & & & & & \ddots \\
	& & & & & & & \ddots \\
	\end{pmatrix}
	$$
	where $A_n\in \text{Mat}_n(F)$ and $d\in J\subseteq F$.
	\begin{enumerate}
	\item $R$ is a ring.
	\item The center of $R$ is the set of all matrices of the form
	$$
	\begin{pmatrix}
	& d & & & & & & \\
	& & d & & & & O \\
	& & & d & \\
	& & & & & \ddots \\
	& & O & & & & \ddots \\
	& & & & & & & \ddots \\
	\end{pmatrix}
	$$
	with $d\in J$ and hence is isomorphic to $J$.
	\item $R$ is primitive (and hence prime by Exercise {IX.4.8}).
	\end{enumerate}
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	The nil radical $N(R)$ of $R$ is the ideal generated by the set of all nil ideals of $R$.
	\begin{enumerate}
	\item $N(R)$ is a nil ideal.
	\item $N(N(R))=N(R)$.
	\item $N(R/N(R))=0$.
	\item $P(R)\subseteq N(R)\subseteq J(R)$.
	\item If $R$ is left Artinian, $P(R)=N(R)=J(R)$.
	\item If $R$ is commutative $P(R)=N(R)$.
	\end{enumerate}
	\end{enumerate}
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\section{Algebras}
	{\it Note:} $K$ is always a commutative ring with identity and $A$ a $K$-algebra.
	\begin{enumerate}
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	The $\mathbb{Q}$-algebra $A$ of Exercise {IV.7.4} is a left Artinian $\mathbb{Q}$-algebra that is {\it not} a left Artinian ring.
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	A finite dimensional algebra over a field $K$ satisfies both the ascending and descending chain conditions on left and right algebra ideals.
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	\begin{enumerate}
	\item If $M$ is a left algebra $A$-module,
	then $\mathfrak{a}(M)=\{r\in A\mid rc=0\text{ for all }c\in M\}$ is an algebra ideal of $A$.
	\item An algebra ideal $P$ of $A$ is said to be primitive if the quotient algebra $R/P$ is primitive (that is,
	has a faithful simple algebra $R/P$-module).
	Show that every primitive algebra ideal is a primitive ideal of the ring $A$ and vice versa.
	\end{enumerate}
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	Let $M$ be a simple algebra $A$-module.
	\begin{enumerate}
	\item $D=\text{Hom}_A(M, M)$ is a division algebra over $K$,
	where $\text{Hom}_A(M, M)$ denotes all endomorphisms of the algebra $A$-module $M$.
	\item $M$ is a left algebra $D$-module.
	\item The ring $\text{Hom}_D(M, M)$ of all $D$-algebra endomorphisms of $M$ is a $K$-algebra.
	\item The map $A\mapsto \text{Hom}_D(M, M)$ given by $r\mapsto \alpha_r$ (where $\alpha_r(x)=rx$) is a $K$-algebra homomorphism.
	\end{enumerate}
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	Let $A$ be the set of all denumerably infinite matrices over a field $K$ (that is,
	matrices with rows and columns indexed by $\mathbb{N}^*$) which have only a finite number of nonzero entries.
	\begin{enumerate}
	\item $A$ is a simple $K$-algebra.
	\item $A$ is an infinite dimensional algebraic $K$-algebra.
	\end{enumerate}
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	The radical $J$ of an algebraic algebra $A$ over a field $K$ is nil.
	[{\it Hint:} if $r\in J$ and $k_n r^n+k_{n-1}r^{n-1}+\cdots+k_t r^t=0$ ($k_t\neq 0$),
	then $r^t=r^t u$ with $u=-k_t^{-1}k_n r^{n-t}-\cdots-k_t^{-1}k_{t+1}r$,
	whence $-u$ is right quasi-regular,
	say $-u+v-uv=0$.
	Show that $0=r^t(-u-v-uv)=-r^t$.]
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	Let $A$ be a $K$-algebra and $C$ the center of the ring $A$.
	\begin{enumerate}
	\item $C$ is a $K$-subalgebra of $A$.
	\item If $K$ is an algebraically closed field and $A$ is finite dimensional semisimple,
	then the number $t$ of simple components of $A$ (as in Theorem {IX.5.7}) is precisely $\dim_K{C}$.
	\end{enumerate}
	\end{enumerate}
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\section{Division Algebras}
	\begin{enumerate}
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	If $A$ is a finite dimensional central simple algebra over the field $K$,
	then $A\otimes_K A^{op}\cong \text{Mat}_n K$,
	where $n=\dim_K{A}$ and $A^{op}$ is defined in Exercise {III.1.17}.
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	If $A$ and $B$ are central simple algebras over a field $K$,
	then so is $A\otimes_K B$.
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	Let $D$ be a division ring and $F$ a subfield.
	If $d\in D$ commutes with every element of $F$,
	then the subdivision ring $F(d)$ generated by $F$ and $d$ (the intersection of all subdivision rings of $D$ containing $F$ and $d$) is a subfield. [See Theorem {V.1.3}.]
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	If $D$ is a division ring,
	then $D$ contains a maximal subfield.
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	If $A$ is a finite dimensional central simple algebra over a field $K$,
	then $\dim_K{A}$ is a perfect square.
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	If $A$ and $B$ are left Artinian algebras over a field $K$,
	then $A\otimes_K B$ need not be left Artinian.
	[{\it Hint:} let $A$ be a division algebra with center $K$ and maximal subfield $B$ such that $\dim_B{A}$ is infinite.]
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	If $D$ is finite dimensional division algebra over its center $K$ and $F$ is a maximal subfield of $D$,
	then there is a $K$-algebra isomorphism $D\otimes_K F\cong \text{Mat}_n F$,
	where $n=\dim_F{D}$.
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	If $A$ is a simple algebra finite dimensional over its center,
	then any automorphism of $A$ that leaves the center fixed elementwise is an inner automorphism.
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	(Dickson) Let $D$ be a division ring with center $K$.
	If $a, b\in D$ are algebraic over the field $K$ and have the same minimal polynomial,
	then $b=dad^{-1}$ for some $d\in D$.
	\end{enumerate}
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\chapter{Categories}
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\section{Functors and Natural Transformations}
	{\it Note:} In these exercises $\mathcal{S}$ is the category of sets and functions;
	$\mathcal{R}$ is the category of rings and ring homomorphisms;
	$R$ is a ring;
	$\mathfrak{M}$ is the category of left $R$-modules and $R$-module homomorphisms;
	$\mathcal{G}$ is the category of groups and group homomorphisms.
	\begin{enumerate}
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	Construct functors as follows:
	\begin{enumerate}
	\item A covariant functor $\mathcal{G}\to \mathcal{S}$ that assigns to each group the set of all its subgroups.
	\item A covariant functor $\mathcal{R}\to \mathcal{R}$ that assigns to each ring $N$ the polynomial ring $N[x]$.
	\item A functor, covariant in both variables $\mathfrak{M}\times \mathfrak{M}\to \mathfrak{M}$ such that
	$$(A, B)\mapsto A\oplus B.$$
	\item A covariant functor $\mathcal{G}\to \mathcal{G}$ that assigns to each group $G$ its commutator subgroup $G'$ (Definition {II.7.7}).
	\end{enumerate}
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	\begin{enumerate}
	\item If $T:\mathcal{C}\to \mathcal{D}$ is a covariant functor,
	let $\text{Im }T$ consist of the objects $\{T(C)\mid C\in \mathcal{C}\}$ and the morphisms $\{T(f):T(C)\to T(C')\mid f:C\to C'\text{ a morphism in }\mathcal{C}\}$.
	Then show that $\text{Im }T$ need {\it not} be a category.
	\item If the object function of $T$ is injective,
	then show that $\text{Im }T$ is a category.
	\end{enumerate}
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	\begin{enumerate}
	\item If $S:\mathcal{C}\to \mathcal{D}$ is a functor,
	let $\sigma(S)=1$ if $S$ is covariant and $-1$ if $S$ is contravariant.
	If $T:\mathcal{D}\to \mathcal{E}$ is another functor,
	show that $TS$ is a functor from $\mathcal{C}$ to $\mathcal{E}$ whose variance is given by $\sigma(TS)=\sigma(T)\sigma(S)$.
	\item Generalize part (a) to any finite number of functors,
	$
	S_1:\mathcal{C}_1\to \mathcal{C}_2,
	S_2:\mathcal{C}_2\to \mathcal{C}_3, ...,
	S_n:\mathcal{C}_n\to \mathcal{C}_{n+1}
	$.
	\end{enumerate}
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	\begin{enumerate}
	\item If $A, B, C$ are sets,
	then there are natural bijections: $A\times B\to B\times A$ and $(A\times B)\times C\to A\times (B\times C)$.
	\item Prove that the isomorphisms of Theorem {IV.4.9}, Theorem {IV.5.8}, Theorem {IV.5.9}, and Theorem {IV.5.10} are all natural.
	\end{enumerate}
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	Let $\mathcal{V}$ be the category whose objects are all finite dimensional vector spaces over a field $F$ (of characteristic $\neq 2, 3$)
	and whose morphisms are all vector-space {\it isomorphisms}.
	Consider the dual space $V^*$ of a left vector space $V$ as a left vector space (see the Remark after Proposition {VII.1.10}).
	\begin{enumerate}
	\item If $\phi:V\to V_1$ is a vector-space isomorphism (morphism of $\mathcal{V}$),
	then so is the dual map $\overline{\phi}:{V_1}^*\to V^*$ (see Theorem {IV.4.10}).
	Hence $\overline{\phi}^{-1}:V^*\to {V_1}^*$ is also a morphism of $\mathcal{V}$.
	\item $D:\mathcal{V}\to \mathcal{V}$ is a covariant functor,
	where $D(V)=V^*$ and $D(\phi)=\overline{\phi}^{-1}$.
	\item For each $V$ in $\mathcal{V}$ choose a basis $\{x_1, ..., x_n\}$ and let $\{f_{x_1}, ..., f_{x_n}\}$ be the dual bases of $V^*$ (Theorem {IV.4.11}).
	Then the map $\alpha_V:V\to V^*$ defined by $x_i\mapsto f_{x_i}$ is an isomorphism.
	Thus $\alpha_V:V\cong D(V)$.
	\item The isomorphism $\alpha_V$ is {\it not} natural;
	that is,
	the assignment $V\mapsto \alpha_V$ is {\it not} a natural isomorphism from the identity functor $I_{\mathcal{V}}$ to $D$.
	[{\it Hint:} consider a one dimensional space with basis $\{x\}$ and let $\phi(x)=cx$ with $c\neq 0, \pm 1_F$.]
	\end{enumerate}
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	\begin{enumerate}
	\item Let $S:\mathcal{C}\to \mathcal{D}$ and $T:\mathcal{C}\to \mathcal{D}$ be covariant functors and $\alpha:S\to T$ a natural isomorphism.
	Then there is a natural isomorphism $\beta:T\to S$ such that $\beta \alpha=I_S$ and $\alpha\beta=I_T$,
	which $I_S:S\to S$ is the identity natural isomorphism and similarly for $I_T$.
	[{\it Hint:} for each $C$ of $\mathcal{C}$,
	$\alpha_C:S(C)\to T(C)$ is an equivalence and hence has an inverse morphism $\beta_C:T(C)\to S(C)$.]
	\item Extend (a) to functors of several variables.
	\end{enumerate}
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	Covariant representable functors from $\mathcal{S}$ to $\mathcal{S}$ preserve surjective maps.
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	\begin{enumerate}
	\item The forgetful functor $\mathfrak{M}\to \mathcal{S}$ (see the Example preceding Definition {X.1.2}) is representable.
	\item The forgetful functor $\mathcal{G}\to \mathcal{S}$ is representable.
	\end{enumerate}
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	\begin{enumerate}
	\item Let $P:\mathcal{S}\to \mathcal{S}$ be the functor that assigns to each set $X$ its power set (set of all subsets) $P(X)$ and to each function $f:A\to B$ the map $P(f):P(B)\to P(A)$ that sends a subset $X$ of $B$ onto $f^{-1}(X)\subseteq A$.
	Then $P$ is a representable contravariant functor.
	\item Let the object function of $Q:\mathcal{S}\to \mathcal{S}$ be defined by $Q(A)=P(A)$.
	If $f:A\to B$,
	let $Q(f):Q(A)\to Q(B)$ be given by $X\mapsto f(X)$.
	Then $Q$ is a covariant functor.
	Is $Q$ representable?
	\end{enumerate}
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	Let $(A, \alpha)$ and $(B, \beta)$ be representations of the covariant functors $S:\mathcal{C}\to \mathcal{S}$ and $T:\mathcal{C}\to \mathcal{S}$ respectively.
	If $\tau:S\to T$ is a natural transformation,
	then there is a unique morphism $f:A\to A$ in $\mathcal{C}$ such that the following diagram is commutative for every object $C$ of $\mathcal{C}$:
	$$
	\xymatrix{
	\text{hom}_{\mathcal{C}}(A, C) \ar[r]^-{\alpha_C} \ar[d]_{\text{hom}(f, 1_C)}& S(C) \ar[d]^{\tau_C} \\
	\text{hom}_{\mathcal{C}}(B, C) \ar[r]_-{\beta_C} & T(C)
	}
	$$
	\end{enumerate}
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\section{Adjoint Functors}
	{\it Note:} $\mathcal{S}$ denotes the category of sets.
	\begin{enumerate}
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	If $T:\mathcal{C}\to \mathcal{S}$ is a covariant functor that has a left adjoint,
	then $T$ is representable.
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	Let $\mathcal{C}$ be a concrete category and $T:\mathcal{C}\to \mathcal{S}$ the forgetful functor.
	If $T$ has a left adjoint $F:\mathcal{S}\to \mathcal{C}$,
	then $F$ is called a free-object functor and $F(x)$ ($X\in \mathcal{S}$) is called a free $F$-object on $X$.
	\begin{enumerate}
	\item The category of groups has a free-object functor.
	\item The category of commutative rings with identity and identity preserving homomorphisms has a free-object functor.
	[If $X$ is finite,
	use Exercise {III.5.11} to define $F(X)$.]
	\end{enumerate}
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	Let $X$ be a fixed set and define a functor $S:\mathcal{S}\to \mathcal{S}$ by $Y\mapsto X\times Y$.
	Then $S$ is a left adjoint of the covariant hom functor $h_X=\text{hom}_{\mathcal{S}}(X, -)$.
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	Let $\mathcal{G}$ be the category of groups,
	$\mathfrak{a}$ the category of abelian groups,
	$\mathcal{F}$ the category of fields,
	$\mathcal{D}$ the category of integral domains,
	$\mathfrak{M}$ the category of unitary left $K$-modules,
	and $\mathcal{B}$ the category of unitary $K$-$K$ bimodules ($K, R$ rings with identity).
	In each of the following cases let $T$ be the appropriate forgetful functor (for example, $T:\mathcal{F}\to \mathcal{D}$ sends each field $F$ to itself,
	considered as an integral domain).
	Show that $(S, T)$ is an adjoint pair.
	\begin{enumerate}
	\item $T:\mathfrak{a}\to \mathcal{G}, S:\mathcal{G}\to \mathfrak{a}$,
	where $S(G)=G/G'$ with $G'$ the commutator subgroup of $G$ (Definition {II.7.7}).
	\item $T:\mathcal{F}\to \mathcal{D}, S:\mathcal{D}\to \mathcal{F}$,
	where $S(D)$ is the field of quotients of $D$ (Section {III.4}).
	\item $T:\mathfrak{M}\to \mathfrak{a}, S:\mathfrak{a}\to \mathfrak{M}$,
	where $S(A)=K\otimes_{\mathbb{Z}} A$ (see Theorem {IV.5.5}).
	\item $T:\mathcal{B}\to \mathfrak{M}, S:\mathfrak{M}\to \mathcal{B}$,
	where $S(M)=M\otimes_{\mathbb{Z}} R$.
	\end{enumerate}
	\end{enumerate}
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\section{Morphisms}
	\begin{enumerate}
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	A morphism in the category of sets is monic [resp. epic] if and only if it is injective [resp. surjective].
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	A morphism $f:G\to H$ in the category of groups is epic if and only if $f$ is a surjective homomorphism (that is, an epimorphism in the usual sense).
	[{\it Hint:} If $f$ is epic,
	$K=\text{Im }f$, and $j:K\to H$ is the inclusion map,
	then $j$ is epic by Proposition {X.3.2}.
	Show that $f$ is surjective (that is, $K=H$) as follows.
	Let $S$ be the set of left cosets of $K$ in $H$;
	let $T=S\cup \{u\}$ with $u\notin S$.
	Let $A$ be the group of all permutations of $T$.
	Let $t:H\to A$ be given by $t(h)(h'K)=hh'K$ and $t(h)(u)=u$.
	Let $s:H\to A$ be given by $\sigma t(h)\sigma$,
	where $\sigma\in A$ is the transposition interchanging $u$ and $K$.
	Show that $s$ and $t$ are homomorphisms such that $sj=tj$,
	whence $s=t$.
	Show that $hK=K$ for all $h\in H$;
	therefore $K=H$.]
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	A commutative diagram
	$$
	\xymatrix{
	B \ar[r]^{g_1} \ar[d]_{g_2} & C_1 \ar[d]^{f_1} \\
	C_2 \ar[r]_{f_2} & D
	}
	$$
	of morphisms of a category $\mathcal{C}$ is called a pullback for $f_1$ and $f_2$ if for every pair of morphisms $h_1:B'\to C_1, h_2:B'\to C_2$ such that $f_1 h_1=f_2 h_2$ there exists a unique morphism $t:B'\to B$ such that $h_1=g_1 t$ and $h_2=g_2 t$.
	\begin{enumerate}
	\item If there is another pullback diagram for $f_1, f_2$ with $B_1$ in the upper left-hand corner,
	then $B$ and $B_1$ are equivalent.
	\item In the pullback diagram above,
	if $f_2$ is a monomorphism,
	then so is $g_1$.
	\item Every pair of functions $f_1:C_1\to D, f_2:C_2\to D$ in the category of sets has a pullback.
	\end{enumerate}
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	Show that every pair of functions $f, g:C\to D$ has a difference cokernel in the category of sets.
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	Let $f, g:C\to D$ be morphisms of a category $\mathcal{C}$.
	For each $X$ in $\mathcal{C}$ let
	$$\text{Eq}(X, f, g)=\{h\in \text{hom}(X, C)\mid fh=gh\}.$$
	\begin{enumerate}
	\item $\text{Eq}(-, f, g)$ is a contravariant functor from $\mathcal{C}$ to the category of sets.
	\item A morphism $i:K\to C$ is a difference kernel of $(f, g)$ if and only if $\text{Eq}(-, f, g)$ is representable with representing object $K$ (that is, there is a natural isomorphism $\tau:\text{hom}_{\mathcal{C}}(-, K)\to \text{Eq}(-, f, g)$).
	[{\it Hint:} show that for $h:X\to K, \tau_X(h)=ih$,
	where $i=\tau_K(1_K)$.]
	\end{enumerate}
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	If each square in the following diagram is a pullback and $B'\to B$ is a monomorphism,
	then the outer rectangle is a pullback.
	[{\it Hint:} See Exercise {X.3.3}.]
	$$
	\xymatrix{
	P \ar[r] \ar[d] & Q \ar[r] \ar[d] & B' \ar[d] \\
	A \ar[r] & I \ar[r] & B.
	}
	$$
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	\item
	In a category with a zero object,
	the kernel of a monomorphism is a zero morphism.
	\end{enumerate}
	\end{document}

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