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\begin{document}

\begin{center}
\textbf{Algebra}

\textit{Mark Steinberger}

PDF dated August 31, 2006

\url{http://www.math.hawaii.edu/~tom/algebra.pdf}

\textbf{Errata and comments} by Darij Grinberg
\end{center}

\noindent The page numbers below refer to the printed page numbers in the PDF
dated August 31, 2006. This list is not claimed to be exhaustive. Items marked
\textquotedblleft substantive\textquotedblright\ change a mathematical
assertion or supply a missing hypothesis; other items are local corrections,
proof gaps, or clarifications.

Almost all of the errata below were found by GPT-5.6 Sol; I have merely
verified them (and, in some cases, improved the wording).

\appendix


\section{Corrections and comments}

\begin{enumerate}
\item \textbf{Page 5, proof of Proposition 1.2.5:} The parenthetical assertion
that ``if $Z$ has more than one element, then $g^{\prime}$ is not unique'' is
false when $f$ is surjective: in that case, $g^{\prime}$ is forced to be
$g\circ f^{-1}$. Replace the assertion by the following:

\textquotedblleft If $Z$ has more than one element and $f$ is not surjective,
then $g^{\prime}$ is not unique: its values on $Y\setminus\operatorname{im}f$
can be changed\textquotedblright.

Likewise, the paragraph preceding Proposition 1.2.5 should say that the
factorization is \emph{not always} unique.

\item \textbf{Page 8, proof of Corollary 1.4.3:} The proof gives only the
forward implication of Corollary 1.4.3. For the converse, if $E(x)=E(y)$, then
reflexivity gives $x\in E(x)=E(y)$, and hence $x\in E(x)\cap E(y)$; thus this
intersection is nonempty.

\item \textbf{Page 10, proof of Proposition 1.4.11:} \emph{Proof gap.} The
argument after the commutative diagram does not handle the case where
$X=\varnothing$ and $Y$ has one element: then $f$ is not surjective, but there
is only one map $Y\rightarrow Y$, so $g^{\prime}=1_{Y}$ cannot be altered.

Here is a uniform argument for the converse direction (i.e., for statement 2
$\Longrightarrow$ statement 1): Let $Z=\{0,1\}$ and let $g:X\rightarrow Z$ be
the constant-$0$ map. By statement 2, there is a unique map $g^{\prime
}:Y\rightarrow Z$ such that $g^{\prime}\circ f=g$. If $f$ were not surjective,
choose $y_{0}\in Y\setminus\operatorname{im}f$ and change the value of
$g^{\prime}$ at $y_{0}$. This would give a second such factorization, a contradiction.

\item \textbf{Page 19, first sentence of Section 2.2:} Delete one occurrence
of \textquotedblleft about\textquotedblright\ in \textquotedblleft about about
a group.\textquotedblright

\item \textbf{Page 20, proof of Lemma 2.2.6:} The proof of
\[
\left(  g^{m}\right)  ^{n}=g^{mn}%
\]
does not justify the assertion that \textquotedblleft we may assume $n\geq
0$\textquotedblright. Let us do so here: If $n=-k<0$, then the already-proved
first power law gives $g^{m}\cdot g^{-m}=g^{m+\left(  -m\right)  }=g^{0}=e$,
so that $\left(  g^{m}\right)  ^{-1}=g^{-m}$, and the positive-$n$ case gives
$\left(  g^{-m}\right)  ^{k}=g^{-mk}=g^{mn}$ (since $-mk=m\underbrace{\left(
-k\right)  }_{=n}=mn$), so that%
\[
\left(  g^{m}\right)  ^{n}=\left(  g^{m}\right)  ^{-k}=\left(  \left(
g^{m}\right)  ^{-1}\right)  ^{k}=\left(  g^{-m}\right)  ^{k}=g^{mn}.
\]


\item \textbf{Page 21, proof of Lemma 2.2.7:} After \textquotedblleft%
$(-1)^{k}m\cdot(-1)^{l}n=(-1)^{k+l}mn$\textquotedblright, add
\textquotedblleft for $k,l\in\mathbb{Z}$\textquotedblright.

\item \textbf{Page 21, paragraph after Definition 2.2.10:} Insert
\textquotedblleft the\textquotedblright\ in \textquotedblleft Cyclic groups
are simplest kind of group there is.\textquotedblright

\item \textbf{Page 22, Definitions 2.2.11:} Replace \textquotedblleft$k\geq
1$\textquotedblright\ by \textquotedblleft$k\geq0$\textquotedblright\ (and
interpret the empty product $g_{1}^{n_{1}}\cdots g_{0}^{n_{0}}$ as $e$).
Otherwise, $\left\langle S\right\rangle $ comes out empty when $S=\varnothing
$, but a subgroup of $G$ cannot be empty.

\item \textbf{Page 24, Exercise 2.2.18 (11):} The hypothesis
\[
aba^{-1}\in\langle b\rangle
\]
does not imply that
\[
H=\{a^{i}b^{j}\mid i,j\in\mathbb{Z}\}
\]
is a subgroup. For example, let $a$ and $b$ be the affine transformations of
$\mathbb{R}$ given by
\[
a(t)=2t\qquad\text{and}\qquad b(t)=t+1.
\]
Then $aba^{-1}=b^{2}$, but
\[
ba(t)=2t+1,\qquad a^{i}b^{j}(t)=2^{i}t+2^{i}j,
\]
so $ba$ is not of the form $a^{i}b^{j}$.

There are two ways to fix this. One is to replace the hypothesis $aba^{-1}%
\in\left\langle b\right\rangle $ by%
\[
a\langle b\rangle a^{-1}=\langle b\rangle,
\]
i.e., that $a$ normalizes $\langle b\rangle$. Another way is to add the
additional hypothesis $a^{-1}ba\in\left\langle b\right\rangle $. In either
case, the claim of the exercise becomes true.

\item \textbf{Page 25, proof of Lemma 2.3.9:} Replace \textquotedblleft The
inverse of $rn+sm$ is $(-r)m+(-s)n$\textquotedblright\ by \textquotedblleft
The inverse of $rm+sn$ is $(-r)m+(-s)n$.\textquotedblright

\item \textbf{Page 32, proof of Corollary 2.5.13:} Before invoking the order
of $g$, note that $g$ has finite order. Indeed, since $(m,n)=1$, at least one
of $m$ and $n$ is nonzero; a nonzero exponent of $g$ yields a positive
exponent by changing its sign if necessary.

\item \textbf{Page 34, Exercise 2.5.21 (11):} Require $d$ to be positive. As
divisibility was defined for arbitrary integers, a negative divisor $d$ of $n$
is otherwise included, although neither a subgroup of order $d$ nor
\textquotedblleft exactly $d$ elements\textquotedblright\ in part (d) makes sense.

\item \textbf{Page 37, proof of Proposition 2.7.2:} The polar-coordinate
argument does not cover the zero vector, for which the angle $\phi$ is not
defined. Begin by observing that $R_{\theta}\mathbf{0}=\mathbf{0}$, and then
assume that $\mathbf{v}\neq\mathbf{0}$ when choosing $\phi$.

\item \textbf{Page 38, Proposition 2.7.7:} In \textquotedblleft$\theta=2\pi
k/n$ for some integers $k$ and $n$,\textquotedblright\ require $n>0$.
Otherwise $n=0$ makes the expression undefined, while a negative $n$ makes the
displayed order $n/(k,n)$ inappropriate.

\item \textbf{Page 40, Exercises 2.8.5 (5) and (7):} In parts (5) and (7),
require $k>0$, since $\mathrm{D}_{2k}$ was defined only for positive $k$.

\item \textbf{Page 42, Exercises 2.9.7 (4) and (7):} In parts (4) and (7),
require $k>0$, since $\mathrm{Q}_{4k}$ was defined only for positive $k$.

\item \textbf{Page 45, Exercises 2.10.12 (1) and (2):} \emph{Missing
convention.} The order of an element was allowed to be $\infty$, but the least
common multiple of a list containing $\infty$ has not been defined. The best
way to fix this is to explicitly extend the least-common-multiple convention
by declaring that the least common multiple of several \textquotedblleft
numbers\textquotedblright\ is $\infty$ when at least one of the
\textquotedblleft numbers\textquotedblright\ is $\infty$.

\item \textbf{Pages 49--50, proof of Lemma 3.1.12:} Replace each
\textquotedblleft$Y-X$\textquotedblright\ in this proof by \textquotedblleft%
$X-Y$\textquotedblright. (There are four of them in total.) Also, on page 49,
replace \textquotedblleft fixes the elements of of $X-Y$\textquotedblright\ by
\textquotedblleft fixes the elements of $X-Y$\textquotedblright.

\item \textbf{Page 53, Remarks 3.2.7:} \emph{Proof correction.} The displayed
inverse
\[
(x^{-1}y)^{-1}=y^{-1}x
\]
establishes symmetry, not reflexivity. Reflexivity, instead, follows from
$x^{-1}x=e\in H$.

Later in the same paragraph, \textquotedblleft takes each $x\in H$ to its
equivalence class\textquotedblright\ should read \textquotedblleft takes each
$x\in G$ to its equivalence class\textquotedblright.

\item \textbf{Page 63, Exercise 3.3.23 (24) (a):} Both occurrences of
$\mathbf{S}_{n}$ in the parenthetical sentence should be $\mathbf{S}_{p}$. The
displayed $p$-cycle generates a subgroup of $\mathbf{S}_{p}$, and the action
from Exercise 3.3.23 (23) being restricted here is the $\mathbf{S}_{p}$-action
on $G^{p}$.

\item \textbf{Page 67, proof of Proposition 3.5.3:} Replace \textquotedblleft
has the same effect on $Y_{i}$ is $\sigma$ does\textquotedblright\ by
\textquotedblleft has the same effect on $Y_{i}$ as $\sigma$
does\textquotedblright.

\item \textbf{Page 74, proof of Corollary 3.6.20:} Insert a period after
\textquotedblleft Let $\sigma\in\mathbf{S}_{n}$\textquotedblright.

\item \textbf{Page 75, Exercise 3.6.24 (10):} The assertion \textquotedblleft
Suppose that $\mathbf{S}_{n}$ has an element of order $k$. Show that
$\mathrm{D}_{2k}$ embeds in $\mathbf{S}_{n}$\textquotedblright\ is false for
small $k$. For example, $\mathbf{S}_{2}$ has an element of order $2$, but
$\mathrm{D}_{4}\cong\mathbb{Z}_{2}\times\mathbb{Z}_{2}$ does not embed in
$\mathbf{S}_{2}$. Add the hypothesis $k\geq3$. With this hypothesis, one can
choose an involution that reverses each cycle of a permutation of order $k$;
this involution conjugates the permutation to its inverse and is not contained
in the cyclic subgroup it generates.

\item \textbf{Page 77, Corollary 3.7.5:} The statement that the number of
conjugates of $H$ \textquotedblleft divides $[G:H]$\textquotedblright%
\ requires $[G:H]$ to be finite. This finiteness condition appears in the
proof but not in the statement.

\item \textbf{Page 84, Notation 4.1.6:} Replace \textquotedblleft and have
$\overline{g}\cdot\overline{g^{\prime}}=\overline{gg^{\prime}}$%
\textquotedblright\ by \textquotedblleft and we have $\overline{g}%
\cdot\overline{g^{\prime}}=\overline{gg^{\prime}}$\textquotedblright.

\item \textbf{Page 86, Corollary 4.1.11:} This holds only for $n\geq2$. For
$n=1$, the sign homomorphism is not surjective, $\mathbf{A}_{1}=\mathbf{S}%
_{1}$, and the formula $\lvert\mathbf{A}_{n}\rvert=n!/2$ is false.

\item \textbf{Page 87, Example 4.1.15:} \textquotedblleft Let $n$ and $k$ be
positive numbers\textquotedblright\ should be \textquotedblleft Let $n$ and
$k$ be positive integers.\textquotedblright

\item \textbf{Page 87, proof of Lemma 4.1.18:} \textquotedblleft and hence
$KH$ a subgroup of $G$\textquotedblright\ should be \textquotedblleft and
hence $KH$ is a subgroup of $G$\textquotedblright.

\item \textbf{Page 88, Corollary 4.1.21:} The divisibility assertion
\[
\lbrack K:H\cap K]\mid\lbrack G:H]
\]
requires $[G:H]$ to be finite. The proof invokes finite group orders and
Lagrange's theorem for the quotient $G/H$.

\item \textbf{Page 90, Definition 4.2.1:} The definition should begin
\textquotedblleft A \emph{nontrivial} group $G$ is simple if \ldots
\textquotedblright. As printed, the trivial group is simple, contradicting
Lemma 4.2.2 (and later Corollary 5.2.4).

\item \textbf{Page 92, proof of Lemma 4.2.11:} Replace \textquotedblleft
Suppose the first conditions holds\textquotedblright\ by \textquotedblleft
Suppose the first condition holds\textquotedblright.

\item \textbf{Page 101, Proposition 4.4.11:} Replace \textquotedblleft Let $G$
be a finite group\textquotedblright\ by \textquotedblleft Let $G$ be a finite
\emph{abelian} group.\textquotedblright\ For a nonabelian group, the elements
whose orders divide a specified integer need not form a subgroup. The proof
itself uses additive notation, Lemma 4.4.3, and commutativity.

\item \textbf{Page 101, proof of Proposition 4.4.12:} The last sentence only
shows that $p$ divides $o(x)$; it does not yet produce an element of order
$p$. To obtain the latter, we write $o(x)=pq$. Then $qx$ has order $p$, which
completes the induction.

\item \textbf{Page 102, proof of Lemma 4.4.17:} The claim that
\textquotedblleft$y=x-tg$ has order $k$\textquotedblright\ could use some more
explanation. Namely,
\[
ky=k\left(  x-tg\right)  =\underbrace{kx}_{=sg}-\underbrace{kt}_{=s}g=sg-sg=0
\]
shows that $o(y)$ divides $k$. But $k$ also divides $o\left(  y\right)  $,
since the image $\overline{y}=\overline{x}$ in $G/\langle g\rangle$ has order
$k$. These two divisibilities give $o(y)=k$.

\item \textbf{Page 103, proof of Proposition 4.4.18:} The induction omits the
trivial $p$-group, whose identity has order $1=p^{0}$. Begin with $\lvert
G\rvert=1$, where $G$ is the empty direct product, and then treat $\lvert
G\rvert=p$ as the next case.

\item \textbf{Page 104, proof of Proposition 4.4.22:} After quotienting by the
subgroups of order $p$, the proof writes factors $\mathbb{Z}_{p^{r_{i}-1}}$.
If $r_{i}=1$, this is the trivial factor $\mathbb{Z}_{p^{0}}=\mathbb{Z}_{1}$,
whereas the induction hypothesis was stated using positive exponents. Delete
the trivial factors before applying induction, and use the already established
equality $k=l$ to keep track of how many factors disappeared.

\item \textbf{Page 105, Lemma 4.5.1:} Replace \textquotedblleft$n>1$%
\textquotedblright\ by \textquotedblleft$n>0$\textquotedblright. (There is no
need to discriminate against $1$ here.)

\item \textbf{Page 105, Definition 4.5.2:} Replace \textquotedblleft%
$n>1$\textquotedblright\ by \textquotedblleft$n>0$\textquotedblright. (The
case $n=1$ is used in Corollary 4.5.5.)

\item \textbf{Page 105, Lemma 4.5.3:} Again, replace \textquotedblleft%
$n>1$\textquotedblright\ by \textquotedblleft$n>0$\textquotedblright.

\item \textbf{Page 106, Lemma 4.5.6:} Lemma 4.5.6 should assume that $G$ is
finite; otherwise the phrase \textquotedblleft the primes dividing the order
of $G$\textquotedblright\ and the asserted primary decomposition are not
available in the stated form.

\item \textbf{Page 110, Proposition 4.5.18:} \emph{Missing hypothesis.}
Require
\[
1\leq s\leq r;
\]
when $p=2$, retain also the hypothesis $s>1$. This is needed to ensure that
$p^{r-s}$ is an integer.

\item \textbf{Page 116, proof of Proposition 4.6.11:} Lowercase
\textquotedblleft There\textquotedblright\ in \textquotedblleft Thus, by
Proposition 4.6.9, There is a homomorphism\textquotedblright.

\item \textbf{Page 118, Exercise 4.6.13 (6):} \emph{Missing hypothesis.} The
formulas use orders as finite integers, although the book permits
$o(x)=\infty$. In particular, an expression such as $x^{o(f(x))}$ need not
make sense. Add that the elements under discussion have finite order (for
example, assume the groups are finite). Under this hypothesis, if $m\mid
o(x)$, then
\[
o(x)=m\cdot o(x^{m}),
\]
and the subsequent deductions are valid.

\item \textbf{Page 123, Example 4.7.8:} The conjugation homomorphism has
codomain $\operatorname{Aut}(\mathbf{A}_{n})$:
\[
\Gamma\colon\mathbf{S}_{n}\longrightarrow\operatorname{Aut}(\mathbf{A}_{n}),
\]
not $\Gamma\colon\mathbf{S}_{n}\rightarrow\mathbf{A}_{n}$.

\item \textbf{Page 126, Remarks 4.7.19:} In \textquotedblleft The extensions
$f\colon G\rightarrow K$ and $f^{\prime}\colon G\rightarrow K$ are equivalent
\ldots\textquotedblright, replace \textquotedblleft$f^{\prime}\colon
G\rightarrow K$\textquotedblright\ by \textquotedblleft$f^{\prime}\colon
G^{\prime}\rightarrow K$\textquotedblright.

\item \textbf{Page 130, Exercise 4.7.27 (3):} Replace \textquotedblleft%
$H\subset\mathbf{A}_{n}$ of Problem 1\textquotedblright\ by \textquotedblleft%
$H\subset\mathbf{A}_{4}$ of Problem 1\textquotedblright.

\item \textbf{Page 130, Exercise 4.7.27 (8):} Replace \textquotedblleft Show
that $f$ is isomorphic to either $\mathrm{D}_{16}$ or $\mathrm{Q}_{16}%
$\textquotedblright\ by \textquotedblleft Show that $G$ is isomorphic to
either $\mathrm{D}_{16}$ or $\mathrm{Q}_{16}$\textquotedblright.

\item \textbf{Page 133, Exercise 4.7.27 (19) (d):} The expression
\[
\frac{m^{k}-1}{m-1}%
\]
is undefined when the action is trivial, i.e., when $m=1$. Use throughout the
polynomial expression
\[
1+m+\cdots+m^{k-1},
\]
which equals $k$ when $m=1$.

\item \textbf{Page 133, Exercise 4.7.27 (19) (g):} The assertion
\textquotedblleft If $n$ is prime, then $H=\mathbf{Z}_{n}$\textquotedblright%
\ is false without a condition on the action. Let
\[
S=1+m+\cdots+m^{k-1}.
\]
Part (d) shows that $H$ consists of those $b^{i}$ for which $n\mid iS$. Thus,
when $n$ is prime, $H=\mathbf{Z}_{n}$ if and only if $n\mid S$; otherwise $H$
is trivial. A nontrivial action ($m\not \equiv 1\pmod n$) implies $n\mid S$,
but a trivial action need not. For example, with $n=5$, $k=3$, and the trivial
action,
\[
G=\mathbf{Z}_{5}\times\mathbf{Z}_{3}\cong\mathbf{Z}_{15}.
\]
The subgroup $\mathbf{Z}_{5}$ is characteristic, as required by the standing
hypothesis of the exercise, but the subgroup $H$ from part (d) is trivial.

\item \textbf{Page 135, chapter introduction:} In the assertion that the full
converse of Lagrange's Theorem holds for nilpotent groups, replace
\textquotedblleft A nilpotent group $G$\textquotedblright\ by
\textquotedblleft A \emph{finite} nilpotent group $G$\textquotedblright. The
result proved later as Corollary 5.7.10 has this finiteness hypothesis.

\item \textbf{Page 136, proof of Corollary 5.1.2:} Replace \textquotedblleft A
second application of Cauchy's Theorem given an element $a\in G$ of order
$2$\textquotedblright\ by \textquotedblleft A second application of Cauchy's
Theorem gives an element $a\in G$ of order $2$\textquotedblright.

\item \textbf{Page 137, proof of Proposition 5.2.2:} In both appearances of
\textquotedblleft$\lbrack G:\mathcal{C}_{G}(g)]$\textquotedblright, replace
\textquotedblleft$g$\textquotedblright\ by \textquotedblleft$x$%
\textquotedblright.

\item \textbf{Page 139, proof of Proposition 5.2.7:} Replace
\[
\alpha(a^{j})=\overline{m^{j}}%
\]
by
\[
\alpha(f(a^{j}))=\overline{m^{j}}%
\]
(or say directly that conjugation by $a^{j}$ acts as the $j$-th power of
conjugation by $a$), since $\alpha$ has domain $\mathbb{Z}_{p}$, not $G$.
Also, the result that $K(p,2)$ is cyclic of order $p$ generated by $1+p$ is
Corollary 4.5.19, not Corollary 7.3.15.

\item \textbf{Page 140, proof of Lemma 5.2.9:} The representative range in
\[
\mathbf{Z}_{p}\times\mathbf{Z}_{p}=\{a^{i}b^{j}\mid0\leq i,j\leq p\}
\]
should be $0\leq i,j<p$ (equivalently, $0\leq i,j\leq p-1$).

\item \textbf{Page 145, proof of Corollary 5.3.18:} Replace the citation to
Corollary 5.3.16 by a citation to Corollary 5.3.13. Once the $2$-Sylow
subgroup $H$ has been shown to be normal, Corollary 5.3.13 gives
\[
G\cong H\rtimes_{\alpha}\mathbf{Z}_{3}.
\]
Corollary 5.3.16 cannot be applied here: neither choice of the two primes in
$12=2^{2}\cdot3$ satisfies its hypothesis $q^{s}<p$.

\item \textbf{Page 149, proof of Lemma 5.4.5:} Taken literally, the printed
argument proves only
\[
f([G,G])\subseteq\lbrack G,G].
\]
Apply the same argument to the automorphism $f^{-1}$ to obtain
\[
f^{-1}([G,G])\subseteq\lbrack G,G],
\]
or equivalently $[G,G]\subseteq f([G,G])$. These two inclusions give the
equality required for $[G,G]$ to be characteristic.

\item \textbf{Page 151, first paragraph of Section 5.5:} \emph{Wording.} The
reference to Galois theory should say that there is no \emph{general} formula
by radicals solving every polynomial of degree at least $5$. There are, of
course, many individual polynomials of such degrees that are solvable by radicals.

\item \textbf{Page 151, proof of Lemma 5.5.4:} Replace \textquotedblleft the
result follows by induction on $k$\textquotedblright\ by \textquotedblleft the
result follows by induction on $i$\textquotedblright, in agreement with the
index quantified in the statement.

\item \textbf{Page 152, paragraph before Definition 5.5.8:} In the statement
that $G_{i+1}$ need not be normal in $G$, replace \textquotedblleft if
$i>1$\textquotedblright\ by \textquotedblleft if $i\geq1$\textquotedblright.
For $i=1$, the subgroup $G_{2}$ is required to be normal in $G_{1}$, but need
not be normal in $G=G_{0}$.

\item \textbf{Page 152, proof of Proposition 5.5.9:} In the forward direction,
replace \textquotedblleft the quotient group $G_{i}/G_{i+1}$\textquotedblright%
\ by \textquotedblleft the quotient group $G^{(i)}/G^{(i+1)}$%
\textquotedblright. The displayed subnormal series at this point is the
derived series $G^{(k)}\mathrel{\triangleleft}\cdots
\mathrel{\triangleleft}G^{(0)}$, not the separately indexed series
$G_{k}\mathrel{\triangleleft}\cdots\mathrel{\triangleleft}G_{0}$ used in the converse.

\item \textbf{Page 153, paragraph after the Hall Converse:} Replace
\textquotedblleft every order $n$ dividing $G$\textquotedblright\ by
\textquotedblleft every order $n$ dividing $\lvert G\rvert$\textquotedblright.

\item \textbf{Page 154, proof of Lemma 5.6.5:} In \textquotedblleft let
$H_{p}$ be the $p$-torsion subgroup of $G$,\textquotedblright\ replace $G$ by
$H$. The next sentence correctly treats $H_{p}$ as a characteristic subgroup
of $H$.

\item \textbf{Pages 154--155, proof of Theorem 5.6.2:} \emph{Missing boundary
cases.} At the beginning of the induction, handle $n=1$ and $k=1$ separately:
for $n=1$, the desired Hall subgroup is $e$; for $k=1$, it is $G$. This also
supplies the case $\lvert G\rvert=1$.

The subsequent second branch implicitly requires $n,k>1$. Without this
reduction, the inference $k=p^{r}$ from $k\mid p^{r}$ and $(n,k)=1$ fails when
$k=1$, while when $n=1$ the quotient $G/P$ may be trivial and therefore has no
nontrivial minimal normal subgroup $\overline{Q}/P$ to choose.

\item \textbf{Page 155, proof of Theorem 5.6.2:} Remove the comma after
\textquotedblleft Then Lemma 5.6.5\textquotedblright.

\item \textbf{Page 159, paragraph before Corollary 5.8.2:} The map
\[
\iota_{n}\colon\operatorname*{Gl}\nolimits_{n}(A)\longrightarrow
\operatorname{Aut}(A^{n})
\]
is an injective \emph{group} homomorphism, not an injective ring homomorphism;
neither displayed object is a ring in this context.

\item \textbf{Page 160, Proposition 5.8.4:} It is worth reminding the reader
that the groups $A^{0}$ and $\operatorname*{Gl}\nolimits_{0}(A)$ are trivial.
Without this convention, the asserted right-hand side would not be defined
when $n=1$.

\item \textbf{Page 160, proof of Proposition 5.8.4:} \emph{Proof error over
general rings.} The conclusion is correct, but the argument using linear
dependence (\textquotedblleft as otherwise the first $n-1$ rows of the full
matrix would be linearly dependent\textquotedblright) is false in this
generality: Already a one-row matrix $(2)$ over $\mathbb{Z}$ is noninvertible
without its row being linearly dependent. Instead, the easiest way to argue
that $\left(
\begin{array}
[c]{cc}%
M & 0\\
X & 1
\end{array}
\right)  \in\operatorname*{Gl}\nolimits_{n}\left(  A\right)  $ implies
$M\in\operatorname*{Gl}\nolimits_{n-1}\left(  A\right)  $ is using the
determinantal identity
\[
\det%
\begin{pmatrix}
M & 0\\
X & 1
\end{pmatrix}
=\det M.
\]


\item \textbf{Page 165, Definition 6.1.6:} Replace the composition rule
\[
(f\times g)\circ(f^{\prime}\times g^{\prime})=\left(  f\circ f^{\prime
}\right)  \times\left(  g\circ g^{\prime}\right)
\]
by
\[
(f^{\prime}\times g^{\prime})\circ(f\times g)=(f^{\prime}\circ f)\times
(g^{\prime}\circ g).
\]


\item \textbf{Page 166, Exercise 6.1.10 (2):} \emph{Missing exception.} Not
every injection in $\mathfrak{Sets}$ has a left inverse. If $X=\varnothing$
and $Y\neq\varnothing$, the injection $X\rightarrow Y$ has no left inverse
because there is no function $Y\rightarrow\varnothing$. Add the hypothesis
$X\neq\varnothing$ (or separately exclude this case).

The assertion concerning surjections is the Axiom of Choice adopted in Chapter 1.

\item \textbf{Page 169, Exercise 6.2.6 (10) (b):} Replace \textquotedblleft%
$\iota_{Y}(X)=X\times Y$\textquotedblright\ by \textquotedblleft$\iota
_{Y}(X)=(X,Y)$\textquotedblright.

\item \textbf{Page 171, Exercise 6.3.2 (6):} \emph{Substantive false
assertion.} The empty set has a unique $G$-action, and there is exactly one
$G$-map from it to every $G$-set. Hence the empty $G$-set is an initial object
in $G\mathfrak{-}\mathfrak{sets}$ for every group $G$, including every
nontrivial group. The requested assertion is therefore false and should be
deleted (or replaced by the assertion just stated).

\item \textbf{Page 181, proof of Proposition 6.4.5:} Insert the missing right
parenthesis in the equality
\[
h((g(a),-f(a))=f^{\prime}g(a)-g^{\prime}f(a)=0.
\]
Thus, the left-hand side should be $h((g(a),-f(a)))$.

\item \textbf{Page 183, first paragraph of Section 6.5:} Binary products and
coproducts give products and coproducts of any \emph{nonempty} finite family
by induction. The empty cases require a terminal object and an initial object,
respectively, and do not follow merely from the existence of binary products
or coproducts.

\item \textbf{Page 186, proof of Proposition 6.5.10:} Replace
\textquotedblleft$f\colon\bigoplus_{i\in I}\longrightarrow B$%
\textquotedblright\ by \newline\textquotedblleft$f\colon\bigoplus_{i\in
I}A_{i}\longrightarrow B$\textquotedblright.

\item \textbf{Page 188, proof of Lemma 6.6.3:} Replace \textquotedblleft with
$A$ and object of $C$\textquotedblright\ by \textquotedblleft with $A$ an
object of $C$\textquotedblright.

\item \textbf{Page 188, paragraph before Proposition 6.6.4:} The parenthetical
statement that one-point sets are the only sets for which free groups are
abelian overlooks the empty set: the free group and the free abelian group on
$\varnothing$ are both trivial. Replace \textquotedblleft one-point
sets\textquotedblright\ by \textquotedblleft sets of cardinality at most
one\textquotedblright.

\item \textbf{Page 188, Proposition 6.6.5 and the following paragraph:} The
free-object structure maps were denoted by $\eta_{X}$ and $\eta_{Y}$, not by
$\iota_{X}$ and $\iota_{Y}$. Accordingly, replace \textquotedblleft%
$F(f)\circ\iota_{X}=\iota_{Y}\circ f$\textquotedblright\ by \textquotedblleft%
$F(f)\circ\eta_{X}=\eta_{Y}\circ f$\textquotedblright. In the following
paragraph, likewise replace \textquotedblleft its image under $\iota_{X}%
$\textquotedblright\ by \textquotedblleft its image under $\eta_{X}%
$\textquotedblright.

\item \textbf{Page 189, Exercise 6.6.6 (5):} Replace \textquotedblleft$G$ is
generated\textquotedblright\ by \textquotedblleft$A$ is
generated\textquotedblright.

\item \textbf{Page 190, Definition 6.7.1:} In \textquotedblleft induced by
$\{x_{1},\ldots,x_{n}\}\subseteq G$\textquotedblright, replace
\textquotedblleft$n$\textquotedblright\ by \textquotedblleft$k$%
\textquotedblright.

\item \textbf{Page 192, Proposition 6.8.3:} In the displayed description
\[
G=\{(g_{i}\mid i\geq0)\mid h_{i}(g_{i})=g_{i-1}\text{ for all }i\geq0\},
\]
replace the final \textquotedblleft$i\geq0$\textquotedblright\ by
\textquotedblleft$i\geq1$\textquotedblright. The morphism $h_{0}$ and the term
$g_{-1}$ are not defined.

\item \textbf{Page 194, Exercise 6.8.8 (7):} Insert \textquotedblleft
is\textquotedblright\ in \textquotedblleft the image of $\iota_{i}$ the
subgroup\textquotedblright, so that it reads \textquotedblleft the image of
$\iota_{i}$ is the subgroup\textquotedblright.

\item \textbf{Page 196, Definition 6.9.3 and Proposition 6.9.5:}
\emph{Nonstandard terminology.} The definition of a skeleton given here is
highly nonstandard. The standard definition would additionally require that
$D$ have no two distinct isomorphic objects. Thus, for example, any category
$C$ is a skeleton of itself according to Definition 6.9.3, but not according
to the standard definition unless it has no two distinct isomorphic objects.

\item \textbf{Page 204, Definitions 7.1.10:} \textquotedblleft for some
integer $n$\textquotedblright\ should be \textquotedblleft for some positive
integer $n$\textquotedblright\ or \textquotedblleft for some nonnegative
integer $n$\textquotedblright.

\item \textbf{Page 211, paragraph before Exercises 7.1.36:} The sentence
\textquotedblleft$\mathbb{C}$ and $\mathbb{H}$ are the only division rings
that are algebras over $\mathbb{R}$ in such a way that the induced real vector
space structure is finite dimensional\textquotedblright\ omits $\mathbb{R}$
itself. Frobenius' theorem says that the three finite-dimensional associative
real division algebras are
\[
\mathbb{R},\qquad\mathbb{C},\qquad\mathbb{H}.
\]


\item \textbf{Page 218, Definitions 7.2.16:} Replace \textquotedblleft$k\geq
1$\textquotedblright\ by \textquotedblleft$k\geq0$\textquotedblright\ three
times in this definition. (The sums should be allowed to be empty, since
ideals must contain $0$ even when $S$ is empty.)

\item \textbf{Page 219, Definitions 7.2.21:} Again, replace \textquotedblleft%
$k\geq1$\textquotedblright\ by \textquotedblleft$k\geq0$\textquotedblright%
\ three times.

\item \textbf{Page 223, proof of Proposition 7.3.3:} \textquotedblleft And
$\overline{g}_{b}\left(  1\right)  =\iota\left(  1\right)  =1$%
\textquotedblright\ should be \textquotedblleft And $\overline{g}_{b}\left(
1\right)  =g\left(  1\right)  =1$\textquotedblright.

\item \textbf{Page 226, Exercises 7.3.16 (6):} The assertion that the units of
$A[X]$ are precisely the constant units is false for a general ring. For
instance, $1+2X\in\left(  \mathbf{Z}/4\right)  \left[  X\right]  $ is a unit
(in fact, $\left(  1+2X\right)  ^{2}=1$ in $\left(  \mathbf{Z}/4\right)
\left[  X\right]  $) but not constant.

What is true is that if $A$ is commutative, then the units of $A\left[
X\right]  $ are precisely those polynomials whose constant term is a unit and
whose all remaining coefficients are nilpotent. See, e.g.,
\url{https://math.stackexchange.com/questions/19132/characterizing-units-in-polynomial-rings}
for a proof of this fact. Thus, if $A$ is an integral domain (or, more
generally, if the only nilpotent element of $A$ is $0$), then the claim of the
exercise does hold.

\item \textbf{Page 227, Exercise 7.3.16 (20):} Require $n>1$ here (otherwise,
the denominator $\zeta_{n}^{k}-1$ will be zero).

\item \textbf{Page 236, Definition 7.4.6:} This does not define the degree of
the zero polynomial. The standard convention is to define the degree of the
zero polynomial to be $-\infty$ (which is not an actual number but a symbol
that is understood to behave according to the rules $\min\left\{
-\infty,k\right\}  =-\infty$ and $\left(  -\infty\right)  +k=-\infty$ for all
$k\in\mathbb{Z}\cup\left\{  -\infty\right\}  $, so that the standard
properties of degrees still hold for the zero polynomial).

\item \textbf{Page 237, proof of Proposition 7.4.7:} The claim that
\textquotedblleft$h$ must have total degree less than or equal to that of
$f$\textquotedblright\ is slightly underexplained. The reason why it holds is
the following: Consider the polynomial rings $A\left[  X_{1},\ldots
,X_{n}\right]  $ and $A\left[  X_{1},\ldots,X_{n-1}\right]  $ as graded
$A$-algebras with their usual grading by total degree (i.e., each variable
$X_{i}$ is homogeneous of degree $1$). Their subrings $A\left[  X_{1}%
,\ldots,X_{n}\right]  ^{\mathbf{S}_{n}}$ and $A\left[  X_{1},\ldots
,X_{n-1}\right]  ^{\mathbf{S}_{n-1}}$ inherit this grading from them. Consider
the polynomial rings $A\left[  Y_{1},\ldots,Y_{n}\right]  $ and $A\left[
Y_{1},\ldots,Y_{n-1}\right]  $ as graded $A$-algebras as well, but with the
grading now defined by letting each indeterminate $Y_{i}$ be homogeneous of
degree $i$. This is called \textquotedblleft weighted degree\textquotedblright%
. Thus, $A\left[  Y_{1},\ldots,Y_{n-1}\right]  $ is a graded subalgebra of
$A\left[  Y_{1},\ldots,Y_{n}\right]  $. Moreover, the $A$-algebra
homomorphisms $\alpha_{n}:A\left[  Y_{1},\ldots,Y_{n}\right]  \rightarrow
A\left[  X_{1},\ldots,X_{n}\right]  ^{\mathbf{S}_{n}}$ and $\alpha
_{n-1}:A\left[  Y_{1},\ldots,Y_{n-1}\right]  \rightarrow A\left[  X_{1}%
,\ldots,X_{n-1}\right]  ^{\mathbf{S}_{n-1}}$ are graded (since $s_{i}\left(
X_{1},\ldots,X_{n}\right)  $ and $s_{i}\left(  X_{1},\ldots,X_{n-1}\right)  $
are homogeneous of the same degree -- namely, $i$ -- as the variable $Y_{i}$).

Now, the equality%
\[
\pi\left(  f\right)  =g\left(  s_{1}\left(  X_{1},\ldots,X_{n-1}\right)
,\ldots,s_{n-1}\left(  X_{1},\ldots,X_{n-1}\right)  \right)  =\alpha
_{n-1}\left(  g\right)
\]
shows that the total degree of $\pi\left(  f\right)  $ equals the weighted
degree of $g$ (since the homomorphism $\alpha_{n-1}$ is graded). Hence, the
weighted degree of $g$ is the total degree of $\pi\left(  f\right)  $, which
is clearly less than or equal to that of $f$. Since the homomorphism
$\alpha_{n}:A\left[  Y_{1},\ldots,Y_{n}\right]  \rightarrow A\left[
X_{1},\ldots,X_{n}\right]  ^{\mathbf{S}_{n}}$ is also graded, we thus conclude
that the total degree of $\alpha_{n}\left(  g\right)  $ is less than or equal
to that of $f$. But let us now recall that%
\begin{align*}
h  &  =h\left(  X_{1},\ldots,X_{n}\right)  =\underbrace{f\left(  X_{1}%
,\ldots,X_{n}\right)  }_{=f}-\underbrace{g\left(  s_{1}\left(  X_{1}%
,\ldots,X_{n}\right)  ,\ldots,s_{n-1}\left(  X_{1},\ldots,X_{n}\right)
\right)  }_{=\alpha_{n}\left(  g\right)  }\\
&  =f-\alpha_{n}\left(  g\right)  .
\end{align*}
Hence, $h$ is a difference of two polynomials whose total degree is less than
or equal to that of $f$. Therefore, $h$ itself must also have total degree
less than or equal to that of $f$.

\item \textbf{Page 238, Definition 7.4.12:} This definition (which is arguably
somewhat nonstandard) is only well-defined if $a_{0}$ is a unit of $A$ (since
$a_{0}$ appears in denominators).

\item \textbf{Page 239, Exercise 7.4.14 (3):} Again, $a$ must be assumed to be
a unit of $A$ (in order for the fraction to be well-defined).

\item \textbf{Page 239, monoid rings:} The canonical map $M\rightarrow
A\left[  M\right]  $ is injective if the ring $A$ is nontrivial, but is
constant if $A$ is trivial. Thus, strictly speaking, it cannot be regarded as
an inclusion (as you do in Lemma 7.5.3), and you cannot really
\textit{identify} each $m\in M$ with the corresponding element of $A\left[
M\right]  $ (as you do in Definitions 7.5.1). This is a very pedantic point,
since its injectivity is not actually used in practice and the identification
only serves as a convenient abuse of notation. (One almost never derives the
equality of two elements of $M$ from the equality of the respective elements
of $A\left[  M\right]  $; and when one does, one usually has the option of
choosing $A=\mathbb{Z}$ or $A=\mathbb{Q}$ or $A=\mathbb{C}$.)

\item \textbf{Page 242, Exercises 7.5.10 (3) and (4):} These exercises require
$A$ to be nontrivial, since the zero ring has no zero-divisors.

\item \textbf{Page 247:} \textquotedblleft every commutative ring has a
maximal ideal\textquotedblright\ should be \textquotedblleft every nontrivial
commutative ring has a maximal ideal\textquotedblright. The zero ring has no
proper ideals.

\item \textbf{Page 248, proof of Corollary 7.6.14:} The union argument applies
to a nonempty chain (i.e., to the case $S\neq\varnothing$). For the empty
chain, use any element of the already nonempty poset $X$, for example the
original ideal $\mathfrak{a}$, as an upper bound.

\item \textbf{Page 254, Lemma 7.7.19:} \textquotedblleft a two-sided ideal of
$M$\textquotedblright\ should be \textquotedblleft of $A$\textquotedblright.

\item \textbf{Page 254, Lemma 7.7.21:} \textquotedblleft$1\leq i\leq
n$\textquotedblright\ should be \textquotedblleft$1\leq i\leq k$%
\textquotedblright.

\item \textbf{Page 255, Definitions 7.7.25:} Again, replace \textquotedblleft%
$k\geq1$\textquotedblright\ by \textquotedblleft$k\geq0$\textquotedblright.

\item \textbf{Page 256, Exercise 7.7.27 (3):} \textquotedblleft Corollary
7.7.26\textquotedblright\ $\rightarrow$ \textquotedblleft Proposition
7.7.26\textquotedblright.

\item \textbf{Page 256, Exercise 7.7.27 (9):} \textquotedblleft Let $m$
be\textquotedblright\ $\rightarrow$ \textquotedblleft Let $M$
be\textquotedblright.

\item \textbf{Page 256, Exercise 7.7.27 (10):} the displayed matrix should be
labelled $x$, not $a$.

\item \textbf{Page 258, proof of Lemma 7.7.30:} \textquotedblleft from $A^{n}$
to $N$\textquotedblright\ should be \textquotedblleft from $A^{n}$ to
$M$\textquotedblright.

\item \textbf{Page 259, footnote }$^{12}$\textbf{:} Strictly speaking,
\textquotedblleft In the case of a basis, the elements $x_{1},\ldots,x_{n}$
must all be distinct\textquotedblright\ is only true if $A$ is nontrivial. But
as a justification for an abuse of notation, this is fine, and the case of a
trivial ring $A$ is trivial anyway.

\item \textbf{Page 261, Lemma 7.7.38:} This is only true if $A$ is nontrivial.

\item \textbf{Page 261, Lemma 7.7.40:} \textquotedblleft Thus, $m_{1}%
,\ldots,m_{k}$ is a basis\textquotedblright\ $\rightarrow$ \textquotedblleft
Thus, $m_{1},\ldots,m_{n}$ is a basis\textquotedblright.

\item \textbf{Page 262, Corollary 7.7.41:} \textquotedblleft some integer
$n$\textquotedblright\ $\rightarrow$ \textquotedblleft some nonnegative
integer $n$\textquotedblright.

\item \textbf{Page 262, after Definition 7.7.44:} \textquotedblleft Thus,
$M/g\left(  M^{\prime}\right)  $\textquotedblright\ should be
\textquotedblleft Thus, $M/f\left(  M^{\prime}\right)  $\textquotedblright.

\item \textbf{Page 265, proof of Proposition 7.7.51:} Replace
\textquotedblleft$\pi\left(  m_{i}\right)  $\textquotedblright\ by
\textquotedblleft$g\left(  m_{i}\right)  $\textquotedblright\ in the first
sentence of the proof.

\item \textbf{Page 266, Exercise 7.7.52 (3):} This requires $x\neq0$.

\item \textbf{Page 271, proof of Lemma 7.8.7:} Replace \textquotedblleft%
$g\left(  N_{k}\right)  \subset f\left(  N_{k+l}\right)  $\textquotedblright%
\ by \textquotedblleft$g\left(  N_{k}\right)  \subset g\left(  N_{k+l}\right)
$\textquotedblright.

\item \textbf{Page 272, proof of the Hilbert Basis Theorem (Theorem 7.8.14):}
Strictly speaking, the set $\mathfrak{b}$ is not an ideal, since it does not
contain $0$. Moreover, if $f$ and $g$ are two polynomials of equal degrees
$m=n$ whose leading coefficients add to zero, then the leading coefficient of
$f+X^{m-n}g=f+g$ is not $a+b=0$ but something else.

Both of these issues can be fixed by redefining $\mathfrak{b}$ as follows:%
\[
\mathfrak{b}=\left\{  0\right\}  \cup\left\{  \text{leading coefficients of
all nonzero polynomials in }\mathfrak{a}\right\}  .
\]


In the second paragraph of the proof, then, we need to assume that all $x_{i}$
are nonzero (since a zero generator is redundant).

\item \textbf{Page 275, Theorem 7.9.8:} \emph{Missing hypothesis.} Require $A$
to be a nonzero commutative ring. For the zero ring, $A^{m}=A^{n}=0$ for every
$m,n$, so the conclusion $m=n$ is false. The proof's choice of a maximal ideal
also requires $A\neq0$.

\item \textbf{Page 276, proof of Proposition 7.9.11:} Replace
\textquotedblleft$1\leq i\leq k$\textquotedblright\ by \textquotedblleft$1\leq
i\leq l$\textquotedblright.

\item \textbf{Page 279:} On the first two lines of this page, replace both
\textquotedblleft$n$\textquotedblright s by \textquotedblleft$r$%
\textquotedblright s.

\item \textbf{Page 284, Exercise 7.10.23 (3):} Replace \textquotedblleft%
$\Phi_{n,n}$\textquotedblright\ by \textquotedblleft$\Phi_{n+k,n+k}%
$\textquotedblright.

\item \textbf{Page 284, Definition 7.11.1:} The set $S$ should be required to
be nonempty here. Otherwise, $S^{-1}A$ is an empty set and thus cannot be a ring.

\item \textbf{Page 290, proof of Proposition 7.11.16:} Replace
\textquotedblleft Thus, $S^{-1}f$ is onto\textquotedblright\ by
\textquotedblleft Thus, $\ker\left(  S^{-1}g\right)  \subseteq
\operatorname{im}\left(  S^{-1}f\right)  $\textquotedblright.

\item \textbf{Page 306, Exercise 8.1.37 (14) (d):} The element $z$ should be
required to be a nonzero nonunit.

\item \textbf{Page 307, Exercise 8.1.37 (16) (e):} This is false when $m=n$.
For example, the element $2+2\zeta_{3}$ is prime in $\mathbf{Z}\left[
\zeta_{3}\right]  $ (since it equals $-2\zeta_{3}^{2}$ and thus is associate
to $2$), but $\varphi\left(  2+2\zeta_{3}\right)  =\left(  2+2\zeta
_{3}\right)  \overline{\left(  2+2\zeta_{3}\right)  }=4$ is not prime in
$\mathbf{Z}$.

This mistake might be limited to the $m=n$ case, but it needs to be fixed nevertheless.

\item \textbf{Page 315, proof of Lemma 8.3.4:} Replace \textquotedblleft%
$A\left[  X_{1},\ldots,X_{n+1}\right]  $\textquotedblright\ by \newline%
\textquotedblleft$K\left[  X_{1},\ldots,X_{n+1}\right]  $\textquotedblright.

\item \textbf{Page 316, proof of Lemma 8.3.8:} Replace \textquotedblleft whose
total degree is $k$ less than that of $f$\textquotedblright\ by
\textquotedblleft whose total degree $\deg f_{k}$ satisfies $\deg f_{k}%
\leq\deg f-k<\deg f$\textquotedblright.

\item \textbf{Page 316, proof of Proposition 8.3.9:} \textquotedblleft Lemma
8.3.5\textquotedblright\ should be \textquotedblleft Corollary
8.3.5\textquotedblright.

\item \textbf{Pages 318--319, Lemmas 8.4.6 and 8.4.8 and proof of Proposition
8.4.9:} Nonzero constant polynomials do not have roots. Every occurrence of
\textquotedblleft every polynomial has a root\textquotedblright\ should say
\textquotedblleft every polynomial of positive degree has a
root\textquotedblright. In particular, the $f_{i}$ in Lemma 8.4.6 must have
positive degree.

\item \textbf{Page 320, proof of Gauss's Lemma 8.5.4:} Replace
\textquotedblleft in $\mathbf{Q}\left(  X\right)  $\textquotedblright\ by
\textquotedblleft in $\mathbf{Q}\left[  X\right]  $\textquotedblright.

\item \textbf{Page 321, Lemma 8.5.6:} After \textquotedblleft Let
$f(X)\in\mathbf{Q}[X]$\textquotedblright\ add \textquotedblleft be
nonzero\textquotedblright.

\item \textbf{Page 325, Exercise 8.6.8 (2):} Require $\mathbf{k}$ to be a
perfect field of characteristic $p>0$. As written, $p$ is undefined, and
\textquotedblleft perfect field\textquotedblright\ also includes
characteristic-zero fields.

\item \textbf{Page 333, proof of Proposition 8.9.6:} The zero module needs a
separate base case: if $M=0$, take $n=0$. The claim that a one-generated
module $M=Am$ has $\operatorname*{Ann}\left(  m\right)  =0$ fails when $M=0$.

\item \textbf{Page 333, proof of Proposition 8.9.6:} After \textquotedblleft
Let $m_{1},\ldots,m_{n}$ be a generating set for $M$.\textquotedblright, add
\textquotedblleft We WLOG assume that all $m_{i}$ are
nonzero.\textquotedblright. This is used when you apply Lemma 8.9.5 to
$m=m_{1}$.

\item \textbf{Page 333, proof of Proposition 8.9.6:} Replace \textquotedblleft%
$A/Am_{1}$\textquotedblright\ by \textquotedblleft$M/Am_{1}$\textquotedblright.

\item \textbf{Page 334, proof of Proposition 8.9.6:} The claim
\textquotedblleft Since $M$ is torsion-free, $q\neq0$\textquotedblright\ is
unnecessary, and (strictly speaking) fails when $m=0$.

Later, \textquotedblleft generate $m$\textquotedblright\ should read
\textquotedblleft generate $M$\textquotedblright.

\item \textbf{Page 335, Lemma 8.9.8:} There is an exceptional case when
$a=x=0$. Then $c=0$, and the equation $a=bc$ does not determine $b$; choosing
$b=0$, for example, gives the wrong annihilator. Either assume $a\neq0$, or
add: \textquotedblleft If $a=x=0$, take $b=1$\textquotedblright.

\item \textbf{Page 335, Lemma 8.9.11:} Require $a_{1},a_{2},\ldots,a_{k}\neq
0$, in order for the weights to be defined.

\item \textbf{Page 336, proof of Corollary 8.9.13:} The indices are
inconsistent: The \textquotedblleft$k$\textquotedblright\ in \textquotedblleft%
$a=p_{1}^{r_{1}}\cdots p_{k}^{r_{k}}$\textquotedblright\ is not the same
number as the \textquotedblleft$k$\textquotedblright\ in Corollary 8.9.13.
Also an \textquotedblleft$n$\textquotedblright\ and an \textquotedblleft%
$l$\textquotedblright\ appear without definition in the last paragraph of the proof.

\item \textbf{Page 339, proof of Theorem 8.9.17:} After writing $a_{i}%
=pa_{i}^{\prime}$, some $a_{i}^{\prime}$ may be units, so the displayed
decomposition%
\[
A/\left(  a_{1}^{\prime}\right)  \oplus\cdots\oplus A/\left(  a_{k}^{\prime
}\right)
\]
may contain zero summands $A/(1)$. The same holds for the decomposition%
\[
A/\left(  b_{1}^{\prime}\right)  \oplus\cdots\oplus A/\left(  b_{k}^{\prime
}\right)  .
\]
These zero summands must be deleted before invoking the induction hypothesis,
since Theorem 8.9.17 requires $a_{k}$ and $b_{l}$ to be non-units. Since $k=l$
has already been proved, the number of deleted zero summands can then be recovered.

\item \textbf{Page 340, Theorem 8.9.20:} The summands should be required to be
nonzero. Otherwise one can add arbitrary zero cyclic modules, which are
$p$-torsion for every $p$, and the decomposition is not literally unique.

\item \textbf{Page 340, Exercise 8.9.21 (1):} Replace \textquotedblleft no
element of $\mathbf{Z}$ annihilates all of $\mathbf{Q}/\mathbf{Z}%
$\textquotedblright\ by \textquotedblleft no nonzero element\ldots
\textquotedblright\ Of course, $0$ annihilates every module.

\item \textbf{Page 344, proof of Proposition 9.1.8:} \textquotedblleft since
since\textquotedblright\ $\rightarrow$ \textquotedblleft
since\textquotedblright.

\item \textbf{Page 344, proof of Lemma 9.1.9:} Replace \textquotedblleft for
$n,m\geq0$\textquotedblright\ by \textquotedblleft for $n,m\geq1$%
\textquotedblright. Otherwise $n+m-1$ can be negative.

\item \textbf{Page 345, Exercise 9.1.17 (2):} This needs an additional
requirement that $A$ is not a field. Indeed, a field is a P.I.D with Jacobson
radical $0$ but only one prime ideal.

\item \textbf{Page 347, Examples 9.2.6 (3):} Replace \textquotedblleft a
nonzero ideal $(a)$\textquotedblright\ by \textquotedblleft a nonzero proper
ideal $(a)$\textquotedblright. Primary ideals must be proper by definition.

\item \textbf{Page 351, Lemma 9.3.1:} \textquotedblleft where $A$
Noetherian\textquotedblright\ $\rightarrow$ \textquotedblleft where $A$ is
Noetherian\textquotedblright.

\item \textbf{Page 352, proof of Corollary 9.3.3:} In the final two sentences,
take $\alpha$ to be an arbitrary \emph{nonzero} element of $A/\mathfrak{p}$.
Then $K[\alpha]$ is a field, so $\alpha$ is invertible. It follows that every
\emph{nonzero} element of $A/\mathfrak{p}$ is invertible and hence that
$A/\mathfrak{p}$ is a field. (As printed, the assertion that $\alpha$ is
invertible fails for $\alpha=0$.)

\item \textbf{Page 353, Definitions 9.3.7:} Replace \textquotedblleft%
$\mathfrak{m}_{a_{1},\ldots,a_{n}}\in\operatorname*{Max}\left(  A\right)
$\textquotedblright\ by \textquotedblleft$\mathfrak{m}_{a_{1},\ldots,a_{n}}%
\in\operatorname*{Max}\left(  K\left[  X_{1},\ldots,X_{n}\right]  \right)
$\textquotedblright.

\item \textbf{Page 355, paragraph after proof of Lemma 9.3.12:} Merely as
topological spaces, neither $\operatorname{Spec}(A)$ nor $\operatorname{Max}%
(A)$ is a scheme. The spectrum $\operatorname{Spec}(A)$ equipped with its
structure sheaf is the affine scheme associated to $A$; $\operatorname{Max}%
(A)$ is generally not a scheme. The sentence should refer here simply to
spectra and their Zariski closed subsets, reserving \textquotedblleft affine
scheme\textquotedblright\ for the locally ringed space
\[
(\operatorname{Spec}(A),\mathcal{O}_{\operatorname{Spec}(A)}).
\]


\item \textbf{Page 356, Corollary 9.3.16:} Require $A$ to be commutative;
otherwise $\operatorname*{Spec}\left(  A\right)  $ has not been defined in the
book's sense.

\item \textbf{Page 356, Examples 9.3.17 (1):} Require the PID $A$ not to be a
field. For a field, $\operatorname*{Spec}\left(  A\right)
=\operatorname*{Max}\left(  A\right)  =\left\{  \left(  0\right)  \right\}  $,
so the claimed description of the proper closed subsets fails.

\item \textbf{Page 356, Remarks 9.3.18:} For functions $f_{i}\colon
\mathbf{R}^{m}\rightarrow\mathbf{R}$, the definition of $f\colon\mathbf{R}%
^{m}\rightarrow\mathbf{R}^{k}$ should say \textquotedblleft for $x\in
\mathbf{R}^{m}$\textquotedblright, not \textquotedblleft for $x\in
\mathbf{R}^{n}$\textquotedblright. Later, if $f\colon\mathbf{C}^{n}%
\rightarrow\mathbf{C}^{k}$ has component functions $f_{i}$, the index range is
$i=1,\ldots,k$, not $i=1,\ldots,n$.

\item \textbf{Page 356, Remarks 9.3.18, second paragraph:} The statement
\textquotedblleft Every smooth manifold is diffeomorphic \ldots\ to a manifold
obtained by the above procedure\textquotedblright\ is false. For instance,
non-orientable manifolds cannot be obtained in this way. A safe replacement
is: \textquotedblleft Every smooth manifold is \emph{locally} of this form,
and many embedded manifolds arise globally as regular level
sets\textquotedblright.

\item \textbf{Page 360, proof of Corollary 9.3.31:} \textquotedblleft
representing $f_{i}$ for $i=1,\ldots,n$\textquotedblright\ should be
\textquotedblleft representing $f_{i}$ for $i=1,\ldots,m$\textquotedblright.

\item \textbf{Page 361:} \textquotedblleft whose objects and
are\textquotedblright\ $\rightarrow$ \textquotedblleft whose objects
are\textquotedblright.

\item \textbf{Page 365, Exercise 9.3.39 (5):} To be maximally unambiguous,
replace \textquotedblleft the upper half plane in $\mathbf{C}$%
\textquotedblright\ by \textquotedblleft the closed upper half plane in
$\mathbf{C}$\textquotedblright. Indeed, the maximal ideals $(X-a)$ for
$a\in\mathbf{R}$ correspond to points on the real axis. Under this
identification, the map
\[
\operatorname{Max}(\mathbf{C}[X])\longrightarrow\operatorname{Max}%
(\mathbf{R}[X])
\]
identifies each $z\in\mathbf{C}$ with its complex conjugate.

\item \textbf{Page 366, Proposition 9.4.3:} Replace \textquotedblleft%
$\overline{f}:M\otimes_{A}N$\textquotedblright\ by \textquotedblleft%
$\overline{f}:M\otimes_{A}N\rightarrow G$\textquotedblright.

\item \textbf{Page 367, Lemma 9.4.6:} Replace \textquotedblleft left and
right\textquotedblright\ by \textquotedblleft right and left\textquotedblright%
\ (both times).

\item \textbf{Page 368, Proposition 9.4.7:} Replace \textquotedblleft%
$\iota:M\times N$\textquotedblright\ by \textquotedblleft$\iota:M\times
N\rightarrow M\otimes_{A}N$\textquotedblright.

\item \textbf{Page 369, Proposition 9.4.11:} \textquotedblleft$B$-bilinear in
$M$\textquotedblright\ should be \textquotedblleft$B$-linear in $M$%
\textquotedblright.

\item \textbf{Page 376, proof of Corollary 9.4.26:} \textquotedblleft every
maximal ideal of $M$\textquotedblright\ should be \textquotedblleft every
maximal ideal $\mathfrak{m}$ of $A$\textquotedblright.

\item \textbf{Page 381, paragraph after proof of Proposition 9.5.14:} The
claim that \textquotedblleft the commutative ring $A$ maps to the field
$A/\mathfrak{m}$ if $\mathfrak{m}$ is a maximal ideal of $A$\textquotedblright%
\ holds only if the ring $A$ is nonzero (i.e., nontrivial).

\item \textbf{Page 381, Proposition 9.5.15:} Replace \textquotedblleft
commutative ring\textquotedblright\ by \textquotedblleft nonzero commutative
ring\textquotedblright. Over a zero ring $A$, all modules are isomorphic, so
the existence of a surjective homomorphism $A^{m}\rightarrow A^{n}$ does not
imply anything about $m$ and $n$.

\item \textbf{Page 382, Lemma 9.5.17:} Add \textquotedblleft
nonzero\textquotedblright\ before \textquotedblleft commutative
ring\textquotedblright, since a zero ring has no prime ideals.

\item \textbf{Page 383, Theorem 9.5.20:} \emph{Missing hypothesis.} Require
$A$ to be a nonzero commutative ring. Over the zero ring there is an
injection
\[
A^{m}=0\longrightarrow0=A^{n}%
\]
for arbitrary $m,n$, so $m\leq n$ need not hold. The proof's reduction to
$A/\mathfrak{p}$ also presupposes the existence of a prime ideal.

\item \textbf{Page 383, Exercise 9.5.21 (5):} \textquotedblleft a left
$A$-module $M$\textquotedblright\ should be \textquotedblleft a left
$A$-module $N$\textquotedblright.

\item \textbf{Page 384, Proposition 9.6.1:} The \textquotedblleft only
if\textquotedblright\ part of the assertion \textquotedblleft$B\otimes_{A}C$
is commutative if and only if both $B$ and $C$ are
commutative\textquotedblright\ is false, because the canonical maps
\[
B\longrightarrow B\otimes_{A}C,\qquad C\longrightarrow B\otimes_{A}C
\]
need not be injective. For example, take
\[
A=\mathbb{Z},\qquad B=M_{2}(\mathbf{Z}/2\mathbf{Z}),\qquad C=\mathbf{Z}%
/3\mathbf{Z}.
\]
Then $B$ is noncommutative but $B\otimes_{\mathbb{Z}}C=0$, which is commutative.

The forward implication (i.e., \textquotedblleft only if\textquotedblright%
\ part) is valid if both canonical maps are injective. Without additional
hypotheses, retain only the \textquotedblleft if\textquotedblright\ part of
the proposition (i.e., the implication that commutativity of $B$ and $C$
implies commutativity of $B\otimes_{A}C$). (Fortunately, this is the one
implication that is commonly used.)

\item \textbf{Page 388, proof of Proposition 9.7.5:} The displayed equation%
\[
\psi:\operatorname*{Hom}\nolimits_{A}\left(  M_{1}\otimes_{B}M_{2}%
,M_{3}\right)  \rightarrow\operatorname*{Hom}\nolimits_{A}\left(
M_{1},\operatorname*{Hom}\nolimits_{A}\left(  M_{2},M_{3}\right)  \right)
\]
should be%
\[
\psi:\operatorname*{Hom}\nolimits_{A}\left(  M_{1}\otimes_{B}M_{2}%
,M_{3}\right)  \rightarrow\operatorname*{Hom}\nolimits_{B}\left(
M_{1},\operatorname*{Hom}\nolimits_{A}\left(  M_{2},M_{3}\right)  \right)
\]
instead.

\item \textbf{Page 391, paragraph after proof of Proposition 9.7.12:} The
claim that \textquotedblleft Infinitely generated free modules are not
reflexive\textquotedblright\ is false as written. Replace it by
\textquotedblleft Infinitely generated free modules are not always
reflexive\textquotedblright. For instance, the infinitely generated free
$\mathbb{Z}$-module $\mathbb{Z}^{\left(  \mathbb{N}\right)  }$ is reflexive by
\href{https://mathoverflow.net/questions/453688/double-dual-of-free-mathbbz-p-modules}{Specker's
theorem}.

\item \textbf{Page 397, proof of Proposition 9.8.16:} Require $P$ to be
projective in the first sentence.

\item \textbf{Page 408, Examples 9.10.8 (2):} \textquotedblleft gives an
isomorph of the following\textquotedblright\ is garbled; probably
\textquotedblleft gives an example of the following\textquotedblright.

\item \textbf{Page 408, paragraph after Lemma 9.10.10:} Replace
\textquotedblleft the elements $m_{1}\otimes m_{2}-m_{1}\otimes m_{1}%
$\textquotedblright\ by \textquotedblleft the elements $m_{1}\otimes
m_{2}-m_{2}\otimes m_{1}$\textquotedblright.

\item \textbf{Page 418, Proposition 10.1.9 (3):} Replace
\[
\operatorname{tr}(f\oplus g)=\operatorname{tr}(f)+\operatorname{tr}(g)
\]
by
\[
\operatorname{tr}(f_{1}\oplus f_{2}) =\operatorname{tr}(f_{1}%
)+\operatorname{tr}(f_{2}).
\]
The maps introduced in the hypothesis are $f_{1}$ and $f_{2}$, not $f$ and $g$.

\item \textbf{Page 419, proof of Proposition 10.2.3:} In the first displayed
calculation, replace $\Delta(x_{1},\ldots,x_{j})$ by $\Delta(x_{1}%
,\ldots,x_{n})$.

\item \textbf{Page 423, Remarks 10.2.14:} Near the bottom of the page,
\textquotedblleft the natural map from the tensor algebra $T_{A}%
(M)$\textquotedblright\ should be \textquotedblleft the natural map from the
tensor algebra $T_{A}(N)$\textquotedblright.

\item \textbf{Page 424, proof of Proposition 10.3.1:} The phrase
\textquotedblleft the nonzero summands above are in one-to-one correspondence
with elements of $\mathbf{S}_{n}$\textquotedblright\ is not to be taken fully
literally: even a summand indexed by a permutation can vanish. Replace
\textquotedblleft the nonzero summands\textquotedblright\ by \textquotedblleft
the summands not forced to vanish by the preceding argument\textquotedblright.

\item \textbf{Page 426, Proposition 10.3.9:} In the displayed short exact
sequence of multiplicatively written groups, replace the two endpoint $0$'s by
$1$'s:
\[
1\longrightarrow\operatorname{Sl}_{n}(A)\overset{\subset}{\longrightarrow
}\operatorname{Gl}_{n}(A)\overset{\det}{\longrightarrow}A^{\times
}\longrightarrow1
\]
(since the groups are multiplicative).

\item \textbf{Page 426, paragraph after Definition 10.3.10:} \textquotedblleft
a choice of basis for $n$\textquotedblright\ should be \textquotedblleft a
choice of basis for $N$\textquotedblright.

\item \textbf{Page 428, Exercise 10.3.12 (10) (d):} Replace \textquotedblleft
of degree $n-1$\textquotedblright\ by \textquotedblleft of degree at most
$n-1$\textquotedblright. For example, if all the $y_{i}$ are zero, the unique
interpolating polynomial subject to the corrected bound is the zero
polynomial, which does not have degree $n-1$.

\item \textbf{Page 430, paragraph preceding Lemma 10.4.5:} Replace
\textquotedblleft$XI_{n}-(M^{\prime}\oplus M^{\prime\prime})$ is the block sum
of $XI_{n}-M^{\prime}$ and $XI_{n}-M^{\prime\prime}$\textquotedblright\ by
\textquotedblleft$XI_{n+k}-(M^{\prime}\oplus M^{\prime\prime})$ is the block
sum of $XI_{n}-M^{\prime}$ and $XI_{k}-M^{\prime\prime}$\textquotedblright.

\item \textbf{Page 432, Exercise 10.4.11 (4):} \textquotedblleft Let
$\mathfrak{a}$ and $\mathfrak{b}$ ideals\textquotedblright\ should be
\textquotedblleft Let $\mathfrak{a}$ and $\mathfrak{b}$ be
ideals\textquotedblright.

\item \textbf{Page 433, Lemma 10.5.2:} Insert \textquotedblleft
a\textquotedblright\ in \textquotedblleft$a\in K$ is root of the
characteristic polynomial\textquotedblright, so that it reads
\textquotedblleft$a\in K$ is a root of the characteristic
polynomial\textquotedblright.

\item \textbf{Page 436, second paragraph:} Replace \textquotedblleft every
torsion module over $K[X]$\textquotedblright\ by \textquotedblleft every
finitely generated torsion module over $K[X]$\textquotedblright. (The module
under discussion is indeed finitely generated, since its underlying $K$-vector
space is finite-dimensional.)

\item \textbf{Page 437, proof of Proposition 10.6.7:} In the first sentence
and the following display, replace \textquotedblleft$XI_{n}-C(f)$%
\textquotedblright\ by \textquotedblleft$XI_{m}-C(f)$\textquotedblright. (The
companion matrix $C(f)$ is $m\times m$, not $n\times n$.)

\item \textbf{Page 440, Exercise 10.6.14 (7):} Require $f_{1}$ and $f_{2}$ to
be \emph{distinct} monic irreducible polynomials.

\item \textbf{Page 441, Definition 10.7.2:} \textquotedblleft Jordan block
with eigenvalue $m$\textquotedblright\ should be \textquotedblleft Jordan
block with eigenvalue $a$\textquotedblright.

\item \textbf{Page 442, proof of Lemma 10.7.6:} In \textquotedblleft%
$\operatorname{ch}_{M}(X)=\operatorname{ch}_{C((X-a)^{k})}(X)=(X-a)^{k}%
$\textquotedblright, replace both occurrences of \textquotedblleft%
$k$\textquotedblright\ by \textquotedblleft$m$\textquotedblright. The matrix
$M=J(a,m)$ is similar to $C((X-a)^{m})$ (by Lemma 10.7.3) and has
characteristic polynomial $(X-a)^{m}$.

\item \textbf{Page 442, proof of Lemma 10.7.6:} Insert \textquotedblleft
is\textquotedblright\ in \textquotedblleft if $b\neq a$, then $bI_{m}-M$
invertible\textquotedblright.

\item \textbf{Page 443, Exercise 10.7.8 (4):} Insert \textquotedblleft
is\textquotedblright\ in \textquotedblleft such that $\rho(a)$ not
diagonalizable if $a\neq0$\textquotedblright.

\item \textbf{Page 446, proof of Proposition 10.8.9:} \textquotedblleft the
first $k-1$ entries of the $i$-th row are empty\textquotedblright\ should be
\textquotedblleft the first $k-1$ entries of the $i$-th row are
zero\textquotedblright.

\item \textbf{Page 446, proof of Proposition 10.8.9:} The final sentence,
\textquotedblleft The result now follows by induction on $n$\textquotedblright%
, names the wrong induction variable. Replace it by, for example,
\textquotedblleft Continuing this induction through $k=n$ proves the
result\textquotedblright.

\item \textbf{Page 446, Exercise 10.8.13 (2):} Insert \textquotedblleft
has\textquotedblright\ before \textquotedblleft order $3$\textquotedblright.

\item \textbf{Page 448, Lemma 10.9.6:} As on page 426, replace the endpoint
$0$'s in this short exact sequence of multiplicatively written groups by
$1$'s:
\[
1\longrightarrow\operatorname{Sl}(A)\overset{\subset}{\longrightarrow
}\operatorname{Gl}(A)\overset{\det}{\longrightarrow}A^{\times}\longrightarrow
1.
\]


\item \textbf{Page 450, sixth paragraph:} \textquotedblleft the subfields of
the algebraic closure of $L$\textquotedblright\ should be \textquotedblleft
the subfields of $L$\textquotedblright.

\item \textbf{Page 454, Example 11.2.2 (3):} \textquotedblleft The complex
numbers $\mathbf{C}$ is a splitting field\textquotedblright\ is a somewhat
ungrammatical shorthand for \textquotedblleft The field $\mathbf{C}$ of
complex numbers is a splitting field\textquotedblright.

\item \textbf{Page 456, proof of Proposition 11.2.9, last paragraph:}
\textquotedblleft the minimal polynomials of the $\alpha_{i}$ over
$\mathbf{K}$\textquotedblright\ should be \textquotedblleft the minimal
polynomials of the $\alpha_{i}$ over $\mathbf{k}$\textquotedblright.

\item \textbf{Page 457, Corollary 11.2.11:} Require the collection
$\{\mathbf{K}_{i}\mid i\in I\}$ to be nonempty. For the empty collection, the
intersection inside the ambient field $\mathbf{K}$ is $\mathbf{K}$, which has
not been assumed normal over $\mathbf{k}$.

\item \textbf{Page 457, Proposition 11.2.13:} Insert \textquotedblleft
of\textquotedblright\ in \textquotedblleft the collection of all roots in
$\mathbf{K}$ the polynomials $f_{1},\ldots,f_{n}$\textquotedblright.

\item \textbf{Page 457, Corollary 11.2.14:} Insert \textquotedblleft
be\textquotedblright\ after \textquotedblleft let $\mathbf{K}_{1}%
$\textquotedblright.

\item \textbf{Page 458, proof of Proposition 11.3.1:} The formal derivative of
$X^{q^{n}}-X$ is $-1$, not $1$. The conclusion that it has no repeated roots
is unchanged.

\item \textbf{Page 459, Exercise 11.3.5 (1):} Require $n>1$. As written, the
smallest answer is trivially $2$: in $\mathbf{Z}_{2}$, take $\alpha=0$. Even
requiring $\alpha\neq0$ would only change the answer to $3$, since
$1\in\mathbf{Z}_{3}$ generates the prime field but is not a generator of
$\mathbf{Z}_{3}^{\times}$. With the intended requirement $n>1$, the smallest
answer is $9$: an element of order $4$ in $\mathbf{F}_{9}^{\times}$ generates
$\mathbf{F}_{9}$ over $\mathbf{F}_{3}$ but is not primitive in the
multiplicative group of order $8$.

\item \textbf{Page 460, paragraph before Proposition 11.4.2:}
\textquotedblleft a better ideal\textquotedblright\ should be
\textquotedblleft a better idea\textquotedblright.

\item \textbf{Page 463, proof of Proposition 11.4.12:} Near the end of the
proof, \textquotedblleft for $i=1,\ldots,n$\textquotedblright\ should be
\textquotedblleft for $i=1,\ldots,k$\textquotedblright, in accordance with
$\mathbf{K}=\mathbf{k}(\alpha_{1},\ldots,\alpha_{k})$.

\item \textbf{Page 464, proof of Proposition 11.4.14:} Replace both
occurrences of \newline\textquotedblleft$\mathbf{k}(\alpha_{1},\ldots
,\alpha_{n})$\textquotedblright\ by \textquotedblleft$\mathbf{k}(\alpha
_{0},\ldots,\alpha_{n-1})$\textquotedblright. (Recall that $\alpha_{n}=1$; the
printed field omits the possibly nontrivial coefficient $\alpha_{0}$.)

\item \textbf{Page 464, proof of Corollary 11.4.15:} \textquotedblleft a
finite extension $\mathbf{k}(\alpha_{i_{1}},\ldots,\alpha_{i_{k}})$ of
$\mathbf{K}$\textquotedblright\ should end with \textquotedblleft of
$\mathbf{k}$\textquotedblright.

\item \textbf{Page 464, Corollary 11.4.16:} Replace \textquotedblleft any
collection of separable elements of $\mathbf{k}[X]$\textquotedblright\ by
\textquotedblleft any collection of separable polynomials in $\mathbf{k}%
[X]$\textquotedblright.

\item \textbf{Page 465, Exercise 11.4.20 (2):} \emph{Missing finiteness
hypothesis.} Require $\mathbf{K}$ to be a \emph{finite} extension of
$\mathbf{k}$, at least in parts (b) and (c). Otherwise the minimal polynomial
in part (b) need not exist, and the separability and inseparability degrees in
part (c), as defined in this section, need not be defined. For example, the
printed hypotheses allow $\mathbf{K}=\mathbf{k}(t)$.

\item \textbf{Page 465, paragraph before Lemma 11.5.2:} \textquotedblleft the
minimal polynomial of $\alpha$ over $\mathbf{K}$\textquotedblright\ should be
\textquotedblleft the minimal polynomial of $\alpha$ over $\mathbf{k}%
$\textquotedblright.

\item \textbf{Page 466, Lemmas 11.5.5 and 11.5.7:} In each statement,
\textquotedblleft an intermediate field between $\mathbf{K}$ and $\mathbf{k}%
$\textquotedblright\ should be \textquotedblleft an intermediate field between
$\mathbf{k}$ and $\mathbf{K}$\textquotedblright.

\item \textbf{Page 467, proof of Lemma 11.5.10:} Replace \textquotedblleft Let
$\alpha_{1},\ldots,\alpha_{m}$ be a basis for $\mathbf{K}$ as a vector space
over $\mathbf{k}$\textquotedblright\ by \textquotedblleft Let $\alpha
_{1},\ldots,\alpha_{m}$ be $m$ elements of $\mathbf{K}$ that are linearly
independent over $\mathbf{k}$\textquotedblright, since the existence of a
finite basis is not a-priori guaranteed by the conditions of the lemma.

\item \textbf{Page 470, Remarks 11.5.15:} The reduction of $X^{3}+bX^{2}+cX+d$
by substituting $X-b/3$ requires $\operatorname{char}\mathbf{k}\neq3$. Add
this hypothesis to the paragraph; in characteristic $3$, the printed
substitution is not defined (and translation does not in general remove the
quadratic term).

\item \textbf{Page 471, proof of Theorem 11.5.17:} Before invoking the
Fundamental Theorem of Galois Theory, state that $\mathbf{K}/\mathbf{k}$ is
finite by Lemma 11.5.10. This supplies a hypothesis needed for the cited
subgroup--subfield correspondence.

\item \textbf{Page 472, proof of Proposition 11.5.19:} \textquotedblleft
adjoining the roots of $g(X)$ to $\mathbf{k}$\textquotedblright\ should be
\textquotedblleft adjoining the coefficients of $g(X)$ to $\mathbf{k}%
$\textquotedblright.

\item \textbf{Page 475, proof of Corollary 11.6.3:} The $2$-Sylow subgroup
$G_{2}$ of $G=\operatorname{Gal}(\mathbf{L}/\mathbf{R})$ acts on $\mathbf{L}$,
not necessarily on the intermediate field $\mathbf{K}$. Replace
\[
\mathbf{K}_{1}=\mathbf{K}^{G_{2}}%
\]
by
\[
\mathbf{K}_{1}=\mathbf{L}^{G_{2}}.
\]
The following statements should then read
\[
G_{2}=\operatorname{Gal}(\mathbf{L}/\mathbf{K}_{1}),\qquad\lbrack
\mathbf{L}:\mathbf{K}_{1}]=|G_{2}|,\qquad\lbrack\mathbf{K}_{1}:\mathbf{R}%
]=[G:G_{2}].
\]
With these replacements, the argument proves the desired contradiction.

\item \textbf{Page 476, Lemma 11.7.2:} \textquotedblleft
elments\textquotedblright\ should be \textquotedblleft
elements\textquotedblright.

\item \textbf{Page 477, paragraph before Corollary 11.7.4:} \textquotedblleft
the splitting field of $X^{n}-1$ over $\zeta$\textquotedblright\ should be
\textquotedblleft the splitting field of $X^{n}-1$ over $\mathbf{k}%
$\textquotedblright.

\item \textbf{Page 477, Proposition 11.7.7:} \textquotedblleft obtained by
adjoining a primitive $n$-th root of unity to $\mathbf{K}$\textquotedblright%
\ should end with \textquotedblleft to $\mathbf{k}$\textquotedblright.

\item \textbf{Page 480, proof of Lemma 11.8.1:} In the first sentence, let
$\mathbf{L}$ be an algebraic closure of $\mathbf{k}$, not of $\mathbf{K}$; the
proof is constructing the splitting field $\mathbf{K}$, which is then set
equal to $\mathbf{k}(\alpha,\zeta)$.

\item \textbf{Page 481, paragraph after Proposition 11.8.2:} \textquotedblleft
a principal $n$-th root of unity\textquotedblright\ should be
\textquotedblleft a primitive $n$-th root of unity\textquotedblright.

\item \textbf{Page 481, proof of Proposition 11.8.3:} \textquotedblleft
divides is order\textquotedblright\ should be \textquotedblleft divides its
order\textquotedblright.

\item \textbf{Page 484, Exercise 11.8.6 (11):} Replace \textquotedblleft%
$n=p_{1},\ldots,p_{k}$\textquotedblright\ by \textquotedblleft$n=p_{1}\cdots
p_{k}$\textquotedblright.

\item \textbf{Page 486, proof of Proposition 11.9.2:} In the last sentence,
\textquotedblleft a nontrivial dependence relation between $f_{2}%
(x),\ldots,f_{n}(x)$\textquotedblright\ should be \textquotedblleft a
nontrivial dependence relation between the functions $f_{2},\ldots,f_{n}%
$\textquotedblright.

\item \textbf{Page 486, Proposition 11.9.4:} Require $G$ to be finite. Both
sums over $G$ in the proof, as well as the application of Corollary 11.9.3,
require this hypothesis.

\item \textbf{Page 487, proof of Proposition 11.9.5, first sentence:}
\textquotedblleft any element $a\in\mathbf{K}$\textquotedblright\ should be
\textquotedblleft any element $a\in\mathbf{k}$\textquotedblright.

Also, the \textquotedblleft by induction\textquotedblright\ argument in the
second paragraph could use some more explanation:

We argue by induction on $n$. Put
\[
d=[\mathbf{K}:\mathbf{k}]=\lvert\operatorname{Gal}(\mathbf{K}/\mathbf{k}%
)\rvert.
\]
Since $\operatorname{Gal}(\mathbf{K}/\mathbf{k})$ is cyclic and has exponent
$n$, we have $d\mid n$. If $d<n$, then the inductive hypothesis, applied with
$d$ in place of $n$, shows that $\mathbf{K}$ is the splitting field of
$X^{d}-a$ for some $a\in\mathbf{k}$ (since $\operatorname{Gal}(\mathbf{K}%
/\mathbf{k})$ is cyclic and thus has exponent $\lvert\operatorname{Gal}%
(\mathbf{K}/\mathbf{k})\rvert=d$). Writing $n=dk$, we see from the first
paragraph that $X^{d}-a$ and $X^{n}-a^{k}$ have the same splitting field over
$\mathbf{k}$. Thus, we can WLOG assume that $d=n$.

\item \textbf{Page 492, proof of Corollary 11.10.5:} (This was reported by
GPT-5.6; I have not thoroughly verified it.)

\emph{Proof gap.} The printed induction does not justify that $B_{1}\subseteq
B$, yet uses this to compare the two indices and to choose $b_{s+1}\in
B\setminus B_{1}$. A shorter argument avoids this issue. Since $B/(\mathbf{k}%
^{\times})^{r}$ is finite, choose $b_{1},\ldots,b_{s}\in B$ whose residue
classes generate it, and let $\mathbf{K}_{0}$ be the splitting field of
\[
\{X^{r}-b_{i}\mid1\leq i\leq s\}.
\]
For any $b\in B$, write
\[
b=a^{r}b_{1}^{e_{1}}\cdots b_{s}^{e_{s}}%
\]
with $a\in\mathbf{k}^{\times}$ and $e_{i}\in\mathbf{Z}$. If $\gamma_{i}%
^{r}=b_{i}$ in $\mathbf{K}_{0}$, then $a\gamma_{1}^{e_{1}}\cdots\gamma
_{s}^{e_{s}}$ is an $r$-th root of $b$. Since $\boldsymbol{\mu}_{r}%
\subseteq\mathbf{k}$, all the roots of $X^{r}-b$ lie in $\mathbf{K}_{0}$. Thus
$\mathbf{K}_{0}$ is already the splitting field of $\{X^{r}-b\mid b\in B\}$.

\item \textbf{Page 493, first sentence of Section 11.11:} Delete one
occurrence of \textquotedblleft finding\textquotedblright\ in
\textquotedblleft the issue of finding finding roots of
polynomials\textquotedblright.

\item \textbf{Pages 493--494, Example 11.11.1 (Cardano's formula):}
\emph{Substantive degenerate case.} The instruction to choose $\beta$ uniquely
so that $3\alpha\beta=-p$, and the later expression involving $p/(3\alpha)$,
require $\alpha\neq0$. When $p=0$, one of the two choices of the square root
can make $\alpha^{3}=0$, so the printed formula is undefined. Handle $p=0$
separately: the depressed cubic is
\[
X^{3}+q,
\]
whose roots are the three cube roots of $-q$ (with the evident multiplicity
convention when $q=0$). Alternatively, when $q\neq0$, choose the sign of the
square root so that $\alpha^{3}=-q$ and take $\beta=0$. For $p\neq0$, the
printed linked choice of $\alpha$ and $\beta$ is valid.

\item \textbf{Page 493, Example 11.11.1:} Delete one occurrence of
\textquotedblleft in\textquotedblright\ in \textquotedblleft As shown in in
Problem 5 of Exercises 7.4.14\textquotedblright.

\item \textbf{Page 494, first paragraph:} Delete the second \textquotedblleft
that\textquotedblright\ in \textquotedblleft We leave it to the reader to
verify that in this degenerate case, that the procedure \ldots
\textquotedblright.

\item \textbf{Page 497, proof of Theorem 11.11.9:} In the first sentence of
the proof, \textquotedblleft$\mathbf{Q}\subset\mathbf{K}$\textquotedblright%
\ should be \textquotedblleft$\mathbf{Q}\subset\mathbf{k}$\textquotedblright.

\item \textbf{Page 497, Exercise 11.11.10 (1):} \emph{Missing hypothesis.}
Require $f$ to be irreducible over $\mathbf{Q}$. Without irreducibility the
claim is false: for $p=5$, the polynomial $(X^{3}-X)(X^{2}+1)$ has exactly
$3=p-2$ real roots, but its splitting field is $\mathbf{Q}(i)$ and its Galois
group is not $\mathbf{S}_{5}$.

\item \textbf{Page 499, fifth paragraph:} \textquotedblleft the $i$-th
elementary function $s_{i}$ in the variables $X_{1},\ldots,X_{1}%
$\textquotedblright\ should be \textquotedblleft the $i$-th elementary
symmetric function $S_{i}$ in the variables $X_{1},\ldots,X_{n}$%
\textquotedblright.

\item \textbf{Page 500, proof of Proposition 11.12.3:} \textquotedblleft gives
in embedding\textquotedblright\ should be \textquotedblleft gives an
embedding\textquotedblright.

\item \textbf{Page 502, opening paragraph of Section 11.14:} \textquotedblleft
Write $\mu_{\alpha}:K\rightarrow K$\textquotedblright\ should be
\textquotedblleft Write $\mu_{\alpha}:\mathbf{K}\rightarrow\mathbf{K}%
$\textquotedblright. In the following sentence, insert \textquotedblleft
a\textquotedblright\ in \textquotedblleft Note that $\mu_{\alpha}$ is a
transformation of vector spaces\textquotedblright.

\item \textbf{Page 504, Exercise 11.14.5 (1):} \textquotedblleft with respect
to any $\mathbf{k}$-basis of $\mathbf{k}$\textquotedblright\ should be
\textquotedblleft with respect to any $\mathbf{k}$-basis of $\mathbf{K}%
$\textquotedblright.

\item \textbf{Page 508, Example 12.1.2 (2):} \textquotedblleft Its nonzero
ideals are $K\times0$ and $0\times L$\textquotedblright\ should be
\textquotedblleft Its proper nonzero ideals are $K\times0$ and $0\times
L$\textquotedblright. The full ring $K\times L$ is also a nonzero ideal.

\item \textbf{Page 511, paragraph before Theorem 12.1.8:} Capitalize
\textquotedblleft compact lie group\textquotedblright\ as \textquotedblleft
compact Lie group\textquotedblright.

\item \textbf{Page 514, proof of Proposition 12.2.6:} Delete one occurrence of
\textquotedblleft and\textquotedblright\ in \textquotedblleft By Lemmas 12.2.3
and and 12.2.5\textquotedblright.

\item \textbf{Page 514, proof of Corollary 12.2.8:} \textquotedblleft Then $D$
is the direct sum of the ideals $\mathfrak{a}_{i}$\textquotedblright\ should
be \textquotedblleft Then $M_{n}(D)$ is the direct sum of the ideals
$\mathfrak{a}_{i}$\textquotedblright.

\item \textbf{Page 516, paragraph before Proposition 12.2.14:}
\textquotedblleft a hands-on argument in instructive\textquotedblright\ should
be \textquotedblleft a hands-on argument is instructive\textquotedblright.

\item \textbf{Page 516, proof of Proposition 12.2.14:} In the sentence
beginning \textquotedblleft Since $\mathfrak{b}$ is generated\ldots
\textquotedblright, both occurrences of \textquotedblleft$\mathfrak{b}%
$\textquotedblright\ should be \textquotedblleft$\mathfrak{b}_{i}%
$\textquotedblright.

\item \textbf{Page 517, proof of Proposition 12.2.14:} In the directness
argument, replace \textquotedblleft$\mathfrak{a}_{1}\mathfrak{a}%
=0$\textquotedblright\ by \textquotedblleft$\mathfrak{a}_{i}\mathfrak{a}%
=0$\textquotedblright. Indeed, $\mathfrak{b}_{1}+\cdots+\mathfrak{b}_{i-1}$ is
a sum of simple submodules of $A$ non-isomorphic to $\mathfrak{a}_{i}$, so
Corollary 12.2.13 gives $\mathfrak{a}_{i}\left(  \mathfrak{b}_{1}%
+\cdots+\mathfrak{b}_{i-1}\right)  =0$. Since $\mathfrak{a}\subset
\mathfrak{b}_{1}+\cdots+\mathfrak{b}_{i-1}$, this entails $\mathfrak{a}%
_{i}\mathfrak{a}=0$. This contradicts the preceding equality $\mathfrak{a}%
_{i}\mathfrak{a}=\mathfrak{a}$ when $\mathfrak{a}\neq0$.

\item \textbf{Page 524, proof of Theorem 12.3.10:} The first sentence should
cite Corollary 12.2.21 for the equivalence of left and right simplicity.
Theorem 12.2.19 gives the equivalence between being simple and being a matrix
ring over a division ring (conditions 4 and 5 here).

\item \textbf{Pages 526--529, Exercise 12.3.14 (2), Corollaries 12.4.2 and
12.4.4, and Proposition 12.4.8:} \emph{Zero-ring exceptions.} These statements
should require $A$ to be nonzero, in keeping with Definitions 12.1.1 and
12.1.4. If $A$ is the zero ring, then $A/\mathfrak{R}(A)=0$, and every unitary
$A$-module is the zero module and hence projective, so the zero ring has
global left homological dimension $0$ (and thus at most $1$). However, the
book's definitions do not call the zero ring Jacobson semisimple, left
semisimple, or left hereditary.

\item \textbf{Page 526, Exercise 12.3.14 (4):} Insert \textquotedblleft
isomorphism classes of\textquotedblright\ before \textquotedblleft those
$K[G]$-modules\textquotedblright. A choice of basis turns a module structure
on $K^{n}$ into a homomorphism $G\rightarrow\operatorname{Gl}_{n}(K)$;
changing the basis conjugates the homomorphism. Thus, conjugacy classes
correspond to isomorphism classes of $n$-dimensional modules, not to the
individual modules.

\item \textbf{Page 528, Lemma 12.4.7:} On the second exact sequence, replace
the label \textquotedblleft$f$\textquotedblright\ on the arrow $Q\rightarrow
M$ by \textquotedblleft$g$\textquotedblright. The proof and its pullback
diagram use $g$ for this map.

\item \textbf{Page 529, proof of Lemma 12.4.7:} In the second split short
exact sequence, replace \textquotedblleft$P\times_{M}A$\textquotedblright\ by
\textquotedblleft$P\times_{M}Q$\textquotedblright.

\item \textbf{Page 529, proof of Proposition 12.4.8:} In the last short exact
sequence, replace \textquotedblleft$A\mathfrak{a}$\textquotedblright\ by
\textquotedblleft$A/\mathfrak{a}$\textquotedblright, thus making it%
\[
0\longrightarrow\mathfrak{a}\longrightarrow A\longrightarrow A/\mathfrak{a}%
\longrightarrow0.
\]


\item \textbf{Page 530, proof of Proposition 12.5.1:} \textquotedblleft every
ideal is a principal\textquotedblright\ should be \textquotedblleft every
ideal is principal\textquotedblright.

\item \textbf{Page 532, proof of Proposition 12.6.6:} After \textquotedblleft%
$a_{1}b_{1}+\cdots+a_{k}b_{k}=a$\textquotedblright, the range
\textquotedblleft$i=1,\ldots,n$\textquotedblright\ should be \textquotedblleft%
$i=1,\ldots,k$\textquotedblright.

\item \textbf{Page 533, proof of Theorem 12.6.8:} In the implication
(2)$\Rightarrow$(1), the proof treats only $\mathfrak{a}\neq0$. The omitted
case $\mathfrak{a}=0$ is easy (just take $\mathfrak{c}=0$).

\item \textbf{Page 533, proof of Theorem 12.6.8:} \textquotedblleft every
maximal ideal of $\mathfrak{a}$ is invertible\textquotedblright\ should be
\textquotedblleft every maximal ideal of $A$ is invertible\textquotedblright.

\item \textbf{Page 533, proof of Theorem 12.6.8:} In \textquotedblleft%
$\mathfrak{m}_{i}$ is a factor if the principal ideal $(a)$\textquotedblright,
replace \textquotedblleft if\textquotedblright\ by \textquotedblleft
of\textquotedblright.

\item \textbf{Page 534, paragraph after Proposition 12.6.10:} Insert
\textquotedblleft are\textquotedblright\ in \textquotedblleft if
$\mathfrak{a}$ and $\mathfrak{b}$ nonzero ideals of a Dedekind
domain\textquotedblright.

\item \textbf{Page 535, proof of Lemma 12.6.14:} The proof silently assumes
that the ideals to which Lemma 12.6.4 is applied are nonzero. Add that if
either $\mathfrak{a}=\mathfrak{c}=0$ or $\mathfrak{b}=\mathfrak{d}=0$, the
conclusion is immediate; otherwise the printed proof applies.

\item \textbf{Page 536, Proposition 12.6.20:} Require $\mathfrak{a}$ and
$\mathfrak{b}$ to be nonzero ideals. For example, if $\mathfrak{a}=0$, the
asserted isomorphism would give $\mathfrak{b}\cong A$ for every ideal
$\mathfrak{b}$, which is false for a nonprincipal ideal in a Dedekind domain.
The printed proof also uses the invertibility of $\mathfrak{a}$.

\item \textbf{Page 537, proof of Theorem 12.6.21:} \textquotedblleft an
$k$-dimensional one\textquotedblright\ should be \textquotedblleft a
$k$-dimensional one\textquotedblright.

\item \textbf{Page 538, Exercise 12.6.23 (2) (a):} Replace \textquotedblleft
an $A$-submodule $\mathfrak{i}$ of $A$\textquotedblright\ by \textquotedblleft
an $A$-submodule $\mathfrak{i}$ of $K$\textquotedblright, and require the
element $a\in A$ satisfying $a\mathfrak{i}\subset A$ to be nonzero.

\item \textbf{Page 538, Exercise 12.6.23 (2) (c):} The groups in the displayed
exact sequence are written multiplicatively, so replace the two endpoint $0$'s
by $1$'s:
\[
1\longrightarrow A^{\times}\longrightarrow K^{\times}\longrightarrow
\mathfrak{I}(A)\overset{p}{\longrightarrow}\operatorname{Cl}(A)\longrightarrow
1.
\]


\item \textbf{Page 540, proof of Corollary 12.7.6:} \textquotedblleft a monic
polynomial over $B$ which has $c$ is a root\textquotedblright\ should be
\textquotedblleft a monic polynomial over $B$ which has $c$ as a
root\textquotedblright. Also, all three occurrences of \textquotedblleft%
$b_{1},\ldots,b_{n}$\textquotedblright\ should be \textquotedblleft%
$b_{0},\ldots,b_{k-1}$\textquotedblright.

\item \textbf{Page 540, Proposition 12.7.7:} \textquotedblleft there is an
$a\in A$\textquotedblright\ should be \textquotedblleft there is a nonzero
$a\in A$\textquotedblright.

\item \textbf{Page 542, Lemma 12.7.13:} Delete the comma in \textquotedblleft
for a prime number, $p$\textquotedblright.

\item \textbf{Page 543, proof of Lemma 12.7.15:} In the noninjective case,
\textquotedblleft$g(bt)$ gives an element of $A$ that lies in the kernel of
$\eta$\textquotedblright\ should say \textquotedblleft$g\left(  bt\right)  $
gives an element of $B$ that lies in the kernel of $\eta:B\rightarrow S^{-1}%
B$\textquotedblright.

\item \textbf{Page 547, proof of Theorem 12.7.30:} The converse argument
immediately assumes that $A$ has Krull dimension $1$, whereas the theorem
allows dimension $0$. Add first that if $A$ is a field, then it is a Dedekind
domain. Otherwise, a domain of Krull dimension at most $1$ that is not a field
has Krull dimension exactly $1$, and the printed argument applies.
\end{enumerate}


\end{document}