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\begin{document}

\begin{center}
\textbf{MA377 Rings and Modules}

\textit{Samir Siksek}

Notes taught in 2019 and 2021; 96-page PDF

\url{https://samirsiksek.github.io/siksek.github.io/teaching/rm/rmversion9.pdf}

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

\noindent The page numbers below refer to the printed page numbers in the
supplied PDF (thus the first page of Chapter~1 is page~1). This list is not
claimed to be exhaustive. Items marked ``substantive'' correct a mathematical
assertion or repair a proof gap. Items marked ``pedagogical'' are not
necessarily errors, but seem likely to cause avoidable difficulty for a
student meeting rings and modules for the first time. The remaining items are
local corrections or clarifications.

Most of the errata below were found with assistance from GPT-5.6 Sol.

\appendix


\section{Corrections and comments}

\begin{enumerate}
\item \textbf{Page 1, list of references and last sentence:} \textquotedblleft
Dummitt\textquotedblright\ should be \textquotedblleft
Dummit\textquotedblright. Also hyphenate the compound adjective in
\textquotedblleft our approach will be more hands-on\textquotedblright.

\item \textbf{Page 3, paragraph after Example 2:} Replace ``subtlties'' by ``subtleties''.

\item \textbf{Page 4, end of Example 6:} Replace ``and that it contained in
$\mathbb{R}^{2}$'' by ``and that it is contained in $\mathbb{R}^{2}$''.

\item \textbf{Page 5, proof of Lemma 12:} \emph{Pedagogical point:} From
\[
a^{n}\bigl(a^{m-n}-1\bigr)=0
\]
the proof concludes that $a^{m-n}=1$. It would help to justify this in more
detail: From $a\neq0$, we get $a^{n}\neq0$, since an integral domain has no
zero divisors. Merely saying \textquotedblleft as $a\neq0$\textquotedblright%
\ skips the small induction (or repeated cancellation) that justifies this.

\item \textbf{Page 5, Section 4:} Replace \textquotedblleft such
at\textquotedblright\ by \textquotedblleft such as\textquotedblright\ and
\textquotedblleft powerseries\textquotedblright\ by \textquotedblleft power
series\textquotedblright. Later, replace \textquotedblleft an multiplicative
inverse\textquotedblright\ by \textquotedblleft a multiplicative
inverse\textquotedblright.

\item \textbf{Page 7, Example 22:} Replace \textquotedblleft$f^{\prime}$
denote the derivate of $f$\textquotedblright\ by \textquotedblleft$f^{\prime}$
denotes the derivative of $f$\textquotedblright. Later in the same sentence,
insert \textquotedblleft a\textquotedblright\ in \textquotedblleft it is
homomorphism\textquotedblright.

\item \textbf{Page 7, Example 24:} Replace \textquotedblleft as $1\notin%
\mathbb{Z}$\textquotedblright\ by \textquotedblleft as $1\notin2\mathbb{Z}%
$\textquotedblright.

\item \textbf{Page 8, definition of $AB$:} Replace \textquotedblleft the empty
sum with $n=0$ as the $0$\textquotedblright\ by \textquotedblleft the empty
sum with $n=0$ as $0$\textquotedblright.

\item \textbf{Page 8, Example 28:} Replace \textquotedblleft as following
exercise shows\textquotedblright\ by \textquotedblleft as the following
exercise shows\textquotedblright.

\item \textbf{Page 9, Exercise 30(ii):} In the description of the summands,
replace \textquotedblleft$y_{j}\in J$\textquotedblright\ by \textquotedblleft%
$y_{i}\in J$\textquotedblright.

\item \textbf{Page 9, Exercise 31:} Replace \textquotedblleft$\mathfrak{b}%
=(b_{1},\ldots,b_{m})$\textquotedblright\ by \textquotedblleft$\mathfrak{b}%
=(b_{1},\ldots,b_{n})$\textquotedblright.

\item \textbf{Page 9, Example 33:} Insert a full stop in \textquotedblleft So
$\mathfrak{a}=\mathbb{Z}[X]$ Hence $1\in\mathfrak{a}$\textquotedblright.

\item \textbf{Page 11, first line:} Replace \textquotedblleft
Workout\textquotedblright\ by \textquotedblleft Work out\textquotedblright.

\item \textbf{Page 11, paragraph after Theorem 38:} The customary spelling is
\textquotedblleft B\'{e}zout identities\textquotedblright.

\item \textbf{Page 12, proof of Theorem 40(b):} In the forward implication,
replace \textquotedblleft Hence by (b), $\overline{a}\in(\mathbb{Z}%
/m\mathbb{Z})^{\ast}$\textquotedblright\ by \textquotedblleft Hence by (a),
\ldots\textquotedblright; part (b) is the statement currently being proved. In
the next sentence, delete one \textquotedblleft and\textquotedblright\ from
\textquotedblleft and and so\textquotedblright.

\item \textbf{Page 13, proof of Theorem 44:} After writing $a=qb+r$, it should
perhaps be said that%
\[
r=a-qb\in\mathfrak{a}.
\]
Only then does the minimality of $\partial(b)$ among the nonzero elements of
$\mathfrak{a}$ imply that $r=0$.

\item \textbf{Page 13, Example 47 and definition of a coset:} Replace
\textquotedblleft It turns that\textquotedblright\ by \textquotedblleft It
turns out that\textquotedblright. A set $r+\mathfrak{a}$ is a
\textquotedblleft coset of $\mathfrak{a}$ in $R$\textquotedblright, rather
than a \textquotedblleft coset of $R$\textquotedblright.

\item \textbf{Page 14, Theorem 50:} Replace \textquotedblleft the addition and
multiplicative identity elements\textquotedblright\ by \textquotedblleft the
additive and multiplicative identity elements\textquotedblright.

\item \textbf{Page 15, first remark after Theorem 53:} Replace
\textquotedblleft corollaries for the first one\textquotedblright\ by
\textquotedblleft corollaries of the first one\textquotedblright.

\item \textbf{Page 16, Example 54 and proof of Theorem 56:} Replace
\textquotedblleft In otherwords\textquotedblright\ by \textquotedblleft In
other words\textquotedblright, and \textquotedblleft a ideal\textquotedblright%
\ by \textquotedblleft an ideal\textquotedblright.

\item \textbf{Pages 17--18, Lemma 60 and its proof:} Clarify that
\textquotedblleft$\deg(r)<\deg(f)$\textquotedblright\ uses the conventions
that $\deg\left(  0\right)  =-\infty$ and that $-\infty$ is smaller than any integer.

\item \textbf{Page 19, proof of Theorem 65:} Replace \textquotedblleft lets
show\textquotedblright\ by \textquotedblleft let's show\textquotedblright, and
replace \textquotedblleft if $f$ is composite\textquotedblright\ by
\textquotedblleft if $f$ is reducible\textquotedblright.

\item \textbf{Page 19, same proof:} The equality \textquotedblleft%
$\gcd(f,f_{1})=f_{1}$\textquotedblright\ need not be literally true under the
stated convention that polynomial gcds are monic, since $f_{1}$ was not chosen
monic. What is needed is simply that $\gcd(f,f_{1})$ has positive degree and
hence is not $1$.

\item \textbf{Page 20, \S 14:} Again: \textquotedblleft
composite\textquotedblright\ should be \textquotedblleft
reducible\textquotedblright.

\item \textbf{Page 22, reduction of $\theta^{n+1}$:} On the third line of the
long-ish computation, replace \textquotedblleft$-a_{1}\theta-a_{1}\theta
^{2}-\cdots--a_{n-2}\theta^{n-1}$\textquotedblright\ by \textquotedblleft%
$-a_{0}\theta-a_{1}\theta^{2}-\cdots-a_{n-2}\theta^{n-1}$\textquotedblright.

\item \textbf{Page 22, summary after that computation:} Insert
\textquotedblleft of\textquotedblright\ in \textquotedblleft as linear
combinations $1,\theta,\ldots,\theta^{n-1}$\textquotedblright.

\item \textbf{Page 25, Theorem 73(i):} Insert \textquotedblleft
a\textquotedblright\ in \textquotedblleft is 2-sided ideal\textquotedblright.

\item \textbf{Pages 25--26, proof of Theorem 73:} In each of the two subgroup
arguments, the proof establishes that the set contains $0$ and is closed under
addition, then calls it a subgroup without checking additive inverses.
Fortunately, the latter is indeed unnecessary if the ideal property
($ra\in\mathfrak{a}$ for all $r\in R$ and $a\in\mathfrak{a}$) has been proved
(since $-a=\left(  -1\right)  a$ for all $a$), but this should be said somewhere.

\item \textbf{Page 26, Example 76:} Replace \textquotedblleft%
$\operatorname{Ann}_{\mathbb{Z}/15\mathbb{Z}}=$\textquotedblright\ by
\textquotedblleft$\operatorname{Ann}_{\mathbb{Z}/15\mathbb{Z}}(10)=$%
\textquotedblright.

\item \textbf{Page 27, definition of group-ring multiplication:} Replace
\textquotedblleft where $gh$ denotes multiplication in $g$\textquotedblright%
\ by \textquotedblleft where $gh$ denotes multiplication in $G$%
\textquotedblright.

\item \textbf{Page 29, second paragraph:} Replace \textquotedblleft maybe
expressed\textquotedblright\ by \textquotedblleft may be
expressed\textquotedblright.

\item \textbf{Page 33, Example 95:} Add a full stop after \textquotedblleft%
$\mathbb{R}$-algebra\textquotedblright.

\item \textbf{Page 33, Example 96:} Replace \textquotedblleft
indentifying\textquotedblright\ by \textquotedblleft
identifying\textquotedblright.

\item \textbf{Page 33, proof of Lemma 98:} Replace \textquotedblleft Let
take\textquotedblright\ by \textquotedblleft Let's take\textquotedblright.

\item \textbf{Page 34, Theorem 99:} In the evaluation map, replace
\textquotedblleft$f(X)\mapsto f(a)$\textquotedblright\ by \textquotedblleft%
$f(X)\mapsto f(\alpha)$\textquotedblright.

\item \textbf{Page 35, first bullet about $\chi_{\alpha}$:} The expression
\textquotedblleft$\det(XI_{n}-\phi_{\alpha})$\textquotedblright\ subtracts a
linear map from a matrix. Either choose a basis and write \textquotedblleft%
$\chi_{\alpha}(X)=\det\bigl(XI_{n}-[\phi_{\alpha}]\bigr)$\textquotedblright,
or use the basis-free operator notation \textquotedblleft$\det(XI_{A}%
-\phi_{\alpha})$\textquotedblright.

\item \textbf{Page 35, Lemma 101:} Replace \textquotedblleft
possitive\textquotedblright\ by \textquotedblleft positive\textquotedblright.

\item \textbf{Page 35, proof of Lemma 101:} Replace \textquotedblleft applying
the above with $\beta=1$ show\textquotedblright\ by \textquotedblleft applying
the above with $\beta=1$ shows\textquotedblright.

\item \textbf{Page 36, Example 102:} The characteristic-polynomial display
concerns $\mathbf{i}$, so replace \textquotedblleft$\chi_{\alpha}%
$\textquotedblright\ and \textquotedblleft$M_{\alpha}$\textquotedblright\ by
\textquotedblleft$\chi_{\mathbf{i}}$\textquotedblright\ and \textquotedblleft%
$M_{\mathbf{i}}$\textquotedblright.

\item \textbf{Page 37, proof of Theorem 107:} Insert \textquotedblleft
you\textquotedblright\ in \textquotedblleft it is easier to do this if think
of quaternions as $2\times2$ matrices\textquotedblright.

\item \textbf{Page 38, Theorem 111:} Require $R\neq0$. In fact, the zero ring
has no nonzero proper left ideals, but it is not a division ring under the
definition on page~37.

\item \textbf{Page 39, sentence before Lemma 115:} Replace \textquotedblleft
The following lemma say no\textquotedblright\ by \textquotedblleft The
following lemma says no\textquotedblright.

\item \textbf{Page 41, Exercise 119:} Replace \textquotedblleft
satisfes\textquotedblright\ by \textquotedblleft satisfies\textquotedblright.

\item \textbf{Page 43, proof of Lemma 128:} \textquotedblleft for some
positive integer $n$\textquotedblright\ should be \textquotedblleft for some
positive integer $r$\textquotedblright.

\item \textbf{Page 43, paragraph before Lemma 130:} Close the parenthesis in
\textquotedblleft(recall $\dim(V)=n-1$, where $n=\dim(A)$\textquotedblright.

\item \textbf{Page 45, penultimate paragraph:} Replace \textquotedblleft%
$\mathbb{R}$ belongs to the centre\textquotedblright\ by \textquotedblleft%
$\mathbb{R}$ is contained in the centre\textquotedblright.

\item \textbf{Page 47, proof of Lemma 137:} Replace the malformed
\textquotedblleft As $D\supseteq C_{s}$ is a division ring\textquotedblright%
\ by \textquotedblleft As $D$ is a division ring\textquotedblright. (The
centralizer in this lemma is $C_{x}$, not $C_{s}$.)

\item \textbf{Page 48, proof of Lemma 137:} Replace \textquotedblleft are
required\textquotedblright\ by \textquotedblleft as required\textquotedblright.

\item \textbf{Page 48, \S 4:} Replace \textquotedblleft
orbit-stablizer\textquotedblright\ by \textquotedblleft
orbit-stabilizer\textquotedblright.

\item \textbf{Page 49, proof of Lemma 139:} Replace \textquotedblleft%
$\operatorname*{Orb}\nolimits_{x}$\textquotedblright\ by \textquotedblleft%
$\operatorname*{Orb}\left(  x\right)  $\textquotedblright.

\item \textbf{Page 50, proof of Theorem 140:} Delete one \textquotedblleft
the\textquotedblright\ from \textquotedblleft all the the
orbits\textquotedblright.

\item \textbf{Page 50, proof of Corollary 141:} In the stabilizer calculation,
replace \textquotedblleft$gy_{i}g^{-1}=y$\textquotedblright\ by
\textquotedblleft$gy_{i}g^{-1}=y_{i}$\textquotedblright.

\item \textbf{Page 51, definition of $\Phi_{n}$:} To make the displayed
quotient unambiguous, specify that the least common multiple is chosen monic.
For $n=1$, also state the convention that the lcm of the empty family is $1$;
otherwise the first answer in Exercise 143 is not determined by the definition.

\item \textbf{Pages 51--52, Theorem 148:} The product
\[
\prod_{\substack{1\leq r<n, \\(r,n)=1}}(X-\zeta_{n}^{r})
\]
is empty when $n=1$, although $\Phi_{1}(X)=X-1$. Either state the theorem for
$n\geq2$, or replace the \textquotedblleft$1\leq r<n$\textquotedblright\ under
the product sign by \textquotedblleft$1\leq r\leq n$\textquotedblright\ (or
\textquotedblleft$0\leq r<n$\textquotedblright).

\item \textbf{Page 52, proof of Theorem 148:} Replace ``enough ot prove'' by
``enough to prove'' and ``Coversely'' by ``Conversely''.

\item \textbf{Page 52, proof of Wedderburn's Little Theorem:} Replace ``But
the definition (13) of $\Phi_{n}(X)$, we know'' by ``By definition (13), we
know''. Later replace ``contradiciton'' by ``contradiction''.

\item \textbf{Page 55, definition of a right module and Example 153:} Replace
\textquotedblleft multipliction\textquotedblright\ by \textquotedblleft
multiplication\textquotedblright, and insert \textquotedblleft
an\textquotedblright\ in \textquotedblleft This is $n\times1$
matrix\textquotedblright.

\item \textbf{Page 56, continuation of Example 153:} The sentence says that
for column vectors, multiplication by matrices on the left is not defined,
immediately after using precisely that multiplication. Replace it by either
\textquotedblleft for column vectors, multiplication by matrices on the right
is not defined\textquotedblright\ or \textquotedblleft for row vectors,
multiplication by matrices on the left is not defined\textquotedblright.

\item \textbf{Page 56, end of Example 154:} Replace ``one-one correspondence''
by ``one-to-one correspondence''.

\item \textbf{Page 57, Example 158:} Insert \textquotedblleft
a\textquotedblright\ in \textquotedblleft the same as subspace of
$V$\textquotedblright.

\item \textbf{Page 58, Lemma 163 and the following paragraph:} Replace
``addition and scalar multiplication is defined'' by ``addition and scalar
multiplication are defined'', and hyphenate ``$R$-module''.

\item \textbf{Page 59, Example 165:} Insert the missing \textquotedblleft
if\textquotedblright\ twice: \textquotedblleft if and only they are $K$-vector
spaces\textquotedblright\ and \textquotedblleft if and only it is a linear
transformation\textquotedblright.

\item \textbf{Page 59, Exercise 169:} Replace ``submodules of the
$R$-submodule $M/N$'' by ``submodules of the $R$-module $M/N$''.

\item \textbf{Page 60, Lemma 170:} Replace \textquotedblleft
decomposted\textquotedblright\ by \textquotedblleft
decomposed\textquotedblright.

\item \textbf{Page 61, definition of span:} Insert \textquotedblleft
is\textquotedblright\ in \textquotedblleft This the set of all finite linear
combinations\textquotedblright, and (optionally) insert \textquotedblleft
an\textquotedblright\ in \textquotedblleft spans (or generates) $M$ as
$R$-module\textquotedblright.

\item \textbf{Page 61, Example 175:} Replace ``for any $M$-module $R$'' by
``for any $R$-module $M$''.

\item \textbf{Page 61, definition of linear independence:} Require
$x_{1},\ldots,x_{m}$ to be \emph{distinct} elements of $X$. If repetitions are
allowed, every nonempty set is declared dependent by writing $1x+(-1)x=0$.

That said, it is probably better to define linear independence for families or
tuples rather than for sets.

\item \textbf{Page 62, Example 180:} Delete \textquotedblleft
a\textquotedblright\ from \textquotedblleft an $R$-basis for a $M$%
\textquotedblright.

\item \textbf{Page 63, Example 183:} Replace \textquotedblleft
indepdendent\textquotedblright\ by \textquotedblleft
independent\textquotedblright.

\item \textbf{Page 64, Exercise 188 and Section 5:} Hyphenate
\textquotedblleft$\mathbb{R}[T]$-module\textquotedblright. In the definition
of $\operatorname{Hom}_{R}(M,N)$, delete the duplicated \textquotedblleft%
$h:$\textquotedblright\ from
\[
\{h:h:M\rightarrow N\text{ is a homomorphism}\}.
\]
Also insert \textquotedblleft the\textquotedblright\ in \textquotedblleft the
additive identity is trivial homomorphism\textquotedblright.

\item \textbf{Page 65, Exercise 191:} \textquotedblleft Check that
$rf\in\operatorname{Hom}_{R}(M,M)$\textquotedblright\ should be
\textquotedblleft Check that $rf\in\operatorname{Hom}_{R}(M,N)$%
\textquotedblright.

\item \textbf{Page 65, Example 192:} Replace \textquotedblleft
contruct\textquotedblright\ by \textquotedblleft construct\textquotedblright.
More importantly, in \textquotedblleft This will be a homomorphism as
$\phi,f,\phi^{-1}$ are isomorphisms\textquotedblright, replace
\textquotedblleft are isomorphisms\textquotedblright\ by \textquotedblleft are
homomorphisms\textquotedblright; $f\in\operatorname{End}(M)$ need not be
invertible. A comma after the subordinate clause \textquotedblleft as
$\phi:M\rightarrow N$ is an isomorphism\textquotedblright\ would also smooth
the preceding paragraph.

\item \textbf{Page 65, Exercise 195:} In both module actions, replace
\[
a_{0}+a_{1}+a_{2}X^{2}+\cdots+a_{r}X^{r}\quad\text{by}\quad a_{0}+a_{1}%
X+a_{2}X^{2}+\cdots+a_{r}X^{r}.
\]


\item \textbf{Page 66, first paragraph of Section 6:} Replace
\textquotedblleft in terms of the $M$\textquotedblright\ by \textquotedblleft
in terms of $M$\textquotedblright, and \textquotedblleft beyond know that it
is a homomorphism\textquotedblright\ by \textquotedblleft beyond knowing that
it is a homomorphism\textquotedblright.

\item \textbf{Page 67, paragraph after (18):} Replace \textquotedblleft the
product of the matrices to two homomorphism\textquotedblright\ by
\textquotedblleft the product of the matrices of two
homomorphisms\textquotedblright.

\item \textbf{Page 68, associativity calculation:} Two closing parentheses are
missing in the left-hand calculation. It should be
\[
(f\circ(g\circ h))(\mathbf{v})=f\bigl((g\circ h)(\mathbf{v}%
)\bigr)=f\bigl(g(h(\mathbf{v}))\bigr).
\]
Also, the right calculation should perhaps go on its separate line, since it
currently bleeds out into the margins.

\item \textbf{Page 68, discussion of Theorem 197:} Insert \textquotedblleft
you\textquotedblright\ in \textquotedblleft to convince of the
truth\textquotedblright.

\item \textbf{Page 68, proof of Theorem 197:} In the last bullet, replace
\textquotedblleft given a matrix how to you write down\textquotedblright\ by
\textquotedblleft given a matrix, how do you write down\textquotedblright.

\item \textbf{Page 69, Example 200:} After \textquotedblleft If $\#A\geq
2$\textquotedblright, the \textquotedblleft the\textquotedblright\ should be a
\textquotedblleft then\textquotedblright.

\item \textbf{Pages 70--71, proof of Theorem 205:} To prove that
$\mathfrak{b}=\bigcup_{i}\mathfrak{b}_{i}$ is an ideal, the proof checks $0$,
addition, and multiplication by ring elements, but not additive inverses.
Again, this is legitimate, but the reason should be mentioned.

\item \textbf{Page 71, Exercise 206:} This holds only if $R$ is nonzero.

\item \textbf{Page 71, Exercise 209:} Replace \textquotedblleft
containining\textquotedblright\ by \textquotedblleft
containing\textquotedblright.

\item \textbf{Page 72, Exercise 210:} In Exercise 210(v), delete one copy of
\textquotedblleft that\textquotedblright\ before \textquotedblleft$\mathbb{Q}$
has no maximal $\mathbb{Z}$-submodules\textquotedblright.

\item \textbf{Page 72, Theorem 211 and Corollary 212:} Use \textquotedblleft
a\textquotedblright\ rather than \textquotedblleft an\textquotedblright%
\ before consonant sounds: \textquotedblleft a $D$-module\textquotedblright,
\textquotedblleft a $D$-basis\textquotedblright, \textquotedblleft a
$D$-linearly independent set\textquotedblright, and similarly
\textquotedblleft a $K$-basis\textquotedblright.

\item \textbf{Page 73, proof of Theorem 211(ii):} Delete one copy of
\textquotedblleft combination\textquotedblright\ in \textquotedblleft a finite
linear combination combination of elements of $T$\textquotedblright.

In the next sentence, after \textquotedblleft Since $T$ is maximal,
$T\cup\left\{  \mathbf{v}\right\}  $ does not belong to $\mathcal{P}%
$\textquotedblright, add \textquotedblleft(unless $\mathbf{v}\in T$, in which
case our claim is obvious)\textquotedblright.

\item \textbf{Page 75, Example 218:} Delete \textquotedblleft
the\textquotedblright\ from \textquotedblleft the submodules of the
$\mathbb{Z}/m\mathbb{Z}$\textquotedblright, and insert \textquotedblleft
if\textquotedblright\ in \textquotedblleft if and only $m$ is
prime\textquotedblright.

\item \textbf{Page 76, Example 221:} Replace \textquotedblleft extended to a
$K$-basis \ldots{} for $K$\textquotedblright\ by \textquotedblleft extended to
a $K$-basis \ldots{} for $K^{n}$\textquotedblright.

\item \textbf{Page 76, Theorem 222:} Delete \textquotedblleft
a\textquotedblright\ from \textquotedblleft$D^{n}$ is a
simple\textquotedblright.

\item \textbf{Page 76, proof of Theorem 222:} Replace \textquotedblleft Let
$A\in M_{n}(K)$\textquotedblright\ by \textquotedblleft Let $A\in M_{n}\left(
D\right)  $\textquotedblright.

\item \textbf{Page 76, proof of Theorem 224:} Insert \textquotedblleft
is\textquotedblright\ in \textquotedblleft Thus there some isomorphism
$g=f^{-1}$\textquotedblright.

\item \textbf{Page 77, proof of Lemma 228:} Replace \textquotedblleft$R$ does
not have any left $R$-submodules\textquotedblright\ and \textquotedblleft$R$
has no left ideals\textquotedblright\ by \textquotedblleft$R$ has no nonzero
proper left $R$-submodules\textquotedblright\ and \textquotedblleft$R$ has no
nonzero proper left ideals\textquotedblright. Both $0$ and $R$ itself are
always left ideals.

\item \textbf{Page 78, proof of Theorem 227:} Theorem 73 concerns 2-sided
ideals of a quotient ring. Here $\mathfrak{m}$ is only a left ideal and
$R/\mathfrak{m}$ is a quotient module, so cite the module Correspondence
Theorem, Exercise 169, instead.

\item \textbf{Page 79, Example 229:} Replace each of the three occurrences of
\textquotedblleft$V$\textquotedblright\ by \textquotedblleft$W$%
\textquotedblright.

\item \textbf{Page 79, Example 230:} Replace \textquotedblleft has not
complementary ideal\textquotedblright\ by \textquotedblleft has no
complementary ideal\textquotedblright.

\item \textbf{Page 80, Exercise 234:} Replace \textquotedblleft the
$\mathbb{R}[X]$-module were\textquotedblright\ by \textquotedblleft the
$\mathbb{R}[X] $-module where\textquotedblright.

\item \textbf{Page 81, Exercise 240:} Replace \textquotedblleft let
$\mathfrak{a},\mathfrak{b},\mathfrak{c}$ are\textquotedblright\ by
\textquotedblleft let $\mathfrak{a},\mathfrak{b},\mathfrak{c}$
be\textquotedblright.

\item \textbf{Page 81, proof of Theorem 241:} The induction begins with
$\dim_{K}(V)=1$, but the theorem also allows $V=0$. Add that when $\dim
_{K}(V)=0$, $V$ is the empty direct sum of simple modules. (On the other hand,
the case $\dim_{K}\left(  V\right)  =1$ does not require separate treatment.)

\item \textbf{Page 81, proof of Theorem 241:} Replace \textquotedblleft there
is a $A$-submodule\textquotedblright\ by \textquotedblleft there is an
$A$-submodule\textquotedblright.

\item \textbf{Page 81, end of the same proof:} Replace \textquotedblleft the
$U_{i}$ and $V_{j}$ are simple\textquotedblright\ by \textquotedblleft the
$U_{i}$ and $W_{j}$ are simple\textquotedblright.

\item \textbf{Page 82, Lemma 243:} Replace \textquotedblleft$=\psi
h$\textquotedblright\ by \textquotedblleft$=\psi$\textquotedblright.

\item \textbf{Page 83, Lemma 245 and Example 246:} Since the subject is
plural, write \textquotedblleft$\phi_{1},\ldots,\phi_{r}$ form a
$K$-basis\textquotedblright\ and \textquotedblleft$\phi_{1},\phi_{2},\phi_{3}$
form a basis\textquotedblright. In the proof, insert \textquotedblleft
We\textquotedblright\ before \textquotedblleft can write\textquotedblright.

\item \textbf{Page 85, proof of Lemma 249:} Replace \textquotedblleft We
compute $Ae_{u,v}$ and $Ae_{v,u}$ and compare\textquotedblright\ by
\textquotedblleft We compute $Ae_{u,v}$ and $e_{u,v}A$ and
compare\textquotedblright, as in the calculation that follows.

\item \textbf{Page 86, Section 6 and Examples 252--253:} Replace
\textquotedblleft otherwords\textquotedblright\ by \textquotedblleft other
words\textquotedblright\ and hyphenate \textquotedblleft$K[G]$%
-module\textquotedblright. In both examples, insert \textquotedblleft
we\textquotedblright\ in \textquotedblleft so can apply Maschke's
theorem\textquotedblright.

\item \textbf{Page 87, proof of Maschke's Theorem:} Replace the semicolon
after claim (iii) by a full stop, use the plural in \textquotedblleft prove
claim (i), (ii), and (iii)\textquotedblright, and replace \textquotedblleft is
is easy\textquotedblright\ by \textquotedblleft it is easy\textquotedblright.

\item \textbf{Page 88, proof of Theorem 256:} The $m$ in the isomorphism
\textquotedblleft$\mathbb{C}\left[  G\right]  \cong\prod\limits_{i=1}%
^{m}M_{n_{i}}\left(  D_{i}\right)  $\textquotedblright\ (which is obtained
from Artin--Wedderburn) is a-priori only some nonnegative integer; that it
equals the $m$ from Theorem 256 (that is, the number of conjugacy classes of
$G$) will only become clear at the end of the proof. Thus, a different letter
(or a warning) is warranted.

\item \textbf{Page 88, same proof:} The direct result saying that a
finite-dimensional complex division algebra is $\mathbb{C}$ is Theorem 118,
not Frobenius' real classification in Theorem 123. Citing Theorem 118 is both
shorter and clearer.

\item \textbf{Page 92, proof of Theorem 268 and Exercise 270:} Replace
\textquotedblleft an 2-sided ideal\textquotedblright\ by \textquotedblleft a
2-sided ideal\textquotedblright, replace \textquotedblleft the kernel \ldots{}
contain\textquotedblright\ by \textquotedblleft the kernel \ldots{}
contains\textquotedblright, and replace \textquotedblleft
endormphism\textquotedblright\ by \textquotedblleft
endomorphism\textquotedblright.
\end{enumerate}


\end{document}