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\begin{document}

\begin{center}
\textbf{Algebra: Abstract and Concrete}

\textit{Frederick M. Goodman}

Edition 2.6 (last revised May 1, 2015)

\url{https://homepage.divms.uiowa.edu/~goodman/algebrabook.dir/download.htm}

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

\noindent The page numbers below refer to the printed page numbers in Edition
2.6. This list is not claimed to be exhaustive. Items marked ``substantive''
change a mathematical assertion or supply a missing hypothesis; other items
are local corrections or clarifications. Formula-sensitive items have been
checked against rendered pages of the PDF, since text extraction from this
file occasionally loses minus signs.

The majority of errata in this list were found and written up by GPT-5.6 Sol.
All have been verified by Darij Grinberg.

\appendix
\setcounter{section}{7}

\section{Corrections and mathematical clarifications}

\begin{enumerate}
\item \textbf{Page 2, sentence below Figure 1.1.1:} Replace \textquotedblleft
An object is symmetric if it has symmetries\textquotedblright\ by
\textquotedblleft An object is symmetric in a nontrivial sense if it has a
symmetry other than the nonmotion.\textquotedblright\ Once the nonmotion is
counted as a symmetry on page 3, every object has at least this trivial symmetry.

\item \textbf{Page 20, paragraph defining the order of a permutation:} Replace
\textquotedblleft no lower power of $\pi$ is the identity\textquotedblright%
\ by \textquotedblleft no lower positive power of $\pi$ is the
identity\textquotedblright. The zeroth power is always the identity.

\item \textbf{Page 22, Theorem 1.5.3 (uniqueness of disjoint-cycle notation):}
\emph{Clarification.} This theorem, as stated, relies on the convention that
each cycle has length $\geq2$ (that is, it permutes at least two elements). If
$1$-cycles (i.e., cycles of length $1$) were allowed, then the representation
of a permutation as a product of disjoint cycles would no longer be unique,
since $1$-cycles are just the identity permutation (after all, cycling through
a single element obviously leaves this element fixed!) and thus can be
inserted into a product at will without affecting the product.

Later on -- e.g., in Example 5.1.17 and Exercise 5.1.17 -- this convention is
abandoned; there, you allow cycles of length $1$, and in fact require them in
certain cases. Namely, in Chapter 5, you include a $1$-cycle $\left(
i\right)  $ for each fixed point $i$ of the permutation, so as to ensure that
each element of $\left\{  1,2,\ldots,n\right\}  $ appears exactly once in the
cycle notation. For example, the permutation in $S_{9}$ that would be written
as $\left[  \left(  2 \ 5\right)  \left(  4 \ 6 \ 9 \ 8\right)  \right]  $
according to the convention of Chapter 1 would be written as $\left[  \left(
1\right)  \left(  2 \ 5\right)  \left(  3\right)  \left(  4 \ 6 \ 9
\ 8\right)  \left(  7\right)  \right]  $ in Chapter 5 (with the redundant
$1$-cycles being included in order to ensure that each element of $\left\{
1,2,\ldots,n\right\}  $ appears exactly once). With this convention, the
representation is again unique.

\item \textbf{Page 22, proof of Theorem 1.5.3:} Replace ``If $|X|=1$, there is
nothing to do'' by ``If $|X|\leq1$, there is nothing to do.'' Also, after the
induction hypothesis, replace ``Let $\pi$ be a nonidentity permutation of
$X$'' by ``If $\pi$ is the identity, there is nothing to prove. Otherwise, let
$\pi$ be a nonidentity permutation of $X$.'' The theorem includes the empty
set and the identity permutation on sets of cardinality greater than $1$.

\item \textbf{Page 26, paragraph before Definition 1.6.3:} Replace
\textquotedblleft Natural numbers that are not prime are called
\textit{composite}\textquotedblright\ by \textquotedblleft Natural numbers
greater than $1$ that are not prime are called \textit{composite}%
\textquotedblright. Otherwise, the sentence incorrectly classifies $1$ as composite.

\item \textbf{Page 30, Euclidean algorithm:} Replace \textquotedblleft after
no more than $n$ steps\textquotedblright\ by \textquotedblleft after no more
than $|n|$ steps\textquotedblright. The integer $n$ has not been assumed positive.

\item \textbf{Page 40, Example 1.7.6:} Replace \textquotedblleft the sum of a
positive number and a negative number can be either positive or
negative\textquotedblright\ by \textquotedblleft the sum of a positive number
and a negative number can be negative or nonnegative\textquotedblright. The
sum can also be $0$.

\item \textbf{Page 52, proof of Theorem 1.8.16:} Replace \textquotedblleft
after at most $\deg(f)$ steps\textquotedblright\ by \textquotedblleft after at
most $\deg(f)+1$ steps\textquotedblright. For example, when $f$ is a nonzero
constant, one division is needed although $\deg(f)=0$.

\item \textbf{Page 54, Definition 1.8.23:} After \textquotedblleft if
$x-\alpha$ appears exactly $k$ times in the irreducible factorization of
$p$\textquotedblright, add \textquotedblleft(or, to be more precise, exactly
$k$ scalar multiples of $x-\alpha$ appear in this
factorization)\textquotedblright. For example, the multiplicity of $3$ in
$\left(  x-2\right)  \left(  x-3\right)  \left(  2x-6\right)  $ is $2$, since
$2x-6$ is a scalar multiple of $x-3$.

\item \textbf{Page 55, Exercise 1.8.13 (b):} Replace \textquotedblleft is an
element of $I$ of smallest degree\textquotedblright\ by \textquotedblleft has
smallest degree among the nonzero elements of $I$\textquotedblright. The zero
polynomial belongs to $I$ and has degree $-\infty$ under the convention of
Definition 1.8.3.

\item \textbf{Page 56, Exercise 1.8.18 (b):} Replace \textquotedblleft is
irreducible if and only if it has no integer root\textquotedblright\ by
\textquotedblleft is irreducible in $\mathbb{Q}[x]$ if and only if it has no
integer root\textquotedblright. Irreducibility has so far been defined only
for polynomial rings over a field.

\item \textbf{Page 60, Corollary 1.9.6:} Parts (b) and (c) are only true for
$n\geq1$.

\item \textbf{Page 61, Example 1.9.7:} Replace \textquotedblleft while
evaluating at $x=-1$ gives\textquotedblright\ by \textquotedblleft while, if
$n>1$, evaluating at $x=-1$ gives\textquotedblright. For $n=1$, the left-hand
side of the differentiated identity evaluates to $1$, not to $0$.

\item \textbf{Page 61, proof of Proposition 1.9.9:} Replace \textquotedblleft
Reducing modulo $n$ gives $[a][s]=[0]$, so $[a]$ is a zero divisor in
$\mathbb{Z}_{n}$, and therefore not invertible\textquotedblright\ by
\textquotedblleft Reducing modulo $n$ gives $[a][s]=[0]$. Since $1\leq s<n $,
we have $[s]\neq\lbrack0]$; hence $[a]$ cannot be invertible\textquotedblright%
. If $[a]=[0]$, then $[a]$ is not a zero divisor according to the definition
on page 41, so the printed intermediate assertion need not hold.

\item \textbf{Page 63:} It is worth mentioning that $1$ means the function
$\mathbf{1}_{U} : U \to\{0,1\}$, which sends every $u \in U$ to $1$.

\item \textbf{Page 65, approximation of the derangement number:} Replace both
occurrences of ``$\leq$'' in the two displayed remainder estimates by ``$<$''.
The strict alternating-series estimate makes the conclusion that $D_{n}$ is
the integer closest to $n!/e$ immediate also when $n=1$.

\item \textbf{Page 65, Definition 1.9.15:} Replace \textquotedblleft natural
numbers $k<n$\textquotedblright\ by \textquotedblleft natural numbers $k\leq
n$\textquotedblright. This gives the standard value $\varphi(1)=1$ and makes
the definition agree with the group of units of $\mathbb{Z}_{1}$ used later.

\item \textbf{Page 74, Exercise 1.10.7:} Replace \textquotedblleft A $C^{1}$
diffeomorphism of $\mathbb{R}^{3}$ is a bijective map $T:\mathbb{R}%
^{3}\rightarrow\mathbb{R}^{3}$ having continuous first-order partial
derivatives\textquotedblright\ by \textquotedblleft A $C^{1}$ diffeomorphism
of $\mathbb{R}^{3}$ is a bijective map $T:\mathbb{R}^{3}\rightarrow
\mathbb{R}^{3}$ such that both $T$ and $T^{-1}$ have continuous first-order
partial derivatives\textquotedblright. Without the condition on the inverse,
the asserted closure under inverses is false; for example, $x\mapsto x^{3}$ on
$\mathbb{R}$ is bijective and $C^{1}$, but its inverse is not $C^{1}$ at $0$.

\item \textbf{Pages 74--75, Exercise 1.10.10:} As it stands, this exercise
relies on the reader correctly inferring how to interpret the definition of
$T$ in the presence of infinities -- e.g., understanding $\dfrac{a\infty
+b}{c\infty+d}$ as $\dfrac{a}{c}$. Also, there is an unwritten condition that
not all of $a,b,c,d$ are $0$, since otherwise nothing can make $T$ well-defined.

\item \textbf{Page 75, matrix-ring examples:} After \textquotedblleft In the
first three examples, multiplication is commutative, but in the fourth
example, it is not\textquotedblright, add \textquotedblleft(unless $n\leq1
$)\textquotedblright.

\item \textbf{Page 76, Example 1.11.2 (d):} After \textquotedblleft The set of
$n$-by-$n$ matrices with integer entries is a noncommutative
ring\textquotedblright, add \textquotedblleft(unless $n\leq1$, in which case
it is a commutative ring)\textquotedblright.

\item \textbf{Page 77, Example 1.11.6 (b):} Replace
\[
a_{0}+a_{1}x+\cdots+a_{s}x^{s}\longmapsto a_{0}+a_{1}T+\cdots+a_{s}T^{s}%
\]
by
\[
a_{0}+a_{1}x+\cdots+a_{s}x^{s}\longmapsto a_{0}I_{n}+a_{1}T+\cdots+a_{s}%
T^{s},
\]
where $I_{n}$ is the $n$-by-$n$ identity matrix. The scalar constant term has
to be interpreted as a scalar matrix.

\item \textbf{Page 81, Lemma 1.12.1:} Replace \textquotedblleft For all
integers $a$ and $h$\textquotedblright\ by \textquotedblleft For every integer
$a$ and every positive integer $h$\textquotedblright. Negative powers of an
arbitrary integer $a$ are not defined in the setting of the lemma.

\item \textbf{Page 83, Exercise 1.12.1 (b):} Replace \textquotedblleft Let
$n=pq$ the product of two primes\textquotedblright\ by \textquotedblleft Let
$n=pq$ be the product of two distinct primes.\textquotedblright\ The displayed
isomorphism $\Phi(pq)\cong\Phi(p)\times\Phi(q)$ requires $p$ and $q$ to be
relatively prime.

\item \textbf{Page 91, proof of Proposition 2.1.19:} Replace \textquotedblleft%
$(a_{1}\cdots a_{k+1})(a_{k+1}\cdots a_{n})$\textquotedblright\ by
\textquotedblleft$(a_{1}\cdots a_{k+1})(a_{k+2}\cdots a_{n})$%
\textquotedblright. The former repeats $a_{k+1}$ and therefore does not equal
$p_{k+1}$.

\item \textbf{Page 96, constructive description of $\langle S\rangle$:} After
\textquotedblleft all possible products $g_{1}g_{2}\cdots g_{n}$, where
$g_{i}\in S$ or $g_{i}^{-1}\in S$\textquotedblright, add \textquotedblleft%
(including the empty product, which is understood to be $e$)\textquotedblright.

\item \textbf{Page 96, footnote 1:} The definition of a partial order needs a
third requirement: \textit{reflexivity} ($x\leq x$ for all $x$). Also, the
second requirement should be called \textit{antisymmetry}, not
\textit{asymmetry}.

\item \textbf{Page 100, proof of Proposition 2.2.21 (a):} Replace
\textquotedblleft by part (a), $d$ and $d^{\prime}$ divide one
another\textquotedblright\ by \textquotedblleft by part (c), $d$ and
$d^{\prime}$ divide one another\textquotedblright.

\item \textbf{Page 101, proof of Corollary 2.2.26 (a):} Replace
\textquotedblleft the least of positive integers $s$ such that $s\in
H$\textquotedblright\ by \textquotedblleft the least of positive integers $s$
such that $[s]\in H$\textquotedblright. Here $H$ is a subgroup of
$\mathbb{Z}_{n}$, so its elements are congruence classes.

\item \textbf{Page 114, proof of Proposition 2.4.11:} Replace the first
reference to \textquotedblleft Proposition 2.1.1(a)\textquotedblright\ by
\textquotedblleft Corollary 2.1.7\textquotedblright, and replace the reference
to \textquotedblleft Proposition 2.1.1(b)\textquotedblright\ by
\textquotedblleft Proposition 2.1.2\textquotedblright.

\item \textbf{Page 117, Example 2.4.20:} This is literally true if all
elements of $G$ have finite order. Otherwise, it requires the understanding
that $\infty$ divides $0$ but does not divide any nonzero integers.

\item \textbf{Page 117, Corollary 2.4.23:} Replace \textquotedblleft The set
of odd permutations in $S_{n}$ is $(1\,2)A_{n}$\textquotedblright\ by
\textquotedblleft For $n\geq2$, the set of odd permutations in $S_{n}$ is
$(1\,2)A_{n}$\textquotedblright. For $n=1$, the transposition $(1\,2)$ is not
an element of $S_{n}$.

\item \textbf{Page 120, Exercise 2.4.15:} Replace \textquotedblleft Show that
$\epsilon$ is the unique homomorphism from $S_{n}$ onto $\{1,-1\}$%
\textquotedblright\ by \textquotedblleft For $n\geq2$, show that $\epsilon$ is
the unique homomorphism from $S_{n}$ onto $\{1,-1\}$\textquotedblright. For
$n=1$, no such surjective homomorphism exists.

\item \textbf{Page 121, Exercise 2.4.21:} Replace \textquotedblleft if $n$ is
relatively prime to the order of $G$\textquotedblright\ by \textquotedblleft
if $G$ is finite and $n$ is relatively prime to $\lvert G\rvert$%
\textquotedblright.

\item \textbf{Page 127, Exercise 2.5.16:} Replace \textquotedblleft the
symmetric group $S_{n}$ has a unique subgroup of index $2$\textquotedblright%
\ by \textquotedblleft for $n\geq2$, the symmetric group $S_{n}$ has a unique
subgroup of index $2$\textquotedblright. The assertion is false for $n=1$.

\item \textbf{Page 131, Example 2.6.10 (b):} Replace \textquotedblleft The
equivalence relation $x\sim y$ for all $x,y\in X$ has just one equivalence
class, namely $X$\textquotedblright\ by \textquotedblleft If $X\neq
\varnothing$, the equivalence relation $x\sim y$ for all $x,y\in X$ has just
one equivalence class, namely $X$; if $X=\varnothing$, it has no equivalence
classes\textquotedblright.

\item \textbf{Page 134, first paragraph of Section 2.7:} Replace
\textquotedblleft Consider the permutation group $S_{n}$ with its normal
subgroup of even permutations\textquotedblright\ by \textquotedblleft Let
$n\geq2$, and consider the permutation group $S_{n}$ with its normal subgroup
of even permutations\textquotedblright. The displayed coset $O=(1\,2)E$ is not
defined for $n=1$.

\item \textbf{Page 136, Example 2.7.3:} The domain $\mathbb{Z}_{n}$ is written
additively, while the target group $G$ is written multiplicatively. Make the
following replacements.

Replace \textquotedblleft$\varphi(a)-\varphi(b)=\varphi(a-b)=0$%
\textquotedblright\ by \textquotedblleft$\varphi(a)\varphi(b)^{-1}%
=\varphi(a-b)=e_{G}$\textquotedblright.

Replace
\[
\widetilde{\varphi}([a][b])=\widetilde{\varphi}([ab])=\varphi(ab)=\varphi
(a)\varphi(b)
\]
by
\[
\widetilde{\varphi}([a]+[b])=\widetilde{\varphi}([a+b])=\varphi(a+b)=\varphi
(a)\varphi(b).
\]


Finally, replace \textquotedblleft If $\widetilde{\varphi}([k])=0$, then
$\varphi(k)=0$, so $k\in n\mathbb{Z}=[0]$, so $[k]=[0]$\textquotedblright\ by
\textquotedblleft If $\widetilde{\varphi}([k])=e_{G}$, then $\varphi(k)=e_{G}%
$, so $k\in n\mathbb{Z}$, and hence $[k]=[0]$\textquotedblright.

\item \textbf{Page 153, Example 3.1.11:} There are three Dalmatian dogs, so
their permutation group $D$ is isomorphic to $S_{3}$, not to $S_{4}$.

Moreover, the displayed copy of $A$ inside
\[
P=A\times B\times C\times D\times E
\]
should be
\[
\widetilde{A}=A\times\{e\}\times\{e\}\times\{e\}\times\{e\},
\]
with four trivial factors $\times\{e\}$, not three.

\item \textbf{Page 155, Definition 3.1.16:} In both formulas defining the
coordinatewise operations, replace the final entry $r_{s}^{\prime}$ of the
second tuple by $r_{n}^{\prime}$.

\item \textbf{Page 155, Proposition 3.1.18:} ``For any integers $x_{1}%
,x_{2},\ldots,x_{s}$'' should be ``For any integers $x_{1},x_{2},\ldots,x_{n}$''.

\item \textbf{Page 162, Exercise 3.2.2:} The automorphism
\[
j:[x]\longmapsto[-x]
\]
does not always have order $2$: it is the identity for $n=1$ and $n=2$. It
does satisfy $j^{2}=\operatorname{id}$ for every $n$, which is all that is
needed to define the indicated homomorphism from $\mathbb{Z}_{2}$. Thus,
either replace ``an order $2$ automorphism'' by ``an automorphism satisfying
$j^{2}=\operatorname{id}$'', or add the hypothesis $n>2$ to the assertion
about its order.

\item \textbf{Page 172, proof of Proposition 3.3.25:} ``there is a nonzero
$\boldsymbol{\alpha}=%
\begin{bmatrix}
\alpha_{1}\\
\vdots\\
\alpha_{n}%
\end{bmatrix}
\in K^{n}$'' should be ``there is a nonzero $\boldsymbol{\alpha}=%
\begin{bmatrix}
\alpha_{1}\\
\vdots\\
\alpha_{s}%
\end{bmatrix}
\in K^{s}$''.

\item \textbf{Page 172, proof of Corollary 3.3.26:} The second application of
Proposition 3.3.25, with the roles of $X$ and $Y$ reversed, gives $|X|\leq
|Y|$, not a second copy of $|Y|\leq|X|$.

\item \textbf{Page 175, proof of Proposition 3.3.33:} The proof cites
Proposition 3.5.1, which has not yet appeared. It would be natural to switch
the order of these two results.

\item \textbf{Page 176, Proposition 3.3.35:} The preceding argument assumes
that $T:V\to W$ is surjective, whereas the proposition is stated for an
arbitrary linear map. To obtain the stated result, apply the preceding
argument to the surjective map $T:V\longrightarrow\operatorname{range}(T)$.

\item \textbf{Page 176, warning after Corollary 3.3.37:} The word ``never'' in
``Complements of a subspace are never unique'' is an overstatement: For
example, the zero subspace has the unique complement $V$, and $V$ has the
unique complement $\{0\}$. Replace ``are never unique'' by ``need not be
unique'' or ``are usually not unique''.

\item \textbf{Page 177, Exercise 3.3.10 (a):} The row $[v_{1},\ldots,v_{n}]$
belongs to $V^{n}$, not to $K^{n}$.

\item \textbf{Page 182, paragraph after Theorem 3.4.5:} The assertion that the
canonical map $V\to V^{**}$ is not surjective when $V$ is infinite-dimensional
is true, but is nontrivial and is neither proved nor assigned as an exercise.

\item \textbf{Pages 184--185, Proposition 3.4.10:} The convention introduced
immediately before the proposition denotes the matrix of $T:V\to W$ by
$[T]_{C,B}$, where $B$ is the basis of $V$ and $C$ the basis of $W$. Thus, in
part (a), replace ``$T\longmapsto[T]_{B,C}$'' by ``$T\longmapsto[T]_{C,B}$''.
Make the same correction in the final sentence of the proof of part (a).

\item \textbf{Page 188, Exercise 3.4.4:} In the displayed inner product, the
second vector should be
\[%
\begin{bmatrix}
\beta_{1}\\
\beta_{2}\\
\beta_{3}%
\end{bmatrix}
,
\]
not
\[%
\begin{bmatrix}
\beta_{2}\\
\beta_{2}\\
\beta_{3}%
\end{bmatrix}
.
\]


\item \textbf{Page 189, Exercise 3.4.10 (b):} The ``integration'' map
$P_{6}\to P_{7}$ is not determined until a choice of integration constant has
been specified. For example, one can define it to be $f(x)\longmapsto\int%
_{0}^{x} f(t)\,dt$, equivalently the unique antiderivative with constant term
$0$.

\item \textbf{Page 190, first paragraph of Section 3.5:} After ``the order of
an element $x$ is the smallest natural number $s$ such that $sx=0$'', add
``(or $\infty$ if no such $s$ exists)''.

\item \textbf{Page 194, Proposition 3.5.9:} The integer $s$ in
\[
\operatorname{diag}(d_{1},\ldots,d_{s},0,\ldots,0)
\]
is not introduced. One should specify that $0\leq s\leq\min\{m,n\}$ and that
$d_{1},\ldots,d_{s}$ are the nonzero diagonal entries.

\item \textbf{Page 200, derivation of the invariant-factor decomposition:} In
the direct product
\[
\mathbb{Z}/d_{i}\mathbb{Z}\times\cdots\times\mathbb{Z}/d_{s}\mathbb{Z}%
\times\mathbb{Z}^{\,n-s},
\]
the first factor should be $\mathbb{Z}/d_{1}\mathbb{Z}$, not $\mathbb{Z}%
/d_{i}\mathbb{Z}$.

\item \textbf{Page 203, proof of uniqueness in Theorem 3.6.2:} The proof
should dispose separately of the case $G_{\mathrm{tor}}=\{0\}$ before writing
\[
a_{s}=b_{t}=a.
\]
In that case $s=t=0$, so neither $a_{s}$ nor $b_{t}$ exists, and uniqueness is
immediate. Add, for example: ``If $G_{\mathrm{tor}}=\{0\}$, then $s=t=0$, and
there is nothing left to prove. Hence assume $G_{\mathrm{tor}}\neq\{0\}$.''

\item \textbf{Page 204, proof of uniqueness in Theorem 3.6.2:} After ``Let
$k^{\prime}$ be the last index such that $a_{k^{\prime}}=p$'', add ``(or $0$
if no such index exists)''. Make a similar change when $k^{\prime\prime}$ is defined.

\item \textbf{Page 204, Example 3.6.6:} Under the convention in Theorem 3.6.2
that the invariant factors satisfy $a_{1}\mid a_{2}\mid\cdots$, the final
decomposition should be written
\[
\mathbb{Z}_{30}\times\mathbb{Z}_{24}\cong\mathbb{Z}_{6}\times\mathbb{Z}%
_{120},
\]
and the invariant factors should be listed as $6,120$, rather than $120,6$.
(The two displayed direct products are of course isomorphic; the issue is the
ordering convention.)

\item \textbf{Page 205, Definition 3.6.9 (b):} If the partition is denoted
$(n_{1},n_{2},\ldots,n_{k})$, then the displayed decomposition should end in
$\mathbb{Z}_{p^{n_{k}}}$, not in $\mathbb{Z}_{p^{n_{s}}}$.

\item \textbf{Page 207, proofs of Theorem 3.6.15 and Corollary 3.6.16:} In
both prime factorizations, replace
\[
p_{1}^{k_{1}}p_{s}^{k_{2}}\cdots p_{s}^{k_{s}}%
\]
by
\[
p_{1}^{k_{1}}p_{2}^{k_{2}}\cdots p_{s}^{k_{s}}.
\]


\item \textbf{Pages 210--212, Examples 3.6.22--3.6.24:} The invariant factors
are systematically listed in the reverse of the order stipulated in Theorem
3.6.2, where $a_{1}\mid a_{2}\mid\cdots\mid a_{s}$.

For instance, the decomposition on page 210 should be written
\[
\mathbb{Z}_{2}\times\mathbb{Z}_{10}\times\mathbb{Z}_{2100},
\]
rather than
\[
\mathbb{Z}_{2100}\times\mathbb{Z}_{10}\times\mathbb{Z}_{2}.
\]
Likewise, the algorithm on pages 210--211 produces the invariant factors from
largest to smallest, so the resulting list should be reversed before it is put
into the convention of Theorem 3.6.2.

On page 212, the six decompositions should accordingly be written as
\[
\begin{gathered} \mathbb Z_{4200},\\ \mathbb Z_5\times\mathbb Z_{840},\\ \mathbb Z_2\times\mathbb Z_{2100},\\ \mathbb Z_{10}\times\mathbb Z_{420},\\ \mathbb Z_2\times\mathbb Z_2\times\mathbb Z_{1050},\\ \mathbb Z_2\times\mathbb Z_{10}\times\mathbb Z_{210}. \end{gathered}
\]
The displayed groups are of course unchanged up to isomorphism; the problem is
only the stated ordering convention.

\item \textbf{Page 213, proof of Proposition 3.6.27 (b):} Exercise 2.2.30
correctly shows that the order of $[3]$ in $\Phi(2^{n})$ is $2^{n-2}$, not
$2^{n-1}$. Consequently, replace ``order $2^{n-1}$'' by ``order $2^{n-2}$'',
and replace each of the three ``$\mathbb{Z}_{2^{n-1}}$''s by ``$\mathbb{Z}%
_{2^{n-2}}$.

\item \textbf{Page 215, Exercise 3.6.17:} If $(m_{1},\ldots,m_{k})$ are
invariant factors with $m_{1}\mid m_{2}\mid\cdots\mid m_{k}$, then a finite
abelian group $G$ has an element of order $s$ if and only if $s\mid m_{k}$,
not if and only if $s\mid m_{1}$. The exponent of $G$ is the largest invariant
factor $m_{k}$.

\item \textbf{Page 249, Exercise 5.1.20:} The proposed description
\[
\mathbb{Z}_{4}=\left\langle (1\ 2\ 3\ 4),(1\ 2)(3\ 4)\right\rangle
\]
is wrong: the displayed generators generate the dihedral group $D_{4}$,
exactly as in the preceding bullet point. Replace this by
\[
\mathbb{Z}_{4}=\left\langle (1\ 2\ 3\ 4)\right\rangle .
\]


\item \textbf{Page 250, proof of Proposition 5.2.2:} The characteristic
function $\mathbf{1}_{F}$ was defined on $G\times X$, not on $X \times G$.
Thus, in both displayed double sums, replace $\mathbf{1}_{F}(x,g)$ by
$\mathbf{1}_{F}(g,x)$.

\item \textbf{Page 259, paragraph following the proof of Theorem 5.4.11:} In
the summary of the Sylow theorems, replace ``the number of such subgroups
divides $|G|$ and is conjugate to $1$ mod $p$'' by ``the number of such
subgroups divides $|G|/p^{n}$ and is congruent to $1$ modulo $p$.'' The
statement that it divides $|G|$ is true but weaker than the third Sylow
theorem (although easily seen to be equivalent in view of the $\equiv1 \mod
p$ property), whereas the word ``conjugate'' is simply wrong.

\item \textbf{Page 260, Example 5.4.14:} ``Hence $n_{r}\in\{1,10\}$'' should
be ``Hence $n_{3}\in\{1,10\}$''.

\item \textbf{Page 261, Example 5.4.14:} The assertion that the map
\[
[k]\longmapsto[-k]
\]
is an automorphism of order $2$ of $\mathbb{Z}_{n}$ ``for any $n$'' is false
for $n=1$ and $n=2$; in these cases it is the identity. Replace ``for any
$n$'' by ``for any $n>2$''. Only $n=3,5$ is needed here.

\item \textbf{Page 263, Exercise 5.4.10, converse assertion:} The relations
\[
a^{7}=b^{4}=1,\qquad bab^{-1}=a^{-1}%
\]
do not by themselves force a group generated by $a,b$ to be the indicated
semidirect product; they only make it a quotient of that group. For example,
one may take $a=1$. Add, for instance, that $a$ and $b$ have orders exactly
$7$ and $4$, respectively (or that the generated group has order $28$), or
rephrase the assertion as a presentation of the semidirect product.

\item \textbf{Page 264, first non-italicized paragraph of Section 5.5:} The
subgroup
\[
\left\langle (1\ 2\ 3\ 4\ 5),(1\ 2)\right\rangle
\]
is $S_{5}$, not $D_{5}$. For example, conjugating $(1\ 2)$ by powers of
$(1\ 2\ 3\ 4\ 5)$ produces enough transpositions to generate $S_{5}$. Replace
$(1\ 2)$ by, for example,
\[
(2\ 5)(3\ 4),
\]
as in Proposition 5.5.3.

\item \textbf{Page 265, paragraph immediately before Proposition 5.5.3:} Since
the common normalizer of the three displayed subgroups has order $20$, its
index in $S_{5}$ is
\[
[S_{5}:\langle\sigma,\rho\rangle]=120/20=6,
\]
not $5$. Thus replace ``the number of conjugates of each of the groups is
$5$'' by ``\ldots is $6$''. In Proposition 5.5.3, ``and its five conjugates''
is consistent with this if it means five \emph{other} conjugates; ``and its
five other conjugates'' would be clearer.

\item \textbf{Page 266, Definition 5.6.1:} Replace ``if for all $x\in X$,'' by
``if for all $g\in G$ and all $x\in X$, we have''.

\item \textbf{Page 267, Exercise 5.6.6:} Add the hypothesis $n\geq3$. For
$n=2$, the point stabilizer $H=\operatorname{Stab}(2)$ is the trivial subgroup
of $S_{2} $, whose normalizer is all of $S_{2}$, not $H$.

\item \textbf{Page 268, Exercise 5.6.18:} This exercise is false. Let
\[
V=\left\{  (x_{1},\ldots,x_{5})\in\mathbb{F}_{2}^{5} \ \middle|\ x_{1}%
+\cdots+x_{5}=0\right\}  \cong(\mathbb{Z}_{2})^{4},
\]
and let $\mathbb{Z}_{5}$ act nontrivially on $V$ by cyclically permuting the
five coordinates. Then
\[
G=V\rtimes\mathbb{Z}_{5}%
\]
has order $2^{4}\cdot5=80$, but the displayed complement $\mathbb{Z}_{5}$ is
not normal. Indeed, $G$ has multiple $5$-Sylow subgroups (since any $(v, c)
\in V \rtimes\mathbb{Z}_{5} = G$ for which $c$ is nonzero has order $5$ in
$G$), which are all conjugate by the Second Sylow Theorem, and thus none of
them is normal.
%OLD PROOF: if both $V$ and this complement were normal,
%then their commutator would lie in their trivial intersection, so the
%action would be trivial.


A simple repair is to assume $1\leq n\leq3$. In that case the number of
$5$-Sylow subgroups divides $2^{n}$ and is congruent to $1$ modulo $5$, and
hence must equal $1$.

\item \textbf{Page 270, etc.:} It might also be worthwhile to point out that
polynomials usually ``know'' the names of their variables; i.e., the
polynomial $x+1 \in K[x]$ is not identified with $y+1 \in K[y]$ (or else
things would go awry when $K[x]$ and $K[y]$ are embedded into $K[x,y]$). The
definition of polynomials as sequences (or families, in the multivariate case)
of coefficients obscures this point, as it suggests that the monomials $x^{2}
y$ in $K[x,y]$ and $y^{2} z$ in $K[y,z]$ both are just the multi-index $(2,
1)$. To avoid this ambiguity, it should be clarified that a monomial is not
\textit{just} the multiindex, but also the list of names of variables involved
-- e.g., the monomial $x^{2} y$ in $K[x,y]$ is the multi-index $(2, 1) $ with
the list of names $(``x^{\prime\prime},``y^{\prime\prime})$, whereas the
monomial $y^{2} z$ in $K[y,z]$ is the multi-index $(2, 1)$ with the list of
names $(``y^{\prime\prime},``z^{\prime\prime})$.

\item \textbf{Page 274, Exercise 6.1.12:} The claim that ``$R$ is isomorphic
to the subring of constant functions on $X$'' is only true if $X$ is nonempty.

\item \textbf{Page 284, Proposition 6.2.27 (a):} The displayed description of
$R\mathcal{S}R$ should specify all three families of conditions:
\[
R\mathcal{S}R=\left\{  \sum_{i=1}^{n}a_{i} s_{i} b_{i}:\ n \geq0,\ a_{i}%
,b_{i}\in R, \ s_{i}\in\mathcal{S}\right\}  .
\]
As printed, only $a_{n},b_{n}\in R$ are mentioned, and no condition at all is
placed on the $s_{i}$.
%If the empty subset $S=\varnothing$ is allowed, one
%must also allow the empty sum (or separately declare $R\varnothing R=\{0\}$).


\item \textbf{Page 295, Definition 6.4.2:} \emph{Missing hypothesis.} To agree
with the standard convention and with subsequent uses, require an integral
domain to be nontrivial, equivalently to satisfy $1\neq0$. Otherwise the zero
ring satisfies the printed condition vacuously.

\item \textbf{Page 298, Definition 6.4.10 and Proposition 6.4.11:}
\emph{Missing hypothesis.} A prime ideal is required to be proper. Thus
Definition 6.4.10 should add the requirement $J\neq R$. Correspondingly,
Proposition 6.4.11 should say that $J$ is prime if and only if $R/J$ is a
nonzero ring with no zero divisors (and, in the unital commutative case, an
integral domain in the corrected sense).

\item \textbf{Page 298, Example 6.4.13:} Every ideal of $\mathbb{Z}$ has the
form $d\mathbb{Z}$ for a unique $d\geq0$, not necessarily for $d>0$. With the
usual definition of prime ideal, $(0)$ is also a prime ideal of $\mathbb{Z}$.
Thus the positive ideals $d\mathbb{Z}$ are prime exactly when $d$ is a prime
number, and $(0)$ is the additional prime ideal.

\item \textbf{Page 300, definition of a Euclidean domain:} The condition
``$d(fg) \geq\max\{d(f), d(g)\}$'' is not used until Lemma 8.4.7 (and possibly
the solution to Exercise 6.5.3, where it allows for a simpler proof). Several
authors omit it from the definition of a Euclidean domain.

\item \textbf{Page 317, Lemma 6.7.4 and its proof:} First dispose separately
of the case $J=(0)$, since otherwise the ``minimum of the degrees of nonzero
elements of $J$'' is undefined. In the induction step, ``choose $g\in J_{0}$
such that $\deg(f-g)<\deg(g)$'' should read ``choose $g\in J_{0}$ such that
$\deg(f-g)<\deg(f)$''.
%This is both what the hypothesis supplies and what is needed for the
%induction.


\item \textbf{Page 324, display (7.2.1):} The first elementary-symmetric
relation should be
\[
\alpha_{1}+\alpha_{2}+\alpha_{3}=0,
\]
not $\alpha_{1}+\alpha_{3}+\alpha_{3}=0$.

\item \textbf{Page 324, Cardano formula near the bottom of the page:} The
second root is labeled $a_{2}$; it should be $\alpha_{2}$. More importantly,
the claim that the square-root choice is immaterial needs an exception: When
$p=0 $, one of the two square-root choices can make $A^{3}=0$, after which the
displayed expression $p/(3A)$ is undefined. Choose the sign so that $A\neq0$
when possible, and handle the case $p=q=0$ separately (where the sole root is
$0$ with multiplicity $3$).

\item \textbf{Page 356, Definition 8.1.28:} In my view, this is a bad
definition, as it lacks the flexibility that is later needed (and silently
assumed). It makes a lot more sense to define linear independence for
\textit{families} or \textit{tuples} (rather than \textit{sets}), and
similarly to say that a \textit{family} or a \textit{tuple} (not a
\textit{set}) is a basis of an $R$-module. This does not mean that it is
always wrong to speak of the linear independence of a set (it is a particular
case of the linear independence of a family: namely, the family $(s)_{s \in
S}$, where $S$ is the set in question), but \textit{most of the time} when
people say ``the set $\{x_{1}, \ldots, x_{n}\}$ is linearly independent'',
they really mean ``the tuple $(x_{1}, \ldots, x_{n})$ is linearly
independent''. This sort of language often causes statements to mean different
things than the author meant to say. For example, on page 328, in the proof of
Proposition 7.3.4, you say that the powers $1, \alpha, \alpha^{2}, \alpha^{3},
\ldots$ cannot be linearly independent over $K$. You definitely mean that the
family $(\alpha^{n})_{n \geq0}$ is not linearly independent, rather than the
set $\{1, \alpha, \alpha^{2}, \alpha^{3}, \ldots\}$. (The latter would be
false for $\alpha= 1$, because in this case the set $\{1, \alpha, \alpha^{2},
\alpha^{3}, \ldots\} = \{1\}$ is linearly independent.) On page 384, in
Exercise 8.4.1 (b), the set $\{v_{1}, v_{2}, \ldots, v_{n}\}$ does not change
if we duplicate some of the $v_{i}$'s, but the claim of the exercise becomes
wrong. In Example 8.5.4, you speak of ``the set of $n+1$ elements $\{v, T(v),
T^{2}(v), T^{3}(v), \ldots, T^{n}(v)\}$'', but this doesn't have to be a set
of $n+1$ elements; it is a \textit{tuple} of $n+1$ elements. On page 398, you
say that ``the $n^{2}+1 $ linear transformations $\operatorname{id}, T, T^{2},
\ldots, T^{n^{2}}$ are not linearly independent'', and again this only holds
if you take their tuple rather than their set. (On the other hand, Proposition
9.5.7 is one of the few places where it really matters that the set is a set.
Here I'd replace ``collection'' by ``set'' to stress this.)

\item \textbf{Page 375, Lemma 8.4.1:} Here you should assume that $R$ is not
the zero ring. Otherwise, the zero ring over itself is a counterexample,
having a basis of any size (as I said, bases are naturally tuples or families,
not sets; but even if you restrict yourself to sets, then there is still a
basis of size $0$ and a basis of size $1$). In
\href{https://doi.org/10.1090/S0002-9939-1988-0954974-5}{his paper
\textit{Nontrivial uses of trivial rings} (Proc. Amer. Math. Soc. \textbf{103}
(1988), pp. 1012--1014)}, Fred Richman makes a highly convincing case that the
``right'' way to state the correct lemma is to say that if $R$ is a
commutative ring with unity, and $M$ is an $R$-module equipped with two bases
of distinct finite cardinalities, then $R = 0$.

\item \textbf{Page 387, Example 8.5.3:} Replace ``there is an $n \in
\mathbb{Z}$'' by ``there is a nonzero $n \in\mathbb{Z}$''.

\item \textbf{Page 448, Definition 9.6.5:} Algebraic independence means that
there is no \emph{nonzero} polynomial $f\in K[x_{1},\ldots,x_{n}]$ satisfying
$f(u_{1},\ldots,u_{n})=0$. The zero polynomial always gives such a relation,
so the word ``nonzero'' is essential.

Also, it is better to define this notion over rings instead of fields; i.e.,
let $K$ be a commutative ring with unity, and let $u_{1}, \ldots, u_{n}$ lie
in a commutative $K$-algebra with unity. This way, the first statement in
Theorem 9.6.6 applies to every commutative ring $K$. (Also, again, I am not
keen on speaking of ``algebraically independent sets'' as opposed to
``algebraically independent families''.)

\item \textbf{Page 449:} Your definition of partitions misses one ingredient:
two partitions that only differ in trailing zeroes should be identified. For
example, $(3, 2, 2) = (3, 2, 2, 0) = (3, 2, 2, 0, 0) = \cdots$. Otherwise, the
conjugate partition $\lambda^{\ast}$ would not be uniquely defined (or would
fail to satisfy $(\lambda^{\ast})^{\ast}= \lambda$) due to too much freedom in
choosing its length.

\item \textbf{Page 450, definition of the monomial symmetric function
$m_{\lambda}$:} The formula
\[
m_{\lambda}=(1/f)\sum_{\sigma\in S_{n}}\sigma(x^{\lambda})
\]
is not defined over an arbitrary field $K$ when the stabilizer size $f$ is
zero in $K$. Instead, $m_{\lambda}$ should be defined as
\[
m_{\lambda}=\sum_{u\in S_{n}\cdot x^{\lambda}}u,
\]
the sum of the distinct monomials in the orbit of the monomial $x^{\lambda}$
under the symmetric group $S_{n}$ permuting the variables.
%(equivalently, sum over a set
%of coset representatives for the stabilizer).


\item \textbf{Pages 450--451, Lemma 9.6.7 (c):} The inverse coefficients
$S_{\lambda\mu}$ need not be nonnegative. For example,
\[
m_{(2)}=\epsilon_{1}^{2}-2\epsilon_{2}.
\]
Replace ``nonnegative integer'' in part (c) by ``integer''. The proof by
inverting a unitriangular integer matrix establishes integrality, not positivity.

\item \textbf{Page 450, proof of Lemma 9.6.7:} ``number of $\lambda_{k}$''
should be ``number of $k$'' (we must count the $k$'s, not the $\lambda_{k}$'s;
for example, if $\lambda_{2} = \lambda_{3}$, then both $2$ and $3$ are counted).

\item \textbf{Page 453, Exercise 9.6.15:} This needs a requirement that
$\operatorname{char} K \neq2$. (Alternatively, retain the condition
$\sigma(f)=\epsilon(\sigma)f$ and additionally require $f$ to vanish whenever
two of its indeterminates are set equal. With these two conditions, the
conclusion holds in every characteristic.)

\item \textbf{Page 461, beginning of Section 9.8:} \emph{Missing hypothesis.}
The substitution $x=y-a/4$ and formulas (9.8.3) require at least
$\operatorname{char}K\neq2$. This assumption should be made before the
substitution, rather than only later in the discussion.

\item \textbf{Page 461, formula (9.8.2):} Replace both ``$x$''s on the
right-hand side of (9.8.2) by $y$'s. That is, the formula should be
\[
f(x)=g(y)=y^{4}+py^{2}+qy+r.
\]


\item \textbf{Page 462, Case 1A:} If the resolvent cubic $h$ is irreducible
and its discriminant is a square, then its splitting field has degree $3$, not
$6$. Indeed its transitive Galois group is a subgroup of $A_{3}$, hence is
$A_{3}\cong C_{3}$. Thus the argument should say that $3$ divides $|G|$, which
still distinguishes $A_{4}$ from the Klein four group and yields $G=A_{4}$.

\item \textbf{Page 463, Lemma 9.8.1, first two sentences:} The variables are
inconsistent. Write, for example,
\[
g(x)=x^{4}+px^{2}+qx+r
\]
and let $h(y)$ denote its resolvent cubic (or consistently use $y$ for the
quartic variable throughout), rather than writing $g(x)=y^{4}+py^{2}+qy+r$.

\item \textbf{Page 479, alternative proof of simplicity of $A_{n}$:}
\emph{Substantive terminology and notation.} The stabilizer of a subgroup $N$
under conjugation is its \emph{normalizer}, not its centralizer. Thus
``centralizer'' should be replaced by ``normalizer'' throughout this page, and
``$\operatorname{Cent}_{S_{n}}(N)$'' should be likewise replaced by
``$N_{S_{n}}(N)$''.

\item \textbf{Page 485, Lemma 10.6.3:} This is false unless the extension
$L/K$ is separable. (The classical example $K = \mathbb{F}_{p}\left(
x^{p}\right)  $ and $L = \mathbb{F}_{p}\left(  x\right)  $ of an inseparable
finite extension is a counterexample.)

\item \textbf{Pages 485--487, proof of Theorem 10.6.2:} Even aside from the
problem with Lemma 10.6.3 noted above, this proof does not work in arbitrary
positive characteristic. If $p=\operatorname{char}K$ divides $n=n_{1}%
n_{2}\cdots n_{r}$, then no primitive $n$-th root of unity exists in any
extension field of $K$. Thus the instruction to adjoin such a root cannot be
carried out.
%Parts of the proof do work if a primitive $n$-th root of unity is
%understood in the weaker sense ``an element $u$ of the field such that
%all roots of $x^n - 1$ \textbf{in a splitting field} of the polynomial
%$x^n - 1$ are powers of $u$''.
%(This requires a slight generalization to Lemma 10.5.1 and
%Proposition 10.5.2, because as these two facts are stated, they require
%$n$ to be relatively prime to the characteristic of $K$.)


\item \textbf{Page 492, definition of the Euclidean norm:} The displayed
formula
\[
\|\boldsymbol{a}\|=\langle\boldsymbol{a},\boldsymbol{a}\rangle=\sum_{i}
a_{i}^{2}%
\]
should read
\[
\|\boldsymbol{a}\|^{2}=\langle\boldsymbol{a},\boldsymbol{a}\rangle=\sum_{i}
a_{i}^{2}, \qquad\text{or equivalently}\qquad\|\boldsymbol{a}\|=\sqrt
{\langle\boldsymbol{a},\boldsymbol{a}\rangle}.
\]


\item \textbf{Page 493, proof of Lemma 11.1.1:} Both displayed equalities have
the wrong sign in front of the inner product. They should be
\[
\|\boldsymbol{a}-\boldsymbol{b}\|^{2} =\|\boldsymbol{a}\|^{2}+\|\boldsymbol{b}%
\|^{2} -2\langle\boldsymbol{a},\boldsymbol{b}\rangle
\]
and the analogous identity for $\tau(\boldsymbol{a}),\tau(\boldsymbol{b})$.
The desired conclusion still follows after making this correction.

\item \textbf{Page 497, Lemma 11.1.9:} The formula ``$\sigma\tau
_{\boldsymbol{b}} \sigma^{-1} = \tau_{\sigma(\boldsymbol{b})}$'' should
instead be
\[
\sigma\tau_{\boldsymbol{b}}\sigma^{-1}=\tau_{\boldsymbol{A}\boldsymbol{b}},
\qquad\text{ where } \sigma\text{ is given by } \sigma(\boldsymbol{x}) =
\boldsymbol{A} \boldsymbol{x} + \boldsymbol{c}.
\]


\item \textbf{Page 501, proof of Theorem 11.2.6:} The induction begins with
graphs having exactly one edge, but the theorem also includes the connected
graph with one vertex and no edges. Add this base case: $v=1$, $e=0$, $f=1$,
so $v-e+f=2$.

The tree step also invokes the existence of a vertex of valence $1$, a fact
deferred to Exercise 11.2.4; it would be clearer to state this as a preceding lemma.

\item \textbf{Page 502, proof of Euler's theorem for polyhedra:} If
``polyhedron'' denotes the solid convex body, the radial projection
$\boldsymbol{x}\mapsto\boldsymbol{x}/\|\boldsymbol{x}\|$ does not map the
whole polyhedron bijectively onto the sphere. It maps the \emph{boundary} of a
convex polyhedron containing the origin bijectively onto the sphere. Replace
``maps the polyhedron bijectively'' by ``maps the boundary of the polyhedron
bijectively'' (or state explicitly that ``polyhedron'' here means its boundary surface).

\item \textbf{Page 550:} The notation $M^{j}$ for the $j$-th column of a
matrix $M$ is rather confusing, even if the matrix $M$ is (in general)
rectangular and thus has no $j$-th power.

\item \textbf{Page 551, Definition E.9:} Replace ``$\{T(\boldsymbol{x}%
):\boldsymbol{x}\in K^{n}\}$'' by ``$\{T(\boldsymbol{x}):\boldsymbol{x}\in
V\}$''.
\end{enumerate}

\section{Minor typographical and editorial corrections}

\begin{enumerate}
\item \textbf{Page 9, paragraph following Figure 1.3.5:} Replace ``in this
table, order in which symmetries are multiplied'' by ``in this table, the
order in which symmetries are multiplied''.

\item \textbf{Page 16:} Replace ``Let $\tau$ be symmetry'' by ``Let $\tau$ be
a symmetry'', and replace ``the third in the place of the first; There are''
by ``the third in the place of the first; there are''.

\item \textbf{Page 19:} Replace \textquotedblleft sends $1$ to $4$, $2$, to
$3$\textquotedblright\ by \textquotedblleft sends $1$ to $4$, $2$ to
$3$\textquotedblright.

\item \textbf{Page 25, first paragraph:} Replace \textquotedblleft probably
most familiar algebraic system\textquotedblright\ by \textquotedblleft
probably the most familiar algebraic system\textquotedblright.

\item \textbf{Page 26, first paragraph:} Replace \textquotedblleft every
natural numbers has\textquotedblright\ by \textquotedblleft every natural
number has\textquotedblright.

\item \textbf{Page 27, proof of Theorem 1.6.6:} Replace \textquotedblleft We
show than\textquotedblright\ by \textquotedblleft We show
that\textquotedblright.

\item \textbf{Page 28, Proposition 1.6.7:} Replace \textquotedblleft such
$a=qd+r$\textquotedblright\ by \textquotedblleft such that $a=qd+r$%
\textquotedblright.

\item \textbf{Page 32, paragraph after proof of Proposition 1.6.18:}
\textquotedblleft it is a only a short way\textquotedblright\ should be
\textquotedblleft it is only a short way\textquotedblright.

\item \textbf{Page 35, paragraph after proof of Proposition 1.6.24:} Replace
\textquotedblleft integers $s_{1},s_{2},\ldots,s_{n}$ such\textquotedblright%
\ by \textquotedblleft integers $s_{1},s_{2},\ldots,s_{n}$ such
that\textquotedblright.

\item \textbf{Page 41, Proposition 1.7.7 (f):} Replace \textquotedblleft The
distributive law hold\textquotedblright\ by \textquotedblleft The distributive
law holds\textquotedblright.

\item \textbf{Page 42, Example 1.7.8 (b):} Replace \textquotedblleft the
conjugacy class of $237$ modulo $3$\textquotedblright\ by \textquotedblleft
the congruence class of $237$ modulo $3$\textquotedblright.

\item \textbf{Page 44, Exercise 1.7.16:} Replace \textquotedblleft Find and
integer $x$\textquotedblright\ by \textquotedblleft Find an integer
$x$\textquotedblright.

\item \textbf{Page 45, paragraph following Example 1.8.1:} Replace
\textquotedblleft$K$ can be regarded as subset of $K[x]$\textquotedblright\ by
\textquotedblleft$K$ can be regarded as a subset of $K[x]$\textquotedblright.

\item \textbf{Page 55, Exercise 1.8.12:} There is an extraneous closing
parenthesis at the end of this exercise.

\item \textbf{Page 61, proof of Lemma 1.9.8:} Replace \textquotedblleft%
$(n-k)!$\textquotedblright\ by \textquotedblleft$(p-k)!$\textquotedblright.

\item \textbf{Page 62, Inclusion--Exclusion:} \textquotedblleft but then
uncounted once in $\left\vert A\cap B\right\vert -\left\vert A\cap
C\right\vert -\left\vert B\cap C\right\vert $\textquotedblright\ should be
\textquotedblleft but then uncounted once in $-\left\vert A\cap B\right\vert
-\left\vert A\cap C\right\vert -\left\vert B\cap C\right\vert $%
\textquotedblright.

\item \textbf{Page 64, Example 1.9.13:} Replace \textquotedblleft%
$A_{1}^{\prime}\cap A_{2}\cap\cdots\cap A_{n}^{\prime}$\textquotedblright\ by
\textquotedblleft$A_{1}^{\prime}\cap A_{2}^{\prime}\cap\cdots\cap
A_{n}^{\prime}$\textquotedblright\ (mind the prime over the $A_{2}$).

\item \textbf{Page 65, simplification of $D_{n}$:} Replace \textquotedblleft%
$D_{n}=\ =$\textquotedblright\ by \textquotedblleft$D_{n}=$\textquotedblright.

\item \textbf{Page 65, Example 1.9.14:} Replace \textquotedblleft$D_{10}%
=1{,}333{,}961$\textquotedblright\ by \textquotedblleft$D_{10}=1{,}334{,}%
961$\textquotedblright. (See \href{https://oeis.org/A000166}{the OEIS} for
more values of $D_{n}$.)

\item \textbf{Page 66, computation of $\varphi(n)$:} Replace \textquotedblleft%
$A_{1}^{\prime}\cap A_{2}^{\prime}\cap\cdots\cap A_{n}^{\prime}$%
\textquotedblright\ by \textquotedblleft$A_{1}^{\prime}\cap A_{2}^{\prime}%
\cap\cdots\cap A_{s}^{\prime}$\textquotedblright.

\item \textbf{Page 66, proof of Corollary 1.9.17:} Replace \textquotedblleft
the number of natural numbers $j\leq n$\textquotedblright\ by
\textquotedblleft the number of natural numbers $j\leq p^{k}$%
\textquotedblright.

\item \textbf{Page 67, Example 1.9.21:} Replace
\[
7^{8}-1=5764801
\]
by
\[
7^{8}-1=5764800.
\]


\item \textbf{Page 68, Exercise 1.9.8 (b):} Replace \textquotedblleft In how
many ways the men and women form pairs\textquotedblright\ by \textquotedblleft
In how many ways can the men and women form pairs\textquotedblright.

\item \textbf{Pages 71 and 74, comparison of }$\mathcal{R}$\textbf{\ with the
fourth roots of unity:} Replace both occurrences of \textquotedblleft%
$H$\textquotedblright\ by \textquotedblleft$C_{4}$\textquotedblright.

\item \textbf{Page 72, homomorphism paragraph:} Replace \textquotedblleft if
$f$ take products to products\textquotedblright\ by \textquotedblleft if $f$
takes products to products\textquotedblright.

\item \textbf{Page 73, Exercises 1.10:} Replace \textquotedblleft Show that
set of symmetries\textquotedblright\ by \textquotedblleft Show that the set of
symmetries\textquotedblright, and replace \textquotedblleft1.10.3. .
Consider\textquotedblright\ by \textquotedblleft1.10.3.
Consider\textquotedblright.

\item \textbf{Page 74, Exercise 1.10.3:} Again, replace \textquotedblleft%
$H$\textquotedblright\ by \textquotedblleft$C_{4}$\textquotedblright.

\item \textbf{Page 75, first paragraph of Section 1.11:} Replace
\textquotedblleft the set is group under the operation of
addition\textquotedblright\ by \textquotedblleft the set is a group under the
operation of addition\textquotedblright.

\item \textbf{Page 76:} Replace \textquotedblleft with with polynomial
entries\textquotedblright\ by \textquotedblleft with polynomial
entries\textquotedblright, and replace \textquotedblleft\textit{inverses}; A
multiplicative inverse\textquotedblright\ by \textquotedblleft%
\textit{inverses}: A multiplicative inverse\textquotedblright.

\item \textbf{Page 78, Proposition 1.11.7:} Replace \textquotedblleft$Z_{b}%
$\textquotedblright\ by \textquotedblleft$\mathbb{Z}_{b}$\textquotedblright.
Make the same change in the proof of the proposition.

\item \textbf{Page 78, proof of Proposition 1.11.7:} Replace \textquotedblleft
we have to check that If\textquotedblright\ by \textquotedblleft we have to
check that if\textquotedblright.

\item \textbf{Page 79, Exercise 1.11.2:} Replace \textquotedblleft the only
units are the constant polynomials\textquotedblright\ by \textquotedblleft the
only units are the nonzero constant polynomials\textquotedblright.

\item \textbf{Page 79, Exercise 1.11.3:} Replace the mismatched quotation
marks around \textquotedblleft polynomial\textquotedblright\ by matching
quotation marks.

\item \textbf{Page 86, Proposition 2.1.4:} Insert a period after the formula
$(ab)^{-1}=b^{-1}a^{-1}$.

\item \textbf{Page 90, discussion of the general associative law:} Replace
\textquotedblleft three different product of four elements\textquotedblright%
\ by \textquotedblleft three different products of four
elements\textquotedblright, and replace \textquotedblleft four or less
elements\textquotedblright\ by \textquotedblleft four or fewer
elements\textquotedblright. In the list of four candidate products of five
elements, insert a comma after $(ab)(cde)$.

\item \textbf{Page 91:} Replace \textquotedblleft no more that $n-1$
elements\textquotedblright\ by \textquotedblleft no more than $n-1$
elements\textquotedblright.

\item \textbf{Page 91, Exercise 2.1.3:} delete the redundant sentence
\textquotedblleft Suppose $e^{\prime}$ and $g$ are elements of $G$%
\textquotedblright.

\item \textbf{Page 95, proof of Example 2.2.4:} Replace \textquotedblleft the
set of complex number of modulus $1$\textquotedblright\ by \textquotedblleft
the set of complex numbers of modulus $1$\textquotedblright.

\item \textbf{Page 96:} Replace \textquotedblleft The family of subgroups of a
group $G$ are partially ordered\textquotedblright\ by \textquotedblleft The
family of subgroups of a group $G$ is partially ordered\textquotedblright.

\item \textbf{Page 96, proof of Proposition 2.2.9:} Replace \textquotedblleft
is always is closed\textquotedblright\ by \textquotedblleft is always
closed\textquotedblright.

\item \textbf{Page 97, proof of Example 2.2.13:} Replace \textquotedblleft
root or unity\textquotedblright\ by \textquotedblleft root of
unity\textquotedblright.

\item \textbf{Page 101, Example 2.2.27:} Replace \textquotedblleft The
inclusion relations among the subgroups of $\mathbb{Z}_{12}$ is
pictured\textquotedblright\ by \textquotedblleft The inclusion relations among
the subgroups of $\mathbb{Z}_{12}$ are pictured\textquotedblright.

\item \textbf{Page 105, Exercise 2.2.11:} Replace \textquotedblleft the
product of a $2$-cycle and a $3$-cycle (disjoint) is $6$%
,while\textquotedblright\ by \textquotedblleft the order of the product of a
$2$-cycle and a $3$-cycle disjoint from it is $6$, while\textquotedblright.

\item \textbf{Page 105, Exercise 2.2.23:} \textquotedblleft therein
$\mathbb{Z}_{200000}$\textquotedblright\ should be \textquotedblleft there in
$\mathbb{Z}_{200000}$\textquotedblright.

\item \textbf{Page 113, Example 2.4.7:} Replace \textquotedblleft the cyclic
subgroup of $G$ generated by $g$\textquotedblright\ by \textquotedblleft the
cyclic subgroup of $G$ generated by $a$\textquotedblright.

\item \textbf{Page 117, paragraph preceding Definition 2.4.21:} Replace
\textquotedblleft there is a homomorphisms\textquotedblright\ by
\textquotedblleft there is a homomorphism\textquotedblright.

\item \textbf{Page 118, proof of Corollary 2.4.24:} Replace \textquotedblleft
a $k$ cycle\textquotedblright\ by \textquotedblleft a $k$%
-cycle\textquotedblright.

\item \textbf{Page 119, Exercise 2.4.10:} Insert a period after
\textquotedblleft for all $\sigma$ and $\tau\in S_{n}$\textquotedblright.

\item \textbf{Page 119, Exercise 2.4.14:} Replace \textquotedblleft$k$
cycle\textquotedblright\ and \textquotedblleft$k$ cycles\textquotedblright\ by
\textquotedblleft$k$-cycle\textquotedblright\ and \textquotedblleft%
$k$-cycles\textquotedblright, respectively.

\item \textbf{Page 120, Exercise 2.4.16:} Replace \textquotedblleft Show that
two answers always agree\textquotedblright\ by \textquotedblleft Show that the
two answers always agree\textquotedblright.

\item \textbf{Page 121, Definition 2.5.1:} Replace \textquotedblleft Let $H$
be subgroup of a group $G$\textquotedblright\ by \textquotedblleft Let $H$ be
a subgroup of a group $G$\textquotedblright.

\item \textbf{Page 127, first paragraph of Section 2.6:} Replace
\textquotedblleft$a\sim b\operatorname{mod}H)$\textquotedblright\ by
\textquotedblleft$a\sim b\operatorname{mod}H$\textquotedblright, and replace
\textquotedblleft if, and only if, if $b^{-1}a\in H$\textquotedblright\ by
\textquotedblleft if, and only if, $b^{-1}a\in H$\textquotedblright\ (in the
same sentence).

\item \textbf{Page 128, Example 2.6.3 (g):} Replace \textquotedblleft an
bijective map\textquotedblright\ by \textquotedblleft a bijective
map\textquotedblright.

\item \textbf{Page 129, last paragraph:} Replace \textquotedblleft we can
define an relation on $X$\textquotedblright\ by \textquotedblleft we can
define a relation on $X$\textquotedblright.

\item \textbf{Page 131, Example 2.6.10 (e):} Replace \textquotedblleft an
bijective map\textquotedblright\ by \textquotedblleft a bijective
map\textquotedblright.

\item \textbf{Page 132, Proposition 2.6.13:} Replace \textquotedblleft
determine the same equivalence relation $X$\textquotedblright\ by
\textquotedblleft determine the same equivalence relation on $X$%
\textquotedblright.

\item \textbf{Page 133, paragraph preceding Definition 2.6.14:} Replace ``a
canonical surjective map on $G$'' by ``a canonical surjective map from $G$''.

\item \textbf{Page 135, proof of Theorem 2.7.1:} Replace ``the definition of
the product on $G/H$ makes sense'' by ``the definition of the product on $G/N
$ makes sense''.

\item \textbf{Pages 136--137, Example 2.7.4:} Replace \textquotedblleft in the
same coset modulo $\mathbb{Z}$\textquotedblright\ by \textquotedblleft in the
same coset of $\mathbb{Z}$ in $\mathbb{R}$\textquotedblright. Replace each
occurrence of \textquotedblleft cosets of $\mathbb{R}$ modulo $\mathbb{Z}%
$\textquotedblright\ by \textquotedblleft cosets of $\mathbb{Z}$ in
$\mathbb{R}$\textquotedblright. On page 137, replace \textquotedblleft
bijections between set $\mathbb{R}/\mathbb{Z}$\textquotedblright\ by
\textquotedblleft bijections between the set $\mathbb{R}/\mathbb{Z}%
$\textquotedblright.

\item \textbf{Page 138, Example 2.7.5:} Replace \textquotedblleft cosets of
$\operatorname{Aff}(n)$ modulo $N$\textquotedblright\ by \textquotedblleft
cosets of $N$ in $\operatorname{Aff}(n)$\textquotedblright.

\item \textbf{Page 141, Example 2.7.12:} Replace \textquotedblleft
Morover\textquotedblright\ by \textquotedblleft Moreover\textquotedblright,
and replace \textquotedblleft homorphism theorem\textquotedblright\ by
\textquotedblleft homomorphism theorem\textquotedblright.

\item \textbf{Page 144, Example 2.7.18:} Replace ``The cyclic subgroup group
generated by $[a]$'' by ``The cyclic subgroup generated by $[a]$''.

\item \textbf{Page 146, Example 2.7.21:} Replace ``then we have $X=\lambda
X^{\prime}$'' by ``Then we have $X=\lambda X^{\prime}$''.

\item \textbf{Page 147, Exercise 2.7.6 (a):} Replace ``Show that
$\operatorname{Aut}(G)$ of $G$ is also a group'' by ``Show that
$\operatorname{Aut}(G)$ is also a group''.

\item \textbf{Page 157, Exercise 3.1.2:} ``$\{e_{a}\}$'' should be
``$\{e_{A}\}$''.

\item \textbf{Page 158, Exercise 3.1.7:} ``Show that $\pi_{i}$ surjective
homomorphism'' should be ``Show that $\pi_{i}$ is a surjective homomorphism''.

\item \textbf{Page 164:} ``The \emph{kernel} of linear transformation'' should
be ``The \emph{kernel} of a linear transformation''. Also, in Lemma 3.3.3,
``then'' at the beginning of the second sentence should be capitalized.

\item \textbf{Page 165, quotient-vector-space paragraph:} Delete the repeated
word in ``check that this this is well-defined'', and replace ``if
$v+W=v^{\prime}+W$ and, then'' by ``if $v+W=v^{\prime}+W$, then''.

\item \textbf{Page 166:} ``straighforward'' should be ``straightforward'', and
``indclude'' should be ``include''.

\item \textbf{Page 169, discussion of the empty set:} The statement that the
empty set is linearly independent because there are no sequences of its
elements is correct if ``sequence'' means a sequence of positive length, as in
the book's convention. If sequences of length $0$ are allowed (which is the
standard convention in mathematics), there is exactly one such sequence,
namely the empty sequence; its coefficient vector is the unique element of
$K^{0}$, so the conclusion about linear independence remains correct.

\item \textbf{Page 172, Proposition 3.3.25:} ``Let $V$ a finite dimensional
vector space'' should be ``Let $V$ be \ldots''.

\item \textbf{Page 172, proof of Corollary 3.3.26:} ``Propostion'' should be ``Proposition''.

\item \textbf{Page 174:} ``several vectors spaces'' should be ``several vector spaces''.

\item \textbf{Page 178:} ``linear maps from $V$ and $W$'' should be ``linear
maps from $V$ to $W$'', and ``preceeding'' should be ``preceding''.

\item \textbf{Pages 179--180, dual-basis discussion and Proposition 3.4.2
(b):} Replace both occurrences of ``is a a basis'' by ``is a basis''.

\item \textbf{Page 183:} ``subpace'' should be ``subspace'', and the stray
period after ``Corollary 3.4.9'' should be deleted.

\item \textbf{Page 187:} ``Similarly is an equivalence relation'' should be
``Similarity is an equivalence relation''.

\item \textbf{Page 190:} ``ususal'' should be ``usual''.

\item \textbf{Page 194:} ``post--mulitplication'' should be
``post--multiplication''.
%and
%``$d_i0s$'' should be ``the $d_i$'s''.


\item \textbf{Page 195:} ``it's own inverse'' should be ``its own inverse''.

\item \textbf{Page 195, Smith-normal-form algorithm:} ``If there is a element
$\beta$'' should be ``If there is an element $\beta$''.

\item \textbf{Page 198, proof of Theorem 3.5.13:} ``a basis of $Z^{n}$''
should be ``a basis of $\mathbb{Z}^{n}$''.

\item \textbf{Pages 202--204:} Replace ``is a also an $\mathbb{Z}_{p}$-vector
space'' by ``is also a $\mathbb{Z}_{p}$-vector space''; replace ``an
$\mathbb{Z}_{p}$-vector space'' by ``a $\mathbb{Z}_{p}$-vector space'';
replace ``irreducible'' by ``prime''; and replace ``occuring in an prime
factorization'' by ``occurring in a prime factorization''.

\item \textbf{Page 209:} ``We have already have observed'' should be ``We have
already observed'', and ``decompostion'' should be ``decomposition''.

\item \textbf{Page 212, Remark 3.6.26:} ``the group of units of $K^{*}$''
should be ``the multiplicative group $K^{*}$'' or ``the group of units of $K$''.

\item \textbf{Page 213:} Replace both references to ``Lemma 3.6.25'' by
references to ``Theorem 3.6.25''.

\item \textbf{Page 216, first paragraph:} The paper models are in Appendix F,
not Appendix E. Also replace ``patterns for making paper model'' by ``patterns
for making paper models''.

\item \textbf{Page 239, first paragraph of Section 4.5:} Delete one occurrence
of ``by'' in ``implemented by by a linear isometry''.

\item \textbf{Page 243, Definition 5.1.5:} The two proposed formulations of
transitivity are equivalent only when $X$ is nonempty. If $X=\varnothing$,
there are no orbits, whereas ``for any two elements $x,x^{\prime}\in X$'' is
vacuously true. Add the hypothesis $X\neq\varnothing$, or choose one of the
two formulations as the definition.

\item \textbf{Page 245, Definition 5.1.16:} Replace ``$\operatorname{Cent}%
_{G}(x)$'' by ``$\operatorname{Cent}_{G}(g)$''.

\item \textbf{Page 253, Remark 5.3.2:} ``The rather unexpected answer that it
is true'' should be ``The rather unexpected answer is that it is true''.

\item \textbf{Page 254, Exercise 5.3.5:} ``determined by rational $2$-by-$2$
matrix'' should be ``determined by a rational $2$-by-$2$ matrix''.

\item \textbf{Page 258, Theorem 5.4.9:} ``there is a $a\in G$'' should be
``there is an $a\in G$''.

\item \textbf{Page 265, proof of Lemma 5.5.1:} ``the intersections of the
stabilizers'' should be ``the intersection of the stabilizers''.

\item \textbf{Page 265, Remark 5.5.2:} ``$A_{n}$ is the only nontrivial normal
subgroup of $S_{n}$'' should read ``$A_{n}$ is the only nontrivial
\emph{proper} normal subgroup of $S_{n}$'', since $S_{n}$ itself is also a
nontrivial normal subgroup.

\item \textbf{Page 284, Proposition 6.2.27 (b):} ``interesection'' should be ``intersection''.

\item \textbf{Page 289, quotient-ring discussion:} Delete one occurrence of
``this'' in ``check that this this is well defined'', and insert a space in
``degree of $f$,the product''.

\item \textbf{Page 305, Lemma 6.5.18:} ``properties of an nonzero nonunit
element'' should be ``properties of a nonzero nonunit element''.

\item \textbf{Page 310, Corollary 6.6.5 and Example 6.6.6:} Replace ``unique
factorization domain domain'' by ``unique factorization domain'', ``an
nonzero'' by ``a nonzero'', and ``In an UFD'' by ``In a UFD''.

\item \textbf{Page 355, Definition 8.1.26:} The subscript ``$s$'' on the left
hand side of the first equality should be ``$n$''.

\item \textbf{Page 363, Lemma 8.3.3:} ``acts $S_{n}$'' should be ``$S_{n}$ acts''.

\item \textbf{Page 363, proof of Lemma 8.3.3:} In the long computation,
replace \newline``$x_{\sigma(\tau(1))}, \ldots, x_{\sigma(\tau(1))}$'' by
``$x_{\sigma(\tau(1))}, \ldots, x_{\sigma(\tau(n))}$''.

\item \textbf{Page 375, proof of Lemma 8.4.1:} In ``Each $w_{j}$ has a unique
expression as an $R$\^{a}\euro``linear combination of the basis elements
$v_{j}$'', it would be better to replace``$v_{j}$'' by ``$v_{i}$''. Replace
``as an $R$\^{a}\euro``linear combinations'' by ``as an $R$\^{a}\euro``linear
combination''. In the formula for $v_{j}$ (between (8.4.1) and (8.4.2)),
replace ``$b_{n,j}w_{n}$'' by ``$b_{m,j}w_{m}$''. Finally, in the last
sentence of the proof, ``two basis'' should be ``two bases''.

\item \textbf{Page 379, definition of length:} The period before ``where $u$
is a unit and the $p_{i}$'s are irreducibles'' should be a comma.

\item \textbf{Page 382, proof of Lemma 8.4.11:} ``as an $R$\^{a}\euro``linear
combinations'' should be ``as an $R$\^{a}\euro``linear combination''.

\item \textbf{Page 399, paragraph after the block-diagonal display:} Delete
one occurrence of ``subspace'' in ``the invariant subspace subspace $V_{i}$''.

\item \textbf{Page 410, Definition 8.6.14:} ``an \textit{nonzero} vector''
should be ``a \textit{nonzero} vector''.

\item \textbf{Page 417, Definition 8.7.7:} ``similar $A$'' should be ``similar
to $A$''.

\item \textbf{Page 435, proof of Proposition 9.4.4:} There is a closing
parenthesis too much in ``$k_{n}\sigma(\alpha^{n}))$''.

\item \textbf{Page 453, Exercise 9.6.15:} The ``g'' should be a mathmode ``$g
$''.

\item \textbf{Page 457:} Replace ``$(b_{j}/b_{n})$'' by ``$(b_{j}/b_{m})$''
(shortly after (9.7.2)). In the next sentence, replace ``and the total degree
as a polynomial in the $(b_{j}/b_{m})$ is $m$'' by ``and the total degree as a
polynomial in the $(b_{j}/b_{m})$ is $n$''.

\item \textbf{Page 466, proof of Lemma 9.8.5:} ``By by the Galois
correspondence'' should be ``By the Galois correspondence''.

\item \textbf{Page 498, Exercise 11.1.1:} ``there is an most one point''
should be ``there is at most one point''.

\item \textbf{Page 547, Definitions E.1 and E.2:} ``A linear combination of
set $S$'' should be ``A linear combination of a set $S$'', and ``A set $S$
vectors'' should be ``A set $S$ of vectors''.

\item \textbf{Page 548, first sentence:} ``a linear independent set'' should
be ``a linearly independent set''.

\item \textbf{Page 548, second sentence:} ``The empty set is linearly
independent, since there are no sequences of its elements''. There are! But
only the empty one, so the conclusion still holds.

\item \textbf{Page 550, first paragraph after Definition E.6:} The map is
printed as $x\mapsto=Mx$; delete the extraneous equals sign.

\item \textbf{Page 551, first line:} The identity matrix $E_{n}$ acts on
$K^{n}$, not on the undefined $K^{N}$.

\item \textbf{Page 551, second paragraph:} ``$\operatorname{Hom}_{k}%
(K^{n},K^{m})$'' should be ``$\operatorname{Hom}_{K}(K^{n},K^{m})$''.

\item \textbf{Page 552, second paragraph:} Replace ``$T(\boldsymbol{x}) =
\sum\alpha_{i} T(\boldsymbol{e}_{i})$'' by ``$T(\boldsymbol{x}) = \sum
\alpha_{i} T(\widehat{\boldsymbol{e}}_{i})$''.

\item \textbf{Page 555, second paragraph:} ``A matrix has a left inverse''
should be ``A matrix $M$ has a left inverse''.

\item \textbf{Page 556, definition of an inner product:} The third property
``$\left<  \boldsymbol{x}, \boldsymbol{x}\right>  \geq0$ and $\left<
\boldsymbol{x}, \boldsymbol{x}\right>  = 0$ if, and only if $\boldsymbol{x} =
\boldsymbol{0}$'' is confusingly worded. Of course, what is meant is that
$\left<  \boldsymbol{x}, \boldsymbol{x}\right>  \geq0$ holds for each vector
$\boldsymbol{x}$, whereas $\left<  \boldsymbol{x}, \boldsymbol{x}\right>  = 0$
holds if and only if $\boldsymbol{x} = \boldsymbol{0}$.

\item \textbf{Page 556, Section E.3:} ``Cauchy--Schwartz'' should be
``Cauchy--Schwarz'' (twice).

\item \textbf{Page 557, first paragraph:} ``$\boldsymbol{v}_{i},\ldots
,\boldsymbol{v}_{s}$'' should be ``$\boldsymbol{v}_{1},\ldots,\boldsymbol{v}%
_{s}$''.

\item \textbf{Page 557, third paragraph:} ``$\{\boldsymbol{v}_{i}%
,\ldots,\boldsymbol{v}_{n}\}$'' should be ``$\{\boldsymbol{v}_{1}%
,\ldots,\boldsymbol{v}_{n}\}$''.

\item \textbf{Page 557, fourth paragraph:} Delete one occurrence of ``to'' in
``$\boldsymbol{w}_{j}$ is orthogonal to to $B_{j}$''.
\end{enumerate}


\end{document}