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\begin{document}

\begin{center}
\textbf{Abstract Algebra}

\textit{John A. Beachy and William D. Blair}

Fourth edition, Waveland Press, 2019

ISBN 978-1-4786-3869-8

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

\noindent The page numbers below refer to the printed page numbers in the
565-page PDF. This list is not claimed to be exhaustive. It combines
mathematical corrections, repairs to references and notation, and local
copyedits into a single list ordered by page. Broader bibliography, index, and
editorial recommendations appear separately at the end.

Almost all the errors in this list were found by GPT-5.6 Sol. I have verified
(and occasionally reworded) them.

\appendix\setcounter{section}{2}

\section{Corrections and comments}

\begin{enumerate}
\item \textbf{Page ix, Preface to the Second Edition:} Replace ``text books''
by ``textbooks''.

\item \textbf{Page xix, \textquotedblleft Writing Proofs\textquotedblright:}
After specializing Euclid's lemma to $p=2$, $a=n$, and $b=n$, the text still
concludes \textquotedblleft if $p$ is a factor of $n^{2}$, then $p$ must be a
factor of $n$\textquotedblright. Replace both occurrences of \textquotedblleft%
$p$\textquotedblright\ in that conclusion by \textquotedblleft$2$%
\textquotedblright.

\item \textbf{Page xx, \textquotedblleft Writing Proofs\textquotedblright:}
Replace \textquotedblleft descendents\textquotedblright\ by \textquotedblleft
descendants\textquotedblright\ for a more standard spelling.

\item \textbf{Page 7, Looking ahead:} In \textquotedblleft and again in
Chapter 5. when \ldots\textquotedblright, replace the period after
\textquotedblleft5\textquotedblright\ by a comma.

\item \textbf{Page 12, Example 1.1.5:} Replace \textquotedblleft%
$-5x+81y=0$\textquotedblright\ by \textquotedblleft$-5x+18y=0$%
\textquotedblright.

\item \textbf{Page 16, Exercise 23:} The exercise first says that $a$ and $b$
are not both zero, but then calls them \textquotedblleft nonzero
integers\textquotedblright. Delete \textquotedblleft nonzero\textquotedblright%
\ before \textquotedblleft integers\textquotedblright.

\item \textbf{Page 20, paragraph after proof of Theorem 1.2.7:} In
\textquotedblleft If $a=p_{1}^{\alpha_{1}}p_{2}^{\alpha_{2}}\cdots
p_{n}^{\alpha_{n}}$, then $b$ is a divisor of $a$\textquotedblright, add
\textquotedblleft a positive integer\textquotedblright\ before
\textquotedblleft$b$\textquotedblright. Moreover, it is worth clarifying that
the $\beta_{i}$ should be nonnegative integers, so in particular are allowed
to be $0$ (so $b$ does not have to use all prime factors of $a$).

\item \textbf{Page 22, Example 1.2.4, first paragraph of the proof:} The
number $2^{d}-1$ is called a \textquotedblleft proper nontrivial
divisor\textquotedblright\ of both $2^{dq}-1$ and $2^{dp}-1$. It need not be
proper when $q=1$ or $p=1$. Say instead that it is a common divisor greater
than $1$ (or merely a nontrivial common divisor).

\item \textbf{Page 23, paragraph before Proposition 1.2.10:} Replace
\textquotedblleft as in next proposition\textquotedblright\ by
\textquotedblleft as in the next proposition\textquotedblright.

\item \textbf{Page 26, Exercise 23:} The notion of \textquotedblleft twin
primes\textquotedblright\ has not been defined. It means two primes that have
a difference of $2$.

\item \textbf{Page 34, alternative proof of the Chinese remainder theorem:}
After correctly starting with $x\equiv a\pmod n$, the recap says
\textquotedblleft the first congruence $x\equiv a\pmod m$\textquotedblright.
Change the modulus in the recap from \textquotedblleft$m$\textquotedblright%
\ to \textquotedblleft$n$\textquotedblright.

\item \textbf{Page 37, Exercise 24, Hint:} Replace \textquotedblleft How you
can use the digits \ldots?\textquotedblright\ by \textquotedblleft How can you
use the digits \ldots?\textquotedblright

\item \textbf{Page 42, Corollary 1.4.6:} The hypothesis \textquotedblleft%
$n>0$\textquotedblright\ admits $n=1$. For $n=1$, conditions (2) and (3) hold
vacuously because $\mathbf{Z}_{1}$ has no nonzero elements, whereas condition
(1) fails because $1$ is not prime. Replace \textquotedblleft$n>0$%
\textquotedblright\ by \textquotedblleft$n>1$\textquotedblright, or strengthen
conditions (2) and (3) to disallow $\mathbf{Z}_{n}$ being trivial.

\item \textbf{Page 43, paragraph before Example 1.4.2:} What is called
\textquotedblleft zero divisor\textquotedblright\ here has been previously
called \textquotedblleft divisor of zero\textquotedblright\ in Definition
1.4.3. The terminology should be made uniform (ideally towards the more
commonly used \textquotedblleft zero divisor\textquotedblright).

\item \textbf{Page 43, paragraph before Example 1.4.2:} The claim that the set
of powers $[1],[a],[a]^{2},\ldots$ \textquotedblleft must contain fewer than
$n$ distinct elements\textquotedblright\ fails for $n=1$. Replace
\textquotedblleft fewer than $n$\textquotedblright\ by \textquotedblleft at
most $n$\textquotedblright\ (which is all that the repetition argument needs),
or assume $n>1$ here.

\item \textbf{Page 44, second paragraph:} The paragraph says that
\textquotedblleft the Chinese remainder theorem is used to show how to
determine the solutions modulo a prime power $p^{\alpha}$ (for integers
$\alpha\geq2$) from the solutions modulo $p$\textquotedblright. The Chinese
remainder theorem combines solutions for coprime moduli; lifting a solution
from $p$ to $p^{\alpha}$ is a separate problem and is not always possible
(Hensel's lemma is commonly used for this). Say that one first solves modulo
the relevant prime powers and then uses the Chinese remainder theorem to
combine those solutions.

\item \textbf{Page 46, second proof of Fermat's little theorem:} ``Writing $a$
as $(1+1+\cdots+1)$'' only directly covers nonnegative integers, although the
theorem is stated for every integer $a$. Choose a nonnegative representative
of $a\bmod p$, or add a sentence treating negative $a$.

\item \textbf{Page 49:} In ``the basic the basic building blocks'', delete one
occurrence of ``the basic''.

\item \textbf{Pages 54--55, Definition 2.1.1 and the following discussion:}
Definition 2.1.1 identifies a function with a subset $F\subseteq S\times T$,
but page 55 then distinguishes two functions with the same graph and domain
when their codomains differ. The subset $F$ alone does not encode its
codomain, so these two conventions are incompatible. If the latter convention
is intended, define a function as a triple $(S,T,F)$ consisting of the
specified domain $S$, the specified codomain $T$ and the graph $F$.

\item \textbf{Page 55:} Replace ``use the more familiar notation to their
definitions'' by ``use the more familiar notation to give their definitions''.

\item \textbf{Page 62, proof of Proposition 2.1.7:} Replace ``$f$ maps $g(y)$
onto $y$'' by ``$f$ maps $g(y)$ to $y$''. The word ``onto'' is being used in
its technical sense in the surrounding discussion, not as a substitute for ``to''.

\item \textbf{Page 65, Exercise 16:} For arbitrary sets, the implication from
``$f$ is onto'' to the existence of a function $g:B\to A$ satisfying $f\circ
g=1_{B}$ uses the axiom of choice; indeed, the assertion that every surjection
has a right inverse is equivalent to the axiom of choice. State that choice is
being assumed, or restrict the exercise (for example, to finite $B$).

\item \textbf{Page 72, factorization of a function after Theorem 2.2.7:}
\textquotedblleft for each $x\in X$\textquotedblright\ should be
\textquotedblleft for each $x\in S$\textquotedblright.

\item \textbf{Page 73, Definition 2.2.8:} \textquotedblleft Let $S\rightarrow
T$ be a function\textquotedblright\ should be \textquotedblleft Let
$f:S\rightarrow T$ be a function\textquotedblright.

\item \textbf{Page 75, Exercise 12(h)--(j):} Literally, the embedding
$f(x,y)=[x,y,1]$ does not take an affine line onto a projective line: it omits
the line's point at infinity. Consequently, images of parallel affine lines
remain disjoint as subsets, although their projective closures meet. Ask about
the \emph{projective closure} of each image, or explicitly say that the image
is a projective line with one point removed.

\item \textbf{Page 82, proof of Theorem 2.3.5:} It should be explained why all
the cycles constructed by this procedure are disjoint. For example, why cannot
some $\sigma^{i}\left(  1\right)  $ equal some $\sigma^{j}\left(  a\right)  $
? This is not hard to prove (if $\sigma^{i}\left(  1\right)  =\sigma
^{j}\left(  a\right)  $, then $a=\left(  \sigma^{j}\right)  ^{-1}\left(
\sigma^{i}\left(  1\right)  \right)  =\sigma^{i-j}\left(  1\right)  $, and
thus $a$ must belong to the set $\left\{  1,\sigma\left(  1\right)
,\ldots,\sigma^{r-1}\left(  1\right)  \right\}  $ because the latter set is
closed under both $\sigma$ and $\sigma^{-1}$), but some argument along these
lines needs to be made.

Furthermore, the theorem asserts the uniqueness of the disjoint cycles of
length $\geq2$, but the proof establishes only existence; the next paragraph
explicitly leaves uniqueness as an exercise. Either remove uniqueness from the
theorem statement or add the short orbit argument: the support of the cycle
containing $i$ is $\{\sigma^{k}(i)\mid k\in\mathbf{Z}\}$, and its cyclic
ordering is forced by the action of $\sigma$.

\item \textbf{Page 89, Exercise 12:} Replace \textquotedblleft the number
cycles\textquotedblright\ by \textquotedblleft the number of
cycles\textquotedblright.

\item \textbf{Page 90, Notes:} The phrases \textquotedblleft combinations of
the coefficients\textquotedblright\ (twice) and \textquotedblleft sums,
differences, products, and quotients of the coefficients\textquotedblright%
\ are misleading. The coefficients are rational and do not get permuted. The
relevant condition is preservation of all polynomial relations over
$\mathbf{Q}$ among the roots (equivalently, of all rational expressions in the
roots whose values lie in $\mathbf{Q}$). Rephrase all three occurrences accordingly.

\item \textbf{Page 107:} Replace ``a abelian group'' by ``an abelian group''.

\item \textbf{Page 111, paragraph before Corollary 3.2.4:} Replace ``If the
subset in question known to be finite'' by ``If the subset in question is
known to be finite''.

\item \textbf{Page 130, proof of Proposition 3.3.9:} Replace \textquotedblleft
if $x$ is a word in $S$, the so \ldots\textquotedblright\ by \textquotedblleft
if $x$ is a word in $S$, then so \ldots\textquotedblright.

\item \textbf{Page 135, Example 3.4.1:} Replace \textquotedblleft it is useful
rearrange the table\textquotedblright\ by \textquotedblleft it is useful to
rearrange the table\textquotedblright.

\item \textbf{Page 136, Example 3.4.3:} \textquotedblleft defined in Example
3.2.14\textquotedblright\ should be \textquotedblleft defined in Example
3.2.13\textquotedblright.

\item \textbf{Page 145, proof of Proposition 3.5.3:} The comma before
\textquotedblleft The order\textquotedblright\ should be a period.

\item \textbf{Page 149, Exercise 12:} Replace ``Proposition 3.5.5'' by
``Theorem 3.5.5''.

\item \textbf{Page 158, Exercise 1(c):} The first cycle is printed as
$(1,3,\ )$, with its third entry missing. Restore the omitted entry, or remove
the comma.

\item \textbf{Page 166, Proposition 3.7.6:} Replace \textquotedblleft If
$H_{2}$ is a normal in $G_{2}$\textquotedblright\ by \textquotedblleft If
$H_{2}$ is normal in $G_{2}$\textquotedblright\ or \textquotedblleft If
$H_{2}$ is a normal subgroup of $G_{2}$\textquotedblright.

\item \textbf{Page 175, Example 3.8.1:} The proof has no end-of-proof symbol.
Add $\square$ after its final sentence, for consistency with the other proofs
in the chapter.

\item \textbf{Page 187, Exercise 33:} The exercise is stated as a declarative
sentence without an instruction. Insert ``Prove that'' or ``Show that'' before
``the left coset $aH$ is a right coset \ldots''.

\item \textbf{Page 187, Notes:} Replace \textquotedblleft the use groups in
the theory of equations\textquotedblright\ by \textquotedblleft the use of
groups in the theory of equations\textquotedblright.

\item \textbf{Pages 203--204, proof of Theorem 4.2.1:} The theorem allows
$f(x)=0$, but the proof immediately writes $f(x)=a_{m}x^{m}+\cdots+a_{0}$ with
$a_{m}\ne0$ and never treats the zero polynomial. Begin by observing that if
$f(x)=0$, one may take $q(x)=r(x)=0$; then assume $f(x)\ne0$ for the argument
that follows.

\item \textbf{Page 209, last sentence:} The irreducibility criterion for
$x^{4}+x^{3}+x^{2}+x+1$ is not developed in the next section (Section 4.3),
but in Section 4.4 (Theorem 4.4.6). Replace \textquotedblleft in the next
section\textquotedblright\ by \textquotedblleft in Section
4.4\textquotedblright.

\item \textbf{Page 210, proof of Proposition 4.2.11:} The proposition is
stated for polynomials over $\mathbf{R}$, but the proof suddenly places $a(x)$
and $b(x)$ in $F[x]$. Use $\mathbf{R}[x]$, or restate the proposition over a
general field $F$ if that generality is intended.

\item \textbf{Page 213, Exercise 20:} Replace \textquotedblleft Your answer
should have degree $1$\textquotedblright\ by \textquotedblleft Your answer
should have degree at most $1$\textquotedblright, since the products can have
constant (or even zero) remainders for particular values of $a,b,c,d$.

\item \textbf{Page 213, Exercise 22(a)--(c):} If $a=b=0$, the congruence
$(a+bx)q(x)\equiv1$ has no solution. Add the hypothesis that $a$ and $b$ are
not both zero.

\item \textbf{Page 217, paragraph after Example 4.3.2:} Replace
\textquotedblleft the cosets $[a]$ coming the from constant
polynomials\textquotedblright\ by \textquotedblleft the cosets $[a]$ coming
from the constant polynomials\textquotedblright.

\item \textbf{Page 218, proof of Corollary 4.3.9:} After all linear factors
have been removed, the remaining factor $f_{1}(x)$ may be constant. In that
case the desired factorization is already complete, whereas the next sentence
incorrectly tries to find a root of $f_{1}(x)$. Insert ``If $f_{1}(x)$ is
constant, we are done; otherwise'' before applying Kronecker's theorem.

\item \textbf{Page 221, Exercise 23(a)--(b):} The zero class, obtained when
$a=b=0$, has no multiplicative inverse. Add the hypothesis that $a$ and $b$
are not both zero.

\item \textbf{Page 227, proof of Corollary 4.4.10:} Replace ``Corollary
4.1.1'' by ``Corollary 4.1.11''.

\item \textbf{Page 233, historical note on the cubic:} After introducing the
new variable $y=x+b/(3a)$, the depressed cubic and its three displayed cases
are still written in $x$. Replace $x$ by $y$ in those four displays, or
explicitly state that the new variable is being renamed $x$.

\item \textbf{Page 241, subfield discussion:} The reference \textquotedblleft
as in Definition 4.4.1\textquotedblright\ should be \textquotedblleft as in
Definition 4.3.1\textquotedblright.

\item \textbf{Page 244, integral-domain discussion:} \textquotedblleft as
given in Definition 3.5.6\textquotedblright\ should be \textquotedblleft as
given in Definition 3.3.5\textquotedblright\ (or \textquotedblleft as given in
Definition 4.1.1\textquotedblright\ for the later expanded definition).

\item \textbf{Page 246, Exercise 5:} The assertion fails for $R=\{0\}$, which
is a ring under the operations of the integral domain $D$ but is not a subring
of $D$, because its identity is not the identity of $D$. Add the hypothesis
that $R$ is not the zero ring.

\item \textbf{Page 248, Exercise 25:} Replace \textquotedblleft except that it
does not have the multiplicative identity of $R$\textquotedblright\ by
\textquotedblleft except that it does not contain the multiplicative identity
of $R_{1}$\textquotedblright.

\item \textbf{Page 252, Example 5.2.4:} Replace \textquotedblleft not
onto\textquotedblright\ by \textquotedblleft not onto unless $R$ is the zero
ring\textquotedblright. (For the zero ring $R$, the natural inclusion
$\iota:R\rightarrow R[x]$ is onto, since $R[x]$ is again the zero ring.)

\item \textbf{Page 253, Example 5.2.6:} Replace \textquotedblleft$\phi
(1)=1$\textquotedblright\ by \textquotedblleft$\phi_{u}(1)=1$%
\textquotedblright.

\item \textbf{Page 253:} Replace \textquotedblleft The previous example can
generalized\textquotedblright\ by \textquotedblleft The previous example can
be generalized\textquotedblright.

\item \textbf{Page 261, Exercise 15:} Replace \textquotedblleft Exercise 6 (c)
of Section 4.3\textquotedblright\ by \textquotedblleft Exercise 6 (c) of
Section 4.4\textquotedblright.

\item \textbf{Page 261, Exercise 15:} \textquotedblleft no nonzero root in
$\mathbf{Z}_{4}[x]$\textquotedblright\ should be \textquotedblleft no nonzero
root in $\mathbf{Z}_{4}$\textquotedblright. (Roots are elements of the
coefficient ring, not polynomials.)

\item \textbf{Page 268, first line:} Insert a period after \textquotedblleft%
$g(x)\in\langle p(x)\rangle$\textquotedblright\ before the sentence beginning
\textquotedblleft Thus\textquotedblright.

\item \textbf{Page 268, paragraph after Definition 5.3.8:} Replace
\textquotedblleft This observations shows\textquotedblright\ by
\textquotedblleft This observation shows\textquotedblright.

\item \textbf{Page 271, Exercise 7:} Replace \textquotedblleft any
ideal\textquotedblright\ by \textquotedblleft any nonzero
ideal\textquotedblright. (If $I=\left\{  0\right\}  $, then the claim of part
(a) is false, since $\left\{  0\right\}  $ contains no monic polynomial.)

\item \textbf{Page 271, Exercise 9:} \textquotedblleft Let $R$ be a
commutative ring\textquotedblright\ should be \textquotedblleft Let $R$ be a
nontrivial commutative ring\textquotedblright. (The zero ring has no proper
ideals, so the hypothesis that every proper ideal is prime holds vacuously,
but the zero ring is not a field under the book's definition.)

\item \textbf{Page 282, last paragraph:} Replace \textquotedblleft then
$\{f(x)\in F[x]\mid f(u)=0\}$ is an ideal of the ring $F\left[  x\right]
$\textquotedblright\ by \textquotedblleft then $\{f(x)\in K[x]\mid f(u)=0\}$
is an ideal of the ring $K\left[  x\right]  $\textquotedblright.

\item \textbf{Page 286, middle of the page:} The displayed minimal-polynomial
relation \textquotedblleft$c_{0}+c_{1}u+\ldots+c_{n-1}u^{n-1}+x^{n}%
=0$\textquotedblright\ should end with \textquotedblleft$+u^{n}=0$%
\textquotedblright\ rather than with \textquotedblleft$+x^{n}=0$%
\textquotedblright.

\item \textbf{Page 287, Example 6.1.5:} In \textquotedblleft$1+I=\left(
1+x^{2}+I\right)  \left(  \dfrac{1}{2}+\dfrac{2}{5}x-\dfrac{1}{5}%
x^{2}+I\right)  $\textquotedblright, replace \textquotedblleft$\dfrac{1}{2}%
$\textquotedblright\ by \textquotedblleft$\dfrac{1}{5}$\textquotedblright.

\item \textbf{Page 292, Example 6.2.5:} The displayed computation ends with an
extra closing bracket. Delete the stray bracket.

\item \textbf{Page 295, introductory paragraph to Section 6.3:} Replace
``passing though a point'' by ``passing through a point''.

\item \textbf{Page 298, proof of Lemma 6.3.5:} Replace ``Given two circles in
$F$'' by ``Given two circles over $F$''.

\item \textbf{Page 298, proof of Theorem 6.3.6:} \textquotedblleft then
$\sqrt{u}$ is constructible for all $u\in F$\textquotedblright\ should be
\textquotedblleft then $\sqrt{u}$ is constructible for all nonnegative $u\in
F$\textquotedblright\ (since constructible numbers have to be real).
Similarly, \textquotedblleft Given $u\in F$\textquotedblright\ should be
\textquotedblleft Given a positive $u\in F$\textquotedblright.

\item \textbf{Page 301, after Definition 6.4.1:} \textquotedblleft if $E$
contains the splitting field of $F$\textquotedblright\ should be
\textquotedblleft if $E$ contains a splitting field of $f\left(  x\right)  $
over $K$\textquotedblright\ (or, more explicitly: \textquotedblleft if
$f\left(  x\right)  $ splits into constant and linear factors in $E\left[
x\right]  $\textquotedblright).

\item \textbf{Page 303, Example 6.4.4:} Replace \textquotedblleft can be
expressed as element of $F$\textquotedblright\ by \textquotedblleft can be
expressed as an element of $F$\textquotedblright.

\item \textbf{Page 307, last paragraph:} In \textquotedblleft identity
element1is\textquotedblright, insert spaces to obtain \textquotedblleft
identity element $1$ is\textquotedblright.

\item \textbf{Page 310, Example 6.5.2:} Replace \textquotedblleft%
$\operatorname*{GF}(p)$\textquotedblright\ by \textquotedblleft%
$\operatorname*{GF}(2)$\textquotedblright.

\item \textbf{Page 311, Example 6.5.3:} In the last sentence, replace
\textquotedblleft$\operatorname*{GF}(3)$\textquotedblright\ by
\textquotedblleft$\operatorname*{GF}(2)$\textquotedblright.

\item \textbf{Page 312, Corollary 6.5.12:} \textquotedblleft For each positive
integer $n$\textquotedblright\ should be \textquotedblleft For each prime $p$
and each positive integer $n$\textquotedblright.

\item \textbf{Page 316, proof of Proposition 6.6.6:} Replace \textquotedblleft
follows form\textquotedblright\ by \textquotedblleft follows
from\textquotedblright.

\item \textbf{Page 316, proof of Theorem 6.6.7:} Delete the stray opening
quotation mark before the fourth equality in the displayed computation.

\item \textbf{Page 316, paragraph before Theorem 6.6.8:} In the last paragraph
of the page, replace \textquotedblleft$F(n)=\log f(n)$\textquotedblright\ by
\textquotedblleft$F\left(  n\right)  =\log G\left(  n\right)  $%
\textquotedblright. Also, this reduction requires the values of $g$ (and hence
of $G$) to be positive.

\item \textbf{Page 317, Theorem 6.6.8:} The multiplicative M\"{o}bius
inversion formula is stated for an arbitrary function into a commutative ring
$R$, but the exponents $\mu(m/n)=-1$ require multiplicative inverses and are
not defined for arbitrary elements of $R$. Require $g:\mathbf{Z}%
^{+}\rightarrow R^{\times}$ (and hence $G:\mathbf{Z}^{+}\rightarrow R^{\times
}$), or more generally let $g$ take values in an abelian group written multiplicatively.

\item \textbf{Page 318, proof of Corollary 6.6.12:} Replace \textquotedblleft%
$q^{n-1}+\cdots+q+1$\textquotedblright\ by \textquotedblleft$q^{m-1}%
+\cdots+q+1$\textquotedblright\ in the last sentence.

\item \textbf{Page 319, Exercise 4:} Specify that the asserted roots
$u^{ip^{k}}$ are obtained for all integers $k\geq0$.

\item \textbf{Page 320, Exercise 12(a)--(b):} Several mathematical symbols run
into adjacent words. Insert spaces in \textquotedblleft that$\mathcal{R}$
is\textquotedblright, \textquotedblleft operations$+$ and $\ast$%
\textquotedblright, \textquotedblleft$f\in\mathcal{R}$ has\textquotedblright,
and \textquotedblleft$f(1)$ is\textquotedblright.

\item \textbf{Page 321, Proposition 6.7.2:} Bad word spacing strikes again.
Insert the missing spaces in \textquotedblleft$a\in\mathbf{Z}$ with $p\nmid
a$\textquotedblright.

\item \textbf{Page 322, proof of Theorem 6.7.3:} Replace \textquotedblleft
multiplication by$a$defines\textquotedblright\ by \textquotedblleft
multiplication by $a$ defines\textquotedblright.

\item \textbf{Page 327, Exercise 6:} As written, the hypotheses permit $p=q$
and $t=0$, in which case both displayed Legendre symbols are undefined.
Require $p$ and $q$ to be \emph{distinct} odd primes (equivalently, add
$t\neq0$), or extend the definition of the Legendre symbol to $\left(
\dfrac{n}{p}\right)  =0$ for $p\mid n$.

\item \textbf{Page 328, historical note on constructible polygons:} The
historical directions are reversed. Gauss proved the \emph{sufficiency} of the
Fermat-prime form for constructibility; Wantzel later proved the
\emph{necessity}. The criterion itself should read
\[
n=2^{\alpha}p_{1}\cdots p_{k},
\]
where the $p_{i}$ are \emph{distinct} Fermat primes of the form $p_{i}%
=2^{2^{m_{i}}}+1$. Thus change Gauss's result to \textquotedblleft if $n$ has
the form \ldots\textquotedblright, identify the later result as the converse
(the necessity), replace the displayed product beginning with $p_{2}$ by one
beginning with $p_{1}$, and do not reuse the polygon variable $n$ in the
Fermat-prime exponent.

\item \textbf{Page 329, Chapter 7 introduction:} The text proposes finding
\textquotedblleft polynomial equations over $\mathbf{C}$ of degree $5$ that
cannot be solved by radicals\textquotedblright. Over the base field
$\mathbf{C}$, the roots already lie in the base field. Say \textquotedblleft
polynomial equations over $\mathbf{Q}$\textquotedblright\ or \textquotedblleft
equations with rational coefficients (and complex roots)\textquotedblright.

\item \textbf{Page 330, first paragraph:} \textquotedblleft which have no
nontrivial normal subgroups\textquotedblright\ should be \textquotedblleft
which have no nontrivial proper normal subgroups\textquotedblright. After all,
every nontrivial group is itself a nontrivial normal subgroup.

\item \textbf{Page 330, discussion of projective special linear groups:} The
assertion that \textquotedblleft the factor group $N/Z$ is simple except for
the cases $n=2$ and $F=\operatorname*{GF}\left(  2\right)  $ or
$\operatorname*{GF}\left(  3\right)  $\textquotedblright\ needs the hypothesis
$n\geq2$. For $n=1$, this factor group (which is later called
$\operatorname{PSL}_{n}(F)$) is trivial, and thus is not simple under
Definition 3.8.10.

\item \textbf{Page 331, Example 7.1.2:} Insert the missing spaces in
\textquotedblleft let$H$\textquotedblright\ and \textquotedblleft
Since$N$\textquotedblright, obtaining \textquotedblleft let $H$%
\textquotedblright\ and \textquotedblleft Since $N$\textquotedblright.

\item \textbf{Page 338, paragraph preceding Example 7.2.1:} Replace
\textquotedblleft each elements\textquotedblright\ by \textquotedblleft each
element\textquotedblright.

\item \textbf{Page 339, Example 7.2.1, last paragraph:} The three order-$2$
subgroups of $S_{3}$ are listed as $\{e,b\}$, $\{e,ab\}$, and $\{e,a^{2}\}$.
The last set is not one of them. Replace $\{e,a^{2}\}$ by $\{e,a^{2}b\}$.

\item \textbf{Page 340, Example 7.2.2:} It is said that $a^{i}=a^{-i}$
\textquotedblleft holds if and only if $i=2$\textquotedblright. In truth, this
holds if and only if $i$ is even (including $i=0$).

\item \textbf{Page 341, Example 7.2.3:} Insert the missing space in
\textquotedblleft conjugate in$S_{n}$\textquotedblright, obtaining
\textquotedblleft conjugate in $S_{n}$\textquotedblright.

\item \textbf{Page 345, Exercise 20 (and the selected answer on page 525):}
The book elsewhere defines $D_{n}$ as the dihedral group of order $2n$, with
$a$ of order $n$. Here $D_{12}$ is explicitly given with $a$ of order $6$, so
the group has order $12$; the selected answer repeats the inconsistent
notation. Rename this group $D_{6}$, or change $a$ to have order $12$ and
adjust the exercise and answer.

\item \textbf{Page 349, paragraph after proof of Proposition 7.3.4:}
\textquotedblleft Applying the above proposition to Example
7.3.6\textquotedblright\ should be \textquotedblleft Applying the above
proposition to Example 7.3.7\textquotedblright.

\item \textbf{Page 350, paragraph after proof of Theorem 7.3.6:}
\textquotedblleft If we apply Theorem 7.3.6 to Example 7.3.5\textquotedblright%
\ should be \textquotedblleft If we apply Theorem 7.3.6 to Example
7.3.6\textquotedblright.

\item \textbf{Page 351, Exercise 4:} \textquotedblleft Let $G$ act on the
subgroup $H$ by conjugation\textquotedblright\ is not well-defined unless $H$
is normal. The next clause introduces the correct $G$-set, namely the set $S$
of conjugates of $H$. Replace the opening \textquotedblleft Let $G$ act on the
subgroup $H$ by conjugation, let $S$ be the set of all conjugates of
$H$\textquotedblright\ by \textquotedblleft Let $H$ be a subgroup of a group
$G$. Let $S$ be the set of all conjugates of $H$ in $G$. Let $G$ act by
conjugation on the set $S$\textquotedblright.

\item \textbf{Page 353, Exercise 18(b) from Section 7.3:} Replace
\textquotedblleft Given an example\textquotedblright\ by \textquotedblleft
Give an example\textquotedblright.

\item \textbf{Page 353, proof of Theorem 7.4.1:} The theorem allows $\alpha
=0$, but the proof assumes $p\mid\lvert G\rvert$ and in Case 1 invokes a
subgroup of order $p^{\alpha-1}$. Begin by observing that when $\alpha=0$ the
trivial subgroup has the required order, and then assume $\alpha\geq1$ for the
two-case argument.

\item \textbf{Page 363, paragraph preceding Lemma 7.5.11:} Replace
\textquotedblleft cyclic of order of $2$\textquotedblright\ by
\textquotedblleft cyclic of order $2$\textquotedblright.

\item \textbf{Page 364, proof of Theorem 7.5.12:} Replace \textquotedblleft%
$\varphi(p^{k})$\textquotedblright\ by \textquotedblleft$\varphi(2^{k}%
)$\textquotedblright.

\item \textbf{Page 376, discussion before Example 7.7.1:} The claim
\textquotedblleft With two exceptions, when $n=2$ and either $\left\vert
F\right\vert =2$ or $\left\vert F\right\vert =3$, the groups
$\operatorname*{PSL}\nolimits_{n}\left(  F\right)  $ are
simple\textquotedblright\ holds only for $n\geq2$. For $n=1$, the group
$\operatorname*{PSL}\nolimits_{n}\left(  F\right)  $ is trivial and thus not
simple under the book's definition.

\item \textbf{Page 377, Example 7.7.2:} Replace \textquotedblleft It can be
shown the action\textquotedblright\ by \textquotedblleft It can be shown that
the action\textquotedblright.

\item \textbf{Page 388, proof of Theorem 8.1.8:} In the first paragraph,
\textquotedblleft Since $f(x)$ has no repeated roots\textquotedblright\ should
be \textquotedblleft Since $x^{p^{n}}-x$ has no repeated
roots\textquotedblright.

\item \textbf{Page 391, proof of Proposition 8.2.3:} In the last paragraph,
the factorization%
\[
f(x)=(x-r_{1})(x-r_{2})\cdots(x-r_{t})
\]
of $f\left(  x\right)  $ omits its leading coefficient. Write
\[
f(x)=c(x-r_{1})(x-r_{2})\cdots(x-r_{t}),\qquad c\neq0.
\]


\item \textbf{Page 396, proof of Lemma 8.3.4:} Replace \textquotedblleft
Dividing each of the solutions by $a_{1}$ still gives us a set of
solutions\textquotedblright\ by \textquotedblleft Dividing each entry of this
solution by $a_{1}$ still gives us a solution\textquotedblright.

\item \textbf{Page 396, proof of Lemma 8.3.4:} Replace \textquotedblleft We
can again relabel the indices on $x_{2},\ldots,x_{n}$ and $u_{2},\ldots,u_{n}%
$\textquotedblright\ by \textquotedblleft We can again relabel the indices on
$x_{2},\ldots,x_{n+1}$ and $u_{2},\ldots,u_{n+1}$\textquotedblright.

\item \textbf{Page 397, Example 8.3.1:} Replace \textquotedblleft Corollary
8.1.8\textquotedblright\ by \textquotedblleft Theorem 8.1.8\textquotedblright\ (twice).

\item \textbf{Page 397, Example 8.3.1, last paragraph:} In \textquotedblleft
for any integer we have $\phi^{t}(x)=\phi^{t+rs}(x)$\textquotedblright, add
\textquotedblleft$s$\textquotedblright\ after \textquotedblleft
integer\textquotedblright.

\item \textbf{Page 399, proof of Theorem 8.3.8(c):} Insert the missing spaces
in \textquotedblleft If$u\in E$\textquotedblright, \textquotedblleft
polynomial$p(x)$\textquotedblright, and \textquotedblleft then$\phi
(u)$\textquotedblright.

\item \textbf{Page 406, proof of Theorem 8.4.3:} The theorem permits $a=0$,
but the proof does not quite work in this case (since \textquotedblleft the
exponent $i$ of $\zeta$ in $\phi\left(  u\right)  =\zeta^{i}u$%
\textquotedblright\ is not uniquely determined when $a=0$). Thus, this case
needs to be handled first (it is easy: $a=0$ entails $F=K$, and the Galois
group is trivial). Then assume $a\neq0$ for the rest of the proof.

\item \textbf{Page 407, Theorem 8.4.4:} Add the hypothesis
$\operatorname{char}(K)\neq p$ (in particular, characteristic zero suffices).
In characteristic $p$, the field has only one $p$th root of unity, the
Vandermonde matrix in the proof is not invertible, and the conclusion fails
for nontrivial cyclic Artin--Schreier extensions of degree $p$.

\item \textbf{Page 408, proof of Lemma 8.4.5:} Replace \textquotedblleft for
$i=1,\ldots,n$\textquotedblright\ by \textquotedblleft for $i=1,\ldots
,m$\textquotedblright\ on line 4 of the proof.

\item \textbf{Page 408, proof of Theorem 8.4.6:} The proof uses $F_{i-1}$ when
$i=1$ without defining $F_{0}$, and defines $N_{i}$ only for $i=1,\ldots,m-1$
although the displayed chain ends with $N_{m}=\{e\}$. Set
\[
F_{i}=K(\zeta,u_{1},\ldots,u_{i}),\qquad N_{i}=\operatorname{Gal}%
(F(\zeta)/F_{i})
\]
for $0\leq i\leq m$. Then $F_{0}=K(\zeta)$ and $N_{0}=N$ and $N_{m}=\{e\}$.

\item \textbf{Page 409, converse proof of Theorem 8.4.6:} The first group in
the \textquotedblleft finite chain of subgroups\textquotedblright\ and the
first field in the \textquotedblleft ascending chain of
subfields\textquotedblright\ are not indexed, yet the factor groups
$N_{i}/N_{i+1}$ and fields $F_{i}$ are used immediately. Define
\[
N_{0}=\operatorname{Gal}(F/K(\zeta)),\quad N_{m}=\{e\},\quad F_{i}=F^{N_{i}%
}\quad(0\leq i\leq m),
\]
so that $F_{0}=K(\zeta)$, $F_{m}=F$, and $\operatorname{Gal}(F_{i+1}%
/F_{i})\cong N_{i}/N_{i+1}$ for $0\leq i<m$.

\item \textbf{Page 409, converse proof of Theorem 8.4.6:} The words around the
formula $f(x)$ are run together. Insert spaces before and after the formula,
yielding \textquotedblleft show that $f(x)$ is solvable\textquotedblright.

\item \textbf{Page 412, proof of Proposition 8.5.2(c):} Replace
\textquotedblleft$\Phi_{d}(x)$ with $d>n$\textquotedblright\ by
\textquotedblleft$\Phi_{d}(x)$ with $d<n$\textquotedblright.

\item \textbf{Page 412, proof of Proposition 8.5.2(c):} \textquotedblleft from
Exercise 1 of Section 4.3\textquotedblright\ should be \textquotedblleft from
Exercise 1 of Section 4.4\textquotedblright.

\item \textbf{Page 414, Corollary 8.5.5:} As stated for positive $n$, the
classification omits the case $n=1$: The polynomial $\Phi_{1}(x)=x-1$ has
trivial, hence cyclic, Galois group. Include $1$ in the list (as in the
standard classification) or state the corollary only for $n>1$. This is also
the omitted edge case in the cited Corollary 7.5.13 unless $p^{0}$ is
explicitly intended.

\item \textbf{Page 415, Example 8.5.2:} In the Gauss--Wantzel form for a
constructible regular polygon, the product begins with $p_{2}$. Replace
\textquotedblleft$n=2^{k}p_{2}\cdots p_{m}$\textquotedblright\ by
\textquotedblleft$n=2^{k}p_{1}\cdots p_{m}$\textquotedblright.

\item \textbf{Page 415, proof of Theorem 8.5.6:} Replace \textquotedblleft if
it has dimension is $n$\textquotedblright\ by \textquotedblleft if it has
dimension $n$\textquotedblright.

\item \textbf{Page 416, proof of Theorem 8.5.6:} From $\lvert\Phi_{n}%
(q)\rvert\geq(q-1)^{\varphi(n)}$ and $\Phi_{n}(q)\mid(q-1)$, the text
immediately concludes $n=1$. The displayed weak inequality alone does not
exclude equality for $n>1$. Instead, argue that for $n>1$, every primitive
$n$th root $\alpha^{j}$ is different from $1$, whence $\lvert q-\alpha
^{j}\rvert>q-1$; the resulting strict inequality $\lvert\Phi_{n}%
(q)\rvert>(q-1)^{\varphi(n)}$ does give the contradiction.

\item \textbf{Page 417, Exercises 7 and 8:} In both exercises, replace
\textquotedblleft Let $D$ be division ring\textquotedblright\ by
\textquotedblleft Let $D$ be a division ring\textquotedblright.

\item \textbf{Page 418, list of transitive subgroups of $S_{5}$:} The list of
possible Galois groups for an irreducible polynomial $f\left(  x\right)  $ of
degree $5$ given here includes $\mathbf{Z}_{4}$. But $\mathbf{Z}_{4}$ cannot
act transitively on five points (since $\left\vert \mathbf{Z}_{4}\right\vert
=4$ is not divisible by $5$), hence does not belong into this list. The
correct list is $S_{5},A_{5},F_{20},D_{5},\mathbf{Z}_{5}$.

\item \textbf{Page 419, proof of Lemma 8.6.4:} The statement that
\textquotedblleft each subgroup in the series is a transitive
subgroup\textquotedblright\ incorrectly includes the last subgroup
$N_{k}=\{(1)\}$. Say instead that repeated use of Lemma 8.6.3 shows that each
\emph{nontrivial} $N_{i}$ is transitive; in particular $N_{k-1}$ is
transitive, which is what the conclusion needs.

\item \textbf{Page 420, proof of Proposition 8.6.5:} The inequality $q<p$
could use some justification. Here is why it holds: We have $q=\left[
N_{i-1}:N_{i}\right]  \mid\left[  G:N_{k-1}\right]  $ (since $N_{i-1}$ is a
subgroup of $G$, while $N_{k-1}$ is a subgroup of $N_{i}$), but $\left[
G:N_{k-1}\right]  =\dfrac{\left\vert G\right\vert }{\left\vert N_{k-1}%
\right\vert }=\dfrac{p!}{p}=\left(  p-1\right)  !$, so that $q\mid\left[
G:N_{k-1}\right]  =\left(  p-1\right)  !=1\cdot2\cdot\cdots\cdot\left(
p-1\right)  $. Since $q$ is prime, this entails that $q$ divides one of
$1,2,\ldots,p-1$. Thus, $q<p$.

\item \textbf{Page 420, Example 8.6.3:} Replace \textquotedblleft The order of
the Galois group is order $p(p-1)$\textquotedblright\ by \textquotedblleft The
order of the Galois group is $p(p-1)$\textquotedblright.

\item \textbf{Page 421, definition of the discriminant:} The definition
\[
\Delta=\prod_{i<j}(r_{i}-r_{j})^{2}%
\]
of the discriminant $\Delta$ omits the leading-coefficient factor. It is
therefore inconsistent with the next assertion that the discriminant of
$ax^{2}+bx+c$ is $b^{2}-4ac$.

A correct definition of $\Delta$ proceeds as follows: For $f(x)=a_{n}\prod
_{i}(x-r_{i})$, define
\[
\Delta=a_{n}^{\,2n-2}\prod_{i<j}(r_{i}-r_{j})^{2}.
\]


\item \textbf{Page 421, Proposition 8.6.6:} The \textquotedblleft
if\textquotedblright\ direction requires the additional hypothesis
$\operatorname{char}(K)\neq2$. Indeed, in characteristic $2$, changing the
order of two roots does not change the sign of $\prod_{i<j}(r_{i}-r_{j})$, so
that $\prod_{i<j}(r_{i}-r_{j})$ always belongs to $K$ and therefore $\Delta$
is always a square in $K$.

\item \textbf{Page 421, Example 8.6.4:} Replace \textquotedblleft%
$-216$\textquotedblright\ by \textquotedblleft$-108$\textquotedblright.

\item \textbf{Page 424, Example 8.6.7:} In the last sentence, replace
\textquotedblleft the only transitive subgroup that contains a $4$-cycle and a
$3$-cycle is $S_{4}$\textquotedblright\ by \textquotedblleft... is $S_{5}%
$\textquotedblright.

\item \textbf{Page 425, Exercise 2:} This needs additional requirements that
$f(x)$ be separable and $\operatorname{char}(K)\neq2$. Without separability,
an irreducible cubic in characteristic $3$ can have trivial Galois group and
zero discriminant; in characteristic $2$, Proposition 8.6.6 itself does not
apply (see above).

\item \textbf{Page 431, proof of Proposition 9.1.6:} Replace all three
occurrences of \textquotedblleft$R$\textquotedblright\ by \textquotedblleft%
$D$\textquotedblright.

\item \textbf{Page 434, Exercise 5:} Replace \textquotedblleft for all $x,y\in
D$\textquotedblright\ by \textquotedblleft for all nonzero $x,y\in
D$\textquotedblright.

More importantly, the conclusion $\delta(1)=1$ is false under Definition
9.1.1, whose codomain includes zero: if $D$ is a field, then $\delta(x)=0$ for
every $x\neq0$ is a multiplicative Euclidean norm. Add the assumption that
$\delta$ is not identically zero, assume more strongly that $D$ is not a
field, or require Euclidean norms to take positive-integer values. Under any
of these corrections, the intended conclusions do follow.

\item \textbf{Page 436, Definition 9.2.4:} Replace \textquotedblleft
nonconstant\textquotedblright\ by \textquotedblleft nonzero\textquotedblright.
There is no reason to exclude nonzero constant polynomials from primitivity.

\item \textbf{Page 437, Lemma 9.2.6(a):} Require $a,b,c,d$ to be nonzero. As
written, the statement fails when $D$ is a field: for example, $a=b=0$ and
$c=d=1$ satisfy the stated relative-primality and associate product hypotheses
(indeed, Definition 9.2.3 makes any two elements of a field relatively prime,
since the field has no irreducible elements), but $a$ and $c$ are not associates.

\item \textbf{Page 437, Lemma 9.2.6(b):} Require $f\neq0$. (Otherwise, the
uniqueness part becomes false.)

\item \textbf{Page 438, proof of Proposition 9.2.7:} Replace \textquotedblleft
gives rise to a factorization $f(x)=tg^{\ast}(x)h^{\ast}(x)$\textquotedblright%
\ by \textquotedblleft gives rise to a factorization $f(x)=\dfrac{s}{t}%
g^{\ast}(x)h^{\ast}(x)$\textquotedblright.

\item \textbf{Page 438, proof of Theorem 9.2.8, first paragraph:} Restore the
missing spaces in \textquotedblleft Let$f(x)$be a nonzero element
of$D[x]$\textquotedblright, \textquotedblleft that$f(x)$is\textquotedblright,
and \textquotedblleft degree of$f(x)$\textquotedblright.

\item \textbf{Page 441, Exercise 6(c):} Replace \textquotedblleft and
$x+\left\langle s\right\rangle =a+\left\langle s\right\rangle $%
\textquotedblright\ by \textquotedblleft$x+\langle s\rangle=b+\langle
s\rangle$\textquotedblright.

\item \textbf{Page 441, Exercise 9(c):} Replace \textquotedblleft$\cup
_{k=1}^{n}$\textquotedblright\ by \textquotedblleft$\cup_{k=0}^{n}%
$\textquotedblright. (It is well possible that $n=0$.)

\item \textbf{Page 444, proof of Theorem 9.3.3:} Replace \textquotedblleft
must differ from $m+ni$ by a unit\textquotedblright\ by \textquotedblleft must
differ from either $m+ni$ or $m-ni$ by a unit\textquotedblright.

\item \textbf{Page 449, proof of Lemma 9.3.7(c):} Replace \textquotedblleft or
$z=[0]$\textquotedblright\ by \textquotedblleft or $[z]=[0]$\textquotedblright.

\item \textbf{Page 449, proof of Lemma 9.3.7(c):} Replace \textquotedblleft It
follows from by part (b)\textquotedblright\ by \textquotedblleft It follows
from part (b)\textquotedblright.

\item \textbf{Page 451, end of the proof of Theorem 9.3.8:} To complete the
minimal-descent contradiction, we need not just some new solution $\left(
x^{\ast},y^{\ast},z^{\ast}\right)  $ whose third entry $z^{\ast}$ is divisible
by $1-\omega$ but not by $\left(  1-\omega\right)  ^{k}$, but we need such a
solution that additionally satisfies conditions (i) and (ii) from page 450.
Lemma 9.3.7(c) says that one of $x_{\ast}^{3},y_{\ast}^{3},z_{\ast}^{3}$ is
divisible by $(1-\omega)^{4}$; because exactly one of $x_{\ast},y_{\ast
},z_{\ast}$ is divisible by $1-\omega$ (since exactly one of $u,v,w$ is
divisible by $1-\omega$), it must be $z_{\ast}^{3}$. Hence $(1-\omega)^{4}\mid
z_{\ast}^{3}$, so the exponent of $1-\omega$ in $z_{\ast}$ is at least $2$.
Moreover, if our solution $\left(  x^{\ast},y^{\ast},z^{\ast}\right)  $ does
not satisfy condition (i) from page 450, then we can divide $x^{\ast},y^{\ast
},z^{\ast}$ by their gcd without breaking condition (ii). The resulting
solution then satisfies both conditions (i) and (ii) and is therefore among
those over which the minimal exponent $k$ was chosen.

\item \textbf{Page 452, Exercise 6(b):} Replace \textquotedblleft show this it
\textit{suffices}\textquotedblright\ by \textquotedblleft show that it
\textit{suffices}\textquotedblright.

\item \textbf{Page 453, opening description of Chapter 10:} Insert
\textquotedblleft finite\textquotedblright\ before \textquotedblleft groups in
this class\textquotedblright. (The claim that nilpotent groups are direct
products of groups of prime-power order applies only to finite nilpotent
groups. For example, the infinite cyclic group is nilpotent but is not such a
direct product.)

\item \textbf{Page 455, proof of Theorem 10.1.4, (2) implies (3):} Replace
\textquotedblleft This implies the $P$ is normal\textquotedblright\ by
\textquotedblleft This implies that $P$ is normal\textquotedblright.

\item \textbf{Page 455, proof of Theorem 10.1.4, (3) implies (4):}
\textquotedblleft for some integers $k_{1},\ldots,k_{n}$\textquotedblright%
\ should be \textquotedblleft for some integers $k_{1},\ldots,k_{i}%
$\textquotedblright.

\item \textbf{Page 456, proof of Lemma 10.1.6:} \textquotedblleft The second
Sylow theorem (Theorem 7.7.4)\textquotedblright\ should be \textquotedblleft
The second Sylow theorem (Theorem 7.4.4(a))\textquotedblright.

\item \textbf{Page 458, first paragraph:} In \textquotedblleft Let $N^{\prime
}=\phi(N)$ and $K=\phi(K)$\textquotedblright, replace \textquotedblleft%
$K=\phi\left(  K\right)  $\textquotedblright\ by \textquotedblleft$K^{\prime
}=\phi(K)$\textquotedblright.

\item \textbf{Page 459, Example 10.2.3:} For arbitrary additive $G_{1}\leq F$
and multiplicative $G_{2}\leq F^{\times}$, the set $G$ is not necessarily a
subgroup of $\operatorname*{GL}\nolimits_{2}\left(  F\right)  $: Indeed,
closure requires $ax\in G_{1}$ for all $a\in G_{2}$ and $x\in G_{1}$. So this
requirement needs to be added in order to obtain a subgroup $G$.

\item \textbf{Page 459, Example 10.2.3:} In \textquotedblleft shows that shows
that\textquotedblright, delete one occurrence of \textquotedblleft shows
that\textquotedblright.

\item \textbf{Page 462, Exercise 6(b):} The displayed set $K$ includes the
singular matrix obtained when $a_{22}=0$, so as written it is not even a
subset of $\operatorname{GL}_{2}(\mathbf{R})$. Add the condition $a_{22}\neq0$
to the definition of $K$.

\item \textbf{Page 462, Exercises 11(a) and 11(b):} In \textquotedblleft show
that that $G_{1}$\textquotedblright\ and \textquotedblleft show that that
$G_{2}$\textquotedblright, delete the second \textquotedblleft
that\textquotedblright.

\item \textbf{Page 463, definition of Heisenberg group:} Replace
\textquotedblleft$G$\textquotedblright\ by \textquotedblleft$H$%
\textquotedblright\ in \textquotedblleft If $F=\operatorname*{GF}\left(
p\right)  $, then $G$ is called\textquotedblright.

\item \textbf{Page 463, semidirect-product setup:} Replace \textquotedblleft%
$g_{2}=h_{2}k_{2}$ in the\textquotedblright\ by \textquotedblleft$g_{2}%
=n_{2}k_{2}$ in the\textquotedblright.

\item \textbf{Page 465, construction of $N^{\prime}$ in an external semidirect
product:} On the penultimate line of the calculation of $\left(  n,k\right)
\left(  n^{\prime},e\right)  \left(  n,k\right)  ^{-1}$, the second coordinate
is printed as \textquotedblleft$3$\textquotedblright. It must be the identity
\textquotedblleft$e$\textquotedblright.

\item \textbf{Page 465, same paragraph:} The calculation is meant to prove
that the conjugate lies in $N^{\prime}$, but the text only says that it lies
in $N\rtimes_{\alpha}K$, which is automatic. Replace the conclusion by
\textquotedblleft is in $N^{\prime}$, so $N^{\prime}$ is
normal\textquotedblright.

\item \textbf{Page 465, Example 10.3.2:} \textquotedblleft external direct
product\textquotedblright\ should be \textquotedblleft external semidirect
product\textquotedblright.

\item \textbf{Page 467, paragraph after Example 10.3.3:} Replace
\textquotedblleft external direct product $N\rtimes_{\alpha}K$%
\textquotedblright\ by \textquotedblleft external semidirect product
$N\rtimes_{\alpha}K$\textquotedblright.

\item \textbf{Page 468, Proposition 10.3.7:} \textquotedblleft for all $x,y\in
X$\textquotedblright\ should be \textquotedblleft for all $x,y\in
N$\textquotedblright.

\item \textbf{Page 468, final sentence of Example 10.3.4:} Replace
\textquotedblleft an linear action\textquotedblright\ by \textquotedblleft a
linear action\textquotedblright.

\item \textbf{Page 469, Example 10.3.6, second sentence:} Replace
\textquotedblleft$\operatorname*{GL}_{n}(F)$. and let $N$\textquotedblright%
\ by \textquotedblleft$\operatorname*{GL}_{n}(F)$, and let $N$%
\textquotedblright.

\item \textbf{Page 472, proof of Proposition 10.4.3:} Replace
\textquotedblleft in unambiguous\textquotedblright\ by \textquotedblleft is
unambiguous\textquotedblright.

\item \textbf{Page 473, proof of Proposition 10.4.4:} Add a full-stop after
\textquotedblleft isomorphic to $C_{4}\cong\mathbf{Z}_{5}^{\times}$ or
$V$\textquotedblright.

\item \textbf{Page 474, paragraph before Proposition 10.4.5:} Replace
\textquotedblleft if $|G|$ is $p^{n}$, $n=pq^{m}$, or $n=p^{2}q$%
\textquotedblright\ by \textquotedblleft if $|G|$ is $p^{n}$, $pq^{m}$, or
$p^{2}q$\textquotedblright.

\item \textbf{Page 475, opening of the proof of Proposition 10.4.6:} Cauchy's
theorem does not by itself rule out a nonabelian simple group of prime-power
order. Replace \textquotedblleft Cauchy's theorem\textquotedblright\ by
\textquotedblleft Theorem 7.2.8\textquotedblright\ (the theorem that a finite
$p$-group has nontrivial center).

\item \textbf{Page 475, same proof:} Replace \textquotedblleft Examples 7.4.2
provides\textquotedblright\ by \textquotedblleft Example 7.4.2
provides\textquotedblright.

\item \textbf{Page 475, Exercise 1:} The displayed matrices do not form a
subgroup for an arbitrary subgroup $H\leq\mathbf{C}^{\times}$. For example, if
$H=\{1\}$, the second matrix
\[%
\begin{bmatrix}
0 & 1\\
-1 & 0
\end{bmatrix}
\]
belongs to the displayed set, but its square $-I$ does not. It is enough to
assume that $H$ is stable under complex conjugation and contains $-1$ (or,
more restrictively, that $H$ is a subgroup of the unit circle containing
$-1$). The subsequent choice $H=C_{2n}$ has these properties.

\item \textbf{Page 477, proof of Lemma 10.5.1:} Replace \textquotedblleft$0$
is all other rows\textquotedblright\ by \textquotedblleft$0$ in all other
rows\textquotedblright.

\item \textbf{Page 477, last line:} \textquotedblleft plane $R^{2}%
$\textquotedblright\ should be \textquotedblleft plane $\mathbf{R}^{2}%
$\textquotedblright.

\item \textbf{Page 479, half-angle calculation for the reflection line:} The
equations
\[
\cos(\theta/2)=\sqrt{\frac{1+\cos\theta}{2}},\qquad\sin(\theta/2)=\sqrt
{\frac{1-\cos\theta}{2}}%
\]
lose signs when $0\leq\theta<2\pi$; in particular, the first one fails for
$\pi<\theta<2\pi$. The subsequent replacement of $\sqrt{1-\cos^{2}\theta}$ by
$\sin\theta$ has the same problem.

Instead, the slope $m=\tan\left(  \theta/2\right)  $ of the $x$-axis rotated
by $\theta/2$ in a counterclockwise direction can be computed using the signed
half-angle identity
\[
\tan(\theta/2)=\frac{1-\cos\theta}{\sin\theta}=\frac{1-c}{s}.
\]
This gives the claimed reflection line without the invalid square roots.

\item \textbf{Page 480, proof of Theorem 10.5.4, Case 1:} Replace
\textquotedblleft where $k\in\mathbf{Z}^{+}$\textquotedblright\ by
\textquotedblleft where $k\in\mathbf{Z}_{\geq0}$\textquotedblright.

Replace \textquotedblleft Since $A_{\theta},A_{\rho}\in G$\textquotedblright%
\ by \textquotedblleft Since $A_{\theta},A_{\mu}\in G$\textquotedblright.

\item \textbf{Page 481, Theorem 10.5.4:} This is not literally true, since the
book defines $D_{n}$ for $n\geq3$ only, but a finite subgroup of
$\operatorname*{O}\nolimits_{2}\left(  \mathbf{R}\right)  $ can also be the
dihedral group $D_{2}$ (aka the Klein four-group).

To fix this, define the dihedral group $D_{n}$ for each positive integer $n$
as the semidirect product of the cyclic group $\mathbf{Z}_{n}$ by
$\mathbf{Z}_{2}$, where $\mathbf{Z}_{2}$ acts on $\mathbf{Z}_{n}$ by
$x\mapsto-x$ (or $x\mapsto x^{-1}$, if you write $\mathbf{Z}_{n}$
multiplicatively). Note that $D_{1}$ is isomorphic to $\mathbf{Z}_{2}$ (so it
is not really necessary), but $D_{2}$ is isomorphic to the Klein four-group.

\item \textbf{Page 481, final sentence of Theorem 10.5.4:} Replace
\textquotedblleft that $B\cong D_{n}$\textquotedblright\ by \textquotedblleft
that $G\cong D_{n}$\textquotedblright.

\item \textbf{Page 482, proof of Proposition 10.6.2:} Replace
\textquotedblleft$\left(  f(X-Y)\right)  \cdot\left(  f(X-Y)\right)
$\textquotedblright\ by \textquotedblleft$\left(  f(X)-f(Y)\right)
\cdot\left(  f(X)-f(Y)\right)  $\textquotedblright.

\item \textbf{Page 484, Proposition 10.6.9 and Theorem 10.6.10:} Replace
\textquotedblleft$O_{2}(\mathbf{R}^{2})$\textquotedblright\ by
\textquotedblleft$O_{2}(\mathbf{R})$\textquotedblright\ throughout these two
results and their proofs.

\item \textbf{Page 484, paragraph after Theorem 10.6.10:} Replace
\textquotedblleft straight forward\textquotedblright\ by \textquotedblleft
straightforward\textquotedblright.

\item \textbf{Page 485:} Replace \textquotedblleft H. S. M,
Coxeter\textquotedblright\ by \textquotedblleft H. S. M.
Coxeter\textquotedblright.

\item \textbf{Page 486, paragraph before the exercises:} Replace
\textquotedblleft a translation on a line parallel to the line of
reflection\textquotedblright\ by \textquotedblleft a translation along a line
parallel to the line of reflection\textquotedblright.

\item \textbf{Page 492, construction of $\mathbf{Z}$, $\mathbf{Q}$, and
$\mathbf{R}$:} The constructions specify equivalence relations but then speak
as though the raw pairs or sequences themselves were the numbers. Define
$\mathbf{Z}$ and $\mathbf{Q}$ as sets of \emph{equivalence classes} of the
indicated pairs, and $\mathbf{R}$ as the set of equivalence classes of Cauchy
sequences. In particular, the sentence identifying $\mathbf{N}$ with ``the set
of ordered pairs $(a,b)$ such that $a\geq b$'' is false literally: that set
has many representatives of each nonnegative integer. Instead embed
\[
n\longmapsto[(n,0)],
\]
or say that the nonnegative integer classes are precisely those having a
representative $(a,b)$ with $a\geq b$.

\item \textbf{Page 493:} Replace ``various text books'' by ``various textbooks''.

\item \textbf{Page 494, Proposition A.3.6:} The labels jump from (h) to (k),
then to (m) and (n). Relabel the final three parts consecutively as (i), (j),
and (k).

\item \textbf{Page 496, proof of Theorem A.4.3:} Replace \textquotedblleft
natural numbers\textquotedblright\ by \textquotedblleft positive
integers\textquotedblright. (The natural numbers include $0$ by definition.)

\item \textbf{Page 498, Exercise A.4.15:} The asserted identity fails for
$n=1$: its left-hand side is $1$, not $0$. Add the hypothesis $n\geq2$.

\item \textbf{Page 498, Exercise A.4.19:} The strict inequality $(1+h)^{n}%
>1+nh$ fails when $h=0$. Either replace $>$ by $\geq$, or retain the strict
inequality and add $h\neq0$.

\item \textbf{Page 500, polar form of a complex number:} The definitions
$\cos\theta=a/r$ and $\sin\theta=b/r$ divide by $r$ and therefore apply only
when $a+bi\neq0$. Add this hypothesis. The zero complex number may instead be
represented by $r=0$ with arbitrary angle $\theta$.

\item \textbf{Page 500, proof of Corollary A.5.3:} The proof exhibits $n$
distinct roots of $z^{n}-1$, but it does not say why there are no others. Add
an invocation of Corollary 4.1.12, which says that a nonzero polynomial of
degree $n$ has at most $n$ distinct roots.

\item \textbf{Page 505, Section A.6.2:} For a real quadratic with discriminant
zero, the assertion
\[
ax^{2}+bx+c=(mx+k)^{2}\qquad(m,k\in\mathbf{R})
\]
is false when $a<0$. Say instead that the quadratic is a real scalar multiple
of the square of a linear polynomial (indeed, $a(x+b/(2a))^{2}$), or add the
hypothesis $a>0$ to the displayed square characterization.

\item \textbf{Pages 505--509, discriminants of quadratics and cubics:} Replace
``imaginary'' by ``nonreal complex'' when describing roots and the quantities
whose cube roots occur. For example, when a real cubic has one real root and a
conjugate pair, the nonreal roots need not be purely imaginary; likewise the
two radicands on page 509 generally have nonzero real parts.

\item \textbf{Page 506, solution of the reduced cubic:} The substitution
$y=z-p/(3z)$ requires $z\neq0$. Handle the degenerate case $p=q=0$ separately:
then the reduced equation is simply $y^{3}=0$, while its resolvent supplies
only $z=0$, for which the displayed substitution is undefined.

\item \textbf{Page 506, discriminant of a cubic:} Replace \textquotedblleft in
F.4\textquotedblright\ by \textquotedblleft in Section A.6.4\textquotedblright.

\item \textbf{Page 510, general quartic equation:} Remove \textquotedblleft%
$a\neq0$\textquotedblright\ from the first displayed equation in Section A.6.5.

\item \textbf{Page 511, Descartes's solution of the quartic:} The formulas for
$m$ and $n$ involve division by $k$, although the resolvent in $k^{2}$ can
have the root $k^{2}=0$ (in particular when $q=0$). Require $k\neq0$ for this
derivation and treat $k=0$ separately: then $q=0$, and the biquadratic can be
factored by choosing $m+n=p$ and $mn=r$.

\item \textbf{Page 513, Definition A.7.1:} Replace \textquotedblleft A vector
space over the field $F$ is set $V$\textquotedblright\ by \textquotedblleft A
vector space over the field $F$ is a set $V$\textquotedblright.

\item \textbf{Pages 513--515, Definitions A.7.3--A.7.5:} Span and linear
independence are defined only for a displayed finite set $S=\{\mathbf{v}%
_{1},\mathbf{v}_{2},\ldots,\mathbf{v}_{n}\}$, but the later results use
arbitrary subsets. Extend the definitions to arbitrary $S$ using finite linear
combinations, and define independence by requiring every finite linear
relation to be trivial.

\item \textbf{Page 525, selected answer to Exercise 7.7.15:} The exercise asks
for the orders of $\operatorname{GL}_{n}(F)$, $\operatorname{SL}_{n}(F)$, and
$\operatorname{PSL}_{n}(F)$, but the selected answer gives only the first. For
$\lvert F\rvert=q$, complete it with
\[
\lvert\operatorname{SL}_{n}(F)\rvert=\frac{\lvert\operatorname{GL}%
_{n}(F)\rvert}{q-1}, \qquad\lvert\operatorname{PSL}_{n}(F)\rvert=\frac
{\lvert\operatorname{GL}_{n}(F)\rvert} {(q-1)\gcd(n,q-1)}.
\]


\item \textbf{Page 526, selected answer to Exercise 8.1.2:} The exercise asks
for an explicit formula for each element of the Galois group, but the answer
only names a generator. Add that the three automorphisms of $\operatorname{GF}%
(2^{3})/\operatorname{GF}(2)$ are
\[
x\longmapsto x,\qquad x\longmapsto x^{2},\qquad x\longmapsto x^{4}.
\]

\end{enumerate}

\section{Additional bibliography, index, and editorial recommendations}

\subsection{Bibliography}

\begin{itemize}
\item \textbf{Page 516:} Add terminal periods to the Jacobson and Halmos entries.

\item \textbf{Page 516:} In ``Niven, I. Irrational Numbers'', add a comma
after ``I.''.

\item \textbf{Page 517:} Add a terminal period to the Hall entry.

\item Consider standardizing publisher punctuation (``Dover Publications
Inc.'' versus ``Dover Publications, Inc.''), edition typography, and
capitalization of titles and volume designations.
\end{itemize}

\subsection{Index}

\begin{itemize}
\item \textbf{Page 534, ``group(s), exponent of'':} Remove the stray trailing
comma after ``148''.

\item \textbf{Page 538, ``permutation(s), even'':} Add the missing comma after ``even''.

\item A final mechanical pass should standardize entries that alternate
between noun-first and inverted forms (for example, ``transitive subgroup''
versus ``subgroup, transitive'').
\end{itemize}

\subsection{Editorial recommendations}

\begin{itemize}
\item \textbf{Use a single convention for dihedral groups.} Most of the book
uses $D_{n}$ for the group of order $2n$, but Exercise 7.2.20 and its selected
answer use $D_{12}$ for a group of order $12$. This is especially risky in a
textbook because both conventions occur in the literature.

\item \textbf{Standardize theorem labels in references.} The most common
reference errors are correct numbers paired with the wrong label
(\textquotedblleft Corollary\textquotedblright\ instead of \textquotedblleft
Theorem\textquotedblright) and references to the right topic in the wrong section.

\item \textbf{Give Chapter 10 an additional notation pass.} This chapter
contains a noticeably higher concentration of duplicated words,
primed/unprimed symbols, and mismatched ambient sets than the earlier chapters.

\item \textbf{Preserve the leading coefficient in discriminant discussions.}
The monic convention in the appendix is coherent, but Chapter 8 immediately
discusses nonmonic quadratics. Stating the full discriminant definition once
will keep both discussions consistent.
\end{itemize}

\bigskip

\begin{noncompile}
My (GPT's) overall verdict is: a good and unusually thoughtful undergraduate
text, but not a sufficiently reliable one to use without an errata sheet.

Its greatest strength is pedagogical architecture. The authors deliberately
postpone the abstract definition of a group until students have accumulated
examples from integers, congruences, functions, permutations, and matrices.
Likewise, polynomials precede the general theory of rings. This works
especially well for students who have had linear algebra but relatively little
experience with proofs. The motivational passages, \textquotedblleft Looking
ahead\textquotedblright\ discussions, and recurring number-theoretic
applications are also genuinely useful rather than ornamental.

The exposition is generally humane and concrete. It often explains why a
definition is needed, not merely what it says. The chapter introductions and
\textquotedblleft To the Student\textquotedblright\ material show a realistic
understanding of where undergraduates struggle. There is enough material for a
two-semester course, with later chapters suitable for a stronger or
honors-level continuation.

There are, however, three qualifications.

The error density is high for a fourth-edition textbook. Many of our 201
main-list entries are harmless copyedits or incorrect cross-references, but
there are also false boundary cases, missing hypotheses, incomplete proofs,
divisions by quantities that may be zero, and occasional genuinely false statements.

The mistakes are particularly unfortunate for inexperienced students. A mature
reader can repair something like the $n=1$ exception in an induction exercise
or the omitted $a>0$ in a quadratic-discriminant statement. A beginning proof
writer may instead assume that their own reasoning is defective.

The increase in difficulty is not entirely gradual. The early chapters are
expansive and supportive, while some later sections become markedly terse. The
authors themselves acknowledge this, but the transition can still be abrupt.

The good news is that the defects are overwhelmingly local. I did not find
evidence that the central development of groups, rings, fields, or Galois
theory is fundamentally unsound. The book's conceptual backbone is solid; its
sentence-by-sentence reliability and copyediting are not.

So I would recommend it as follows:

For a year-long undergraduate course with an attentive instructor: yes,
certainly---especially for students new to proof-based algebra.

For a one-semester course: yes, but the first two chapters would need
substantial selection or acceleration.

For independent study by a beginner: with reservations; the errata should be
kept open alongside the book.

For very strong or already abstractly minded students: the opening may feel
slow, though the later chapters contain worthwhile material.

In one sentence: it is pedagogically better than its error count suggests, and
mathematically less dependable than its polished appearance suggests. With the
errata available, I would be quite comfortable teaching from it; without them,
I would hesitate.
\end{noncompile}

\section{Corrections to the student solutions booklet}

\noindent In this section, page numbers refer to the printed page numbers in
the booklet \textit{Selected Solutions for Students to Accompany Abstract
Algebra, Fourth Edition} (downloadable from
\url{https://www.johnabeachy.com/abstractalgebra} ); the roman-numbered front
matter is cited in the same way.

\begin{enumerate}
\item \textbf{Page iii, Contents:} Change the second occurrence of
\textquotedblleft3.6\textquotedblright\ (before \textquotedblleft Cosets,
Normal Subgroups, and Factor Groups\textquotedblright) to \textquotedblleft%
3.8\textquotedblright. Also replace \textquotedblleft
Homomophisms\textquotedblright\ by \textquotedblleft
Homomorphisms\textquotedblright.

\item \textbf{Page v, To the Student:} Replace \textquotedblleft
everthing\textquotedblright\ by \textquotedblleft everything\textquotedblright.

\item \textbf{Page 1, solution to Exercise 1.1.17:} Replace \textquotedblleft%
$k=n^{2}-m^{2}$\textquotedblright\ by \textquotedblleft$k=m^{2}-n^{2}%
$\textquotedblright\ in the first sentence.

\item \textbf{Page 5, solution to Exercise 2.1.3(d):} In \textquotedblleft for
all $x\in\mathbf{R}$, This shows\textquotedblright, replace the comma by a period.

\item \textbf{Page 6, solution to Exercise 2.2.11(d):} Replace
\textquotedblleft the horizontal line through $y$\textquotedblright\ by
\textquotedblleft the horizontal line at height $y$\textquotedblright.

\item \textbf{Page 7, solution to Exercise 2.2.12(d):} Replace
\textquotedblleft then $\left(  \lambda a_{1}\right)  x+\left(  \lambda
b_{1}\right)  y+\left(  \lambda c_{1}\right)  z=0$\textquotedblright\ by
\textquotedblleft then $\left(  \lambda a_{2}\right)  x+\left(  \lambda
b_{2}\right)  y+\left(  \lambda c_{2}\right)  z=0$\textquotedblright.

\item \textbf{Page 8, solutions to Exercise 2.2.12(h)--(j):} These solutions
confuse the image of an affine line with its projective closure. If
$f(x,y)=[x,y,1]$ and
\[
\ell=\{(x,y)\in\mathbf{R}^{2}\mid ax+by+c=0\},\qquad L=\{[x,y,z]\in
\mathbf{P}^{2}\mid ax+by+cz=0\},
\]
then
\[
f(\ell)=L\cap\{[x,y,z]\mid z\neq0\},
\]
not $L$ itself. Thus in part (i) the displayed common point only proves that
the two projective closures meet; the literal equality of the images of the
intersection and the intersection of the images follows instead from the
injectivity of $f$. In part (j), the images of two parallel affine lines are
still disjoint, while their projective closures meet in a point at infinity.
Accordingly, the final sentence should say that all \emph{projective} lines in
$\mathbf{P}^{2}$ intersect.

\item \textbf{Page 11, solution to Exercise 3.1.13:} In the last parenthetical
sentence, replace \textquotedblleft$(2-1)\ast a=1$\textquotedblright\ by
\textquotedblleft$(2-a)\ast a=1$\textquotedblright.

\item \textbf{Page 15, solution to Exercise 3.3.19:} Replace \textquotedblleft%
$ab=e$ implies $b\neq a$\textquotedblright\ by \textquotedblleft$ab=e$ implies
$b=a$\textquotedblright. (This contradicts the choice of distinct $a$ and $b$,
and hence proves $ab\neq e$.)

\item \textbf{Page 15, solution to Exercise 3.3.21:} The label of the last row
of the multiplication table should be $a^{2}b$, not $a^{b}$.

\item \textbf{Page 16, solution to Exercise 3.4.26:} In the computation of
$\phi(a)+\phi(b)$, replace \textquotedblleft$mb-m1$\textquotedblright\ by
\textquotedblleft$mb-m$\textquotedblright.

In the final note, if $\theta(x)=x-1$ and $\psi(x)=mx$, then the required
composition is $\phi=\theta\circ\psi$, not \textquotedblleft$\phi=\psi\theta
$\textquotedblright.

\item \textbf{Page 17, solution to Exercise 3.5.13:} The conclusion
\textquotedblleft$c\in\{e,a^{k},a^{2k},\ldots,a^{(m-1)k}\}$\textquotedblright%
\ could use more explanation: Let $q$ be the remainder of $tr$ modulo $m$.
Thus, $tr\equiv q\pmod m$ and $0\leq q<m$. Now, $o(a)=n=mk$ and $k\cdot
tr\equiv kq\pmod{mk}$ (since $tr\equiv q\pmod m$), so that $a^{k\cdot
tr}=a^{kq}$. Thus, $c=a^{k\cdot tr}=a^{kq}\in\{e,a^{k},a^{2k},\ldots
,a^{(m-1)k}\}$ (since $0\leq q<m$).

\item \textbf{Page 18, solution to Exercise 3.7.6:} Exercise 2.1.11 only shows
that $[x]_{n}\mapsto\lbrack x]_{m}$ is a well-defined map into $\mathbf{Z}%
_{m}$. To see that its displayed codomain is really $\mathbf{Z}_{m}^{\times}$,
add that $\gcd(x,n)=1$ and $m\mid n$ imply $\gcd(x,m)=1$.

\item \textbf{Page 18, alternate solution to Exercise 3.7.14:}
\textquotedblleft determinant homomorphism from $\operatorname{GL}_{n}(R)$ to
$\mathbf{R}$\textquotedblright\ should be \textquotedblleft determinant
homomorphism from $\operatorname{GL}_{n}(\mathbf{R})$ to $\mathbf{R}^{\times}%
$\textquotedblright.

\item \textbf{Page 23, solution to Exercise 4.3.20:} In the line for $[x]^{3}%
$, remove the mismatched closing parenthesis in \textquotedblleft%
$\lbrack4x+2+x)]$\textquotedblright.

In the comment, replace \textquotedblleft$\mathbf{Z}_{3}/\langle
x^{2}+x-1\rangle$\textquotedblright\ by \textquotedblleft$\mathbf{Z}%
_{3}[x]/\langle x^{2}+x-1\rangle$\textquotedblright.

\item \textbf{Page 27, solution to Exercise 5.3.32(c):} The argument proves
only $I\subseteq2^{k}R$. For the reverse inclusion, argue that $u=2^{k}v\in I$
with $v$ odd. Since $1/v\in R$, it follows that $2^{k}=u/v\in I$, and hence
$2^{k}R\subseteq I$.

\item \textbf{Page 27, solution to Exercise 5.3.32(d):} Replace
\textquotedblleft and so $\dfrac{m}{n}=\dfrac{mu+mv2^{k}}{n}$%
\textquotedblright\ by \textquotedblleft and so $\dfrac{m}{n}=mu+\dfrac
{mv2^{k}}{n}$\textquotedblright.

\item \textbf{Page 28, solution to Exercise 5.3.32(d):} After proving that
each coset has a unique representative $r$ with $0\leq r<2^{k}$ and checking
the homomorphism laws, the solution establishes only that $\phi$ is onto. Add
a justification that it is one-to-one: if $\phi(r+2^{k}R)=[r]_{2^{k}}=0$, the
chosen range for $r$ forces $r=0$.

\item \textbf{Page 30, solution to Exercise 6.1.7:} In the last sentence,
replace \textquotedblleft$\sqrt{v}+\sqrt{v}$\textquotedblright\ by
\textquotedblleft$\sqrt{u}+\sqrt{v}$\textquotedblright.

\item \textbf{Page 30, Section 6.2 heading:} Replace \textquotedblleft Finite
and Algebraic Elements\textquotedblright\ by \textquotedblleft Finite and
Algebraic Extensions\textquotedblright.

\item \textbf{Page 30, solution to Exercise 6.2.5:} Replace \textquotedblleft
contracts the assumption\textquotedblright\ by \textquotedblleft contradicts
the assumption\textquotedblright.

\item \textbf{Page 31, solution to Exercise 6.2.7:} The proof notes only that
$n\mid\lbrack F:K]$. It must also use $m\mid\lbrack F:K]$, since
$K(u)\subseteq F$. As $\gcd(m,n)=1$, this gives $mn\mid\lbrack F:K]$; together
with $[F:K]\leq mn$, it yields $[F:K]=mn$.

\item \textbf{Page 33, solution to Exercise 6.5.15:} In the characteristic $2$
case, handle $u=1$ before dividing into the two possibilities for $1+u$. If
$u=1$, then $F^{\times}=\langle u\rangle=\{1\}$ and $F=\operatorname{GF}(2)$
immediately; otherwise the printed argument applies.

\item \textbf{Page 35, solution to Exercise 7.2.22:} Replace \textquotedblleft
a divisor of $G$\textquotedblright\ by \textquotedblleft a divisor of $\lvert
G\rvert$\textquotedblright.

\item \textbf{Pages 39--40, solution to Exercise 7.5.10:} In the decomposition
of $H$, replace \textquotedblleft$\sum_{i=1}^{t}\beta_{i}=k$\textquotedblright%
\ by \textquotedblleft$\sum_{i=1}^{s}\beta_{i}=k$\textquotedblright.

The induction also needs two repairs. Once we know that the number of elements
of order $p$ gives $s=t$, we conclude that multiplication by $p$ has kernel of
size $p^{t}$ on both $G$ and $H$. For $j\geq2$, every element of order
$p^{j-1}$ in $pG$ has exactly $p^{t}$ preimages, all of order $p^{j}$; hence
$pG$ and $pH$ have the same number of elements of each order. Induction
therefore gives $pG\cong pH$. Uniqueness of the cyclic decomposition now shows
that the multisets
\[
\{\alpha_{i}-1\mid\alpha_{i}>1\}\quad\hbox{and}\quad\{\beta_{i}-1\mid\beta
_{i}>1\}
\]
agree. Since $s=t$, the number of exponents $\alpha_{i}$ equal to $1$ also
agrees with the number of exponents $\beta_{i}$ equal to $1$, and therefore we
obtain $G\cong H$. The printed componentwise conclusion $\alpha_{i}%
-1=\beta_{i}-1$ overlooks the cyclic factors that become trivial after
multiplication by $p$.

\item \textbf{Page 41, solution to Exercise 7.6.7:} Replace \textquotedblleft
observing that fact $K_{2}$ has index $2$\textquotedblright\ by
\textquotedblleft observing that $K_{2}$ has index $2$\textquotedblright.

\item \textbf{Pages 41--42, solution to Exercise 7.6.10(b):} In the case
$p=q$, the displayed series proves solvability only after normality is
checked. A subgroup $H$ of order $p^{2}$ has index $p$ in $G$, so
$H\mathrel{\triangleleft}G$; moreover, $H$ is abelian, so every subgroup $K$
of order $p$ is normal in $H$. Thus $G\supset H\supset K\supset\{e\}$ is
indeed a subnormal series with abelian factors.

(Alternatively, the case $p=q$ also follows directly from Theorem 7.6.3.)

\item \textbf{Page 42, solution to Exercise 7.6.14:} After obtaining
$\phi(N)\subseteq N$, add that $\lvert\phi(N)\rvert=\lvert N\rvert$ because
$\phi$ is an automorphism. Hence $\phi(N)=N$, as required for $N$ to be characteristic.

\item \textbf{Page 49, solution to Exercise 8.5.12:} In the congruence for
$\Phi_{105}(x)$ modulo $x^{25}$, the final term \textquotedblleft$-x^{24}%
$\textquotedblright\ is printed twice. Delete one occurrence. (The final
displayed polynomial has the correct coefficient.)

\item \textbf{Pages 49--50, solution to Exercise 8.6.5:} Abstract isomorphism
does not preserve transitivity: the subgroup $V$ is isomorphic to $H_{1}$,
although $V$ is transitive and $H_{1}$ is not. Repair the argument using
conjugacy. Every subgroup of order $4$ lies in a Sylow $2$-subgroup $P$, and
some conjugation carries $P$ to the displayed copy of $D_{4}$ (by the second
Sylow theorem). The conjugate of the subgroup is then one of $H_{1}$, $H_{2}$,
or $H_{3}$, and conjugation preserves transitivity. Here $H_{1}$ is not
transitive, $H_{2}$ is cyclic and transitive, and $H_{3}=V$ is transitive. The
earlier order-$8$ case should likewise use conjugacy of Sylow $2$-subgroups,
rather than merely their abstract isomorphism with $D_{4}$.

\item \textbf{Page 50, solution to Exercise 8.6.8:} Replace \textquotedblleft
a $5$-cyclic\textquotedblright\ by \textquotedblleft a $5$%
-cycle\textquotedblright.

\item \textbf{Page 51, solution to Exercise 9.2.8:} The proof of the converse
begins by choosing a nonzero $x_{1}\in I$, which is impossible for $I=(0)$.
First dispose of this case: the zero ideal is generated by $0$ (or by the
empty set). Then assume $I\neq(0)$ and continue as printed.

\item \textbf{Page 52, solution to Exercise 9.2.9(c):} The proposed set
$\mathcal{B}$ omits the degree-zero generators. Replace \textquotedblleft%
$\cup_{k=1}^{n}$\textquotedblright\ by \textquotedblleft$\cup_{k=0}^{n}%
$\textquotedblright.

In the case $\deg f=k\geq n$, the \textquotedblleft repeat the argument using
the generators of $I_{n}$\textquotedblright\ argument is somewhat
underexplained. In more detail, this argument proceeds as follows: We have
$a\in I_{k}=I_{n}$ (by part (b), since $k\geq n$), and thus $a=\sum
_{j=1}^{m\left(  n\right)  }r_{j}a_{jn}$ for some $r_{j}\in R$, where each
$a_{jn}$ is the $x^{n}$-coefficient of $p_{jn}\left(  x\right)  $. Hence, the
polynomial $f\left(  x\right)  -\sum_{j}r_{j}x^{k-n}p_{jn}(x)$ has lower
degree than $f\left(  x\right)  $, and still cannot be expressed as a linear
combination of elements of $\mathcal{B}$.

\item \textbf{Page 52, Sarges's proof of the Hilbert basis theorem:} Replace
\textquotedblleft for $1\leq i<k-1$\textquotedblright\ by \textquotedblleft
for $1\leq i<k$\textquotedblright\ when describing the polynomials already
chosen before $f_{k}$.

\item \textbf{Page 52, Sarges's proof of the Hilbert basis theorem:} Replace
\textquotedblleft be the degree $f_{i}$\textquotedblright\ by
\textquotedblleft be the degree of $f_{i}$\textquotedblright.

\item \textbf{Page 54, solution to Exercise 9.3.5(i):} Replace
\textquotedblleft such that $x=k^{2}$\textquotedblright\ by \textquotedblleft
such that $s=k^{2}$\textquotedblright.

Replace \textquotedblleft a solution $\left(  a^{2},b^{2},k\right)
$\textquotedblright\ by \textquotedblleft a solution $\left(  a,b,k\right)
$\textquotedblright.

\item \textbf{Page 54, Exercise 9.3.6(b):} Replace \textquotedblleft show this
it \textit{suffices}\textquotedblright\ by \textquotedblleft show that it
\textit{suffices}\textquotedblright.

\item \textbf{Page 55, solution to Exercise 10.1.5(b):} The proof of the
converse reads rather confusingly when $k=0$ or $k=1$; in these cases, the
ascending central series stabilizes either immediately or after one step, and
some of the arguments given do not apply. It is probably better to explain
things more rigorously: Any $j\in\left\{  1,2,\ldots,k\right\}  $ satisfies
\[
D_{2^{j}m}/Z(D_{2^{j}m})\cong D_{2^{j-1}m}%
\]
(by the Lemma and by what was said just before it). Thus, by induction over
$i$, we can easily see that the ascending central series of $D_{n}$ satisfies
\[
D_{n}/Z_{i}\left(  D_{n}\right)  \cong D_{2^{k-i}m}%
\ \ \ \ \ \ \ \ \ \ \text{for each }i\in\left\{  0,1,\ldots,k\right\}  ,
\]
and thus in particular $D_{n}/Z_{k}\left(  D_{n}\right)  \cong D_{m}$. If
$m>1$, then $Z\left(  D_{m}\right)  =\left\{  e\right\}  $, which entails
$Z_{k+1}\left(  D_{n}\right)  =Z_{k}\left(  D_{n}\right)  $. Thus, if $m>1$
(equivalently, if $n$ is not a power of $2$), the ascending central series of
$D_{n}$ stabilizes at $Z_{k}(D_{n})$ and never reaches $D_{n}$.

\item \textbf{Page 57, Exercise 10.2.8:} Replace \textquotedblleft
isomorpic\textquotedblright\ by \textquotedblleft isomorphic\textquotedblright.

\item \textbf{Page 58, solution to Exercise 10.3.7:} Replace both occurrences
of \textquotedblleft identity\textquotedblright\ in \textquotedblleft We can
identity \ldots\textquotedblright\ by \textquotedblleft
identify\textquotedblright.

\item \textbf{Page 59, solution to Exercise 10.3.8(b):} The matrix
\[%
\begin{bmatrix}
1 & 0\\
x & a
\end{bmatrix}
\]
has order $4$ for $a=2,3$, but has order $2$ for $a=4=-1$ (unless it is the
identity, which cannot occur here since $a\neq1$). Thus replace
\textquotedblleft has order $4$ for $a=2,3,4$\textquotedblright\ by
\textquotedblleft has order $4$ for $a=2,3$ and order $2$ for $a=4$%
\textquotedblright. The conclusion that $F_{20}$ has no element of order $10$
remains valid.

\item \textbf{Page 62, solution to Exercise 10.4.10:} Delete the duplicated
\textquotedblleft and\textquotedblright\ in \textquotedblleft and and
$10\cdot2$\textquotedblright; replace \textquotedblleft
sybgroup\textquotedblright\ by \textquotedblleft subgroup\textquotedblright;
replace \textquotedblleft which isomorphic\textquotedblright\ by
\textquotedblleft which is isomorphic\textquotedblright; and replace
\textquotedblleft$G$ is belongs\textquotedblright\ by \textquotedblleft$G$
belongs\textquotedblright.
\end{enumerate}

\section{Corrections to the student study guide}

\noindent In this section, page numbers refer to the printed page numbers in
the booklet \textit{Abstract Algebra: A Study Guide for Beginners}
(downloadable from \url{https://www.johnabeachy.com/abstractalgebra} ); the
roman-numbered front matter is cited in the same way.

\begin{enumerate}
\item \textbf{Page 5:} Replace \textquotedblleft then are certainly
congruent\textquotedblright\ by \textquotedblleft then they are certainly
congruent\textquotedblright.

\item \textbf{Pages 6 and 94, Problem 1.2.39:} Replace \textquotedblleft
module $20$\textquotedblright\ by \textquotedblleft modulo $20$%
\textquotedblright.

\item \textbf{Page 9:} Replace \textquotedblleft if has an
element\textquotedblright\ by \textquotedblleft if it has an
element\textquotedblright.

\item \textbf{Pages 9 and 98, Problem 1.4.45:} In part (a), replace
\textquotedblleft for not\textquotedblright\ by \textquotedblleft but
not\textquotedblright. In part (c), replace \textquotedblleft such $(p-1)/2$
is even\textquotedblright\ by \textquotedblleft such that $(p-1)/2$ is
even\textquotedblright.

\item \textbf{Pages 15 and 108, Problem 2.1.35:} Replace \textquotedblleft
such a that\textquotedblright\ by \textquotedblleft such
that\textquotedblright.

\item \textbf{Pages 19 and 113, Problem 2.3.21:} Delete the trailing comma in
the cycle $(2,4,9,7,)$.

\item \textbf{Page 23:} Replace \textquotedblleft The process the
ended\textquotedblright\ by \textquotedblleft The process that
ended\textquotedblright.

\item \textbf{Pages 26 and 121, Problem 3.1.39:} For $c\neq0$, the expression
$(az+b)/(cz+d)$ has a pole, so it does not define a function $\mathbf{C}%
\rightarrow\mathbf{C}$. Replace $\mathbf{C}$ by the Riemann sphere
$\widehat{\mathbf{C}}$ and interpret the formula at the pole and at infinity,
or reformulate the problem with appropriate domains.

\item \textbf{Page 28, Problem 3.1.50:} Replace \textquotedblleft
definied\textquotedblright\ by \textquotedblleft defined\textquotedblright.

More importantly, the displayed operation is not closed on the stated set: its
first component $ad$ can vanish when $d=0$. Replace \textquotedblleft%
$(a,b)\ast(c,d)=(ad,bc+d)$\textquotedblright\ by \textquotedblleft%
$(a,b)\ast(c,d)=(ac,bc+d)$\textquotedblright\ in order to avoid this issue
(and obtain a more meaningful operation, too). This is the operation obtained
by identifying $(a,b)$ with the matrix $\left[
\begin{smallmatrix}
a & 0\\
b & 1
\end{smallmatrix}
\right]  $.

\item \textbf{Pages 29 and 125, Problem 3.2.38(a):} Replace \textquotedblleft
a subgroup of $G$\textquotedblright\ by \textquotedblleft a subgroup of
$\operatorname{Sym}(S)$\textquotedblright.

In the hint, replace \textquotedblleft$\sigma(V)=V$ for all $\sigma
\in\operatorname{Sym}(S)$\textquotedblright\ by \textquotedblleft$\sigma(V)=V$
for all $\sigma\in H$\textquotedblright.

\item \textbf{Page 36:} Replace \textquotedblleft and the
transferred\textquotedblright\ by \textquotedblleft and then
transferred\textquotedblright.

\item \textbf{Pages 39 and 146, Problem 3.5.34:} Replace \textquotedblleft Use
the the result\textquotedblright\ by \textquotedblleft Use the
result\textquotedblright.

\item \textbf{Pages 44 and 153, Problem 3.7.32:} Replace \textquotedblleft%
$\phi:G_{1}\rightarrow G_{2}$\textquotedblright\ by \textquotedblleft%
$\phi:G\rightarrow G$\textquotedblright.

\item \textbf{Pages 44 and 153, Problem 3.7.33:} Replace \textquotedblleft%
$\phi^{2}=\phi^{2}$\textquotedblright\ by \textquotedblleft$\phi^{2}=\phi
$\textquotedblright.

\item \textbf{Pages 57 and 172, Problem 4.2.33:} Insert the missing closing
parentheses in \textquotedblleft$\deg(f(x)$\textquotedblright\ and
\textquotedblleft$\deg(g(x)$\textquotedblright, and replace \textquotedblleft
such $f$\textquotedblright\ by \textquotedblleft such that $f$%
\textquotedblright.

\item \textbf{Pages 65 and 187, Problem 5.1.37(a):} Replace \textquotedblleft%
$a$ an idempotent\textquotedblright\ by \textquotedblleft$a$ is an
idempotent\textquotedblright.

\item \textbf{Pages 65 and 187, Problems 5.1.38 and 5.1.39:} Replace
\textquotedblleft an unit\textquotedblright\ by \textquotedblleft a
unit\textquotedblright.

\item \textbf{Page 69, Answers and Hints:} Replace \textquotedblleft$R$ is an
field\textquotedblright\ by \textquotedblleft$R$ is a field\textquotedblright.

\item \textbf{Pages 72 and 203, Problem 5.3.48(e):} Replace \textquotedblleft
If $L$ is and ideal\textquotedblright\ by \textquotedblleft If $L$ is an
ideal\textquotedblright.

\item \textbf{Page 72, Problem 5.3.54:} The strict inclusion \textquotedblleft%
$I_{n}\subset I_{m}$\textquotedblright\ cannot hold whenever $n\leq m$, since
equality of the indices is allowed. Either replace $n\leq m$ by $n<m$, or
replace $\subset$ by $\subseteq$.

\item \textbf{Page 73, Problem 5.3.60(b):} Delete the unmatched opening
parenthesis on the left-hand side. The assertion should read
\[
I(J\cap K)\subseteq IJ\cap IK.
\]


\item \textbf{Page 80:} Delete the duplicated \textquotedblleft
to\textquotedblright\ in \textquotedblleft one way to to
construct\textquotedblright, and replace \textquotedblleft passing
though\textquotedblright\ by \textquotedblleft passing
through\textquotedblright.

\item \textbf{Page 80, Theorem:} Replace \textquotedblleft$n=2^{k}p_{2}\cdots
p_{m}$\textquotedblright\ by \textquotedblleft$n=2^{k}p_{1}p_{2}\cdots p_{m}%
$\textquotedblright.

\item \textbf{Page 86, solution to Problem 1.1.38:} Replace \textquotedblleft%
$3=\left(  1\right)  (n^{2}+n+1)+(-n-2)(n+1)$\textquotedblright\ by
\textquotedblleft$3=\left(  1\right)  (n^{2}+n+1)+(-n-2)(n-1)$%
\textquotedblright.

\item \textbf{Page 86, solution to Problem 1.1.40:} After writing $m=2k+1$ and
$n=2q+1$, the expansion should use powers of $k$ and $q$, not powers of $m$
and $n$. In particular,
\[
(2k+1)^{4}=16k^{4}+32k^{3}+24k^{2}+8k+1,
\]
and similarly with $q$.

\item \textbf{Page 90, solution to Problem 1.2.37:} Replace \textquotedblleft
Theorem 2.6\textquotedblright\ by \textquotedblleft Theorem
1.1.6\textquotedblright.

\item \textbf{Page 92, solution to Problem 1.3.32:} Replace \textquotedblleft
if an only if\textquotedblright\ by \textquotedblleft if and only
if\textquotedblright.

\item \textbf{Page 93, alternate solution to Problem 1.3.35:}
\textquotedblleft substitute for $x$ in the second
congruence\textquotedblright\ should be \textquotedblleft substitute for $x$
in the first congruence\textquotedblright.

\item \textbf{Page 94, solution to Problem 1.3.39:} In the sentence
introducing the additive order of $[5]$, replace \textquotedblleft the
solution of $4x\equiv0\pmod{20}$\textquotedblright\ by \textquotedblleft the
solution of $5x\equiv0\pmod{20}$\textquotedblright.

\item \textbf{Page 98, solution to Problem 1.4.45(c):} Replace
\textquotedblleft$x\equiv k\pmod p$\textquotedblright\ by \textquotedblleft%
$x\equiv k!\pmod p$\textquotedblright.

\item \textbf{Page 99, solution to Problem 1.4.46:} The multiplicative
argument takes place in the group of units. Replace each relevant occurrence
of \textquotedblleft$\mathbf{Z}_{35}$\textquotedblright\ by \textquotedblleft%
$\mathbf{Z}_{35}^{\times}$\textquotedblright.

\item \textbf{Page 101, review solutions:} In the solution to Problem 3,
replace \textquotedblleft$\operatorname{lcm}[1275,495)$\textquotedblright\ by
\textquotedblleft$\operatorname{lcm}(1275,495)$\textquotedblright.

In the solution to Problem 5, dividing the congruence by $8$ gives
$3x\equiv21\pmod{25}$, not the original congruence $24x\equiv168\pmod{200}$.

In the solution to Problem 6, delete the duplicated \textquotedblleft
the\textquotedblright.

\item \textbf{Page 104, solution to Problem 2.1.27:} Replace \textquotedblleft%
$\lbrack2x_{1}]_{8}=[x_{2}]_{8}$\textquotedblright\ by \textquotedblleft%
$\lbrack2x_{1}]_{8}=[2x_{2}]_{8}$\textquotedblright.

Also delete the duplicated \textquotedblleft but\textquotedblright.

\item \textbf{Page 105, solution to Problem 2.1.29(c):} Repair the malformed
bracket and extraneous parenthesis in the verification.

\item \textbf{Page 106, solutions to Problems 2.1.31--2.1.32:} In the solution
to Problem 2.1.31, replace $\{f(x_{1}\}$ by $\{f(x_{1})\}$. In parts (a) and
(c) of the solution to Problem 2.1.32, replace $f(x)$ by $f(s)$ where the
element under discussion is denoted by $s$.

\item \textbf{Page 107, solution to Problem 2.1.34:} The composite $KL$ is the
identity transformation on $\mathbf{R}^{n}$, not on $\mathbf{R}^{m}$.

\item \textbf{Page 112, solution to Problem 2.2.22:} Replace \textquotedblleft%
$f^{-1}(P_{\alpha}\cap(P_{\beta})$\textquotedblright\ by \textquotedblleft%
$f^{-1}(P_{\alpha}\cap P_{\beta})$\textquotedblright.

\item \textbf{Page 113, solution to Problem 2.3.23:} Replace \textquotedblleft
cannot be less than $n$\textquotedblright\ by \textquotedblleft cannot be less
than $m$\textquotedblright.

\item \textbf{Page 116, solution to Review Problem 7:} Delete the
parenthetical sentence \textquotedblleft You could use $P=A$ instead of
$P=I$\textquotedblright. This is not valid for a singular matrix $A$.

\item \textbf{Page 118, solution to Problem 3.1.29:} The calculation of
\[
\left(  x_{1},y_{1},z_{1}\right)  \ast\left(  \left(  x_{2},y_{2}%
,z_{2}\right)  \ast\left(  x_{3},y_{3},z_{3}\right)  \right)
\]
has a missing operation symbol $\ast$ (on the second line, between the two triples), a
missing plus sign (on the third line, in \textquotedblleft$z_{3}x_{2}y_{3}%
$\textquotedblright) and a missing closing parenthesis (at the end of the
third line).

\item \textbf{Page 120, solution to Problem 3.1.36:} The formula
\textquotedblleft$x^{-1}=(axa)^{-1}$\textquotedblright\ at the end confuses
the original group inverse with the inverse for the new operation. Replace it
by: \textquotedblleft The inverse of $x$ with respect to $\ast$ is
$(axa)^{-1}$.\textquotedblright

\item \textbf{Page 120, solution to Problem 3.1.37:} Replace \textquotedblleft
Note the $e\cdot e$\textquotedblright\ by \textquotedblleft Note that $e\cdot
e$\textquotedblright.

\item \textbf{Page 121, solution to Problem 3.1.39:} In the displayed matrix
product, replace its lower-right entry \textquotedblleft$c_{2}b_{1}+d_{2}%
d_{2}$\textquotedblright\ by \textquotedblleft$c_{2}b_{1}+d_{2}d_{1}%
$\textquotedblright.

\item \textbf{Page 123, answer to Problem 3.1.42(b):} The stated set
$\mathbf{R}^{2}$ is not a group under
\[
(x_{1},y_{1})\ast(x_{2},y_{2})=(x_{1}x_{2},y_{1}x_{2}+y_{2}),
\]
since an element whose first component is $0$ has no inverse. Restrict the set
to $\{(x,y)\in\mathbf{R}^{2}\mid x\neq0\}$.

\item \textbf{Page 124, solution to Problem 3.2.34:} Replace \textquotedblleft
each nonzero element\textquotedblright\ by \textquotedblleft each nonidentity
element\textquotedblright.

\item \textbf{Page 124, solution to Problem 3.2.35:} In the final paragraph of
the solution to Problem 3.2.35, replace the two erroneous occurrences of
\textquotedblleft$H$\textquotedblright\ by \textquotedblleft$K$%
\textquotedblright.

\item \textbf{Page 125, solution to Problem 3.2.37:} \textquotedblleft If
$\sigma,\tau\in\operatorname{Sym}(S)$\textquotedblright\ should be
\textquotedblleft If $\sigma,\tau\in H$\textquotedblright.

\item \textbf{Page 125, solution to Problem 3.2.38:} After proving
$\sigma(V)=V$ for $\sigma\in H$, verify closure and inverses directly;
\textquotedblleft substitute $V$ for $x$\textquotedblright\ in the preceding
solution is not a valid argument (as the analogy to Problem 3.2.37 is somewhat thin).

\item \textbf{Page 131, solution to Problem 3.3.31:} Replace the final
\textquotedblleft$BA^{3}$\textquotedblright\ by \textquotedblleft$AB^{3}%
$\textquotedblright.

\item \textbf{Page 136, solution to Problem 3.3.42:} Replace the final
\textquotedblleft$\lbrack H:H\cap N]$\textquotedblright\ by \textquotedblleft%
$\lbrack H:H\cap K]$\textquotedblright.

\item \textbf{Page 136, solution to Problem 3.3.50:} Delete the duplicated
\textquotedblleft Answer:\textquotedblright.

\item \textbf{Page 139, solution to Problem 3.4.38:} Replace \textquotedblleft%
$z_{1}+z_{2}+x_{1}y+2$\textquotedblright\ by \textquotedblleft$z_{1}%
+z_{2}+x_{1}y_{2}$\textquotedblright.

\item \textbf{Page 140, solution to Problem 3.4.41:} In the homomorphism
calculation, the second components of all values of $\phi$ lie in
$\mathbf{Z}_{6}$, not in $\mathbf{Z}_{2}$. Replace the relevant subscripts
\textquotedblleft$2$\textquotedblright\ on those second components by
\textquotedblleft$6$\textquotedblright.

\item \textbf{Page 141, solution to Problem 3.4.43:} Replace the false
equality \textquotedblleft$\langle p^{n-1}\rangle=(p^{n}-1)\mathbf{Z}_{p^{n}}%
$\textquotedblright\ by \textquotedblleft$\langle p^{n-1}\rangle
=p^{n-1}\mathbf{Z}_{p^{n}}$\textquotedblright.

\item \textbf{Page 142, solution to Problem 3.4.46:} In the matrix
factorization, replace
\[%
\begin{bmatrix}
a_{1} & 0\\
0 & b_{1}%
\end{bmatrix}%
\begin{bmatrix}
a_{2} & 0\\
0 & b_{2}%
\end{bmatrix}
\]
by
\[%
\begin{bmatrix}
a_{1} & 0\\
0 & a_{2}%
\end{bmatrix}%
\begin{bmatrix}
b_{1} & 0\\
0 & b_{2}%
\end{bmatrix}
.
\]
In the alternate solution, replace \textquotedblleft$\operatorname{GL}%
_{2}(\mathbf{Z}_{p})=HK$\textquotedblright\ by \textquotedblleft the subgroup
of diagonal matrices in $\operatorname{GL}_{2}(\mathbf{Z}_{p})$ is $HK$ and we
have\textquotedblright.

\item \textbf{Page 147, answer to Problem 3.5.40:} Replace \textquotedblleft
One possibility it to\textquotedblright\ by \textquotedblleft One possibility
is to\textquotedblright.

\item \textbf{Page 149, solution to Problem 3.6.38:} Replace \textquotedblleft
elments\textquotedblright\ by \textquotedblleft elements\textquotedblright.

\item \textbf{Page 153, solution to Problem 3.7.31:} Replace
\textquotedblleft for every divisor of $k$\textquotedblright\ by
\textquotedblleft for every divisor $k$ of $m$\textquotedblright.

\item \textbf{Page 153, solution to Problem 3.7.32(b):} \textquotedblleft Let
$y\in G$\textquotedblright\ should be \textquotedblleft Let $g\in
G$\textquotedblright.

\item \textbf{Page 153, solution to Problem 3.7.33:} Replace%
\[%
\begin{bmatrix}
\det\left(  A\right)   & 0\\
0 & 1
\end{bmatrix}%
\begin{bmatrix}
\det\left(  A\right)   & 0\\
0 & 1
\end{bmatrix}
\]
by
\[%
\begin{bmatrix}
\det\left(  A\right)   & 0\\
0 & 1
\end{bmatrix}%
\begin{bmatrix}
\det\left(  B\right)   & 0\\
0 & 1
\end{bmatrix}
.
\]


\item \textbf{Page 154, solution to Problem 3.7.34:} In the alternate
solution to Problem 3.7.34, replace \textquotedblleft$\psi(G)$%
\textquotedblright\ by \textquotedblleft$\psi(G_{2})$\textquotedblright.

\item \textbf{Page 154, solution to Problem 3.7.36:} Replace
\textquotedblleft$gag^{-1}\in G$\textquotedblright\ by \textquotedblleft%
$gag^{-1}\in N$\textquotedblright.

\item \textbf{Page 155, solution to Problem 3.7.37:} In the computation of the
conjugacy class of $a^{2}$, replace \textquotedblleft$xax^{-1}$%
\textquotedblright\ by \textquotedblleft$xa^{2}x^{-1}$\textquotedblright.

\item \textbf{Page 158, solution to Problem 3.8.43:} Replace \textquotedblleft
cosets of $N$\textquotedblright\ by \textquotedblleft cosets of $H$%
\textquotedblright. 

\item \textbf{Page 158, solution to Problem 3.8.45:} Replace \textquotedblleft%
$Nb=\{ab,a^{3}b,a^{6}b,a^{9}b\}$\textquotedblright\ by \textquotedblleft%
$Nb=\{b,a^{3}b,a^{6}b,a^{9}b\}$\textquotedblright.

\item \textbf{Page 160, solution to Problem 3.8.50:} The question concerns
conjugates of $a$, but the proof switches to $x$. Use one fixed element
throughout; for example,
\[
gag^{-1}=hah^{-1}\quad\Longleftrightarrow\quad h^{-1}g\in C(a).
\]


\item \textbf{Page 160, solution to Problem 3.8.52:} Also replace
\textquotedblleft classses\textquotedblright\ by \textquotedblleft
classes\textquotedblright.

\item \textbf{Page 172, alternate solution to Problem 4.2.31:} Replace
\textquotedblleft This give\textquotedblright\ by \textquotedblleft This
gives\textquotedblright.

\item \textbf{Page 174, solution to Problem 4.3.28:} In the multiplication
calculation, replace
\[
(ac-3bd)+(ad+bc)\sqrt{3}%
\]
by
\[
(ac-3bd)+(ad+bc)\sqrt{3}\,i.
\]


\item \textbf{Page 175, solution to Problem 4.3.30:} Delete the final comment.
It refers to $\mathbf{Z}_{3}[x]/\langle x^{2}+x-1\rangle$, whereas the field
just constructed is $\mathbf{Z}_{5}[x]/\langle x^{2}+x+1\rangle$; moreover,
the claim about all nonzero powers of $[x]$ is incompatible with the order-$3$
calculation on the same page.

\item \textbf{Page 176, solution to Problem 4.3.34:} Replace \textquotedblleft%
$x^{2}+bc+1$\textquotedblright\ by \textquotedblleft$x^{2}+bx+1$%
\textquotedblright.

\item \textbf{Page 178, answer to Problem 4.4.35:} Replace \textquotedblleft
Problems 34 and 35\textquotedblright\ by \textquotedblleft Problems 33 and
34\textquotedblright.

\item \textbf{Page 181, solution to Review Problem 7(b):} In the
back-substitution, replace
\[
(x+1)(x^{3}+x^{2}+2x+1)
\]
by
\[
(x+1)(x^{3}+2x^{2}+x+1).
\]


\item \textbf{Page 185, solution to Problem 5.1.35:} In part (a), the list of
units of $\mathbf{Z}_{24}$ omits $17$. The complete list is
\[
\{1,5,7,11,13,17,19,23\}.
\]
In part (b), the conditions modulo $3$ and modulo $8$ must both hold, so
replace \textquotedblleft or else\textquotedblright\ by \textquotedblleft and
furthermore\textquotedblright.

\item \textbf{Page 186, solution to Problem 5.1.36(a):} In the lower-left
entry of the first displayed matrix product, replace \textquotedblleft%
$cd+bd+af$\textquotedblright\ by \textquotedblleft$cd+be+af$\textquotedblright.

\item \textbf{Page 186, solution to Problem 5.1.36(b):} The displayed inverse
of the matrix with diagonal entries $-1$ should be
\[%
\begin{bmatrix}
-1 & 0 & 0\\
b & -1 & 0\\
c & b & -1
\end{bmatrix}
^{-1}=%
\begin{bmatrix}
-1 & 0 & 0\\
-b & -1 & 0\\
-c-b^{2} & -b & -1
\end{bmatrix}
.
\]


\item \textbf{Page 186, solution to Problem 5.1.36(e):} The product in part
(e) has the same erroneous term \textquotedblleft$bd$\textquotedblright\ as
the one in part (a); replace it by \textquotedblleft$be$\textquotedblright.

\item \textbf{Page 187, solution to Problem 5.1.36(e):} This uses the
convention that zero divisors must be nonzero. But the book does allow $0$ to
count as a zero divisor (unless the ring is trivial). Some work is needed to
square this conflict of definitions.

\item \textbf{Page 188, solution to Problem 5.1.40(a):} Delete the extra
closing parenthesis after \textquotedblleft$=ab+ac-2a^{2}bc$\textquotedblright%
, and replace \textquotedblleft$S$ is Boolean ring\textquotedblright\ by
\textquotedblleft$S$ is a Boolean ring\textquotedblright.

\item \textbf{Page 189, solution to Problem 5.1.43:} Delete the extra $x$ in
\textquotedblleft$b_{n-1}x^{n-1}x+\cdots$\textquotedblright.

More importantly, the final comment is false: it is not enough that each
nonzero coefficient separately be a zero divisor. The correct general
criterion is that $f\in R[x]$ is a zero divisor if and only if there is one
nonzero $c\in R$ that annihilates every coefficient of $f$.

\item \textbf{Page 192, solution to Problem 5.2.30:} Replace \textquotedblleft%
$\phi_{1}\neq0$\textquotedblright\ by \textquotedblleft$\phi(1)\neq
0$\textquotedblright.

\item \textbf{Page 193, solution to Problem 5.2.34(a):} Replace
\textquotedblleft$Z_{8}$\textquotedblright\ by \textquotedblleft%
$\mathbf{Z}_{8}$\textquotedblright\ (twice).

\item \textbf{Page 195, solution to Problem 5.2.38:} Delete the extra closing
parenthesis after \textquotedblleft$\left(  m_{1}m_{2}-n_{1}n_{2}\right)
-\left(  m_{1}n_{2}+m_{2}n_{1}\right)  i$\textquotedblright.

\item \textbf{Page 196, solution to Problem 5.2.44(a):} Replace
\textquotedblleft$a^{b}b^{2}$\textquotedblright\ by \textquotedblleft%
$a^{2}b^{2}$\textquotedblright.

\item \textbf{Page 197, solution to Problem 5.2.46:} The second paragraph of
the solution omits the zero homomorphism. Add the case
\[
\phi(1)=0,\qquad\operatorname{im}\phi=\{0\},\qquad\ker\phi=\mathbf{Z}_{120}.
\]


\item \textbf{Page 198, solution to Problem 5.3.34:} Replace the malformed
characterization of $I$ by
\[
I=\{f\in\mathbf{R}[x]\mid f(0)=f^{\prime}(0)=0\}.
\]


\item \textbf{Page 200, solution to Problem 5.3.42(a):} Replace
\textquotedblleft since $n\equiv m\pmod2$\textquotedblright\ by
\textquotedblleft since $a\equiv b\pmod2$\textquotedblright. In more detail:%
\[
\begin{aligned}
m\underbrace{a}_{\equiv b\pmod2}-\,n\underbrace{b}_{\equiv a\pmod2}&\equiv
mb-\underbrace{na}_{\equiv-na\pmod2}\\
&\equiv mb-\left(  -na\right) =na+mb\pmod2.
\end{aligned}
\]


\item \textbf{Page 201, solution to Problem 5.3.44:} Replace \textquotedblleft%
$a\in K$\textquotedblright\ by \textquotedblleft$a\in I$\textquotedblright.

\item \textbf{Page 202, solution to Problem 5.3.46(b):} Replace
\textquotedblleft Suppose that $P$ is a prime ideal of $R$\textquotedblright%
\ by \textquotedblleft Suppose that $P$ is a prime ideal of $S$%
\textquotedblright.

\item \textbf{Page 203, solution to Problem 5.3.48:} In parts (d) and (e),
replace \textquotedblleft The results holds\textquotedblright\ by
\textquotedblleft The result holds\textquotedblright.

\item \textbf{Page 205, solution to Problem 5.4.20:} Replace \textquotedblleft%
$\widehat{\phi_{1}\beta}\,\widehat{\phi_{2}\alpha}:Q(D_{1})\rightarrow
Q(D_{2})$\textquotedblright\ by \textquotedblleft$\widehat{\phi_{1}\beta
}\,\widehat{\phi_{2}\alpha}:Q(D_{1})\rightarrow Q(D_{1})$\textquotedblright.

Replace \textquotedblleft$\widehat{\phi_{2}\alpha}\,\widehat{\phi_{1}\beta
}:Q(D_{2})\rightarrow Q(D_{1})$\textquotedblright\ by \textquotedblleft%
$\widehat{\phi_{2}\alpha}\,\widehat{\phi_{1}\beta}:Q(D_{2})\rightarrow
Q(D_{2})$\textquotedblright.

\item \textbf{Page 206, review solutions:} In Problem 1(c), replace
\textquotedblleft Thus and element\textquotedblright\ by \textquotedblleft
Thus an element\textquotedblright. In Problem 1(e), replace \textquotedblleft
nozero\textquotedblright\ by \textquotedblleft nonzero\textquotedblright.

\item \textbf{Page 208, solution to Review Problem 8(b):} In the comparison of
products, replace \textquotedblleft$\phi\left(  \left(  c+di\right)  \right)
$\textquotedblright\ by \textquotedblleft$\theta\left(  \left(  c+di\right)
\right)  $\textquotedblright.

Also replace \textquotedblleft it is clear the $\theta$\textquotedblright\ by
\textquotedblleft it is clear that $\theta$\textquotedblright.

\item \textbf{Page 210, solution to Problem 6.1.21:} Replace \textquotedblleft
is a just an isomorphism\textquotedblright\ by \textquotedblleft is just an
isomorphism\textquotedblright.

\item \textbf{Page 211, solution to Problem 6.2.19:} From $[\mathbf{Q}%
(u):\mathbf{Q}]\mid7$, it follows only that the degree is $1$ or $7$. To prove
that it is actually $7$ (not $1$), argue as follows: For $\alpha=\sqrt[7]{2}$,
we have $u=\alpha^{4}+3\alpha^{3}\notin\mathbf{Q}$ by linear independence of
$1,\alpha,\ldots,\alpha^{6}$. Hence the degree is indeed $7$.

\item \textbf{Page 213, solution to Problem 6.3.7:} Replace \textquotedblleft
Comment We\textquotedblright\ by \textquotedblleft Comment:
We\textquotedblright.
\end{enumerate}


\end{document}