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\ihead{Errata to ``Mathematics via Problems 1'', 2021}
\ohead{\today}
\begin{document}

\begin{center}
\textbf{Mathematics via Problems, part 1: Algebra}

\textit{Arkadiy Skopenkov}

MSRI/AMS 2021

\textbf{Errata and addenda by Darij Grinberg (version of
%TCIMACRO{\TeXButton{today}{\today}}%
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\end{center}

\section{Errata}

Some of the items below are not corrections of literal mistakes but rather
suggestions written according to my taste and ideology. I hope they are
nevertheless helpful.

\subsection{Introduction}

\begin{itemize}
\item \textbf{page xx:} The definition of $A\sqcup B$ at the very bottom is
not really understandable if one has not seen disjoint unions of non-disjoint
sets before. I would suggest adding the formal definition, as I suspect that
many olympiad-trained students do not know this concept.
\end{itemize}

\subsection{Chapter 1}

\begin{itemize}
\item \textbf{1.1.3:} \textquotedblleft for any $a$ and $b$\textquotedblright%
\ $\rightarrow$ \textquotedblleft for all $a$ and $b$\textquotedblright. The
word \textquotedblleft any\textquotedblright\ is notoriously slippery in
mathematics, as it can mean both $\forall$ and $\exists$ depending on its
position in the sentence. In situations that are not completely clear-cut, I
prefer not to use it at all.

\item \textbf{1.1.3:} The solution to this uses problem 1.1.4. Why not switch
the order of the two exercises then?

\item \textbf{1.1.5:} The word \textquotedblleft or\textquotedblright\ is
unclear: Does it mean \textquotedblleft$2\nmid k$ or $3\nmid k$ or $5\nmid
k$\textquotedblright\ or \textquotedblleft$2\nmid k$ and $3\nmid k$ and
$5\nmid k$\textquotedblright? I suggest writing \textquotedblleft If none of
$2$, $3$ and $5$ divides $k$\textquotedblright.

\item \textbf{1.2.1 (a):} \textquotedblleft$a$\textquestiondown $1$%
\textquotedblright\ should be \textquotedblleft$a>1$\textquotedblright.

\item \textbf{1.2.1 (b):} After \textquotedblleft which are divisible by
$p_{i}$\textquotedblright, add \textquotedblleft and larger than $p_{i}%
$\textquotedblright.

\item \textbf{1.2.5 (c):} This has nothing to do with primes. (The only thing
necessary is that $p_{1}=2$ and $p_{2}<p_{3}<p_{4}<\cdots$ and that
$p_{2},p_{3},p_{4},\ldots$ are odd.) Doesn't this inconsequential puzzle
distract the reader here?

\item \textbf{1.2.8 (d):} This uses a later result (1.5.7 (c)). Why not swap
the order of the sections?

\item \textbf{1.3.3 (a):} This requires $b>0$; otherwise it should be $\left(
a,b\right)  =\left\vert b\right\vert $.

\item \textbf{between 1.3.5 and 1.3.6:} \textquotedblleft smallest
number\textquotedblright\ $\rightarrow$ \textquotedblleft smallest positive
integer\textquotedblright.

\item \textbf{solution to 1.3.3 (c):} Here you use the Euclidean algorithm,
which is only explained later (1.5.9 (b)). Maybe better restate this as an
inductive proof? (This would probably be shorter anyway.)

\item \textbf{hint to 1.4.1 (a):} Replace \textquotedblleft$0\leq
a\leq\left\vert b\right\vert $\textquotedblright\ by \textquotedblleft$0\leq
a<\left\vert b\right\vert $\textquotedblright. Also, replace \textquotedblleft
about $a-b$\textquotedblright\ by \textquotedblleft about $a-\left\vert
b\right\vert $\textquotedblright.

\item \textbf{1.5.5 (a):} Why write $\left(  \left\vert a\right\vert
,\left\vert b\right\vert \right)  $ instead of $\left(  a,b\right)  $ ?

\item \textbf{1.5.9 (b):} \textquotedblleft set $d:=a_{k}$\textquotedblright%
\ $\rightarrow$ \textquotedblleft set $d:=\left\vert a_{k}\right\vert
$\textquotedblright.

\item \textbf{1.5.10 (d):} I find it strange that such a crucial and
nontrivial fact is stated without a hint or proof.

\item \textbf{solution to 1.5.7 (c), \textquotedblleft Another
hint\textquotedblright:} \textquotedblleft so $p\mid ab$\textquotedblright%
\ should be \textquotedblleft so $p\mid a$\textquotedblright.

Next, \textquotedblleft let $p\leq ib$\textquotedblright\ should be
\textquotedblleft let $p<ib$\textquotedblright.

Finally, \textquotedblleft Note that $0\leq ib-p\leq b$\textquotedblright%
\ should be \textquotedblleft Note that $0<ib-p<b$\textquotedblright.

\item \textbf{solution to 1.5.8 (a):} The right hand side should be
$\left\vert n^{\left(  a,b\right)  }-1\right\vert $ (since $n^{\left(
a,b\right)  }-1$ can be negative but a gcd cannot).

\item \textbf{1.6.3:} When the number is called $n$, it is probably better to
use a different letter for the number of its prime factors.

\item \textbf{1.6.6 (a):} This must require $a,b>0$; otherwise, $a=-4$ and
$b=-9$ would be a counterexample.

\item \textbf{1.6.6 (d):} Same as for part (a) (unless $n$ is odd).
\end{itemize}

\subsection{Chapter 2}

\begin{itemize}
\item \textbf{2.1.1 (d), Alternative formulation:} \textquotedblleft$n^{p}%
-1$\textquotedblright\ should be \textquotedblleft$n^{p-1}-1$%
\textquotedblright.

\item \textbf{2.1.5 (c):} \textquotedblleft and $\varphi\left(  m\right)
$\textquotedblright\ $\rightarrow$ \textquotedblleft and if $\varphi\left(
m\right)  $\textquotedblright. (Otherwise, it sounds like $n$ should be
relatively prime to both $m$ and $\varphi\left(  m\right)  $\textquotedblright.

\item \textbf{between 2.1.6 and 2.1.7:} The relation to cryptography is
unlikely to be understood by anyone who does not already know about RSA.

\item \textbf{2.3.1 (d):} \textquotedblleft numbers\textquotedblright%
\ $\rightarrow$ \textquotedblleft positive integers\textquotedblright.

\item \textbf{last line of page 24:} Any references to these texts?

\item \textbf{solution to 2.4.1 (a):} An \textquotedblleft$\equiv
$\textquotedblright\ sign is missing between \textquotedblleft$\cdot\left(
8k+3\right)  $\textquotedblright\ and \textquotedblleft$2^{4k+2}%
$\textquotedblright\ on the second line of the long computation.

\item \textbf{solution to 2.4.2 (a):} An \textquotedblleft$\equiv
$\textquotedblright\ sign is missing between \textquotedblleft$\cdot\left(
8k-1\right)  $\textquotedblright\ and \textquotedblleft$\left(  -1\right)
2^{4k}$\textquotedblright\ on the second line of the long computation.

\item \textbf{solution to 2.4.5 (d):} Replace \textquotedblleft$\left[
\dfrac{py}{p}\right]  $\textquotedblright\ by \textquotedblleft$\left[
\dfrac{qy}{p}\right]  $\textquotedblright\ three times in this solution.

\item \textbf{2.5.6 (b):} Are you sure you want to leave this crucial and
difficult fact without proof or hint?

\item \textbf{solution to 2.5.6 (a):} \textquotedblleft may
by\textquotedblright\ $\rightarrow$ \textquotedblleft may be\textquotedblright.

\item \textbf{2.6.3 (b):} Add \textquotedblleft for $p>2$
prime\textquotedblright.

\item \textbf{page 29, footnote }$^{3}$\textbf{:} \textquotedblleft where $k$
satisfies\textquotedblright\ $\rightarrow$ \textquotedblleft where $u$
satisfies\textquotedblright.

\item \textbf{hint to 2.6.2 (a):} \textquotedblleft\textit{smallest nonzero
degree}\textquotedblright\ $\rightarrow$ \textquotedblleft\textit{smallest
nonzero-degree power}\textquotedblright.

\item \textbf{hint to 2.6.2 (c):} \textquotedblleft\textit{smallest nonzero
degree}\textquotedblright\ $\rightarrow$ \textquotedblleft\textit{smallest
nonzero-degree power}\textquotedblright.
\end{itemize}

\subsection{Chapter 3}

\begin{itemize}
\item \textbf{solution to 3.2.4 (b):} Don't forget to define $\varepsilon_{3}%
$. (I understand you want $\varepsilon_{3}:=\dfrac{-1+i\sqrt{3}}{2}$.)

\item \textbf{solution to 3.2.7 (a):} \textquotedblleft Since $p=2\alpha
-A^{2}$ and $\alpha$ are roots\textquotedblright\ $\rightarrow$
\textquotedblleft Since $p=2\alpha-A^{2}$ and since $\alpha$ is a
root\textquotedblright. There is no reason why $p$ should be a root!

\item \textbf{page 39, third paragraph:} \textquotedblleft function
$\overline{P}:R\rightarrow R$\textquotedblright\ should be \textquotedblleft
function $\overline{P}:\mathbb{R}\rightarrow\mathbb{R}$\textquotedblright.

\item \textbf{3.3.2:} This really needs a solution, to ensure the reader is
not left with a wrong definition.

\item \textbf{3.4.5:} It should be explained what the \textquotedblleft
unique\textquotedblright\ means here (unique up to scalar factors, not just up
to sign as for integers).

\item \textbf{3.5.11:} A whitespace is missing in \textquotedblleft
and$\left\{  y_{n}\right\}  $\textquotedblright.

\item \textbf{solution to 3.6.3:} This needs a definition of \textquotedblleft
lexicographic order\textquotedblright. On a related note, I don't expect many
readers not already familiar with the theory to come up with a useful notion
of \textquotedblleft multi-degree\textquotedblright\ on their own.

\item \textbf{3.7.2 (a):} Again, it should be explained what \textquotedblleft
uniqueness\textquotedblright\ means (unique up to multiplication of any factor
by $\pm1$ or $\pm i$).

\item \textbf{3.7.2 (d):} \textquotedblleft there exists $k$\textquotedblright%
\ $\rightarrow$ \textquotedblleft there exists a Gaussian integer
$k$\textquotedblright\ (the letter $k$ otherwise suggests an integer).

\item \textbf{3.7.3:} \textquotedblleft$Z\left[  \xi\right]  $%
\textquotedblright\ $\rightarrow$ \textquotedblleft$\mathbb{Z}\left[
\xi\right]  $\textquotedblright.

\item \textbf{3.7.4 (b), (c):} Add \textquotedblleft up to permuting the two
addends\textquotedblright.

\item \textbf{3.7.4:} What do you mean by \textquotedblleft section
3\textquotedblright?

\item \textbf{solution to 3.9.1:} I can't make anything out of this. I know
that Igor Pak gives a fairly nontrivial proof in his \textit{Lectures on
Discrete and Polyhedral Geometry} (
\url{https://www.math.ucla.edu/~pak/geompol8.pdf} , Theorem 3.2).
\end{itemize}

\subsection{Chapter 4}

\begin{itemize}
\item \textbf{first paragraph:} \textquotedblleft
subsection3.I\textquotedblright\ needs a whitespace.

\item \textbf{after 4.1.1:} The two definitions of a permutation (either as a
list of the elements in some order, or as a one-to-one mapping) are not
equivalent, unless the set is of the form $\left\{  1,2,\ldots,n\right\}  $
for some $n\in\mathbb{N}$. You use the latter definition.

\item \textbf{page 59:} This would be a good place to define a graph. In a
book that defines divisibility, I wouldn't expect the reader to know what a
graph is! It doesn't help that there are several inequivalent notions of graph
(directed/undirected, simple/multi) around that don't always behave identically.

\item \textbf{4.1.4:} After \textquotedblleft For any permutation
$f$\textquotedblright, add \textquotedblleft of a finite set\textquotedblright.

\item \textbf{4.1.6:} \textquotedblleft order of the
composition\textquotedblright\ $\rightarrow$ \textquotedblleft order of a
composition\textquotedblright.

\item \textbf{4.1.8 (b):} \textquotedblleft cyclic
decomposition\textquotedblright\ has not been defined.

\item \textbf{4.2.1 (c):} The 15-puzzle was invented by Noyes Palmer Chapman
in New York State; I think its only connection to Russia is its popularity in
the Soviet Union.

\item \textbf{4.2.1 (c):} \textquotedblleft by sequentially moving the squares
to an open square\textquotedblright\ is not really a clear description of the
allowed moves.

\item \textbf{4.2.4:} Don't talk about \textquotedblleft a
permutation\textquotedblright\ each time. Rather, say \textquotedblleft Given
a permutation $\sigma$ of a finite set\textquotedblright\ at the beginning,
and then speak of $\sigma$. Otherwise, it sounds like you have an implicit
$\forall\sigma$ quantifier in front of each statement, which is not what you
intend (you want $\sigma$ to be fixed).

\item \textbf{page 65:} \textquotedblleft Distribute trains between stations
so that at each station we place all the different trains that can be obtained
from a single coloring of the carousel (by decoupling two cars in the
carousel). Then the required number $Z$ of colorings is equal to the number of
stations.\textquotedblright: I would be completely lost trying to understand
this if I did not already know what this is supposed to explain (the partition
of the set of all painted trains into rotation-equivalence classes). I think
it would be much clearer to talk about equivalence classes instead of trying
to invent nicknames for them. For the sake of clarity, an example would work wonders...
\end{itemize}

\subsection{Chapter 5}

\begin{itemize}
\item \textbf{page 69, definition of concavity:} This definition of
\textquotedblleft concave up\textquotedblright\ is nonstandard (the standard
definition would require $f\left(  sx+ty\right)  \leq sf\left(  x\right)
+tf\left(  y\right)  $ for any $x,y\in I$ and $s,t\in\lbrack0,\infty)$
satisfying $s+t=1$), and allows for \textquotedblleft Cauchy
monsters\textquotedblright\ (nowhere continuous functions that satisfy
$f\left(  x+y\right)  =f\left(  x\right)  +f\left(  y\right)  $ for all
$x,y\in\mathbb{R}$). I'm not sure if problem 5.1.2 can even be solved with
this definition (the minimum is clear, but the maximum?).

\item \textbf{5.1.7 (c):} \textquotedblleft A function with\textquotedblright%
\ $\rightarrow$ \textquotedblleft A function $f$ with\textquotedblright.

\item \textbf{5.1.9 (a):} \textquotedblleft$f:I\rightarrow R$%
\textquotedblright\ should be \textquotedblleft$f:I\rightarrow\mathbb{R}%
$\textquotedblright.

\item \textbf{5.1.9 (a):} \textquotedblleft convex down\textquotedblright%
\ $\rightarrow$ \textquotedblleft concave up\textquotedblright\ (that's how
you called it when you defined it). And again, this seems to require the
standard definition.

\item \textbf{page 72, line 2:} The word \textquotedblleft
Therefore\textquotedblright\ here is inappropriate: The fact that we get to
$\dfrac{1}{n},\dfrac{1}{n},\ldots,\dfrac{1}{n}$ after no more than $n$
replacements (actually $n-1$ are enough) has nothing to do with the product increasing.

\item \textbf{solution to 5.1.2:} Remove the \textquotedblleft%
(a)\textquotedblright; this problem has no parts.

\item \textbf{5.2.4:} It is important to notice that the variables in this
problem are not assumed by default to be nonnegative, except for the
$y_{1},y_{2},\ldots,y_{n}$ in part (d) (which are assumed to be positive).
Otherwise, part (a) would be false (assuming $at^{2}+2bt+c\geq0$ only for
$t\geq0$ does not guarantee $b^{2}\leq ac$).

\item \textbf{5.2.4 (a):} \textquotedblleft for any\textquotedblright%
\ $\rightarrow$ \textquotedblleft for all\textquotedblright.

\item \textbf{Hint to 5.2.3 (c):} A non-Russian reference would be helpful to
the Anglophone reader. \href{https://arxiv.org/abs/2201.01270}{Melvyn
Nathanson's arXiv:2201.01270 preprint} is one possible reference.

\item \textbf{solution to 5.3.2 (a):} \textquotedblleft The case of integer
values\textquotedblright\ is irrelevant, since all variables are positive by assumption.

\item \textbf{solution to 5.3.7 (a):} On the first line of the computation,
\textquotedblleft$\dfrac{c^{2}}{\left(  d+a\right)  }$\textquotedblright%
\ should be \textquotedblleft$\dfrac{c^{2}}{c\left(  d+a\right)  }%
$\textquotedblright.
\end{itemize}

\subsection{Chapter 6}

\begin{itemize}
\item \textbf{between 6.1.3 and 6.1.4:} \textquotedblleft$\Delta\left(
\Delta\left(  \cdots\right)  \right)  a_{n}$\textquotedblright\ should be
\textquotedblleft$\Delta\left(  \Delta\left(  \cdots\left(  a_{n}\right)
\right)  \right)  $\textquotedblright.

\item \textbf{solution to 6.1.2 (b):} The displayed equation%
\[
\Delta\dfrac{1}{n\left(  n+1\right)  \cdots\left(  n+\left(  k+1\right)
\right)  }=-\dfrac{k+2}{\left(  n+1\right)  \left(  n+2\right)  \cdots\left(
n+k+1\right)  }%
\]
should be%
\[
\Delta\dfrac{1}{n\left(  n+1\right)  \cdots\left(  n+\left(  k+1\right)
\right)  }=-\dfrac{k+2}{n\left(  n+1\right)  \cdots\left(  n+k+2\right)  }.
\]
The error, alas, worms its way through the entire solution. In particular, the
answer, too, is false.

\item \textbf{6.2.6 (c):} You are using the letter $k$ for two different
things here (the degree of the polynomial and the number of \textbf{distinct} roots).

\item \textbf{6.5.10 (b):} The sum should start at $n=0$, not at $n=1$.

\item \textbf{6.6.5 (b):} In the definition of $d_{n}\left(  q\right)  $,
replace \textquotedblleft$w_{1},\ldots,w_{q}>0$\textquotedblright\ by
\textquotedblleft$w_{1},\ldots,w_{q}\geq0$\textquotedblright.

\item \textbf{page 102, proof of Mahler's theorem 6.6.7:} \textquotedblleft
Assume the converse\textquotedblright\ $\rightarrow$ \textquotedblleft Assume
the contrary\textquotedblright.

\item \textbf{page 103, first displayed equation:} Again replace
\textquotedblleft$w_{1},\ldots,w_{q}>0$\textquotedblright\ by
\textquotedblleft$w_{1},\ldots,w_{q}\geq0$\textquotedblright.

\item \textbf{page 103:} After \textquotedblleft It is clear that
$d_{n}\left(  q\right)  =0$ if and only if $n$ has more than $q$ ones in its
binary expansion\textquotedblright, add \textquotedblleft or $n<q$%
\textquotedblright.
\end{itemize}

\subsection{Chapter 7}

\begin{itemize}
\item \textbf{page 108, second paragraph:} \textquotedblleft defined a similar
way\textquotedblright\ $\rightarrow$ \textquotedblleft defined in a similar
way\textquotedblright.

\item \textbf{\S 7.2, first paragraph:} \textquotedblleft called the
\textit{change of sign}\textquotedblright\ $\rightarrow$ \textquotedblleft
called a \textit{change of sign}\textquotedblright.

\item \textbf{7.2.7 (a):} \textquotedblleft the maximum and
minimum\textquotedblright\ $\rightarrow$ \textquotedblleft a maximum and a
minimum\textquotedblright\ (these are neither unique nor identical).

\item \textbf{7.2.7 (b):} \textquotedblleft lies the root\textquotedblright%
\ $\rightarrow$ \textquotedblleft lies a root\textquotedblright.

\item \textbf{Theorem 7.2.8 (b):} It is strange to speak of a $c\in\left[
a,b\right]  $ here, since $f^{\left(  n\right)  }$ is constant and thus
$f^{\left(  n\right)  }\left(  c\right)  $ does not depend on $c$. Perhaps you
don't want to require $f$ to have degree $n$ ?

\item \textbf{7.3.1 (a):} The words \textquotedblleft convex
hull\textquotedblright\ appear here for the first time in the book; the reader
might not be familiar with them.

\item \textbf{page 116, first two paragraphs:} \textquotedblleft A point $x\in
R$\textquotedblright\ $\rightarrow$ \textquotedblleft A point $x\in\mathbb{R}%
$\textquotedblright\ (twice).

\item \textbf{page 116, fourth paragraph:} \textquotedblleft of the
function\textquotedblright\ $\rightarrow$ \textquotedblleft of the function
$f$\textquotedblright.

\item \textbf{7.3.5 (c):} \textquotedblleft polynomial\textquotedblright%
\ $\rightarrow$ \textquotedblleft nonzero polynomial\textquotedblright.

\item \textbf{7.3.10 (b):} \textquotedblleft$p\left(  a\right)  \neq
0$\textquotedblright\ should probably be \textquotedblleft$p\left(  0\right)
\neq0$\textquotedblright.

\item \textbf{7.5.1:} The period in \textquotedblleft$0001.$\textquotedblright%
\ should be outside of the quotation marks.
\end{itemize}

\subsection{Chapter 8}

\begin{itemize}
\item \textbf{\S 1.C, first line:} \textquotedblleft\textit{expressible by
radicals}\textquotedblright\ $\rightarrow$ \textquotedblleft%
\textit{expressible in real radicals}\textquotedblright.

\item \textbf{page 127, bullet point at the top:} This equivalence is not
obvious, because the polynomials $p_{0},p_{1},\ldots,p_{s}$ can emulate
addition, subtraction and multiplication but not obviously division. I
understand that division is unnecessary because all the constructed real
numbers are algebraic over $\mathbb{Q}$, but this is not obvious to the reader
at this point.

\item \textbf{Remark 8.1.9:} \textquotedblleft none of the
roots\textquotedblright\ $\rightarrow$ \textquotedblleft all of the
roots\textquotedblright.

\item \textbf{between Remark 8.1.9 and Theorem 8.1.10:} \textquotedblleft A
polynomial\textquotedblright\ $\rightarrow$ \textquotedblleft A nonconstant
polynomial\textquotedblright.

\item \textbf{Conjecture 8.1.11 (a):} After \textquotedblleft(defined after
problem 3.2.6 (b)\textquotedblright, add a closing parenthesis.

Also, it is worth stating a precise definition of the resolvent cubic. It is
only vaguely hinted at in problem 3.2.6 (b).

\item \textbf{Conjecture 8.1.11 (b):} I. M. Isaacs, in
\href{https://doi.org/10.1080/00029890.1985.11971682}{his paper
\textit{Solution of Polynomials by Real Radicals} (American Mathematical
Monthly \textbf{92} (1985), issue 8)}, shows the following fact: If an
irreducible polynomial $f\in\mathbb{Q}\left[  x\right]  $ (irreducible over
$\mathbb{Q}$) splits over $\mathbb{R}$ (that is, all its complex roots are
real) and has a root that is expressible in real radicals (Isaacs calls this
\textquotedblleft real radical\textquotedblright), then $\deg f$ is a power of
$2$. Applying this to the minimal polynomial of $\cos\dfrac{2\pi}{n}$ (which
has degree $\varphi\left(  n\right)  /2$ because all the $\varphi\left(
n\right)  /2$ numbers $\cos\dfrac{2\pi k}{n}$ where $k$ is coprime to $n$ are
algebraic conjugates of $\cos\dfrac{2\pi}{n}$\ \ \ \ \footnote{Alternatively,
we can argue that this polynomial is the $n$-th cyclotomic Chebyshev
polynomial of the first kind (whose roots are the $\varphi\left(  n\right)
/2$ numbers $\cos\dfrac{2\pi k}{n}$ where $k$ is coprime to $n$).}), we see
that $\varphi\left(  n\right)  /2$ is a power of $2$ if $\cos\dfrac{2\pi}{n}$
is expressible in real radicals. But this quickly yields that $\cos\dfrac
{2\pi}{n}$ is real constructive by Theorem 8.1.5. This proves Conjecture
8.1.11 (b).

\item \textbf{Remark 8.2.1 (b):} What does this mean? Are you just saying that
we can construct any polynomial of the form $P\left(  \sigma_{1},\sigma
_{2},\ldots,\sigma_{n}\right)  $ ?

\item \textbf{Remark 8.2.1 (d):} Replace \textquotedblleft$sigma_{n}%
$\textquotedblright\ by \textquotedblleft$\sigma_{n}$\textquotedblright%
\ (missing backslash).

\item \textbf{8.2.4 (b):} \textquotedblleft mulitiplication\textquotedblright%
\ $\rightarrow$ \textquotedblleft multiplication\textquotedblright.

\item \textbf{page 138, solution to 8.2.7 (a):} On the very last line of the
page, a closing parenthesis is missing after the product.

\item \textbf{page 143, line 2:} \textquotedblleft such that $mx+ny=1$%
\textquotedblright\ $\rightarrow$ \textquotedblleft such that $nx+my=1$%
\textquotedblright.

\item \textbf{page 143, proof of constructibility:} It should be explained
that $\varepsilon$ means $\varepsilon_{n}$ here.

\item \textbf{page 143, proof of constructibility:} At the end of the last
display, \textquotedblleft for any $k$\textquotedblright\ should be
\textquotedblleft for any $k\not \equiv 0\operatorname{mod}n$%
\textquotedblright.

\item \textbf{page 143, proof of constructibility:} \textquotedblleft Consider
the polynomial $x+a_{2}x^{2}+\cdots+a_{n-1}x^{n-1}$\textquotedblright%
\ $\rightarrow$ \textquotedblleft Consider the polynomial $a_{0}+a_{1}%
x+a_{2}x^{2}+\cdots+a_{n-1}x^{n-1}$\textquotedblright.

\item \textbf{page 143, end of \S 2.F:} The same argument that you use to
prove 8.1.5 here also proves 8.1.15 (a). This is worth mentioning, since the
other proof you give on page 146 is much more complicated and hard to follow.

\item \textbf{page 144, sketch of the proof:} This is not a sketch of a proof,
but rather an approach that you only finish later. Calling it a
\textquotedblleft sketch\textquotedblright\ is confusing.

\item \textbf{8.2.18 (c):} \textquotedblleft(in residues modulo $2^{m}%
$)\textquotedblright\ $\rightarrow$ \textquotedblleft(in residues modulo
$2^{m-1}$)\textquotedblright.

\item \textbf{8.2.20 (a):} \textquotedblleft$4k+g^{4l+2}$\textquotedblright%
\ should be \textquotedblleft$g^{4k}+g^{4l+2}$\textquotedblright.

\item \textbf{page 146:} There are several typos in this proof and I cannot
completely follow it (although I understand the idea). I think a lot of it
would be simplified if you gave names to the groups and cosets that you are
summing over; with all the groups being cyclic, this hardly requires any
serious abstraction. Anyway, here is a short (perhaps incomplete) list of typos:

\begin{itemize}
\item There is a $j$ appearing sometimes. Is it a synonym for $k$ ?

\item Replace \textquotedblleft$gb_{0}$\textquotedblright\ by
\textquotedblleft$g^{b_{0}}$\textquotedblright.

\item Replace \textquotedblleft$g^{2j}$\textquotedblright\ by
\textquotedblleft$g^{2^{j}}$\textquotedblright.

\item Replace \textquotedblleft$gb_{0}^{\prime}$\textquotedblright\ by
\textquotedblleft$g^{b_{0}^{\prime}}$\textquotedblright.
\end{itemize}

\item \textbf{page 152, hint to 8.3.4 (d):} \textquotedblleft$\mathbb{Q}%
\left[  \sqrt{2}\right]  $\textquotedblright\ should be \textquotedblleft%
$\mathbb{Q}\left[  r\right]  $\textquotedblright.

\item \textbf{page 153, solution to 8.3.2:} On the last line, remove the
\textquotedblleft$b$\textquotedblright\ in \textquotedblleft$a_{1}%
+b\sqrt{a_{2}^{2}c}$\textquotedblright.

\item \textbf{page 157, solution to 8.3.12:} How exactly? (And I'm not sure
how helpful this is, seeing that you give no hint to 8.4.1.)

\item \textbf{solution to 8.3.15 (d):} \textquotedblleft$x^{2}-2-\sqrt{3}%
$\textquotedblright\ and \textquotedblleft$x^{2}-2+\sqrt{3}$\textquotedblright%
\ should be \textquotedblleft$x^{2}-1-\sqrt{3}$\textquotedblright\ and
\textquotedblleft$x^{2}-1+\sqrt{3}$\textquotedblright, respectively.

\item \textbf{solution to 8.3.18 (c):} \textquotedblleft Divide $x^{3}-r^{3}%
$\textquotedblright\ $\rightarrow$ \textquotedblleft Divide by $x^{3}-r^{3}%
$\textquotedblright.

\item \textbf{8.3.32:} Please don't use the notation $f^{\prime}$ for
something that isn't the derivative of $f$ (or at least explain that it does
not mean the derivative here).

\item \textbf{8.3.32 (c):} There is an unnecessary linebreak before
\textquotedblleft$f_{s}$\textquotedblright.

\item \textbf{8.3.33:} What does \textquotedblleft only one
way\textquotedblright\ mean?

\item \textbf{page 182, proof of Lemma 8.4.13:} In the long equality that
defines $\rho$, should the \textquotedblleft$\varepsilon_{q}^{\left(
1-k\right)  l}$\textquotedblright\ in the numerator perhaps be
\textquotedblleft$\varepsilon_{q}^{\left(  1-q\right)  l}$\textquotedblright?
\end{itemize}


\end{document}