\documentclass[numbers=enddot,12pt,final,onecolumn,notitlepage]{scrartcl}% \usepackage[headsepline,footsepline,manualmark]{scrlayer-scrpage} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsmath} \usepackage{amsthm} \usepackage{color} \usepackage{hyperref} \usepackage[sc]{mathpazo} \usepackage[T1]{fontenc} \usepackage[utf8]{inputenc} \usepackage{enumitem} \usepackage{graphicx}% \setcounter{MaxMatrixCols}{30} %TCIDATA{OutputFilter=latex2.dll} %TCIDATA{Version=5.50.0.2960} %TCIDATA{LastRevised=Tuesday, July 28, 2026 22:06:13} %TCIDATA{} %TCIDATA{} %TCIDATA{BibliographyScheme=Manual} %BeginMSIPreambleData \providecommand{\U}[1]{\protect\rule{.1in}{.1in}} %EndMSIPreambleData \setlength\textheight{22.5cm} \setlength\textwidth{15cm} \setlength\oddsidemargin{0.5cm} \setlength\evensidemargin{0.5cm} \setlength\topmargin{-1cm} \clearpairofpagestyles \ihead{Errata to \textit{MA3D5 Galois Theory}} \ohead{\today} \cfoot{\pagemark} \newcommand{\booktitle}{\textit{MA3D5 Galois Theory}} \newcommand{\ZZ}{\mathbb Z} \newcommand{\QQ}{\mathbb Q} \newcommand{\RR}{\mathbb R} \newcommand{\CC}{\mathbb C} \newcommand{\FF}{\mathbb F} \newcommand{\Aut}{\operatorname{Aut}} \newcommand{\Gal}{\operatorname{Gal}} \newcommand{\id}{\operatorname{id}} \newcommand{\lcm}{\operatorname{lcm}} \begin{document} \begin{center} \textbf{MA3D5 Galois Theory} \textit{Samir Siksek} 65-page PDF generated on 28 October 2019 \url{https://samirsiksek.github.io/siksek.github.io/teaching/gt/whole.pdf} \textbf{Errata and comments} by Darij Grinberg \end{center} \noindent The page numbers below refer to the printed page numbers in the supplied PDF. (The cover is PDF page~1 and the first chapter begins on printed page~3.) This list is not claimed to be exhaustive. Items marked ``substantive'' correct a mathematical assertion or repair a proof gap. Items marked ``pedagogical'' are not necessarily errors, but seem likely to cause avoidable difficulty for a student meeting Galois theory for the first time. The remaining items are local corrections or clarifications. Most of the errata below were found with assistance from GPT-5.6 Sol and have been verified (and occasionally reworded) by myself. \appendix \section{Corrections and comments} \begin{enumerate} \item \textbf{Page 4, Section 2 and the final paragraph:} Replace \textquotedblleft pleny\textquotedblright\ by \textquotedblleft plenty\textquotedblright. Replace \textquotedblleft dip in to\textquotedblright\ by \textquotedblleft dip into\textquotedblright. \item \textbf{Page 5, definition of an ideal:} Replace ``alway commutative with 1'' by ``always commutative with 1''. \item \textbf{Page 6, definition of a quotient:} The display defining $r+I$ should end with a comma, not a full stop, since the sentence continues with ``and the quotient $R/I$ \ldots''. \item \textbf{Page 6, Example 8:} In the surjectivity argument, replace \textquotedblleft$a,b\in\mathbb{Q}$\textquotedblright\ by \textquotedblleft% $a,b\in\mathbb{R}$\textquotedblright. Not every complex number has rational real and imaginary parts. \item \textbf{Page 6, Example 8:} Replace ``In otherwords'' by ``In other words''. \item \textbf{Page 7, Example 10:} At the end of the computation in $\mathbb{C}[x]$, replace \textquotedblleft$J\neq R[x]$\textquotedblright\ by \textquotedblleft$J\neq\mathbb{C}[x]$\textquotedblright. \item \textbf{Page 7, paragraph after Example 12:} The subject of ``Hence it is not maximal'' is accidentally the polynomial $x^{2}+1$. Write instead: ``Hence the ideal $(x^{2}+1)$ is not maximal''. \item \textbf{Page 10, proof of Theorem 18:} After \textquotedblleft so $m_{1}=0$ in $K$ or $m_{2}=0$ in $K$\textquotedblright, I would add \textquotedblleft(since a field has no zero divisors)\textquotedblright. \item \textbf{Page 10, definition of field generation:} Delete one copy of \textquotedblleft the\textquotedblright\ from \textquotedblleft We define the the \textbf{subfield}\textquotedblright. \item \textbf{Page 12, paragraph before Proposition 27:} \textquotedblleft indeterminant\textquotedblright\ is a rather uncommon spelling of \textquotedblleft indeterminate\textquotedblright. \item \textbf{Page 13, Example 30:} The roots of $(x^{2}+1)(x^{2}+2x+2)$ are \[ i,\quad-i,\quad-1+i,\quad-1-i, \] not $i,-i,1+i,1-i$. The stated splitting field $\mathbb{Q}(i)$ is nevertheless correct. \item \textbf{Page 13, Definition of a splitting field:} As stated, this makes no sense when $f$ is the zero polynomial. A trivial case, but worth getting right... \item \textbf{Page 13, Theorem 32:} Also replace \textquotedblleft isomophism\textquotedblright\ by \textquotedblleft isomorphism\textquotedblright. \item \textbf{Page 13, proof of Theorem 32:} The case $\deg f=0$ is also worth briefly mentioning (it is trivial, of course). \item \textbf{Page 15, proof of Lemma 37:} Replace \textquotedblleft a basis for $K/\mathbb{F}_{p}$\textquotedblright\ by \textquotedblleft a basis for $K$ over $\mathbb{F}_{p}$\textquotedblright. Here $K/\mathbb{F}_{p}$ denotes a field extension, not a quotient vector space. \item \textbf{Page 17, Lemma 44(ii):} Replace \textquotedblleft the minimal polynomial $m$ over $\alpha$\textquotedblright\ by \textquotedblleft the minimal polynomial $m$ of $\alpha$ over $K$\textquotedblright. \item \textbf{Page 17, proof of Lemma 44(ii):} After writing $m=f_{1}f_{2}$, replace \textquotedblleft$f_{1}(\alpha)f_{2}(\alpha)=f(\alpha)=0$% \textquotedblright\ by \textquotedblleft$f_{1}(\alpha)f_{2}(\alpha )=m(\alpha)=0$\textquotedblright. In the next sentence, \textquotedblleft the degree of $f$ is minimal\textquotedblright\ should likewise be \textquotedblleft the degree of $m$ is minimal\textquotedblright. Also replace \textquotedblleft stricly\textquotedblright\ by \textquotedblleft strictly\textquotedblright. \item \textbf{Page 19, proof of Theorem 50:} Replace \textquotedblleft% $0\leq\deg\left( r\right) \leq d-1$\textquotedblright\ by \textquotedblleft% $\deg\left( r\right) \leq d-1$\textquotedblright. The polynomial $r$ can be $0$, in which case its degree is $-\infty$. \item \textbf{Page 20, first line:} A parenthesis is missing after \textquotedblleft$\mathbb{Q}(\sqrt[3]{2}$\textquotedblright. \item \textbf{Page 22, calculation in the extended example:} In \textquotedblleft Thus either $a=0,\ b=\sqrt{\dfrac{6}{5}}$ or $b=0,\ a=\sqrt {6}$\textquotedblright, add a \textquotedblleft$\pm$\textquotedblright\ sign in front of both square roots. \item \textbf{Page 23, minimal polynomial of $\sqrt{5}+\sqrt{6}$:} The reference \textquotedblleft from (iii)\textquotedblright\ has no appropriate antecedent. The relevant result is Theorem 50(ii): since $\mathbb{Q}% (\alpha)=M$ and $[M:\mathbb{Q}]=4$, the minimal polynomial of $\alpha$ has degree~$4$. \item \textbf{Page 24, proof of Lemma 53:} Replace \textquotedblleft$L\left( x\right) $\textquotedblright\ by \textquotedblleft$L\left[ x\right] $\textquotedblright\ (both times it appears). \item \textbf{Page 27, Example 61:} Replace \textquotedblleft Examples 48 and 77 before\textquotedblright\ by \textquotedblleft Example 48 above and Example 77 below\textquotedblright. More importantly for a first reader, explain the potentially surprising terminology: $x^{p}-t$ is squarefree as a polynomial in $K[x]$ (it has no repeated irreducible factor there), although over its splitting field it has one root with multiplicity $p$ and is therefore inseparable. \item \textbf{Page 27, Example 62:} Insert a full stop after \textquotedblleft We continue Example 61\textquotedblright. \item \textbf{Page 28, discussion of formal derivatives and Lemma 68:} Replace \textquotedblleft non-contant\textquotedblright\ by \textquotedblleft non-constant\textquotedblright\ and \textquotedblleft Let $K$ has characteristic $0$\textquotedblright\ by \textquotedblleft Let $K$ have characteristic $0$\textquotedblright. \item \textbf{Page 32, proof of Lemma \textbf{73}:} Replace \textquotedblleft Now suppose let $f=m$\textquotedblright\ by \textquotedblleft Now let $f=m$\textquotedblright. \item \textbf{Page 32, proof of Theorem 74:} Replace \textquotedblleft the possibilities for each $\sigma(\alpha_{i})$ is finite\textquotedblright\ by \textquotedblleft the set of possibilities for each $\sigma(\alpha_{i})$ is finite\textquotedblright. \item \textbf{Page 33, proof that Frobenius is an automorphism:} After \textquotedblleft$\phi(\alpha^{-1})=\phi(\alpha)^{-1}$\textquotedblright, add \textquotedblleft(for $\alpha\neq0$)\textquotedblright. \item \textbf{Page 36, proof of Lemma 85:} In \textquotedblleft$\psi (a)=\phi(a)$ for all $a\in K$\textquotedblright, replace \textquotedblleft% $K$\textquotedblright\ by \textquotedblleft$K_{1}$\textquotedblright. Also replace \textquotedblleft isomomorphism\textquotedblright\ by \textquotedblleft isomorphism\textquotedblright. \item \textbf{Page 37, proof of Proposition 86:} In the second sentence, replace \textquotedblleft$n=1$\textquotedblright\ by \textquotedblleft$n\leq 1$\textquotedblright\ (the trivial case $n=0$ is also possible). When choosing the irreducible factor $g$, it is also cleanest to choose it monic before invoking Lemma 85. \item \textbf{Page 37, \S 6, cyclotomic-field discussion:} Replace \textquotedblleft These are the root of $x^{p}-1$\textquotedblright\ by \textquotedblleft These are the roots of $x^{p}-1$\textquotedblright. \item \textbf{Page 39, proof of Lemma 88:} Replace \textquotedblleft As any such $\sigma$ is contained in $\operatorname{Aut}(L/K)$\textquotedblright\ by \textquotedblleft As any such $\sigma$ belongs to $\operatorname{Aut}% (L/K)$\textquotedblright. (\textquotedblleft Contained in\textquotedblright% \ can also mean \textquotedblleft subset of\textquotedblright, and in fact has been used in that exact meaning in the first paragraph of this proof.) \item \textbf{Page 41, proof of Lemma 93:} Replace \textquotedblleft% $\sigma(A)$ has the same kernel at $A$\textquotedblright\ by \textquotedblleft% $\sigma(A)$ has the same kernel as $A$\textquotedblright. \item \textbf{Page 42, proof of Lemma 93:} Write \textquotedblleft The rank of $A$\textquotedblright\ rather than \textquotedblleft The $\operatorname{rank}% (A)$\textquotedblright. \item \textbf{Page 43, Example 95:} Replace \textquotedblleft This a continuation\textquotedblright\ by \textquotedblleft This is a continuation\textquotedblright. \item \textbf{Page 46, proof of Theorem 101:} In the proof of (c) $\Longrightarrow$ (b), \textquotedblleft Moreover $f$ is separable\textquotedblright\ is not true in general: The product $f=m_{1}\cdots m_{n}$ can have equal factors (since two of the $\alpha_{i}$ may have the same minimal polynomial) and thus have repeated roots. Take instead the product of the \emph{distinct} polynomials among $m_{1}% ,\ldots,m_{n}$. It still has splitting field $L$ and is separable. \item \textbf{Page 47, proof of Theorem 102:} Replace \textquotedblleft has distincts roots\textquotedblright\ by \textquotedblleft has distinct roots\textquotedblright. In the last paragraph, replace \textquotedblleft So $K^{\prime}$ is contained in a splitting field $K^{\prime\prime}$ for $f$\textquotedblright\ by \textquotedblleft So $K^{\prime}$ is a splitting field for $f$ (since all $p^{n}$ elements of $K^{\prime}$ are roots of the polynomial $x^{p^{n}}-x$, and the latter polynomial has degree $p^{n}$, so it cannot have any further roots). Thus, $K^{\prime}\cong K$, since the splitting field is unique up to isomorphism\textquotedblright. This renders the (rather murky) next two sentences unnecessary. \item \textbf{Page 47, \S 4, paragraph defining the permutation representation:} \textquotedblleft think of $\operatorname{Aut}(L/\mathbb{Q}% )$\textquotedblright\ should be \textquotedblleft think of $\operatorname*{Aut}\left( L/K\right) $\textquotedblright. \item \textbf{Page 48, Example 104:} Replace \textquotedblleft Letting $\alpha_{1}=\sqrt{p},\ldots$\textquotedblright\ by \textquotedblleft Let $\alpha_{1}=\sqrt{p},\ldots$\textquotedblright. \item \textbf{Page 49, Example 105:} The conventional abbreviation for \textquotedblleft compare\textquotedblright\ is \textquotedblleft cf.\textquotedblright, not \textquotedblleft c.f.\textquotedblright; it would also help to identify the intended Section~2 more precisely. \item \textbf{Page 51, proof of Theorem 106(iii):} Replace \textquotedblleft% $\lbrack L:K]=\#\operatorname{Aut}(L/K)=G$\textquotedblright\ by \textquotedblleft$\lbrack L:K]=\#\operatorname{Aut}(L/K)=\#G$% \textquotedblright. \item \textbf{Page 52, proof of Theorem 106(i):} In \textquotedblleft$L/L^{H}$ is Galois by part (i)\textquotedblright, replace \textquotedblleft% (i)\textquotedblright\ by \textquotedblleft(iii)\textquotedblright. \item \textbf{Page 52, proof of Lemma 108:} Replace \textquotedblleft a basis for $F/K$\textquotedblright\ by \textquotedblleft a basis for $F$ over $K$\textquotedblright. (Once again, the intended meaning is a field extension, not a quotient vector space.) \item \textbf{Page 53, Completing the proof of the Fundamental Theorem:} The assertion that \textquotedblleft$\ast\colon\mathcal{F}\rightarrow\mathcal{H}$ is a bijection\textquotedblright\ comes from part~(i), not part~(ii), of the Fundamental Theorem. \item \textbf{Page 54, worked example:} Replace \textquotedblleft$\zeta$ is the root of the irreducible $x^{2}+x+1$\textquotedblright\ by \textquotedblleft$\zeta$ is a root of the irreducible polynomial $x^{2}% +x+1$\textquotedblright. \item \textbf{Page 56, definition of a normal series:} Replace \textquotedblleft we say call the chain\textquotedblright\ by \textquotedblleft we call the chain\textquotedblright. \item \textbf{Page 56, Example 116:} Write, for example, \textquotedblleft the two quotients $R/1\cong C_{4}$ and $D_{4}/R\cong C_{2}$ are abelian\textquotedblright; the present sentence combines \textquotedblleft with\textquotedblright\ and \textquotedblleft are\textquotedblright\ ungrammatically. \item \textbf{Page 57, proof of Proposition 120(i):} The two subnormal series \textquotedblleft$1\subseteq G_{0}\subseteq\cdots\subseteq G_{n}% =G$\textquotedblright\ and \textquotedblleft$1\subseteq H_{0}\subseteq \cdots\subseteq H_{n}=H$\textquotedblright\ should have equality signs after \textquotedblleft$1$\textquotedblright, not subset signs. The further subnormal series \textquotedblleft$1\subseteq H_{1}\cdots\subseteq H_{n}% =H$\textquotedblright\ should be \textquotedblleft$1=H_{0}\subseteq \cdots\subseteq H_{n}=H$\textquotedblright. \item \textbf{Page 57, proof of Proposition 122:} In the first paragraph of the proof, \textquotedblleft$f$ is separable\textquotedblright\ is false in general: Even in characteristic zero, the product $f=m_{1}\cdots m_{n}$ need not be separable if some $m_{i}$ coincide. Replace it by the product of the distinct minimal polynomials among the $m_{i}$. The splitting field is unchanged and the resulting polynomial is separable. \item \textbf{Page 57, proof of Proposition 122:} Replace \textquotedblleft Let $M$ be the splitting field of $f$ over $K$\textquotedblright\ by \textquotedblleft Let $M$ be a splitting field of $f$ over $K$ that contains $K\left( \alpha_{1},\alpha_{2},\ldots,\alpha_{n}\right) $ as a subfield\textquotedblright\ (such a splitting field indeed exists, since $\alpha_{1},\alpha_{2},\ldots,\alpha_{n}$ are roots of $f$ already). This is what allows you to conclude afterwards that \textquotedblleft$M$ contains the $\alpha_{i}$ and $K$\textquotedblright. \item \textbf{Page 58, proof of Proposition 122:} Replace \textquotedblleft Fundamental Theorem of Galois Theorem\textquotedblright\ by \textquotedblleft Fundamental Theorem of Galois Theory\textquotedblright. \item \textbf{Page 58, proof of Proposition 122:} Replace the displayed equation% \[ \gamma_{i}^{r_{n}}=\sigma_{i}(\alpha_{n}^{r_{n}})=\alpha_{n}^{r_{n}}\in F \] by \[ \gamma_{i}^{r_{n}}=\sigma_{i}(\alpha_{n}^{r_{n}})\in\sigma_{i}(F)=F. \] (There is no reason why $\sigma_{i}(\alpha_{n}^{r_{n}})$ must equal $\alpha_{n}^{r_{n}}$.) \item \textbf{Page 58, proof of Lemma 123:} Replace \textquotedblleft% $\zeta^{a+b}$\textquotedblright\ by \textquotedblleft$\zeta^{ab}% $\textquotedblright. (On the other hand, the additive exponent in Lemma 124 is correct because there the automorphisms fix $\zeta$ and multiply $\sqrt[p]{\alpha}$ by a power of it.) \item \textbf{Page 59, proof of Proposition 126:} The assertion that $L/M$ is radical deserves one line of justification: If $K=L_{0}\subseteq \cdots\subseteq L_{n}=L$ is a radical tower, then \[ M\subseteq ML_{1}\subseteq\cdots\subseteq ML_{n}=L \] is a radical tower after repetitions are removed. \item \textbf{Page 59, proof of Corollary 127:} Replace \textquotedblleft By definition of soluble polynomial\textquotedblright\ by \textquotedblleft By the definition of soluble by radicals\textquotedblright. Incidentally, \textquotedblleft soluble by radicals\textquotedblright\ and \textquotedblleft soluble in radicals\textquotedblright\ are used synonymously, but only the first has been defined. \item \textbf{Page 60, proof of Lemma 130:} With the usual right-to-left composition convention, \[ \left( a,c,d,e\right) ^{2}(a,b)=(a,b,d)(c,e), \] not $(a,d)(b,c,e)$. Both permutations have order~$6$, so the order argument survives. \item \textbf{Page 60, same proof:} From $60\mid\#G$ and $G\leq S_{5}$ one obtains $\#G\in\{60,120\}$. To conclude that $G=A_{5}$ or $S_{5}$, add the standard fact that $A_{5}$ is the unique subgroup of index~$2$ in $S_{5}$ (or prove the needed special case). \item \textbf{Page 60, proof of Theorem 131:} It is not immediately obvious that the polynomial $f=2x^{5}-10x+5$ is irreducible. One way to prove it is using Eisenstein's criterion at~$5$. Also, the existence of a $5$-cycle in $\operatorname{Aut}(L/\mathbb{Q})$ follows from Cauchy's theorem, since $5\mid\#\operatorname{Aut}(L/\mathbb{Q})$ and since the only permutations of order $5$ in $S_{5}$ are $5$-cycles. \item \textbf{Page 62, Exercise 133:} Replace \textquotedblleft non-colinear\textquotedblright\ by \textquotedblleft non-collinear\textquotedblright. \item \textbf{Page 62, proof of Lemma 135:} Replace \textquotedblleft rearragned\textquotedblright\ by \textquotedblleft rearranged\textquotedblright. \item \textbf{Page 63, proof of Lemma 135:} Replace \textquotedblleft the intersection of a non-parallel lines\textquotedblright\ by \textquotedblleft the intersection of two non-parallel lines\textquotedblright. \item \textbf{Page 64, proof of Theorem 137:} There are two intersections of the line $OD$ with the unit circle. Specify that $P$ is the one on the ray $\overrightarrow{OD}$. \item \textbf{Page 64, proof of Theorem 137:} Replace \textquotedblleft% $\lbrack K:\mathbb{Q}]=[\mathbb{Q}(\sqrt{3}):\mathbb{Q}]$\textquotedblright% \ by \textquotedblleft$\lbrack K:\mathbb{Q}]=[\mathbb{Q}(\sqrt{3}% ):\mathbb{Q}]=2$\textquotedblright. This is the value used in the next equality. \item \textbf{Page 64, proof of Theorem 137:} For a less-prepared reader, maybe also justify the final irreducibility assertion: by the rational-root test, $8x^{3}-6x-1$ has no rational root, and a cubic over $\mathbb{Q}$ with no rational root is irreducible. \item \textbf{Page 65, proof of Theorem 138:} Theorem 136 is stated for the field generated by one constructible point, whereas the proof applies it directly to $L=\mathbb{Q}(P,Q)$. Explain that the construction sequences for $P$ and $Q$ can be combined into one sequence; the proof of Theorem 136 then gives $[\mathbb{Q}(P,Q):\mathbb{Q}]=2^{r}$. Alternatively, state the corresponding finite-set version of Theorem 136. \end{enumerate} \end{document}