%! TEX root = Galois.tex
% vim: tw=50
% 05/10/2023 12PM

\setcounter{section}{-1}
\section{Introduction}

\emph{Galois Theory} is named after the French
mathematician Evariste Galois (1811-1832). It is
the study of roots of polynomials (or more
generally field extensions).

It can be used to show that certain classical
problems cannot be solved -- for example there is
no formula (in terms of radicals) for the roots of
a general polynomial of degree $n$ when $n \ge 5$.
(Radicals means $+$, $-$, $\times$, $\div$,
$\sqrt[m]{}$). This is related to the fact that
the alternating group $A_n$ is simple for $n \ge
5$.

More positively, Galois theory is foundational to
the study of Algebraic Number Theory and Algebraic
Geometry.

Prerequisites: Linear Algebra and GRM

\newpage
\section{Field Extensions}

\begin{flashcard}[field-defn]
\begin{definition*}[Field]
  \glsdefn{field}{field}{fields}
  \cloze{
  A \emph{field} $K$ is a ring (commutative with a
  $1$, with $0_K \neq 1_K$) in which every non-zero
  element has an inverse under $\times$.
  }
\end{definition*}
\end{flashcard}

\begin{flashcard}[characteristic-defn]
\begin{definition*}[Characteristic]
  \glsdefn{char}{characteristic}{N/A}
  \cloze{
  The \emph{characteristic} of a \gls{field} $K$ is the least
  positive integer $p$ (necessarily prime) such
  that $p \cdot 1_K = 0_K$, or $0$ \fcemph{if no such
  integer exists}.
  }
\end{definition*}
\end{flashcard}
\vspace{-1em}

\noindent
\glsnoundefn{prime_subfield}{prime subfield}{prime
subfields}
Then $K$ contains a smallest subfield (called its
prime subfield) which is isomorphic to $\FF_p =
\ZZ / p\ZZ$ if $K$ has \gls{char} $p$, and to
$\QQ$ if it has \gls{char} $0$.

\begin{flashcard}[polynomial-has-at-most-deg-roots-lemma]
\begin{lemma}
  % Lemma 1.1
  \label{roots_bound_lemma}
  Let $K$ be a \gls{field} and $0 \neq f \in K[X]$. Then
  $f$ has $\le \deg f$ roots in $K$.
\end{lemma}

\begin{proof}
  \cloze{
  By induction on $n = \deg f$. If $f$ has no
  roots then there is nothing to prove.

  Otherwise let $a \in K$ be a root of $f$. Then
  $f(X) = (X - a) g(X)$ with $g \in K[X]$, where
  $\deg g = n - 1$. If $b \in K$ is a root of $f$,
  then either $b = a$ or $g(b) = 0$. Therefore
  \begin{align*}
    |\{\text{roots of $f$ in $K$}\}|
    &\le 1 + |\{\text{roots of $g$ in K}\}| \\
    &\le 1 + (n - 1)
    &&\text{(by induction)} \\
    &= n
    &&\qedhere
  \end{align*}
  }
\end{proof}
\end{flashcard}

\begin{flashcard}[field-extension-defn]
\begin{definition*}[Field extension]
  \glsdefn{field_ex}{field extension}{field
  extensions}
  \glsalias{ex}{extension}{extensions}
  \cloze{
  Let $L$ be a \gls{field} and $K \subset L$ a subfield
  (i.e. a subset which is a \gls{field} under the same
  operations $+$ and $\times$).

  \glssymboldefn{field_ex}{field extension}{$/$}
  We say $L$ is an \emph{extension} of $K$,
  written $L / K$.
  }
\end{definition*}
\end{flashcard}
\vspace{-1em}

\noindent
We note that $L$ and $K$ necessarily have the same
\gls{char}.

\begin{example*}
  \phantom{}
  \begin{enumerate}[(i)]
    \item $\CC / \RR$, $\QQ(\sqrt{2}) / \QQ$, $\RR
      / \QQ$.
    \item Adjoining a root of an irreducible
      polynomial: Let $K$ be a \gls{field} and $f \in
      K[X]$ an irreducible polynomial. We recall
      from GRM that $K[X]$ is Euclidean, hence a
      PID. Therefore $(f) \subset K[X]$ is a
      maximal ideal and hence $L =
      \frac{K[X]}{(f)}$ is a \gls{field_ex} of
      $K$ and $\alpha = X + (f) \in L$ is a root
      of $f$.

      For example taking $K = \RR$, $f = X^2 + 1$
      gives the first example in (i).
  \end{enumerate}
\end{example*}
\vspace{-1em}

Let $L \extendover K$ be a \gls{field_ex}. Then $+$ in
$L$ and multiplication by elements in $K$ make $L$
a vector space over $K$ (for example $\CC$ is an
$\RR$ vector space).

\begin{flashcard}[extension-degree-defn]
\begin{definition*}[Degree of an Extension]
  \glsdefn{ex_deg}{degree}{N/A}
  \glssymboldefn{ex_deg}{degree of extension}{$[L : K]$}
  \glsdefn{finite_ex}{finite}{N/A}
  \cloze{
  Let $L \extendover K$ be a \gls{field_ex}. We say $L
  \extendover K$ is \emph{finite} if $L$ is finite
  dimensional as a $K$-vector space, in which case
  we write $[L : K] = \dim_K L$ for its dimension,
  which we call the \emph{degree} of $L \mid K$.

  \glsdefn{infinite_ex}{infinite}{N/A}
  If not, we say $L \extendover K$ is an \emph{infinite
  extension} and write $[L : K] = \infty$.
  }
\end{definition*}
\end{flashcard}
\vspace{-1em}

$L \extendover K$ is a quadratic (cubic, quartic, \ldots)
\gls{ex} if $[L : K] = 2 (3, 4, \ldots)$.

\begin{example*}[Continued]
  \phantom{}
  \begin{enumerate}[(i)]
    \item $\degover{\CC}{\RR} = 2$ (basis $1$, $i$),
      $\degover{\QQ(\sqrt{2})}{\QQ} = 2$ (basis $1$,
      $\sqrt{2}$)

      $\degover{\RR}{\QQ} = \infty$ (exercise: use
      countability).
    \item If $L = \frac{K[X]}{(f)}$ where $f \in
      K[X]$ irreducible then $\degover{L}{K} = \deg f$.
      Indeed if $\alpha = X + (f) \in L$ and $n =
      \deg f$ then $1, \alpha, \ldots, \alpha^{n -
      1}$ is a $K$-basis for $L$.
  \end{enumerate}
\end{example*}

\begin{remark*}
  \glsdefn{embedding}{embedding}{embeddings}
  \glsalias{field_embedding}{field
  embedding}{field embeddings}
  Let $K$, $L$ be \glspl{field} and $\phi : K \to L$ a
  ring homomorphism. Then $\ker(\phi)$ is an
  ideal in $K$, but $K$ is a \gls{field}, so
  $\ker(\phi) = \{0\}$ or $\ker(\phi) = K$. But
  by definition, a ring homomorphism must have
  $\phi(1_K) = 1_L$, so can't have $\ker(\phi) =
  K$. So must have $\ker(\phi) = \{0\}$, i.e.
  $\phi$ is injective. We call $\phi$ an
  \emph{embedding} of $K$ in $L$.

  We may use $\phi$ to identify $K$ as a
  subfield of $L$, i.e. we get a \gls{field_ex}
  $L \extendover K$.
\end{remark*}

\begin{example*}
  Taking $K = \FF_2$, $f = X^ + X + 1 \in
  \FF_2[X]$ (which is irreducible) gives $L =
  \frac{\FF_2[X]}{(X^2 + X + 1)}$ a \gls{field} with $4$
  elements.
\end{example*}

\begin{flashcard}[finite-field-sizes-proposition]
\begin{proposition}[Possible sizes of finite \glspl{field}]
  % Proposition 1.2
  \label{finite_field_sizes_prop}
  \cloze{
  Let $K$ be a finite \gls{field} of \gls{char} $p$
  (necessarily $> 0$). Then $|K| = p^n$ where $n =
  \degover{K}{\FF_p}$.
  }
\end{proposition}

\begin{proof}
  \cloze{
  $\degover{K}{\FF_p} = n \implies K \equiv \FF_p^n$ as
  an $\FF_p$-vector space.
  }
\end{proof}
\end{flashcard}

Later we will show that (up to isomorphism) there
is exactly one \gls{field} of order $p^n$ for each prime
power $p^n$.

\glsdefn{mult_group}{multiplicative
group}{N/A}
\glssymboldefn{mult_group}{multiplicative group}{$K^*$}
The multiplicative group of a \gls{field} $K$ is the set
$K^* = K \setminus \{0\}$, which is an abelian group under $\times$.

\begin{flashcard}[cyclic-multiplicative-group-proposition]
\begin{proposition}
  % Proposition 1.3
  \label{cyclic_mult_subgroup}
  If $K$ is a \gls{field} then any finite subgroup $G
  \subset \mult{K}$ is cyclic. In particular if $K$ is
  a finite \gls{field} then $\mult{K}$ is cyclic.
\end{proposition}

\begin{proof}
  \cloze{
  The structure theorem for finite abelian groups
  gives
  \[ G \equiv C_{d_1} \times C_{d_2} \times \cdots
  \times C_{d_t} \]
  where $1 < d_1 \mid d_1 \mid \cdots \mid d_t$.
  If $G$ is not cyclic then picking a prime
  dividing $d_1$ shows that $G$ contains a
  subgroup isomorphic to $C_p \times C_p$. Hence
  the polynomial $X^p - 1$ has $\ge p^2$ roots in
  $K$, which contradicts \cref{roots_bound_lemma}.
  }
\end{proof}
\end{flashcard}