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\begin{document}
\title{6320-001 - Spring 2021 - Weeks 11-12 (3/30, 4/01, 4/06, 4/08)}
\maketitle
\section{The next two weeks in 6320}
{\bf 3/30} -- Galois extensions and the fundamental theorem of Galois theory.
{\bf 4/01} -- Cyclotomic fields.
{\bf 4/06} -- Galois groups of polynomials.
{\bf 4/08} -- Solvability in radicals. \\
\noindent{\bf The basic results and definitions in Galois theory:}
If $L$ is a field and $S \subset \mr{Aut}(L)$ is a subset, we write
\[ L^S := \{\ell \in L\; | \; \sigma(\ell)=\ell \, \forall\, \sigma \in S \}. \]
It is simple to check that it is a subfield, called the fixed field of the collection of automorphisms $S$. We will use this especially when $S$ is a subgroup of $\mr{Aut}(L)$.
If $L$ is a field and $M \subset L$ is a subset, we write
\[ \mathrm{Fix}(M) := \{\sigma \in \Aut(L) \; |\; \sigma(m)=m\, \forall\, m \in M\}.\]
It is simple to check that it is a subgroup; it consists of all automorphisms of $L$ that act trivially (i.e., restrict to the identity) on $M$. We will use this especially when $M$ is a subfield.
\begin{theorem}[Galois extensions and the fundamental theorem of Galois theory]\hfill
For $L/K$ a finite extension, $|\Aut(L/K)|\leq [L:K]$, and the following are equivalent:
\begin{enumerate}
\item $[L:K]=|\Aut(L/K)|$.
\item $K=L^{\mathrm{Aut}(L/K)}$,
\item $L/K$ is separable and normal: that is, if $\alpha \in L$, the minimal polynomial $m_\alpha(x) \in K[x]$ of $\alpha$ over $K$ has $\mr{deg}\,m_\alpha$ distinct roots in $L$.
\item $L/K$ is a splitting field of a separable polynomial in $K[x]$,
\end{enumerate}
When these equivalent conditions hold, $L/K$ is called a \textbf{Galois extension}.
Moreover, if $L/K$ is a Galois extension then there is an inclusion reversing bijection
\begin{eqnarray*}
\{ \textrm{Intermediary field extensions }L \supseteq M \supseteq K \} & \leftrightarrow & \{ \textrm{Subgroups } \{e\} \leq H\leq\Aut(L/K) \} \\
M & \mapsto & \mathrm{Fix}(M) \\
L^H & \leftmapsto & H.
\end{eqnarray*}
Furthermore, for any intermediary extension $L \supseteq M \supseteq K$,
\begin{enumerate}
\item $|\mr{Fix}(M)|=[L:M]$ and $[\Aut(L/K):\mr{Fix}(M)]=[M:K].$
\item $L/M$ is Galois and $\Aut(L/M)=\mathrm{Fix}(M)$,
\item $M/K$ is Galois if and only if $\mathrm{Fix}(M)$ is a normal subgroup, in which case
\[ \Aut(M/K)=\Aut(L/K)/\mathrm{Fix}(M). \]
\end{enumerate}
Finally, if $M_1 \leftrightarrow H_1$, $M_2 \leftrightarrow H_2$, then
\[ M_1 M_2 \leftrightarrow H_1 \cap H_2 \textrm{ and } M_1 \cap M_2 \leftrightarrow \langle H_1, H_2 \rangle. \]
\end{theorem}
For a Galois extension $L/K$, we sometimes write $\mr{Gal}(L/K)$ in place of $\mr{Aut}(L/K)$ and refer to it as the \emph{Galois group} of the extension. If $f(x) \in K[x]$ is a polynomial, the Galois group of $f$ means the Galois group of a splitting field of $f$.
We record a particularly useful computational corollary :
\begin{corollary}\hfill
\begin{enumerate}
\item If $L/K$ is a Galois extension, and $\alpha \in L$ then the roots of the minimal polynomial $m_\alpha(x)\in K[x]$ of $\alpha$ over $K$ are the orbit $\mr{Gal}(L/K)\cdot \alpha$.
\item If $f\in K[x]$ is separable then the irreducible factors of $f$ are in bijection with the orbits of the Galois group of $f$ on the roots of $f$ in the splitting field, with each orbit corresponding to the roots of an irreducible factor.
\end{enumerate}
\end{corollary}
\begin{proof} Left to the reader.
\end{proof}
We also note a corollary when this is combined with the primitive element theorem proved in the problem on your Week 10 homework:
\begin{corollary}
If $L=K(\alpha_1, \ldots, \alpha_m)$ for $\alpha_i\in L$ separable algebraic over $K$, then $L$ is a simple extension of $K$, i.e. $L=K(\alpha)$ for some $\alpha\in L$.
\end{corollary}
\begin{proof} Left to the reader (hint: embed $L$ in the splitting field of a separable polynomial over K).
\end{proof}
The following lemma, which is the main ingredient in the proof, is also very useful on its own:
\begin{lemma}\label{lemma:fixed-index}
If $L$ is a field and $G \subset \Aut(L)$ is a finite subgroup then $[L: L^G]=|G|.$\end{lemma}
In particular, this implies that $L/L^G$ is Galois with group $G$. The proof of Lemma \ref{lemma:fixed-index} uses independence of characters to establish $|G|\leq [L:L^G]$ and a primitive version of \emph{descent} for the other inequality; it is elaborated in one of the problems below. We recall the statement of independence of characters, which was covered in Week 6, here:
\subsection{Independence of characters}
If $G$ is a group and $L$ is a field, recall that a \emph{character} of $G$ with values in $L$ is a group homomorphism $G \rightarrow L^\times$. We can view a character as an element of the set $\mr{Maps}(G,L)$ of all functions on $G$ with values in $L$, which is an $L$-vector space.
\begin{theorem}[Independence of characters] If $X$ is a set of distinct characters of $G$ with values in $L$, then $X$ is a linearly independent set when viewed as a subset of the $L$-vector space $\mathrm{Maps}(G, L)$.
\end{theorem}
\begin{proof}
We sketch the proof: It suffices to consider $X$ finite, for which we can argue by induction. The base case is clear since any character maps $1 \in G$ to $1 \in L$. Suppose it is known for $|X|