\documentclass[12pt]{amsart}
\title{Lecture 29}
\author{}

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\begin{document}
\maketitle

\PP{29.0}{Review}
%---------------
Let $L/K$ be a field extension. Last time we defined

\[
\Gal{L}{K} = \Aut_{\Alg{K}}(L) = \{\sigma:L\isom L \text{ field automorphism, such
that } \sigma|_K=\Id_K\},
\]
called the Galois group of the extension $L/K$.\\

For a group $\Gamma$ acting on $L$ via field automorphisms (that is,
we are given a group homomorphism $\Gamma\to \Aut_{\scriptstyle{\operatorname{field}}}(L)$),
we denote by $L^\Gamma\subset L$, the subfield of $\Gamma$--fixed elements:

\[
L^{\Gamma} = \{z\in L \text{ such that } \sigma(z)=z,\ \forall\ \sigma\in\Gamma\}.
\]
\ \\

A field extension $L/K$ is called a {\em Galois extension} if
$L^{\Gal{L}{K}}=K$.\\

\PP{29.1}{Linear independence of algebra homomorphisms}
%------------------------------------------------------
Let $K$ be a field, $L/K$ a field extension, and $A$ a unital (not necessarily
commutative) algebra over $K$. Meaning, $A$ is a unital ring together with
a ring homomorphism, necessarily injective, $K\hookrightarrow A$.\\

\begin{thm}
$\ds\Hom_{\Alg{K}}(A,L) \subset \Hom_{\vs{K}}(A,L)$
is linearly independent over $L$.
\end{thm}

\begin{rem}
For a $K$--vector space $V$, we view $\Hom_{\vs{K}}(V,L)$ as an $L$--vector space
via the following operations. For every $\xi,\eta:V\to L$, $K$--linear maps, and 
$a,b\in L$, we set:
\[
(a\xi+b\eta)(v) = a\xi(v) + b\eta(v).
\]
Note that, we have the canonical $L$--linear map:
\[
\beta: V^*\otimes_K L \to \Hom_{\vs{K}}(V,L),\qquad
\beta(f\otimes z) : v\mapsto f(v)z.
\]
If $V$ is finite--dimensional, this map is an isomorphism, and we obtain:
\[
\boxed{
\dim_{\vs{L}}\lp\Hom_{\vs{K}}(V,L)\rp = \dim_{\vs{K}}(V^*) = \dim_{\vs{K}}(V)
}
\]
\end{rem}
\ \\

\begin{pf}
Let $\{\xi_1,\ldots,\xi_n\}\subset \Hom_{\Alg{K}}(A,L)$ be a finite set of
$K$--algebra homomorphisms $A\to L$. We will show, by induction
on $n$, that this set of linearly indepedent. For $n=1$, we have
$\{\xi:A\to L\}$ is linearly independent if and only if $\xi\neq 0$,
which is true since $\xi(1)=1$.\\

Now assume that we have a linear relation
$\ds \sum_{i=1}^n a_i\xi_i = 0$, where $a_1,\ldots,a_n\in L$. Note
that for every $x,y\in A$, we get:
\[
a_n\xi_n(xy)- \xi_n(x)(a_n\xi_n(y)) = 0.
\]

Replacing $a_n\xi_n = -\sum_{i=1}^{n-1} a_i\xi_i$, we get:
\[
\sum_{i=1}^{n-1} a_i(\xi_i(x)-\xi_n(x))\xi_i(y) = 0,\text{ for every } y\in A.
\]
Thus, we obtain (by induction) that for every $1\leq i\leq n-1$, and $x\in A$:
$a_i(\xi_i(x)-\xi_n(x))=0$. Since $\xi_i\neq \xi_n$, there must exist some
$x\in A$ such that $\xi_i(x)\neq \xi_n(x)$, implying that $a_i=0$. Now
$a_n\xi_n = -\sum_{i=1}^{n-1}a_i\xi_i = 0$, but $\xi_n\neq 0$, and we conclude
that $a_n=0$. Therefore, the set $\{\xi_1,\ldots,\xi_n\}$ is linearly
independent over $L$.
\end{pf}
\ \\

\PP{29.2}{Application of Theorem 29.1 I: independence of characters}
%------------------------------------
Let $\Gamma$ be a group and $L$ be a field. Let $L[\Gamma]$ be the group
algebra of $\Gamma$ over $L$. As an $L$--vector space, $L[\Gamma]$ has a basis
$\{e(g) : g\in\Gamma\}$, relative to which multiplication is determined
by $e(g)\cdot e(h) = e(gh)$. Note that
\[
\Hom_{\scriptstyle{\mathrm{gp}}}(\Gamma,L^{\times}) = 
\Hom_{\Alg{L}}(L[\Gamma],L).
\]
Therefore, we obtain the following result, due to Dedekind, known as
{\em independence of characters}. An $L$--valued character of a group $\Gamma$,
is just a group homomorphism $\Gamma\to L^\times$.\\

\begin{cor}
Elements of $\Hom_{\scriptstyle{\mathrm{gp}}}(\Gamma,L^\times)$ are
linearly independent over $L$.
\end{cor}
\ \\

\PP{29.3}{Application of Theorem 29.1 II: inequalities}
%------------------------------------------------------
Let $K$ be a field and $E/K$, $L/K$ two field extensions.
By Theorem 29.1, $\Hom_{\Alg{K}}(E,L)\subset\Hom_{\vs{K}}(E,L)$ is 
linearly independent over $L$. Moreover, if $[E:K]<\infty$, 
$\Hom_{\vs{K}}(E,L)$ is an $L$--vector space of dimension $[E:K]$
(see Remark 29.1). Thus,
we get:

\[
\boxed{
\left| \Hom_{\Alg{K}}(E,L)\right| \leq [E:K]
}
\]

Taking $E=L$, and viewing $\Gal{L}{K}\subset\Hom_{\Alg{K}}(L,L)$, we
get:

\[
\boxed{
\left| \Gal{L}{K} \right| \leq [L:K]
}
\]
\ \\

\PP{29.4}{Application of Theorem 29.1 III: Artin's theorem}
%--------------------------------------------------------
Now let $L$ be a field and let $\Gamma\subset\Aut_{\scriptstyle{\mathrm{field}}}(L)$
be a finite subgroup. Let $F=L^{\Gamma}\subset L$.\\

\begin{thm}
$L/F$ is a Galois extension of degree $|\Gamma|$.
\end{thm}

\begin{pf}
Let $n=|\Gamma|$ and $m=[L:F]$.
Since $\Gamma\subset\Hom_{\Alg{F}}(L,L)$, the inequalities from the
previous paragraph imply that $n\leq m$. Assume that $n<m$. Let
$\Gamma=\{\sigma_1,\ldots,\sigma_n\}$ and $\{x_1,\ldots,x_m\}$
be a basis of $L$ as an $F$--vector space. Form an $n\times m$
matrix:
\[
X = (\sigma_i(x_j))_{1\leq i\leq n, 1\leq j\leq m} \in
\mathrm{Mat}_{n\times m}(L).
\]
Since $m>n$, there exists a non--zero vector $\ul{a}\in L^m$,
such that $X\ul{a}=\ul{0}$. That is, for every $1\leq i\leq n$:
\[
\sum_{j=1}^m a_j\sigma_i(x_j) = 0.
\]
We will arrive at a contradiction as follows. Let $p$ be the smallest
positive integer such that there exists $\ul{a}\in\Ker(X)$ with
$p$ non--zero entries. Note that $p=1$ is absurd since it will imply
the existence of $1\leq j\leq m$ such that $\sigma_i(x_j)=0$ for each
$i$. But $\sigma_i$ is an automorphism and $x_j\neq 0$.

Assuming the existence of $\ul{a}\in\Ker(X)$ with $p$ non--zero entries,
we will produce $\ul{b}\in\Ker(X)$ with $p-1$ non--zero entries, thus
a contradiction, proving that $\Ker(X)=\{0\}$.

Upon reordering elements of $\Gamma$, if necessary, we can assume
that $a_1,\ldots,a_p\in L^\times$ and $a_{p+1}=\ldots=a_m=0$. Further,
we can scale $\ul{a}$ to assume that $a_p=1$. Thus,
\[
\text{For every } 1\leq i\leq n,\ \sigma_i(x_p) = - \sum_{j=1}^{p-1} a_j\sigma_i(x_j).
\]

Since $\Gamma$ is a subgroup, there is $i$ such that $\sigma_i=\Id_L$.
We are assuming that $\{x_1,\ldots,x_m\}$ are linearly independent over $F$,
so $x_p = -\sum_{j=1}^{p-1}a_j x_j$ implies that there must be some
$1\leq \ell\leq p-1$ so that $a_\ell\not\in F$. By definition of $F$,
this means there is $1\leq k\leq n$, with $\sigma_k(a_\ell)\neq a_\ell$.

Apply $\sigma_k$ to $\sigma_i(x_p) = -\sum_{j=1}^{p-1} a_j\sigma_i(x_j)$
to get:
\[
(\sigma_k\sigma_i) (x_p) = 
-\sum_{j=1}^{p-1} \sigma_k(a_j) (\sigma_k\sigma_i)(x_j),\qquad
\text{for every } 1\leq i\leq n.
\]
Now left multiplication by $\sigma_k$ is a permutation of $\Gamma$. So we
get:
\[
\sigma_q(x_p) = -\sum_{j=1}^{p-1} \sigma_k(a_j)\sigma_q(x_j),\qquad
\text{for every } 1\leq q\leq n.
\]
Subtracting from the original relation, we get:
\[
0 = \sum_{j=1}^{p-1} (\sigma_k(a_j)-a_j) \sigma_q(x_j),\qquad
\text{for every } 1\leq q\leq n.
\]
Thus we obtain a non--zero (since $\sigma_k(a_\ell)\neq a_\ell$) 
element of $\Ker(X)$ with strictly less than $p-1$ non--zero entries.
\end{pf}
\ \\

\begin{cor}
Let $L/K$ be a finite extension. Then it is a Galois extension if
and only if $|\Gal{L}{K}| = [L:K]$.
\end{cor}
\ \\

\PP{29.5}{Algebraic Galois extensions}
%-------------------------------------
Let $L/K$ be an algebraic extension. The following result gives
algebraic characterization for $L/K$ to be a Galois extension.\\

\begin{thm}
An algebraic extension $L/K$ is Galois if and only if for every $\alpha\in L$,
its minimal polynomial $\minpoly{\alpha}{x}\in K[x]$ has $\deg(\minpoly{\alpha}{x})$
distinct roots in $L$.
\end{thm}

\begin{pf}
Let $\Gamma = \Gal{L}{K}$. Assume that $L/K$ is Galois, that is,
$L^\Gamma=K$. Let $\alpha\in L$ and $\minpoly{\alpha}{x}\in K[x]$
its minimal polynomial. Let $n=\deg(\minpoly{\alpha}{x})$.

Consider the $\Gamma$--orbit of $\alpha$.
\[
\Gamma\alpha = \{\sigma(\alpha) : \sigma\in \Gamma\} \subset L.
\]
Note that for every $\sigma\in\Gamma$, $\sigma(\alpha)$ is another root
of $\minpoly{\alpha}{x}$. Since number of roots of a polynomial
$\leq$ degree of that polynomial, we conclude that $\Gamma\alpha$
is finite and has at most $n$ elements.

Define $\ds f(x) = \prod_{\beta\in\Gamma\alpha} (x-\beta)$. Since $f(x)$
is invariant under $\Gamma$, we have $f(x)\in K[x]$. Moreover, it
divides $\minpoly{\alpha}{x}$ whose irreducibility implies that
$\deg(f)=n$. That is $|\Gamma\alpha|=n$ consists of $n$ distinct
roots of $\minpoly{\alpha}{x}$.\\

Let us prove the converse now. 
Note that the assumption on $L/K$ implies two statements.
\begin{itemize}
\item $L$ is the splitting extension of the following
set of (irreducible, monic) polynomials.
\[
L=\SE{P}{K},\qquad\text{where, } P=\{\minpoly{\alpha}{x} : \alpha\in L\}\subset K[x].
\]
\item Every $f(x)\in P$ has distinct roots in $L$.
\end{itemize}

Assume $\alpha\in L\setminus K$. 
We will exhibit an element $\sigma\in \Gamma$ such that $\sigma(\alpha)\neq \alpha$.
Note that $n=\deg(\minpoly{\alpha}{x})\geq 2$,
therefore there exists $\beta\neq\alpha$ also a root of $\minpoly{\alpha}{x}$.
By Theorem 26.0, there exists an isomorphism $\ol{\sigma} : K(\alpha)\isom K(\beta)$
uniquely determined by $\ol{\sigma}|_K=\Id_K$ and
$\ol{\sigma}(\alpha)=\beta$. By the same theorem, part (2),
$\ol{\sigma}$ extends to an element $\sigma\in\Gal{L}{K}$, since $L/K$
is a splitting extension. Thus, we have shown the existence of
$\sigma\in\Gal{L}{K}$ such that $\sigma(\alpha)=\beta\neq \alpha$.
\end{pf}
\ \\

\PP{29.6}{Separable polynomials and normal extensions}
%----------------------------------------------------
Let us record the two important properties listed in the proof of
the theorem given above.\\

\begin{defn}
Let $K$ be a field and $f(x)\in K[x]$ be a polynomial. We say $f(x)$
is {\em separable} if all its roots (in its splitting extension, for instance)
are distinct.\\

An extension $L/K$ is called {\em separable} if it is algebraic and for every
$\alpha\in L$, its minimal polynomial $\minpoly{\alpha}{x}$ is separable.\\

An extension $L/K$ is called {\em normal} if it is algebraic
and for every $\alpha\in L$, its minimal polynomial $\minpoly{\alpha}{x}$
splits as a product of linear terms in $L[x]$. That is,
$L$ is the splitting extension of $\{\minpoly{\alpha}{x} : \alpha\in L\}$.\\
\end{defn}

Theorem 29.5 is often phrased as {\em Galois if and only if
separable and normal}. Note that there exist fields $F$ over which
irreducible polynomials are not necessarily separable. We will 
discuss such $F$ (called {\em imperfect fields}) next time.\\


\PP{29.7}{Example}
%----------------
Recall that last time we defined $\Q(\mu_n)\subset\C$, where
\[
\mu_n = \left\{e^{\frac{2\pi k\iota}{n}} : 0\leq k\leq n-1\right\}.
\]

We saw that $|\Gal{\Q(\mu_n)}{\Q}|=\phi(n)$, where
\[
\phi(n) = \left|\left\{
1\leq k\leq n-1 : \gcd(k,n)=1\right\}\right|,\ \text{ Euler's }
\phi \text{ function}.
\]

It is immediate that $\Q(\mu_n)/\Q$ is Galois.
By Corollary 29.4, we have:
\[
[\Q(\mu_n) : \Q] = \phi(n).
\]

\end{document}
