\documentclass[12pt]{amsart}
\title{Lecture 35}
\author{}

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\begin{document}
\maketitle

\PP{35.0}{Overview}
%-----------------
Recall that for a Galois extension $L/K$ with Galois group $\Gamma=\Gal{L}{K}$,
we established a bijection between the following two partially ordered
sets (order is inclusion).
\[
\xy
(0,0)*{\boxed{\begin{array}{cc} \scriptstyle{\text{Sub-$K$-extensions}}\\
\scriptstyle{\text{of } L/K} \end{array}}}="a";
(50,0)*{\boxed{\scriptstyle{\text{Closed subgroups of } \Gamma}}}="b";
{\ar@{<->} "a"; "b"};
\endxy
\]

Here, we view $\Gamma$ with the canonical topology of an inverse limit
of finite discrete groups. For the rest of the course, we will narrow
our focus to the case of finite $\Gamma$ (so topological considerations
are no longer necessary).\\

We say a finite Galois extension $L/K$ is {\em cyclic (resp. abelian, solvable,
simple)} if $\Gal{L}{K}$ is cyclic (resp. abelian, solvable, simple).
Recall that a group $G$ is said to be {\em solvable} if
there exists a chain of normal subgroups:
\[
G=G_0 \triangleright G_1 \triangleright \cdots \triangleright
G_\ell = \{e\}\ ,
\]
such that $G_j/G_{j+1}$ is abelian, for every $0\leq j < \ell$. Thus,
every abelian group is solvable. Solvability of an extension $L/K$ is
equivalent to solvability by radicals of any minimal polynomial
$\minpoly{\alpha}{x}$, where $\alpha\in L$.

We say a group $G$ is {\em simple} if there are no non--trivial,
proper normal subgroups of $G$. That is, $H\triangleleft G \Rightarrow
H=\{e\} \text{ or } G$. By convention, we do not consider the trivial
group to be simple. The only abelian, simple, finite groups are
$\Z/p\Z$ where $p\in\Z_{\geq 2}$ is a prime number.\\

Our next topic is to study {\em abelian extensions}. \\

\begin{itemize}
\item For each $n\in\Z_{\geq 3}$, let $\mu_n\in\C^\times$
be the cyclic subgroup consisting of $n^{\sth}$ roots of $1$.
Then $\Gal{\Q(\mu_n)}{\Q} \cong \lp\Z/n\Z\rp^\times$ is abelian.\\

\item Let $k$ be a finite field. Let $K/k$ be the finite
extension of degree $n$. Then $\Gal{K}{k}\cong\Z/n\Z$ is cyclic,
hence abelian.\\
\end{itemize}

We also know that, for instance, if $K$ is the splitting extension
over $\Q$ of $x^5-7$, then $\Gal{K}{\Q}\cong \Z/5\Z \rtimes \Z/4\Z$
is not abelian. Here, $\Z/4\Z \cong \lp\Z/5\Z\rp^\times \cong
\Aut_{\scriptstyle{\text{gp}}}(\Z/5\Z)$ acts naturall on $\Z/5\Z$.
Note that $\Gal{K}{\Q(\mu_5)} \cong \Z/5\Z$ is abelian.
Thus, we will often have to assume that our base field contains
a primitive $n^{\sth}$ root of unity.\\

\begin{defn}
Let $F$ be a field and let $n\in\Z_{\geq 3}$. Consider the subgroup
(which we know to be cyclic):

\[
\mu_n(F) := \{x\in F : x^n=1\} \subset F^\times.
\]

We say that $F$ contains a primitive $n^{\sth}$ root of unity
if $\mu_n(F) \cong \Z/n\Z$. That is, there exists $\zeta\in F$
such that $\langle\zeta\rangle \cong \Z/n\Z$.\\
\end{defn}

\PP{35.1}{Norm and trace}
%-----------------------
Let $F$ be a field and let $A$ be an $F$--algebra, which is finite-dimensional
as an $F$--vector space. Given an element $a\in A$, consider the
$F$--linear endomorphism of left multiplication by $a$,
$\mathcal{L}_a:A\to A$. That is, $\mathcal{L}_a(x) = ax$.\\

\[
\boxed{
\text{Norm of } a = 
\norm{A/K}{a} :=
\det(\mathcal{L}_a)
}
\]

\[
\boxed{
\text{Trace of } a = 
\trace{A/K}{a} :=
\mathrm{Tr}(\mathcal{L}_a)
}
\]

\noindent {\bf Example.} Let $f(x) = x^n+a_{1}x^{n-1}+\cdots+a_n \in K[x]$,
and let $A = K[x]/(f(x))$. Let $\alpha = \ol{x} \in A$. Then multiplication
by $\alpha$, in the basis $\{1,x,x^2,\ldots,x^{n-1}\}$ of $A$ over $K$,
has the following form:
\[
\mathcal{L}_\alpha = \left[
\begin{array}{cccccc}
0 & 0 & \cdots & \cdots & 0 & -a_n \\
1 & 0 & \cdots & \cdots & 0 & -a_{n-1} \\
0 & 1 & \ddots & \cdots & 0 & \vdots \\
\vdots & 0 & \ddots & \ddots & \vdots & \vdots \\
\vdots & \vdots & \ddots & \ddots & 0 & \vdots \\
0 & 0 & \cdots & 0 & 1 & -a_1
\end{array}
\right]
\]

Hence, we get $\norm{A/K}{\alpha} = (-1)^na_n = (-1)^{\deg(f)}f(0)$ and
$\trace{A/K}{\alpha} = -a_1$.\\

\begin{prop}
Let $L/K$ be a finite field extension, $A$ a finite--dimensional $L$--algebra. Let
$m = \dim_L(A)$. For $\alpha\in L$, we have:
\[
\norm{A/K}{\alpha} = \lp\norm{L/K}{\alpha}\rp^m,\qquad
\trace{A/K}{\alpha} = m \trace{L/K}{\alpha}.
\]
\end{prop}

\begin{pf}
Let $\ell = \dim_K(L)$ and let $X = (x_{i,i'})_{1\leq i,i'\leq \ell}
\in \mathrm{Mat}_{\ell\times\ell}(K)$ be the matrix of $\mathcal{L}_\alpha:L\to L$,
in a chosen basis $\{\lambda_i\}_{1\leq i\leq \ell}$.

Let $\{a_j\}_{1\leq j\leq m}$ be a basis of $A$ as an $L$--vector space.
Recall that $\{\lambda_i\alpha_j\}$ is a basis of $A$ as a $K$--vector space.
In this basis, the matrix of left multiplication by $\alpha$, say
$\wt{\mathcal{L}}_\alpha :A\to A$ is a block $m\ell\times m\ell$ size
matrix with $X$ on the diagonals. The proposition follows.
\end{pf}
\ \\

\PP{35.2}{Norm and trace via the Galois group}
%---------------------------------------------
Let $L/K$ be a finite Galois extension. Let $\Gamma = \Gal{L}{K}$.\\

\begin{prop}
For every $\alpha\in L$, we have:
\[
\boxed{
\norm{L/K}{\alpha} = \prod_{\sigma\in\Gamma} \sigma(\alpha)
}
\qquad
\boxed{
\trace{L/K}{\alpha} = \sum_{\sigma\in\Gamma} \sigma(\alpha)
}
\]
\end{prop}

\begin{pf}
Let $f(x)=\minpoly{\alpha}{x}\in K[x]$ and $n=\deg{f}$. Note that $f(x)$ splits over $L$
since $L$ is Galois. That is, we have:
\[
f(x) = (x-\alpha_1)\cdots (x-\alpha_n) = \prod_{\beta\in\Gamma\alpha} (x-\beta)
\]
Since $K(\alpha)\cong K[x]/(f(x))$, by Example from the previous paragraph,
we have:
\[
\norm{K(\alpha)/K}{\alpha} =  (-1)^n\prod_{\beta\in\Gamma\alpha} \beta,
\]
\[
\trace{K(\alpha)/K}{\alpha} = -\sum_{\beta\in\Gamma\alpha}\beta
\]

Let $rn = [L:K]$, where $r=[L:K(\alpha)]$. Note that 
$L/K(\alpha)$ is also Galois, hence $\Gal{L}{K(\alpha)}$ has order
$r$, and is isomorphic to the subgroup
$\operatorname{Stab}_\Gamma(\alpha) \subset \Gamma$. This observation
leads to the following identity:

\[
\prod_{\sigma\in\Gamma} \sigma(\alpha) = \lp\prod_{\beta\in\Gamma\alpha} \beta\rp
^{|\operatorname{Stab}_\Gamma(\alpha)|} = \norm{K(\alpha)/K}{\alpha}^r.
\]
\[
\sum_{\sigma\in\Gamma} \sigma(\alpha) = {|\operatorname{Stab}_\Gamma(\alpha)|}
\sum_{\beta\in\Gamma\alpha} \beta
= r\trace{K(\alpha)/K}{\alpha}.
\]

The right--hand sides of these two equations are respectively
$\norm{L/K}{\alpha}$ and $\trace{L/K}{\alpha}$, by Proposition
35.1, and the result follows.
\end{pf}
\ \\

\PP{35.3}{Hilbert's $90^{\sth}$ problem}
%---------------------------------------
The following result is the key step in classifying cyclic extensions.\\

\begin{thm}
Let $L/K$ be a cyclic extension, with $\Gamma = \Gal{L}{K} \cong \Z/m\Z$.
Choose a generator $\sigma\in\Gamma$.
\begin{enumerate}
\item
For $\beta\in L^\times$, $\norm{L/K}{\beta}=1$ if and only if
there exists $\alpha \in L^\times$ such that $\beta=\frac{\sigma(\alpha)}{\alpha}$. 

\item For $\beta\in L$, $\trace{L/K}{\beta}
=0$ if and only there exists $\alpha\in L$ such that
$\beta=\sigma(\alpha)-\alpha$.
\end{enumerate}
\end{thm}

\begin{pf}
(1). If $\beta=\frac{\sigma(\alpha)}{\alpha}$, then we have:
\[
\norm{L/K}{\beta} = \prod_{j=0}^{m-1} \sigma^j\lp\frac{\sigma(\alpha)}{\alpha}\rp
= \frac{\sigma(\alpha)\sigma^2(\alpha)\cdots \sigma^m(\alpha)}
{\alpha\sigma(\alpha)\cdots \sigma^{m-1}(\alpha)} = 1,
\]
using the fact that $\sigma^m=\Id$.

For the converse, define $u:\Gamma \to L^\times$ by
\[
u_{\sigma^k} = \prod_{j=0}^{k-1} \sigma^j(\beta).
\]
This definition is unambiguous, since 
\[
u_e = u_{\sigma^m} = \prod_{j=0}^{m-1} \sigma^j(\beta) = \norm{L/K}{\beta}=1.
\]
It is also easy to see that $u_{\sigma^{k+1}} = u_{\sigma} \sigma(u_{\sigma^k})$.
This implies, by Corollary 34.5, that there exists $\alpha\in L^\times$
such that $u_\tau = \frac{\tau(\alpha)}{\alpha}$. For $\tau=\sigma$ we get
the claimed result.\\

(2). As in the previous part, it is easy to see that if $\beta=\sigma(\alpha)-\alpha$,
for some $\alpha\in L$, then $\trace{L/K}{\beta}=0$.

Now we prove the converse. Define $u:\Gamma\to L$ by
\[
u_{\sigma^k} = \sum_{j=0}^{k-1} \sigma^j(\beta).
\]
Again, using $\trace{L/K}{\beta}=0$, we can check that this definition
is unambiguous. Moreover, $u_{\sigma^{k+1}} = u_\sigma + \sigma(u_{\sigma^k})$.
Corollary 34.5 again implies the existence of an $\alpha\in L$
so that $\beta = \sigma(\alpha)-\alpha$.
\end{pf}
\ \\

\PP{35.4}{Cyclic extensions}
%--------------------------
Again we assume that $L/K$ is a cyclic extension, with $\Gamma
=\Gal{L}{K}\cong\Z/m\Z$. We choose a generator $\sigma\in\Gamma$.
We further assume that $K$ contains a primitive $m^{\sth}$
root of unity, which we denote by $\zeta$. \\

\begin{thm}
There exists $\alpha\in L^\times$ such that the following assertions
hold.
\begin{enumerate}
\item $a=\alpha^m \in K^\times$.
\item $\sigma(\alpha)=\zeta\alpha$. Hence $\Gamma\alpha = \{\alpha,
\zeta\alpha,\ldots,\zeta^{m-1}\alpha\}$.
\item $L = K(\alpha)$. Hence $L$ is the splitting extension over $K$ of 
$x^m-a\in K[x]$
\[
x^m-a = \prod_{j=0}^{m-1} x-\zeta^j\alpha \ ,\qquad \text{in } L[x].
\]
\end{enumerate}
\end{thm}

\begin{pf}
Note that $\zeta\in K^\times$, and $\zeta^m=1$, which implies:
\[
\norm{L/K}{\zeta} = \zeta^m = 1.
\]
Therefore, by Theorem 35.4, there exists $\alpha\in L^\times$
such that $\zeta = \frac{\sigma(\alpha)}{\alpha}$. (2) is proved.

As norm of any element of $L$ is an element of $K$,
we get 
\[
\norm{L/K}{\alpha} = \zeta^{\frac{m(m-1)}{2}} \alpha^m = (-1)^{m-1}\alpha^m \in K.
\]

Here, we have used that $\ds x^m-1 = \prod_{j=0}^{m-1} (x-\zeta^j)$ in $K[x]$, which
implies (upon setting $x=0$) that $-1 = (-1)^m \zeta^{\frac{m(m-1)}{2}}$.
Thus, $a=\alpha^m\in K$ as claimed in (1).

It remains to show that $L=K(\alpha)$.
Note that the identity $\ds x^m-a = \prod_{j=0}^{m-1} x-\zeta^j\alpha$
implies that $K(\alpha)/K$ is the splitting extension of $x^m-a$,
which has $m$ distinct roots in $K(\alpha)$. Hence $K(\alpha)/K$
is a Galois extension. 
The claim that $K(\alpha)=L$ follows from the fact that
$\operatorname{Stab}_{\Gal{L}{K}}(\alpha)$
is trivial (since $\sigma^j(\alpha)=\zeta^j\alpha\neq \alpha$ for $1\leq j\leq m-1$).
\end{pf}
\end{document}

