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\title{Lecture 34}
\author{}

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\begin{document}
\maketitle

\PP{34.0}{Overview}
%-----------------
In this lecture we will see some equations (Noether's equations)
related to the cohomology of
the Galois group. Theorem 34.4 and Corollary 34.5 are the main results
of these notes.
Theorem 34.4 is proved using some basic facts about {\em rationalily over
the base field} - which was outlined in Homework 10, problems 10 and 11.
The theorem itself, and its proof are not part of the syllabus. In that
sense it is somewhat optional. However, its application (Corollary 34.5)
is needed later in studying cyclic (and more generally abelian)
extensions. For this reason, I have included a ``direct proof" of
this corollary, not assuming Theorem 34.4. In the end these results
rely heavily on {\em Dedekind's independence of characters},
see Lecture 29, Sections 29.1 and 29.2.\\

\PP{34.1}{Rationality over the base field}
%----------------------------------------
Recall the notion of a $K$--{\em structure}
on a vector space. Namely, let $L/K$ be a Galois extension,
and let $V$ be an $L$--vector space.\\

\begin{defn}
A $K$--{\em structure} on $V$ is a sub-$K$-vector space $V^0\subset V$
such that extension of scalars $\varphi: V^0\otimes_K L \to V$
is an isomorphism of $L$--vector spaces.

Assuming $(V,V^0)$ is a $K$--structure on $V$ as above, a vector
$x\in V$ is said to be {\em rational over $K$} if $x\in V^0$. Similarly,
a sub-$L$-vector space $W\subset V$ is said to be rational
over $K$ if $W^0 := W\cap V^0$ is a $K$--structure on $W$.

Let $(V_\ell, V_\ell^0)$, $\ell=1,2$ be two $L$--vector spaces
together with their respective $K$--structures. Let $f:V_1\to V_2$
be an $L$--linear map. We say that $f$ is rational over $K$
if $f(V_1^0)\subset V_2^0$.\\
\end{defn}

\begin{rem}
A $K$--structure on $V$ is essentially same as a choice of a basis
of $V$. Namely, if $\{v_i\}\subset V$ is a basis of $V$ (over $L$),
we can define $V^0$ to be the $K$--span of $\{v_i\}$. Conversely,
if $\{e_i\}$ is a basis of $V^0$, it is clear from the definition
that $\{e_i\}$ will be a basis (over $L$) of $V$.\\
\end{rem}

\PP{34.2}{Rationality via $\Gamma$--action}
%-----------------------------------------------
Now, let $\Gamma = \Gal{L}{K}$.
Given a $K$--structure $V^0$ on an $L$--vector space $V$, we can
define an action of $\Gamma = \Gal{L}{K}$ on $V$, via $K$--linear automorphisms:
\[
u : \Gamma \to \Aut_{\vs{K}}(V)
\]
as follows. If $\{e_i\}_{i\in I}$ is a basis of $V^0$, then every
$x\in V$ can be written as $x = \sum_{i\in I} \lambda_i e_i$, where
$\lambda_i\in L$. Set:
\[
u_\sigma(x) = \sum_{i\in I} \sigma(\lambda_i) e_i,\ \forall\ \sigma\in\Gamma.
\]

Alternately, via the identification $\varphi:V^0\otimes_K L \isom V$,
we have:
\[
\text{For every } \sigma\in\Gamma,\ 
u_\sigma = \Id_{V^0}\otimes \sigma \in \Aut_{\vs{K}}(V^0\otimes_K L).
\]

This way of defining $u_\sigma$ makes it clear that it is independent
of the choice of the basis $\{e_i\}_{i\in I}$ of $V^0$. Moreover, the following
equation follows directly from the definitions\\

\begin{equation}\label{G-linear}
\tag{$\Gamma$--linear}
\boxed{
u_\sigma(\lambda v) = \sigma(\lambda) u_\sigma(v)
}
\ \forall\ 
\sigma\in\Gamma, \lambda\in L, v\in V.
\end{equation}
\ \\

\begin{thm}\hfill
\begin{enumerate}
\item $x\in V$ is rational over $K$ if and only if $u_\sigma(x)=x$ for
every $\sigma\in \Gamma$. In other words, $V^0 = V^\Gamma
:= \{v\in V : u_\sigma(v)=v,\ \forall\ \sigma\in\Gamma\}$.\\

\item Let $f:V_1\to V_2$ be an $L$--linear map between two $L$--vector spaces.
Asssume that $V_\ell^0\subset V_\ell$ is a $K$--structure ($\ell=1,2$).
Then $f$ is rational over $K$ if, and only if it commutes with
the $\Gamma$--action. That is,
\[
f(u_\sigma(x)) = u_\sigma(f(x)),\ \text{ for every }
\sigma\in\Gamma, x\in V_1.
\]

\item With the same notational set up of the previous part, let
$\mathcal{G}(f) \subset V_1\oplus V_2$ be the graph of $f$. Then
$f$ is rational over $K$ if and only if $\mathcal{G}(f)$ is rational
over $K$, where the $K$--structure on $V_1\oplus V_2$ is given
by $V_1^0\oplus V_2^0$.\\

\item Let $W\subset V$ be a sub-$L$-vector space of $V$. Then $W$
is rational over $K$ if and only if $u_\sigma(W)\subset W$
for every $\sigma\in\Gamma$.
\end{enumerate}
\end{thm}

\begin{pf}
(1). Let us choose a basis $\{e_i\}_{i\in I}$ of $V^0$. Given $x\in V$,
we can write it, uniquely, as a finite sum $x = \sum_{i\in I} \lambda_i e_i$,
where $\lambda_i\in L$. The condition $u_\sigma(x)=x$ is then
equivalent to
\[
\sum_{i\in I} (\sigma(\lambda_i)-\lambda_i) e_i = 0,
\]
which implies (since $\{e_i\}$ is a basis of $V$ as an $L$--vector space)
that $\sigma(\lambda_i)=\lambda_i$. Thus, $u_\sigma(x)=x$ for every
$\sigma$, if and only if $\sigma(\lambda_i)=\lambda_i$ for every
$\sigma$ and $i\in I$. As $L/K$ is a Galois extension, this implies
that $\lambda_i\in K$ for every $i\in I$, and hence $x\in V^0$.\\

(2). Assume that $f$ commutes with $\Gamma$--actions. Let $v\in V_1^0$.
Then we have:
\[
f(v) = f(u_\sigma(v)) = u_\sigma(f(v)),
\]
showing that $f(v)\in V_2^0$ by the previous part. Conversely, assume that $f$
is rational over $K$, i.e, $f(V_1^0)\subset V_2^0$. Choose a basis $\{e_i\}_{i\in I}$
of $V_1^0$. Then, for every $v = \sum_{i\in I} \lambda_i e_i \in V$, and
$\sigma\in\Gamma$, we have:
\[
f(u_\sigma(v)) = \sum_{i\in I} \sigma(\lambda_i) f(e_i)
= u_\sigma (\sum_{i\in I} \lambda_i f(e_i)) = u_\sigma(f(v)).
\]
Here, we have used $L$--linearity of $f$, $K$--linearity of
$u_\sigma$ and equation \eqref{G-linear}.\\

(3) is obvious from the definitions.\\

(4) Let $W\subset V$ be a sub-$L$-vector space of $V$.

\noindent {\bf Claim.} There exist two sub-$L$-vector spaces $V_1,V_2\subset V$,
each rational over $K$, an $L$--linear map $f:V_1\to V_2$ such that
$V = V_1\oplus V_2$ and $W\subset V$ is identified with the graph of $f$,
$W = \mathcal{G}(f) \subset V_1\oplus V_2$.

{\em Proof of the claim.} We need to make some choices.
\begin{itemize}
\item Let $\{w_j\}_{j\in J}$ be a basis of $W$ (over $L$).
\item Let $\{e_b\}_{b\in B}\subset V^0$ be such that $\{w_j\}\cup \{e_b\}$
is a basis of $V$.
\item Let $\{e_a\}_{a\in A}\subset V^0$ be such that $\{e_a\}\cup \{e_b\}$
is a basis of $V$.
\end{itemize}
The fact that these choices can be made is a foundational result
of linear algebra.

Now let $V_1=L-\text{Span}\{e_a\}$ and
$V_2=L-\text{Span}\{e_b\}$. These sub-$L$-vector spaces of $V$ are rational
over $K$ by construction. Moreover, $V=V_1\oplus V_2$ and for every
$v_1\in V_1$, there exists a unique $f(v_1)\in V_2$ such that $v_1+f(v_1)\in W$.
This is because each $e_a$ can be written uniquely in the basis
$\{w_j\}\cup \{e_b\}$. In other words, we have:
\[
V_1 \hookrightarrow V \twoheadrightarrow V/W \cong V_2.
\]
Thus, $f:V_1\to V_2$ defined as: $f(v_1)$ is the unique vector in $V_2$
so that $v_1+f(v_1)\in W$ satisfies the requirements of the claim.

Now part (4) follows from parts (2) and (3) (left as an easy exercise).
\end{pf}
\ \\

\PP{34.3}{$K$--structures via $\Gamma$--actions}
%-----------------------------------------------
Now assume that $\Gamma$ is finite. Let $V$ be an $L$--vector space
together with a group homomorphism $u:\Gamma\to \Aut_{\vs{K}}(V)$
satisfying:
\[
u_\sigma(\lambda v) = \sigma(\lambda) u_\sigma(v),\ \forall\ 
\sigma\in\Gamma,\ \lambda\in L,\ v\in V.
\]

\begin{thm}
$V^0 = V^\Gamma$ is a $K$--structure on $V$.
\end{thm}

\begin{pf}
Let $\varphi: V^0\otimes_K L \to V$ be the canonical $L$--linear map.
Let $u'_\sigma = \Id_{V^0}\otimes \sigma \in \Aut_K(V^0\otimes_K L)$
be the $\Gamma$--action on $V^0\otimes_K L$ (associated to the
canonical $K$--structure $V^0 = V^0\otimes_K K \subset V^0\otimes L$).
Let $W=\Ker(\varphi)\subset V^0\otimes_K L$. It is clear
that $u'_\sigma(W)\subset W$, for every $\sigma\in\Gamma$, since $\varphi$
commutes with $\Gamma$--actions. 
Theorem 34.2 (4) implies that $W^0 = W\cap V^0$ spans
$W$ as an $L$--vector space. But for every $w\in W^0$, we have
$w = \varphi(w) = 0$, proving that $W=\{0\}$, i.e, $\varphi$ is injective.
Note that we did not use $|\Gamma|<\infty$ hypothesis for this part.\\

To prove surjectivity, we have to show that $V^0\subset V$ spans
$V$ as an $L$--vector space. If not, we will be able to find a non--zero linear
form $\xi : V\to L$ ($L$--linear) such that $\xi|_{V_0} \equiv 0$.
Let $v\in V$ be such that $\xi(v)\neq 0$. Then, for any
$\alpha\in L$, $\ds \sum_{\sigma\in\Gamma} u_\sigma(\alpha v)$ is in $V^0$,
which implies:
\[
0=\xi(\sum_{\sigma\in\Gamma} u_\sigma(\alpha v))
= \xi(\sum_{\sigma\in\Gamma} \sigma(\alpha) u_\sigma(v))
= \sum_{\sigma\in\Gamma} \sigma(\alpha) \xi(u_\sigma(v))
\]

By Dedekind's independence of characters, this means $\xi(u_\sigma(v))=0$
for every $\sigma\in\Gamma$. In particular, for $\sigma=e$, we get
$\xi(v)=0$ which is a contradiction.
\end{pf}
\ \\

\PP{34.4}{Noether's equations}
%----------------------------
Assume that $L/K$ is a {\em finite} Galois extension, with
$\Gamma = \Gal{L}{K}$. Let $m,n\in\Z_{\geq 1}$. For an $m\times n$
matrix $X=(X_{ij})\in\mathrm{Mat}_{m\times n}(L)$ and $\sigma\in \Gamma$,
we denote by $\sigma(X) = (\sigma(X_{ij}))$.\\

\begin{thm}
Let $U : \Gamma \to \GL_n(L)$ be a set map. Then the following
two conditions are equivalent.
\begin{enumerate}
\item For every $\sigma,\tau\in\Gamma$, we have
\[
\boxed{
U_{\sigma\tau} = U_\sigma \cdot \sigma(U_\tau)
}
\]
These equations for the set map $U$ are often called {\em Noether's equations}.\\

\item There exists $A\in \GL_n(L)$ such that $U_\sigma = A^{-1} \cdot \sigma(A)$
for every $\sigma\in \Gamma$.
\end{enumerate}
\end{thm}

\begin{pf}
(2)$\Rightarrow$(1) is easily verified as follows.
\[
U_{\sigma\tau} = A^{-1}\cdot \sigma(\tau(A)) = A^{-1}\cdot \sigma(A)\cdot
\sigma(A^{-1})\cdot \sigma(\tau(A)) = U_\sigma \cdot \sigma(U_\tau).
\]

Conversely, let $U:\Gamma\to \GL_n(L)$ be a set map satisfying Noether's equations. 
We can define $u_\sigma : L^n
\to L^n$ as $u_\sigma(\xi) = U_\sigma\cdot \sigma(\xi)$. Here, we are
viewing $L^n$ as $n\times 1$ matrices (column vectors).

\begin{itemize}
\item $u_{\sigma\tau} = u_\sigma u_\tau$. This is clear, since for every
$\xi\in L^n$, we have:
\[
u_{\sigma\tau}(\xi) = U_{\sigma\tau} \cdot \sigma(\tau(\xi))
= U_\sigma \cdot \sigma(U_\tau) \cdot \sigma(\tau(\xi))
= U_\sigma \cdot \sigma(U_\tau\cdot \tau(\xi))
= u_\sigma(u_\tau(\xi)).
\]

\item $u_\sigma (\lambda \xi) = \sigma(\lambda) u_\sigma(\xi)$,
for every $\xi\in L^n$ and $\lambda\in L$. This is clear from the definition
of $u_\sigma$.

\item $u_e = \Id$, where $e\in \Gamma$ is the neutral element. 
Using Noether's equations with $\sigma=\tau=e$, we get
$U_e = U_e^2$. Since $U_e$ is invertible, this implies that $U_e$ is the identity
matrix.
\end{itemize}

Thus, we have verified the hypotheses of Theorem 34.3, showing that
$V^0 = \lp L^n\rp^\Gamma$ is a $K$--structure on $L^n$. Unfolding the
definitions, this means that $\dim_K(V^0)=n$. Let $B\in\GL_n(L)$
be an invertible matrix whose columns constitute
a basis of $V^0$. The condition $u_\sigma(B) = B$ translates to
$U_\sigma \cdot \sigma(B) = B$. Taking $A=B^{-1}$ proves our theorem.
\end{pf}
\ \\

\begin{rem}
The statement of this theorem is often written as
\[
\boxed{
H^1(\Gamma, \GL_n(L)) = \{1\}
}
\]

where $H^1(\Gamma, M)$ is the first {\em Galois cohomology} group.
We will learn more about this next week during Will Newman's presentation.
\end{rem}
\ \\

\PP{34.5}{Special cases}
%----------------------
The following result is a corollary of Theorem 34.4. Again, $L/K$
is a finite Galois extension, with $\Gamma=\Gal{L}{K}$.\\

\begin{cor}\hfill
\begin{enumerate}
\item Let $\{c_\sigma\in L^\times\}_{\sigma\in\Gamma} \subset L^\times$.
Then there exists $a\in L^\times$ such that $c_\sigma = \frac{\sigma(a)}{a}$
($\forall\ \sigma\in\Gamma$)
if and only if $c_{\sigma\tau} = c_\sigma \sigma(c_\tau)$
($\forall\ \sigma,\tau\in\Gamma$).\\

\item Let $\{x_\sigma\in L\}_{\sigma\in L}\subset L$. Then there
exists $a\in L$ such that $x_\sigma = \sigma(a) - a$ ($\forall\sigma\in\Gamma$)
if and only if $x_{\sigma\tau} = x_\sigma + \sigma(x_\tau)$
($\forall\ \sigma,\tau\in\Gamma$).
\end{enumerate}
\end{cor}

\begin{pf}
(1) is obtained by taking $n=1$ in Theorem 34.4. (2) is proved using $n=2$
and $\ds U_\sigma = \left[ \begin{array}{cc} 1 & x_\sigma \\ 0 & 1
\end{array} \right]$. We give a direct proof below, since this result
is going to be crucial in our study of abelian extensions.\\

\noindent {\em Proof of (1):} As it was for the proof of Theorem 34.4,
if there exists $a\in L^\times$ such that $c_\sigma = \sigma(a)/a$,
then $c_{\sigma\tau} = c_\sigma \sigma(c_\tau)$. Let us prove the
converse. Using Dedekind's independence of characters,
$\sum_{\tau\in\Gamma} c_\tau \tau : L \to L$ is not identically zero. 
Meaning, there exists $x\in L^\times$ such that
\[
b := \sum_{\tau\in\Gamma} c_\tau \tau(x) \neq 0.
\]
Now, for every $\sigma\in\Gamma$, we have:
\[
\sigma(b) = \sum_{\tau\in\Gamma} \sigma(c_\tau) \sigma(\tau(x))
= \sum_{\tau\in\Gamma} \frac{c_{\sigma\tau}}{c_\sigma} (\sigma\tau)(x)
= \frac{1}{c_\sigma} \sum_{\tau'\in\Gamma} c_{\tau'} \tau'(x) = \frac{b}{c_\sigma}.
\]
Here, we changed variables $\tau' = \sigma\tau$. We conclude
that, for every $\sigma\in\Gamma$, $\ds c_\sigma = \frac{b}{\sigma(b)}$.
(1) is proved by taking $a=b^{-1}$.\\

\noindent {\em Proof of (2):} Again we only show the sufficiency
of Noether's equations. Namely, assuming $x_{\sigma\tau}=x_\sigma + \sigma(x_\tau)$
we will establish the existence of $a\in L$ so that
$x_\sigma = \sigma(a)-a$ for every $\sigma\in\Gamma$.
Using Dedekind's independence of characters, we conclude the existence
of an element $\theta\in L$ so that
\[
y = \sum_{\tau\in\Gamma} \tau(\theta) \neq 0.
\]

Note that $\sigma(y)=y$ for every $\sigma\in\Gamma$, which implies
that $y\in K^\times$. Define:
\[
b = \frac{1}{y} \sum_{\tau\in\Gamma} x_\tau \tau(\theta).
\]
For any $\sigma\in\Gamma$, we get, using $\sigma(x_\tau)
= x_{\sigma\tau} - x_\sigma$:

\[
\sigma(b) = \frac{1}{y} \sum_{\tau} \sigma(x_\tau) \sigma(\tau(\theta))
= \frac{1}{y} \sum_\tau (x_{\sigma\tau} - x_\sigma) (\sigma\tau)(\theta)
\]
\[
=\frac{1}{y} \sum_{\tau'} x_{\tau'} \tau'(\theta)
- \frac{x_\sigma}{y}\sum_{\tau'} \tau'(\theta)
= b - x_\sigma.
\]

Thus, $x_\sigma = b - \sigma(b)$ for every $\sigma\in\Gamma$.
(2) is proved by taking $a=-b$.
\end{pf}

\end{document}

