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09.tex
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\section{22/04/2024}
\subsection{Norm and trace}
\begin{definition}
\index{Trace}
\index{Norm}
Let $E/K$ be a finite extension and $C/K$ be an algebraic closure
that contains $E$. Let $A=\Hom(E/K,C/K)$. For $x\in E$
we define the \textbf{trace} of $x$ in $E/K$
as
\[
\trace_{E/K}(x)=[E:K]_{\operatorname{ins}}\sum_{\varphi\in A}\varphi(x)
\]
and the \textbf{norm} of $x$ in $E/K$ as
\[
\norm_{E/K}(x)=\left(\prod_{\varphi\in A}\varphi(x)\right)^{[E:K]_{\operatorname{ins}}}.
\]
\end{definition}
As an optional exercise, one can show that these definitions do not depend on the algebraic closure.
We collect some basic properties as an exercise:
\begin{exercise}
\label{xca:norm_and_trace}
Let $E/K$ be a finite extension. The following statements hold:
\begin{enumerate}
\item If $E/K$ is not separable, then $\trace_{E/K}(x)=0$ for all $x\in E$.
\item If $x\in K$, then $\trace_{E/K}(x)=[E:K]x$.
\item $\trace_{E/K}(x)\in K$ for all $x\in E$.
\item $\norm_{E/K}(x)=0$ if and only if $x=0$.
\item If $x\in K$, then $\norm_{E/K}(x)=x^{[E:K]}$.
\item $\norm_{E/K}(x)\in K$ for all $x\in E$.
\end{enumerate}
\end{exercise}
One proves, moreover, that
$\trace_{E/K}\colon E\to K$
satisfies
\[
\trace_{E/K}(x+\lambda y)=
\trace_{E/K}(x)+\lambda\trace_{E/K}(y)
\]
for all $x,y\in E$ and $\lambda\in K$, that is to say that
$\trace_{E/K}\colon E\to K$
is a
linear form in $E$ The norm
$\norm_{E/K}\colon E^{\times}\to K^{\times}$
is a group homomorphism.
\begin{exercise}
Let $E/K$ be a finite extension and
$x\in E$ be algebraic. If
\[
f(x,K)=X^n+a_{n-1}X^{n-1}+\cdots+a_1X+a_0,
\]
then
$\norm_{E/K}(x)=\left((-1)^na_0\right)^{[E:K(x)]}$ and
$\trace_{E/K}(x)=-[E:K(x)]a_{n-1}$.
\end{exercise}
\begin{example}
Let $E=\Q(\sqrt{2},\sqrt{3})$. Then
\begin{align*}
&\trace_{E/\Q}(\sqrt{2})=0,
&&
\norm_{E/\Q}(\sqrt{2})=4,\\
&\trace_{E/\Q(\sqrt{2})}(\sqrt{2})=2\sqrt{2},
&&\norm_{E/\Q(\sqrt{2})}(\sqrt{2})=2.
\end{align*}
\end{example}
\begin{example}
If $E/K$ is a finite Galois extension, then
\[
\trace_{E/K}(x)=\sum_{\sigma\in\Gal(E/K)}\sigma(x)
\quad\text{and}\quad
\norm_{E/K}(x)=\prod_{\sigma\in\Gal(E/K)}\sigma(x)
\]
for all $x\in E$. In particular, since $E=K(y)$ for some
$y$ by Proposition \ref{pro:monogenic},
\[
\trace_{E/K}(y)=-a_{n-1}
\quad\text{and}\quad
\norm_{E/K}(y)=(-1)^na_0,
\]
where
$f(y,K)=X^n+a_{n-1}X^{n-1}+\cdots+a_1X+a_0$.
\end{example}
\subsection{Finite fields}
In this section, $p$ will be a prime number.
\begin{proposition}
Let $m$ be a positive integer.
Up to isomorphism, there exists a unique
field $F_m$ of size $p^m$.
\end{proposition}
\begin{proof}
Let $C$ be an algebraic closure of the field $\Z/p$ and
let $F_m=\{x\in C:x^{p^m}=x\}$ be the set of roots of $X^{p^m}-X$. Since
the polynomial $X^{p^m}-X$ has no multiple roots, $|F_m|=p^m$. Moreover,
$F_m$ is the unique subfield of $C$ of size $p^m$.
To prove the uniqueness, it is enough to note that
if $K$ is a field of $p^m$ elements, then
$K$ is the decomposition field of $X^{p^m}-X$ over $\Z/p$.
\end{proof}
Let $K=\Z/p$ and $C$ be an algebraic closure of $K$.
We claim that $C=\cup_k F_k$. If $x\in C$, then $x$ is algebraic over $K$.
Since $K(x)/K$ is finite, $K(x)$ is a finite field, say
$|K|=p^r$ for some $r$. Then $x^{p^r}=x$ and hence $x\in F_r$.
\begin{exercise}
Prove the following statements:
\begin{enumerate}
\item If $x\in F_r$, then $x^{p^{rk}}=x$ for all $k\geq0$.
%\item If $m\mid n$, then $F_m\subseteq F_n$.
\item $F_m\subseteq F_n$ if and only if $m\mid n$.
\item $F_m\cap F_n=F_{\gcd(m,n)}$.
\end{enumerate}
\end{exercise}
\begin{proposition}
Every finite extension of a finite field is cyclic.
\end{proposition}
\begin{proof}
Let $K=\Z/p$. It is enough to show that $F_n/F_m$ is cyclic if $m$ divides $n$.
We first prove that $F_n/K$ is cyclic.
Let
\[
\sigma\colon F_n\to F_n,\quad
x\mapsto x^p.
\]
Then
$\sigma\in\Gal(F_n/K)$ (it is bijective because all field homomorphisms
are injective and $F_n$ is finite).
Note that
$F_n/K$ is a Galois extension, as $F_n$ is the splitting
field over $K$
of the separable polynomial $X^{p^n}-X\in K[X]$.
Thus $|\Gal(F_n/K)|=[F_n:K]=n$.
We claim that $\sigma$ generated $\Gal(F_n/K)$. Since
$\sigma^i(x)=x^{p^i}$ for all $i\geq 0$, in particular,
\[
\sigma^n(x)=x^{p^n}=x.
\]
Thus $\sigma^n=\id$ and hence $|\sigma|$ divides $n$. Let
$s=|\sigma|$. We know that $F_n^{\times}=F_n\setminus\{0\}$ is
cyclic, say $F_n^{\times}=\langle g\rangle$. Since $|g|=p^n-1$,
\[
g=\sigma^s(g)=g^{p^s}
\]
and hence $p^s\equiv 1\bmod (p^n-1)$. Thus $p^n-1$ divides $p^s-1$ and
hence $n$ divides $s$. Therefore $n=s$ and $\Gal(F_n/K)=\langle\sigma\rangle$.
For the general case note that if $m$ divides $n$,
then $\Gal(F_n/F_m)$ is a subgroup of $\Gal(F_n/K)$. Since $\Gal(F_n/K)$ is cyclic,
the claim follows.
\end{proof}
\index{Frobenius automorphism}
If $K=\Z/p$ and
$m$ divides $n$, the subextension $F_m$ corresponds
to the unique
subgroup of index $m$ of $\Gal(F_n/K)=\langle\sigma\rangle$. This subgroup
is $\langle\sigma^m\rangle$, where
\[
\sigma^m(x)=x^{p^m}=x^{|F_m|}.
\]
Note that $\Gal(F_n/F_m)=\langle\sigma^m\rangle$.
The map $\sigma^m$ is known as
the \textbf{Frobenius automorphism}.
\begin{exercise}
Let $E/K$ be an extension of finite fields. Then $E/K$
is cyclic. Moreover, $\Gal(E/K)=\langle\tau\rangle$, where $\tau(x)=x^{|K|}$.
\end{exercise}
% page 96
% number of irreducible polynomials
% Moebius inversion formula in commutative rings
\subsection{Cyclotomic extensions}
For $n\geq1$ let $G_n(K)=\{x\in K:x^n=1\}$ be the
set of $n$-roots of one in $K$. Note that
$G_n(K)$ is a cyclic subgroup of $K^{\times}$ and that
$|G_n(K)|$ divides $n$.
\begin{example}
$G_n(\R)=\{-1,1\}$ if $n$ is odd and $G_{n}(\R)=\{1\}$ if $n$ is even.
\end{example}
\begin{exercise}
Let $K$ be a field of characteristic $p>0$. Let $n=p^sm$ for some $m$ not divisible by $p$.
Then $G_n(K)=G_m(K)$.
\end{exercise}
\begin{exercise}
Let $q$ be a prime number. Then $G_n(\Z/q)\simeq\Z/\gcd(n,q-1)$.
\end{exercise}
Similarly, one can prove that if $K$ is a finite field, then $G_n(K)$ is a cyclic group
of order $\gcd(n,|K^{\times}|)$.
\begin{example}
If $C$ is algebraically closed of characteristic coprime with $n$,
then $G_n(C)$ is cyclic of order $n$, as $X^n-1$
has all its roots in $C$ and does not contain multiple roots.
\end{example}
Let $K$ be an algebraically closed field and $n$ be
such that $n$ is coprime with the characteristic of $K$. The set of
\textbf{primitive $n$-roots} is defined as
\[
H_n(K)=\{x\in G_n(K):|x|=n\}.
\]
\begin{definition}
\index{Cyclotomic polynomial}
Let $K$ be an algebraically closed field and $n$ be
such that $n$ is coprime with the characteristic of $K$. The \textbf{$n$-th cyclotomic
polynomial} is defined as
\[
\Phi_n=\prod_{x\in H_n(K)}(X-x)\in K[X].
\]
\end{definition}
\index{Euler's $\phi$ function}
For $n\geq1$ the Euler's function is defined as
\[
\varphi(n)=|\{k:1\leq k\leq n,\;\gcd(k,n)=1\}|.
\]
For example, $\varphi(4)=2$, $\varphi(8)=\varphi(10)=4$ and $\varphi(p)=p-1$ for every prime $p$.
\begin{proposition}
Let $K$ be an algebraically closed field and $n$ be
such that $n$ is coprime with the characteristic of $K$. Let $A$ be
the ring of integers of $K$.
\begin{enumerate}
\item $\deg\Phi_n=\varphi(n)$.
\item $\Phi_n\in A[X]$.
\end{enumerate}
\end{proposition}
\begin{proof}
The first statement is clear. Let us prove 2) by induction on $n$. The case $n=1$ is
trivial, as $\Phi_1=X-1$. Assume that $\Phi_d\in A[X]$ for all $d$ such that $d<n$.
In particular,
\[
\gamma=\prod_{\substack{d\mid n\\d\ne n}}\Phi_d\in A[X].
\]
Since $\gamma$ is monic, it follows that
$\frac{X^n-1}{\gamma}\in A[X]$. Now the claim follows from
\[
X^n-1=\prod_{d\mid n}\Phi_d=\Phi_n\prod_{\substack{d\mid n\\d\ne n}}\Phi_d=\Phi_n\gamma.\qedhere
\]
\end{proof}
By taking degree in the equality
$X^n-1=\prod_{d\mid n}\Phi_d$
one gets
\[
n=\sum_{d\mid n}\varphi(d).
\]
\begin{definition}
\label{defn:cyclotomic}
\index{Extension!cyclotomic}
Let $n\geq2$ and $K$ be a field of characteristic coprime with $n$. A
\textbf{cyclotomic extension} of $K$ of index $n$ is a
decomposition field of $X^n-1$ over $K$.
\end{definition}
Let $C$ be an algebraic closure of $K$ and $n\geq2$ be coprime with the characteristic of $K$.
If follows from Definition \ref{defn:cyclotomic}
that a cyclotomic extension of index $n$ is of the form
$K(\omega)/K$ for some $\omega\in H_n(K)$.
\begin{proposition}
A cyclotomic extension of index $n$ is abelian and of degree a divisor of $\varphi(n)$.
\end{proposition}
\begin{proof}
Let $C$ be an algebraic closure of $K$ and $n\geq2$ be coprime with the characteristic of $K$.
Let $\omega\in H_n(C)$ and $K(\omega)/K$ be a cyclotomic extension. Then $K(\omega)/K$
is a Galois extension, as it is a decomposition field of a separable polynomial.
Let $U=\mathcal{U}(\Z/n)$ be the group of units of $\Z/n$ and
\[
\lambda\colon \Gal(K(\omega)/K)\to U,
\quad
\sigma\mapsto m_{\sigma},
\]
where $m_{\sigma}$ is such that $\sigma(\omega)=\omega^{m_{\sigma}}$. The map $\lambda$ is well-defined and
it is a group homomorphism, as if $\sigma,\tau\in\Gal(K(\omega)/K)$, then, since
\[
(\tau\sigma)(\omega)=\tau(\sigma(\omega))=\tau(\omega^{m_\sigma})=\left(\omega^{m_\sigma}\right)^{m_\tau}=\omega^{m_\sigma m_\tau},
\]
it follows that $\lambda(\sigma)\lambda(\tau)=\lambda(\sigma\tau)$. Since
$\lambda$ is injective, $\Gal(K(\omega)/K)$ is isomorphic to a subgroup
of the abelian group $U$. Hence $\Gal(K(\omega)/K)$ is abelian. Moreover,
$[K(\omega):K]=|\Gal(K(\omega)/K)|$ is a divisor of $|U|=\varphi(n)$.
\end{proof}
\begin{exercise}
Prove that a cyclotomic extension $K(\omega)/K$ has degree $\varphi(n)$ if and only if
$\Phi_n$ is irreducible over $K$.
\end{exercise}
Note that $\Phi_n$ is irreducible over $\Q$. Some concrete examples:
\[
\Phi_1=X-1,
\quad
\Phi_2=X+1,
\quad
\Phi_3=X^2+X+1,
\quad
\Phi_6=X^2-X+1.
\]
If $p$ is a prime number, then $\Phi_p=X^{p-1}+\cdots+X+1$.
\begin{example}
$\Phi_5$ is irreducible over $\Z/2$. First note that
$\Phi_5=X^{4}+\cdots+X+1$ does not have roots in $\Z/2$. If
$\Phi_5$ is reducible, then, since
$X^2+X+1$ is the unique degree-two
monic irreducible polynomial
over $\Z/2$, it follows that
\[
\Phi_5=(X^2+X+1)(X^2+X+1)=(X^2+X+1)^2=X^4+X^2+1,
\]
a contradiction.
\end{example}
\begin{exercise}
Prove that
$\Phi_{12}=X^4-X^2+1$ is not irreducible over $\Z/5$.
\end{exercise}
\subsection{Hilbert's theorem 90}
\begin{theorem}[Hilbert]
Let $E/K$ be a cyclic extension. Assume that
$\Gal(E/K)$ is generated by $\tau$. For
$a\in E$, $\norm_{E/K}(a)=1$ if and only
if $a=b/\tau(b)$ for some $b\in E\setminus\{0\}$.
\end{theorem}
\begin{proof}
Let $n=|G|$. We first prove $\impliedby$. If $a=b/\tau(b)$ and $b\ne 0$, then
\[
\norm_{E/K}(a)=a\tau(a)\tau^2(a)\cdots\tau^{n-1}(a)
=\frac{b}{\tau(b)}\frac{\tau(b)}{\tau^2(b)}\cdots\frac{\tau^{n-1}(b)}{\tau^n(b)}=1.
\]
Now we prove $\implies$. Let $a\in E$ be such that $\norm_{E/K}(a)=1$. For
$c\in E$ let
\begin{align*}
d_0 &= ac,\\
d_1 &= a\tau(a)\tau(c),\\
d_2 &= a\tau(a)\tau^2(a)\tau^2(c),\\
&\vdots\\
d_{n-1} &= \underbrace{a\tau(a)\cdots\tau^{n-1}(a)}_{=\norm_{E/K}(a)}\tau^{n-1}(c)=\tau^{n-1}(c).
\end{align*}
Then
\[
a\tau(d_j)=a\tau(a)\cdots\tau^{j+1}(a)\tau^{j+1}(c)=d_{j+1}
\]
for all $j\in\{0,\dots,n-2\}$. Let $b=d_0+\cdots+d_{n-1}$. We claim that
$b\ne 0$ for some $c$. Suppose this is not true, say $b=0$ for all $c$. Then
\begin{align*}
0&=ac+(a\tau(a))\tau(c)+\cdots+(a\tau(a)\cdots\tau^{n-1}(a))\tau^{n-1}(c)
\end{align*}
for every $c\in E$. This
implies that $a=0$ by Dedekind's theorem, a contradiction.
So let $c\in E$ be
such that $b\ne 0$. Then
\begin{align*}
\tau(b)&=\tau(d_0)+\cdots+\tau(d_{n-1})\\
&=\tau(ac)+\tau(a\tau(c))+\cdots+\tau(\tau^{n-1}(c))\\
&=\frac{1}{a}(d_1+\cdots+d_{n-1})+\tau^n(c)\\
&=\frac{1}{a}(d_0+\cdots+d_{n-1})\\
&=b/a.\qedhere
\end{align*}
\end{proof}
\begin{exercise}
Let $E/K$ be a cyclic extension. Assume that
$\Gal(E/K)$ is generated by $\tau$. Prove that for
$a\in E$, $\trace_{E/K}(a)=0$ if an only
if $a=b-\tau(b)$ for some $b\in L\setminus\{0\}$.
\end{exercise}