\documentclass[12pt]{amsart}
\title{Lecture 30}
\author{}

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\begin{document}
\maketitle

\PP{30.0}{Galois extensions}
%---------------------------
Let $L/K$ be a field extension, and let $\Gamma=\Gal{L}{K}$ be its Galois group.
Recall that we say $L/K$ is a Galois extension if $L^{\Gamma}=K$.\\

In previous lecture, we showed that an algebraic extension $L/K$ is
Galois if and only if it is {\em normal and separable}.\\

\begin{itemize}
\item An algebraic extension $L/K$ is called {\em normal} if for 
every $\alpha\in L$, $\minpoly{\alpha}{x}$ splits into linear
factors in $L[x]$.\\

\item A polynomial $f(x)\in K[x]$ is said to be {\em separable}
if $f(x)$ has distinct roots in $\SE{f(x)}{K}$, the splitting extension of $f(x)$
over $K$.\\

\item An algebraic extension $L/K$ is said to be {\em separable}
if for every $\alpha\in L$, the minimal polynomial $\minpoly{\alpha}{x}$
is separable.\\
\end{itemize}

We will prove that normal extensions are nothing but splitting extensions
of various subsets of $K[x]$. Theorem 29.5 says that an algebraic extension
is Galois if and only if it is the splitting extension of a set of
separable polynomials. We will discuss how an irreducible polynomial
may fail to be separable.\\

\PP{30.1}{Normal extensions}
%--------------------------
Recall that a field extension $L/K$ is normal if it is algebraic,
and for every $\alpha\in L$, $\minpoly{\alpha}{x}$ splits into a product
of (not necessarily distinct) linear factors in $L[x]$. \\

\begin{thm}
The following are equivalent, for an algebraic extension $L/K$.
\begin{enumerate}
\item $L/K$ is normal.
\item There exists a set $P\subset K[x]$, such that $L=\SE{P}{K}$ is
the splitting extension of $P$.
\item For every extension $E/L$ and $g\in \Gal{E}{K}$, $g(L) = L$.
Thus we have a short exact sequence:
\[
\mathbf{1}\to \Gal{E}{L} \to \Gal{E}{K} \to \Gal{L}{K} \to \mathbf{1},
\]
proving that $\Gal{E}{L}\subset \Gal{E}{K}$ is a normal subgroup.
\item Let $\ol{K}$ be the algebraic closure of $K$. We view $L\subset \ol{K}$
via a fixed embedding. Then, for every $g\in\Gal{\ol{K}}{K}$, we
have $g(L)=L$.
\end{enumerate}
\end{thm}
\begin{pf}
$(1)\Rightarrow (2)$. It
is clear that $L=\SE{P_L}{K}$, where 
\[
P_L(x) = \{\minpoly{\alpha}{x} : \alpha\in L\}\subset K[x].
\]

$(2)\Rightarrow (3)$. Assume that a set $I\subset K[x]$ of irreducible, monic
polynomials is given. It is not a serious restriction, since given
any set $P\subset K[x]$, we may replace $P$ by $I$ consisting of
irreducible polynomials which divide some element of $P$. Then
$\SE{P}{K} = \SE{I}{K}$. Let $L=\SE{I}{K}$. Let $E/L$ be an arbitrary
extension and $g\in \Gal{E}{K}$. To show that $g(L)=L$, it is enough
to prove that $g(\alpha)\in L$ for every $\alpha\in L$ which is a root
of some $f(x)\in P$ (since $L$ is generated over $K$ by such elements).
Now $g(\alpha)$ is another root of $f(x)$, hence is in $L$.

$(3)\Rightarrow (4)$ is obvious. Now we show that $(4)$ implies (1).
Let $\alpha\in L$ and $f(x)=\minpoly{\alpha}{x}\in K[x]$. 
In $\ol{K}[x]$, we have the factorization
$f(x)=(x-r_1)\cdots (x-r_n)$, where $n=\deg(f)$ and $r_1,\ldots,r_n\in \ol{K}$
are not necessarily distinct. Assuming $r_1=\alpha$, let $r_j=\beta$
be different from $\alpha$ (if $r_1=\ldots=r_n=\alpha$, then $f(x)=(x-\alpha)^n$
is already in $L[x]$ as we want to prove). Then $f(x)=\minpoly{\alpha}{x}
=\minpoly{\beta}{x}$ and hence (by Theorem 26.1) there exists
$g\in \Gal{\ol{K}}{K}$ such that $g(\alpha)=\beta$. As
$g(L)\subset L$, we conclude that $\beta\in L$.
We have shown that $r_1,\ldots,r_n\in L$, that is, $f(x)$
splits into a product of linear factors in $L[x]$.
\end{pf}
\ \\

\PP{30.2}{Separable polynomials}
%------------------------------
Theorem 29.5 can now be stated as follows. An algebraic extension $L/K$
is Galois if and only if $L=\SE{I}{K}$ where $I\subset K[x]$
is a set of irreducible, separable polynomials.\\

\begin{thm}
Let $K$ be a field and let $f(x)\in K[x]$ be an irreducible, monic polynomial
of degree $n\geq 1$.
Then the following conditions are equivalent.
\begin{enumerate}
\item $f(x)$ has $n$ distinct roots in its splitting extension
(i.e, $f(x)$ is separable).
\item The ideal generated by $f$ and its derivative $f'$ in $K[x]$ is the unit ideal.
\item $f'(x)\neq 0$.
\item Either $\Char(K)=0$, or $\Char(K)=p$ ($p\in\Z_{\geq 2}$ a prime
number) and $f(x)\not\in K[x^p]\subset K[x]$.
\end{enumerate}
\end{thm}

\begin{pf}
We will first show that (1) and (2) are equivalent. (2) and (3)
are clearly equivalent, since $f(x)$ is irreducible. (4) follows since
\[
f(x)=x^n+\sum_{j=0}^{n-1} c_j x^j \qquad\Rightarrow\qquad
f'(x) = nx^{n-1}+\sum_{j=1}^{n-1} jc_j x^{j-1},
\]
Thus $f'(x)=0$ if and only if $jc_j=0$ for every $1\leq j\leq n$ (with the assumption
that $c_n=1$). If $\Char(K)=0$, this is equivalent to $c_j=0$ for every $1\leq j\leq n$,
i.e, $f$ is a constant. But $\deg(f)\geq 1$. In $\Char(K)=p$ case,
we conclude that either $c_j=0$ or $p|j$. That is, $f(x)\in K[x^p]$.\\

(1)$\iff$(2). Let $L/K$ be the splitting extension of $f(x)$.
Write $\ds f(x)=\prod_{i=1}^n (x-r_i)$. Then:
\[
f'(x) = \sum_{j=1}^n \prod_{i\neq j} (x-r_i).
\]
Thus, for each $1\leq j\leq n$, $f'(r_j)=\prod_{i\neq j} (r_j-r_i)$.
Therefore, $f(x)\in K[x]$ has a repeated root $r\in L$ if and only if
$f'(r)=0$. Now, if $(f,f')=(1)$ in $K[x]$, then there exist $a(x),b(x)\in K[x]$
such that $a(x)f(x) + b(x)f'(x)=1$. Setting $x=r$ gives $0=1$ which
is absurd.

Conversely, assume that $(f,f')=(d(x))$. Since $d(x)$ divides $f(x)$,
it has a root $\gamma\in L$. Since $d(x)$ divides $f'(x)$, $f'(\gamma)=0$.
So $f$ has a repeated root.
\end{pf}
\ \\

\PP{30.3}{Perfect fields}
%-----------------------

\begin{defn}
A field $K$ is said to be {\em perfect} if every irreducible polynomial $f(x)\in K[x]$
is separable. Thus, every field of characteristic zero is perfect (by Theorem 30.2).
Every algebraically closed field is perfect.\\
\end{defn}

\begin{lem}
Let $p\in \Z_{\geq 2}$ be a prime number. Let $K$ be a field
of characteristic $p$. Then $\sigma_p : K\to K$ given by
$\sigma_p(x)=x^p$ is a homomorphism (known as Frobenius endomorphism). 
$K$ is perfect if and only
if $\sigma_p$ is an isomorphism.
\end{lem}

\begin{pf}
It is clear that $\sigma_p(xy)=\sigma_p(x)\sigma_p(y)$. Note that
\[
\sigma_p(x+y) = (x+y)^p = \sum_{i=0}^p \lp\begin{array}{c} p \\ i \end{array}\rp
x^iy^{p-i} = x^p + y^p,
\]
since for every $1\leq i\leq p-1$, the binomial coefficient:
\[
\lp\begin{array}{c} p \\ i \end{array}\rp = \frac{p(p-1)\ldots (p-i+1)}{i!}
\]
is divisible by $p$, hence zero in $K$. Thus $\sigma_p:K\to K$
is a homomorphism. \\

Now assume that $\sigma_p:K\to K$ is an isomorphism. If $K$ is not perfect,
the there would exist $g(x)\in K[x]$ monic irreducible such that $g'(x)=0$.
But that means $g(x)\in K[x^p]$, i.e,
\[
g(x) = \sum_{j=0}^n c_j x^{pj}, \text{ with } c_n=1.
\]
Let $a_j\in K$ be such that $a_j^p = c_j$. Then
\[
g(x) = \lp \sum_{j=0}^n a_j x^j \rp^p \text{ is not irreducible}.
\]

For the converse, we will need the following claim.

\noindent {\bf Claim.} For every $a\in K$, $x^p-a\in K[x]$ is irreducible
if and only if $a\not\in\Im(\sigma_p)$.\\

Let us assume the claim for now. Assume that $\sigma_p$ is not surjective.
That is there exists $a\not\in\Im(\sigma_p)$. By the claim, $x^p-a\in K[x]$
is irreducible, and its derivative is $0$, so it is not separable. Thus,
$K$ is not perfect.\\

{\em Proof of the claim.} Let $L/K$ be the splitting extension of $f(x)=x^p-a\in K[x]$,
and let $b\in L$ be a root of $f(x)$. Then $f(x)=(x-b)^p$.
Now if $f(x)=f_1(x)^{n_1}\cdots f_r(x)^{n_r}$ is the unique factorization
of $f(x)$ into a product of monic irreducible polynomials in $K[x]$, then
in $L$, each $f_j(x)$ can only have one root, namely $b$. But distinct
irreducible polynomials are coprime, so they cannot share a root. This
implies that $r=1$ and $f(x) = g(x)^n$. By degree reasons, either $g(x)=f(x)$
is irreducible, or $g(x)$ is linear (hence necessarily equal to $x-b$) and $n=p$,
proving that $b = a^{1/p}\in K$.
\end{pf}
\ \\

\PP{30.4}{Imperfect fields and purely inseparable extensions}
%-----------------------------------------------------------
According to Lemma 30.3 above, every finite field is perfect. The first
non--trivial example of imperfect fields is thus $K=\FF_p(\lambda)$.
Since $\lambda\not\in \Im(\sigma_p)$, $K$ is imperfect. Moreover
the splitting extension of $x^p-\lambda\in K[x]$ is $K_1=\FF_p(\lambda_1)$,
where $\lambda_1^p = \lambda$. Continuing this way, we obtain
a tower of field extensions:
\[
K\subset K_1 \subset K_2 \subset \cdots 
\]
where $K_n = \FF_p(\lambda_n)$ is the splitting extension of $x^p-\lambda_{n-1}
\in K_{n-1}[x]$. In other words,
\[
K_n = \text{ the splitting extension of } x^{p^n}-\lambda \in K[x].
\]

Note that $\Gal{K_n}{K} = \{\Id\}$, so in a very concrete sense, the Galois
group cannot {\em separate} different $K_n/K$. Such extensions are thus
orthogonal to Galois extensions and are defined to be purely inseparable 
(or $p$--radical) extensions.\\

\begin{defn}
Let $K$ be a field of characteristic $p\in\Z_{\geq 2}$. Let $E/K$
be a field extension. An element $\alpha\in E$ is said to be {\em purely inseparable}
(or $p$--{\em radical}) if there exists $n$ such that $\alpha^{p^n}\in K$.
The smallest such $n$ is often called {\em height of} $\alpha$.
An algebraic extension $E/K$ is called purely inseparable if it is
generated by a set of purely inseparable elements.
\end{defn}
\ \\

\end{document}
