\documentclass[12pt]{amsart}
\title{Lecture 25}
\author{}

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\begin{document}
\maketitle

\PP{25.0}{Galois theory: sketch}
%------------------------------
Last time we started our study of Galois theory. A brief sketch of
the main ideas of this theory is as follows.\\

Let $K$ be a field and $f(x)\in K[x]$ a monic polynomial. Recall that
{\em monic} means that the coefficient of $x^{\deg(f)}$ (called leading
coefficient) is $1$. Assume $n=\deg(f)\in\Z_{\geq 1}$.\\

\begin{itemize}
\item Construct a field $L$ containing $K$ so that $f(x)$ splits
into a product of linear terms:
\[
f(x) = \prod_{j=1}^{n} (x-\alpha_j) \qquad \text{ in } L[x],
\]
and $L = K(\alpha_1,\ldots,\alpha_n)$. Such extensions will be called
{\em splitting extensions} and we will have to show that they exist
and are uniquely determined by $f$. This will be done next week
when we will also establish the existence of an {\em algebraic closure}.\\

\item Define a group (in honour of Galois, we will call them {\em Galois groups})
\[
G = \{\sigma:L\to L \text{ field isomorphism such that } \sigma|_{K} = \Id_K\}.
\]
\end{itemize}

The fundamental theorem of Galois theory will establish a dictionary
between {\em field extensions} and {\em groups}. We will carefully
translate the properties of one to the other.\\

\begin{rem}
It is worth noticing the similarity between the construction of a group
associated to a field extension $L/K$, and the group of {\em deck
transformations of a covering space}.
In fact this is more than
an analogy, and in algebraic geometry fundamental group of algebraic schemes
is defined via Galois theory.\\
\end{rem}

\PP{25.1}{Terms introduced last time}
%-----------------------------------

{\em Prime fields} are $\Q$ and $\FF_p$ where $p\in\Z_{\geq 2}$ is
a prime number. For any field $K$, there is only one prime field
$P$ and only one homomorphism (necessarily injective) $P\hookrightarrow K$.
We say $K$ is of characteristic zero (resp. $p$) if $P=\Q$
(resp. $P=\FF_p$).\\

A {\em field extension} is a pair consisting of a field $L$ and
a subfield $K$. We denote a field extension by $K\subset L$, or
$L/K$. In some textbooks, a field extension is also denoted by
(implying authors' fondness for covering spaces) $\fe{K}{L}$.\\

\begin{itemize}
\item Degree of an extension $L/K$ is defined as the dimension of $L$ as
a $K$--vector space
\[
\boxed{
[L:K] = \dim_K(L)
}
\]
We say $L/K$ is a {\em finite extension} if $[L:K]<\infty$.\\

\item For a set of elements $\{\alpha_i\}_{i\in I} \subset L$, we denote
by $K(\alpha_i:i\in I)\subset L$ the smallest subfield containing $K$
and the set $\{\alpha_i\}_{i\in I}$.\\

\item An element $\alpha\in L$ is said to be {\em algebraic} over $K$
if there exists $p(x)\in K[x]$ such that $p(\alpha)=0$. {\em Transcendental}
elements are the ones which are not algebraic.\\

\item For $\alpha\in L$, we defined a homomorphism (evaluation at $\alpha$)
$\ev_\alpha : K[x]\to L$, by $p(x)\mapsto p(\alpha)$. We observed from
the definition that:
\[
\boxed{
\alpha \text{ is algebraic } \iff \Ker(\ev_\alpha)\neq \{0\}
}
\]

The {\em minimal polynomial} of $\alpha$ is defined as the unique monic
polynomial $\minpoly{\alpha}{x}\in K[x]$ such that
$\Ker(\ev_\alpha) = (\minpoly{\alpha}{x})$.\\
\end{itemize}

\PP{25.2}{Equivalent characterizations of algebraic elements}
%------------------------------------------------------------
Again, let $L/K$ be a field extension, and let $\alpha\in L$.\\

\begin{prop}
The following three statements are equivalent:\\
$(1)$ $\alpha$ is algebraic.
$(2)$ $\Ker(\ev_\alpha)\neq \{0\}$.
$(3)$ $K[\alpha]=K(\alpha)$. \\

In this case, we have:
\[
\boxed{
[K(\alpha):K] = \deg(\minpoly{\alpha}{x})
}
\]
\end{prop}

The last equation was proved in the previous lecture
(see Proposition 24.5 on page 5).\\

\begin{pf}
As observed above, (1)$\iff$ (2) by definition. (2)$\Rightarrow$(3)
is also proved in Proposition 24.5. Let us check (3)$\Rightarrow$(2).
By $K[\alpha]=K(\alpha)$ we know that $\alpha^{-1}$ exists in
$K[\alpha]$. That is, there is a polynomial $p(x)\in K[x]$
such that $p(\alpha) = \alpha^{-1}$. But then, we have
$xp(x)-1\in\Ker(\ev_\alpha)$, hence it is non--zero.
\end{pf}
\ \\

\PP{25.3}{Degree of successive extensions}
%----------------------------------------

\begin{thm} Let us assume that $K_1\subset K_2\subset K_3$ are field extensions.
Then we have:
\[
\boxed{
[K_3:K_1] = [K_3:K_2]\cdot [K_2:K_1]
}
\]
\end{thm}

\begin{pf}
Note that the equation written above is only meaningful when the three degrees
involved are finite. Our proof however will not assume this.

Let $\{\alpha_i\}_{i\in I}$ be a basis of $K_2$ as a $K_1$--vector space, and 
let $\{\beta_j\}_{j\in J}$ be a basis of $K_3$ as a $K_2$--vector space.
The equation claimed in the theorem above is an easy consequence of the following
claim.\\

\noindent {\bf Claim:} $\{\alpha_i\beta_j\}_{i\in I, j\in J}$ is a basis of
$K_3$ as a $K_1$ vector space.\\

\noindent {\em Proof of the claim.} Let $\alpha\in K_3$. Then we can write
it as a (finite) linear combination
\[
\alpha = \sum_{j\in J} \beta_j c_j,
\]
where the sum is finite, and $c_j\in K_2$. Each $c_j$ can be written
as a finite sum $\ds c_j = \sum_{i\in I} \alpha_i d_{ij}$, where $d_{ij}\in K_1$.
This implies:
\[
\alpha = \sum_{i,j\in I\times J}^{\scriptstyle{\text{finite}}}
d_{ij}\cdot \alpha_i\beta_j,
\]
proving that $\{\alpha_i\beta_j\}$ span $K_3$ as a $K_1$--vector space.

Now we will check that this set is linearly independent. Consider a linear
dependence relation:
\[
\sum_{i,j\in I\times J}^{\scriptstyle{\text{finite}}}
x_{ij}\cdot \alpha_i\beta_j = 0,
\]
where $x_{ij}\in K_1$. Collect terms with same $j$ subscript to write it as
\[
\sum_{j\in J} \beta_j \lp \sum_{i\in I} x_{ij} \alpha_i \rp = 0.
\]
Now $\{\beta_j\}_{j\in J}$ are linearly independent over $K_2$, proving that
for each $j\in J$, we have:
\[
\sum_{i\in I} x_{ij} \alpha_i = 0.
\]
We get $x_{ij}=0$ for each $i\in I$ by using linear independence of $\{\alpha_i\}$
over $K_1$.
\end{pf}

\begin{cor}
If $K_1\subset K_2\subset \cdots \subset K_\ell$ are field extensions, then
\[
[K_\ell:K_1] = \prod_{j=1}^{\ell-1} [K_{j+1}:K_j].
\]
\end{cor}
\ \\

\PP{25.4}{Algebraic extensions}
%----------------------------
A field extension $L/K$ is said to be an {\em algebraic extension} if
every $\alpha\in L$ is algebraic over $K$.\\

\begin{thm}\hfill
\begin{enumerate}
\item Every finite extension is algebraic.\\

\item If $L/K$ is any field extension, then:
\[
L^\alg := \{\alpha\in L : \alpha
\text{ is algberaic over } K\} \subset L
\]
is a subfield of $L$. $L^{\alg}/K$ is an algberaic extension.
\end{enumerate}
\end{thm}

\begin{pf}
(1). Let $L/K$ be a finite extension, and let $\alpha\in L$. Since $L$
is a finite--dimensional vector space over $K$, the infinite
set $\{\alpha^n : n\in \N\}$ has to be linearly dependent. A dependence
relation $\sum_{n=0}^N c_n \alpha^n = 0$ gives us a non--zero polynomial
$p(x) = \sum_{n=0}^N c_n x^n \in \Ker(\ev_\alpha)$, proving that
$\alpha$ is algebraic over $K$.\\

(2). Now we assume $L/K$ is an arbitrary extension. 
We have to show that given two (say, non--zero) algebraic
elements $\alpha,\beta\in L^\alg$, $\alpha\beta$, $\alpha+\beta$
and $\alpha^{-1}$ are again algebraic. The last one is obvious,
since:

\[
\sum_{j=0}^n c_j \alpha^j = 0 \iff \sum_{j=0}^n c_j (\alpha^{-1})^{n-j} = 0.
\]

As for the first two, consider the successive extensions:

\[
K=K_1 \subset K(\alpha)=K_2 \subset K(\alpha,\beta)=K_3.
\]

As $\alpha,\beta$ are algebraic, each of the extensions $K_2/K_1$ and $K_3/K_2$ 
is finite by Proposition 25.2 (note that
$\beta$ being algebraic over $K$ implies that it is algebraic over
$K(\alpha)\supset K$). Hence, using Theorem 25.3, we know
$K_3/K_1$ is finite, hence algebraic by (1). So $\alpha+\beta,\alpha\beta
\in K(\alpha,\beta)$ are algebraic over $K$.
\end{pf}
\ \\

\begin{rem}
In general it is a bit non--trivial to get the minimal polynomial
of $\alpha+\beta$ from that of $\alpha$ and $\beta$. For instance,
let $\alpha=2^{\frac{1}{3}}$ and $\beta=\sqrt{3}$. Over $\Q$,
the minimal polynomials are easily computable:
\[
\minpoly{\alpha}{x} = x^3-2,\qquad \minpoly{\beta}{x} = x^2-3.
\]
(I am leaving as an easy exercise to prove that these polynomials
are irreducible over $\Q$).\\

Let $\gamma=\alpha+\beta$ and try to compute the minimal polynomial
of $\gamma$. Our proof of the theorem above implies that its degree
will be $\leq 6$.
\[
(\gamma-\sqrt{3})^3 = 2\ \Rightarrow\ \gamma^3-3\sqrt{3}\gamma^2
+9\gamma-3\sqrt{3} = 2.
\]
We have to eliminate radicals from $3\sqrt{3}(\gamma^2+1)
= \gamma^3+9\gamma-2$ to get:
\[
27(\gamma^2+1)^2 = (\gamma^3+9\gamma-2)^2.
\]
Expanding this out, we get:
\[
\gamma^6-9\gamma^4-4\gamma^3+27\gamma^2-36\gamma-23=0.
\]
(I am also leaving, not so easy, exercise of proving that this is irreducible).\\

\begin{ex}
There exist algebraic extensions which are not finite. A typical example
is obtained by taking $\Q\subset\C$ and looking at all algebraic elements
of $\C$ (called {\em algebraic numbers} and denoted by $\ol{\Q}$):
\[
\ol{\Q} = \C^\alg = \{z\in\C : z \text{ is algebraic over } \Q\}.
\]
(2) of the theorem above implies that $\ol{\Q}\subset\C$ is a field,
clearly algebraic over $\Q$. It is not hard to see that this field
is infinite--dimensional as a $\Q$--vector space.
\end{ex}
\end{rem}
\end{document}
