\documentclass[12pt]{amsart}
\title{Lecture 28}
\author{}

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\begin{document}
\maketitle

\PP{28.0}{Review}
%---------------
Last time we stated and proved the main theorem about symmetric polynomials
(Theorem 27.3). We saw two of its applications (Corollaries 27.5 and
27.6) which stated the following.\\

\begin{itemize}
\item For a field $K$ and a non--negative integer $n$, the field extension:
\[
F(S) = K(T_1,\ldots,T_n)^{\Sym_n} \subset F(R) = K(T_1,\ldots,T_n)
\]
is of degree $n!$.

\item Let $p(x)\in K[x]$ be monic of degree $n$. Let $r_1,\ldots,r_n\in L$
be its roots in an extension $L/K$. Then every symmetric polynomial
in $r_1,\ldots,r_n$ is a polynomial in the cofficients of $p(x)$, hence
an element of $K$.\\
\end{itemize}

\PP{28.1}{Application III: Existence of splitting extensions}
%--------------------------------------------------------
We can use Theorem 27.3 to give another proof of the existence of
splitting extensions.\\ 

\begin{prop}
Let $K$ be a field and let $p(x)\in K[x]$
be a monic polynomial of degree $n$. Then there exists a field
extension $L/K$ and $r_1,\ldots,r_n\in L$ such that 
\begin{enumerate}
\item $\ds p(x)=\prod_{i=1}^n (x-r_i)$ in $L[x]$.
\item $L=K(r_1,\ldots,r_n)$.
\end{enumerate}
\end{prop}

\begin{pf}
Let $R=K[x_1,\ldots,x_n]$ and $S=R^{\Sym_n} = K[e_1,\ldots,e_n]$,
where $e_k$ is the degree $k$ elementary symmetric polynomial in
$x_1,\ldots,x_n$. 
Write $p(x) = x^n + \sum_{j=1}^n a_jx^{n-j}\in K[x]$, and
define the ideal $\ga = (e_k-(-1)^ka_k : 1\leq k\leq n)
\subset R$. Set $\ol{R} := R/\ga$.\\

\noindent {\bf Claim.} $\ga\subsetneq R$. Hence, $\ol{R}\neq \{0\}$.\\

Given the claim, we can choose a maximal ideal $\gm\subsetneq \ol{R}$
and define $L=\ol{R}/\gm$. Let $\pi:R\to L$ be the natural quotient
homomorphism. Composing the ring homomorphisms
$K\hookrightarrow R \twoheadrightarrow L$,
we obtain $K\to L$ which is necessarily injective,
showing that $L$ is a field extension of $K$. Let $r_i = \pi(x_i)\in L$
($1\leq i\leq n$).
As $p(x)=\prod (x-\ol{x_i})$ in $\ol{R}[x]$, we get the following in $L[x]$.
\[
p(x) = \prod_{i=1}^n (x-r_i),\ \text{ in } L[x].
\]
Similarly, since $R$ is generated as an $K$--algebra by $x_1,\ldots, x_n$,
$L=K(r_1,\ldots,r_n)$.\\

{\em Proof of the claim.} Note that $\ga=R\ \iff\ \ol{R}=\{0\}$.
We will show that $\ol{R}$ is $n!$--dimensional $K$ vector space, which will
prove the claim. For this, recall that (Theorem 27.3) $S=R^{\Sym}$
if a polynomial ring $S=K[e_1,\ldots,e_n]$. Hence there exists
a ring homomorphism $\psi:S\to K$ such that $\psi(e_k)=(-1)^ka_k$.
This allows us to view $K\cong S/\Ker(\psi)$ as an $S$--module, and
by definition $\ga = \Ker(\psi)R\subset R$. Viewing $R$ as
$S^{\oplus n!}$, we get:
\[
\ol{R} = R/\ga = R\otimes_S \lp S/\Ker(\psi)\rp \cong
(S\otimes_S S/\Ker(\psi))^{\oplus n!} = K^{n!} \text{ as a } K--\text{vector space}.
\]
This finishes the proof of the claim.
\end{pf}
\ \\

\PP{28.2}{Application III continued}
%----------------------------------
Now let $P\subset K[x]$ consist of monic polynomials. For every
$p(x)\in P$, let $L_p/K$ be the field extension constructed above.
Define:
\[
\mathcal{R} = \bigotimes_{p\in P} L_p,
\]
as an infinite tensor product of finite--dimensional $K$--vector spaces.
Note that we have $\phi: K \to \mathcal{R}$, given by sending
$1\in K$ to $\otimes_p 1_p$, where $1_p\in L_p$ is the unit element.
We consider componenet--wise multiplication on $\mathcal{R}$ which
gives it a structure of a (unital, commutative)
$K$--algebra. Moreover, $\mathcal{R}\neq\{0\}$,
since upon choosing a basis $\{\xi^{(p)}_1,\ldots,\xi^{(p)}_{\ell_p}\}$
of $L_p$ as a $K$--vector space, we get a basis of $\mathcal{R}$:
\[
\left\{\otimes_{p\in P} \xi^{(p)}_{j_p} : 1\leq j_p \leq \ell_p\right\}.
\]

Now we proceed as before. Choose a maximal ideal $M\subsetneq \mathcal{R}$,
and define $L = \mathcal{R}/M$, which is easily seen to be a splitting
extension of $P\subset K[x]$.\\

\PP{28.3}{Fundamental theorem of algebra}
%---------------------------------------
We can now give a proof of the fundamental theorem of algebra. We
view $\C \cong \R[x]/(x^2+1)$ as a degree $2$ extension of $\R$.
Let $\iota := \ol{x} \in \C$ so that $\iota^2 = -1$.\\

The following proof was sketched by Euler in 1749, and completed
by Lagrange in 1776. At the time of its appearance, this proof was considered
{\em incomplete} but these objections were superficial in nature, and the
underlying idea is definitely flawless.\\

\begin{thm}
$\C$ is the algebraic closure of $\R$.
\end{thm}

\begin{pf}
Note that it is sufficient to show that every $f(x)\in\R[x]$ has a root
in $\C$. If the root $\alpha$ lies in $\R$, then $f(x)=(x-\alpha)g(x)$ with
$g(x)\in\R[x]$ of smaller degree. If $\alpha\in\C\setminus\R$ is a root,
then so is $\ol{\alpha}$ and we get
\[
f(x) = (x-\alpha)(x-\ol{\alpha}) h(x) = (x^2-2\Re(\alpha)x+|\alpha|^2) h(x),
\]
so $h(x)\in \R[x]$ and it has smaller degree than $f(x)$.\\

The proof is split into three claims.\\

\noindent {\bf Claim 1.} Every odd degree polynomial in $\R[x]$ has a real root.

{\em Proof.} This is the only topological step and requires the
intemediate value theorem{\footnote{Intermediate value theorem was proved by Bolzano 
in 1817. Bolzano's
argument rested on the fact that every bounded infinite set of real numbers
has a cluster point, which was rigorously proved by Weierstrass in 1872 (now called
Bolzano--Weierstrass theorem).}}. Namely, if $p(x)\in\R[x]$ is of odd degree,
then $\ds\lim_{x\to\pm\infty} p(x) = \pm\infty$, hence we can find two real
numbers $a<b\in\R$ such that $p(a)<0$ and $p(b)>0$. Therefore, there must
be some real number $c\in (a,b)$ such that $p(c)=0$.\\

\noindent {\bf Claim 2.} Every quadratic polynomial with coefficients
from $\C$ splits in $\C$.\\

{\em Proof.} This is easily shown by the well--known formula for
quadratic polynomials. For $b,c\in\C$, we have:
\[
x^2+bx+c=0,\qquad\Rightarrow\qquad x=\frac{-b+\sqrt{b^2-4c}}{2} \in \C.
\]

\noindent {\bf Claim 3.} Every $p(x)\in \R[x]$ has a root in $\C$.

{\em Proof.} Without loss of generality, we may assume that $p(x)$ is monic.
Note that the claim is true for linear and quadratic polynomials.
Let us write $\deg(p)=2^nm$ where $m$ is odd. Our argument is going
to be by induction on $n$. The base case, $n=0$ is settled in the first
claim.\\

Let $L/\C$ be an extension where $p(x)$ splits{\footnote{This is the part
of Euler--Lagrange's proof that was heavily criticized by, for instance,
Gauss, to whom the first {\em complete} proof is often attributed. Gauss
objected that the proof requires the existence of roots in order to
show existence of the roots.}}.
Let $r_1,\ldots,r_N\in L$ be roots of $p(x)$ in $L$ ($N=2^nm=\deg(p)$).
For every $b\in\R$, define:
\[
y_{ij}(b) = r_i+r_j+br_ir_j \in L,\qquad
P^{(b)}(x) = \prod_{1\leq i<j\leq N} (x-y_{ij}(b))
\]
Note that coefficients of $P^{(b)}(x)\in L[x]$ are symmetric under permutation
of $r_1,\ldots,r_N$. Therefore, by Corollary 27.6, $P^{(b)}(x)\in\R[x]$, and
\[
\deg(P^{(b)}(x)) = 2^{n-1}m (2^nm-1),
\]
has smaller exponent of $2$ dividing its degree. By induction hypothesis,
it has a root in $\C$, that is,
there is a pair $i<j$ such that $y_{ij}(b)\in\C$.

Since there are infinitely many real numbers, and for each $b\in\R$
there is a pair $i<j$ with $y_{ij}(b)\in\C$, we can find two different
$b\neq c\in \R$ which have the same pair $(i,j)$,
that is, there exists $i<j$ such that $y_{ij}(b),y_{ij}(c)\in \C$.
Solving the linear system, we conclude that
\[
r_i+r_j+br_ir_j\ \text{ and }\  r_i+r_j+cr_ir_j\in \C\qquad\Rightarrow\qquad
r_i+r_j,r_ir_j\in \C.
\]
Now $x^2-(r_i+r_j)x+r_ir_j\in\C[x]$ is a quadratic polynomial.
By Claim 2, its roots lie in $\C$. But its roots are $r_i$ and $r_j$.
So, $r_i,r_j\in \C$. Hence, $p(x)=0$ has a root in $\C$.
\end{pf}
\ \\

\begin{cor}\hfill
\begin{enumerate}
\item If $p(x)\in\R[x]$ is irreducible, then $\deg(p)=1$ or $2$.

\item $\ol{\Q} := \{z\in\C : z\text{ is algebraic over } \Q\}\subset\C$
is the algebraic closure of $\Q$.
\end{enumerate}
\end{cor}
\ \\

\PP{28.4}{Group of automorphisms: Galois group}
%----------------------------------------------
Let $L/K$ be an arbitrary field extension.\\

\begin{defn}
The {\em group of automorphisms of} $L$ over $K$, denoted by
$\Gal{L}{K}$, also called the {\em Galois group} of $L/K$, is defined as:
\[
\Gal{L}{K} = \left\{
\sigma:L\isom L \text{ field automorphism such that } \sigma|_K = \Id_K
\right\}
\]
\end{defn}

\begin{rem}
Note that $\Gal{L}{K}$ acts on $L$ via field automorphisms. As usual,
we denote by $F=L^{\Gal{L}{K}}\subset L$ the subfield of elements
fixed by $\Gal{L}{K}$:
\[
F=L^{\Gal{L}{K}} = \{r\in L : \sigma(r)=r,\ \forall\ \sigma\in\Gal{L}{K}\}.
\]
It is clear that $K\subset F$, however the reverse inclusion is false
in general. For instance, let $K=\Q$ and $L=\Q(2^{\frac{1}{3}})\subset \R$.
It is not hard to show that $\Gal{L}{K}=\{\Id\}$. Therefore
$K\subsetneq L^{\Gal{L}{K}}=L$.\\
\end{rem}

\PP{28.5}{Galois extensions}
%----------------------------
Next week we will discuss two results (due to Dedekind and Artin),
which will help us find inequalities relating $|\Gal{L}{K}|$ and the degree
of the extension $[L:K]$. For now, we can give a definition.\\

\begin{defn}
A field extension $L/K$ is called a {\em Galois} extension, if
\[
\boxed{
K = L^{\Gal{L}{K}}
}
\]
\end{defn}

As we saw in the last paragraph, $\Q(2^{1/3})/\Q$ is not a Galois extension.

\begin{ex}
$\C/\R$ is a Galois extension. Note that $\Gal{\C}{\R}$ contains complex
conjugation $\sigma:\C\to\C$, given by $\sigma(z)=\ol{z}$. It is a very
easy exercise to show that:
\[
\Gal{\C}{\R} = \{\Id,\sigma\} \cong \Z/2\Z.
\]
We know that $z=\ol{z}$ if and only if the imaginary part of $z$ is $0$,
i.e, $z\in \R$. Hence $\C^{\Gal{\C}{\R}} = \R$.
\end{ex}
\ \\

\PP{28.6}{Example: $n^{\sth}$ roots of unity}
%-----------------------------------------------
Let $n\in\Z_{\geq 2}$.
The roots of $x^n-1$ in $\C$ are often called $n^{\sth}$ roots of unity,
and are easy to list. Let
\[
\omega_n = e^{\frac{2\pi}{n}\iota} \in \C,
\]
then we have:
\[
x^n-1 = \prod_{k=0}^{n-1} \lp x-\omega_n^k\rp.
\]
Thus $\mu_n = \{1,\omega_n,\omega_n^2,\ldots,\omega_n^{n-1}\}\subset\C$
is the set of $n^{\sth}$ roots of unity. As a subgroup of $\nC$,
we have $\mu_n\cong \Z/n\Z$.\\

Consider the field extension $\Q(\mu_n)$ of $\Q$. It is clear that
$\Q(\mu_n)$ is the splitting extension of $x^n-1$ over $\Q$. Note that, for
every field automorphism $\sigma:\Q(\mu_n)\to \Q(\mu_n)$, $\sigma(\omega_n)$
is another root of $x^n-1$, hence given by
$\sigma(\omega_n) = \omega_n^k$ for some $0\leq k\leq n-1$. Moreover,
in order to be surjective, it is necessary and sufficient that
$\gcd(k,n)=1$. Thus we conclude:
\[
\Gal{\Q(\mu_n)}{\Q} = \Aut_{\scriptstyle{\text{gp}}}\lp\Z/n\Z\rp
\]

The complex conjugation is still an element of $\Gal{\Q(\mu_n)}{\Q}$
and can be used to conclude that $\Q(\mu_n)/\Q$ is a Galois extension.

\end{document}
