\documentclass[12pt]{amsart}
\title{Lecture 27}
\author{}

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\begin{document}
\maketitle

\PP{27.0}{Splitting extensions}
%-----------------------------
Recall that last time we proved the existence and uniqueness (up to
a non--unique isomorphism) of splitting extension of a set of polynomials.
More precisely, for a field $K$ and a set of polynomials $P\subset K[x]$,
there exists a unique field extension $L/K$ such that (i) every polynomial
$p(x)\in P$ splits completely in $L[x]$ and (ii) the smallest subfield of
$L$ containing $K$ and the roots of all the polynomials from $P$
is $L$ itself.\\

Recall that the argument for its existence had three main steps:
\begin{itemize}
\item We can adjoin one root of an irreducible polynomial (Kronecker's
theorem).
\item By induction, we proved the existence of the splitting extension
of a finite set of polynomials.
\item Direct limit of the splitting extensions corresponding to 
finite subsets of $P$ is the splitting extension for $P$.\\
\end{itemize}

In this lecture we will go over the theory of {\em symmetric polynomials},
and see two of its applications. Next time we will use the basic results
about symmetric polynomials (Proposition 27.2 and Theorem 27.3) to give
a different proof of the existence of splitting extensions.\\

\PP{27.1}{Symmetric polynomials}
%------------------------------
Let $A$ be a unital commutative ring and let $n\in \N$. We denote by
$\Sym_n$ the symmetric group on $n$ letters. Consider the ring of
polynomials in $n$ variables, with coefficients from $A$, and the
natural action of $\Sym_n$:

\[
\xy
(0,0)*{\Sym_n}; (28,0)*{R = A[x_1,\ldots,x_n]};
(10,-1); (10,1) **\crv{(2,-8)&(2,8)}?(1)*\dir{>};
\endxy
\]
given by $(\sigma\cdot p)(x_1,\ldots,x_n) = p(x_{\sigma(1)},\ldots, x_{\sigma(n)})$.\\

\begin{defn}
A polynomial $p(x_1,\ldots,x_n)\in R$ is symmetric if
$\sigma\cdot p = p$, for every $\sigma\in\Sym_n$. The set of all symmetric
polynomials is denoted by $S=R^{\Sym_n}\subset R$, and is an $A$--subalgebra
of $R$.
\[
S = \{p\in R : \sigma\cdot p = p\ \forall\ \sigma\in\Sym_n\}
\] 
\end{defn}
\ \\

\PP{27.2}{Elementary symmetric polynomials}
%-----------------------------------------
We keep the notations of the previous paragraph.\\

\begin{defn}
For each $k\in \N$, the $k^{\sth}$ elementary symmetric polynomial,
denoted by $e_k(x_1,\ldots,x_n)$ (or just $e_k$ if the number of variables
is clear from the context) is defined as:
\[
\boxed{
e_k(x_1,\ldots,x_n) = \sum_{1\leq i_1<\ldots <i_k\leq n} x_{i_1}\cdots x_{i_k}
}
\]
Note that $e_0=1$ and $e_k=0$ for every $k>n$. Each $e_k$ is homogeneous
of degree $k$ (if we assign degree $1$ to each of the variables $x_1,\ldots,x_n$).\\
\end{defn}

It is clear that $e_k\in S=R^{\Sym_n}$. For instance,
\[
e_1 = x_1+\cdots+x_n,\qquad e_2=\sum_{i<j} x_ix_j,\qquad e_n = x_1\cdots x_n.
\]

\begin{prop}\hfill
\begin{enumerate}
\item We have the following identity:
\[
\boxed{
\prod_{i=1}^n (x-x_i) = x^n + \sum_{k=1}^n (-1)^k e_k(x_1,\ldots,x_n) x^{n-k}
}
\]

\item $\ds (-1)^{n+1}e_n = x_n^n + \sum_{k=1}^{n-1} (-1)^k e_k x_n^{n-k}$.\\

\item Let $e_k' = e_k(x_1,\ldots,x_{n-1})$ and $e_k=e_k(x_1,\ldots,x_n)$.
That is, $e_k' = e_k|_{x_n=0}$. Then we have:
\[
e_k = e_k' + e_{k-1}'x_n,\qquad
e_k' = \sum_{j=0}^k (-1)^j x_n^j e_{k-j}.
\]
Here, $e_{-1}'=0$ (if $k=0$ in the first equation).
\end{enumerate}
\end{prop}

\begin{pf}
(1) is obtained by expanding the left--hand side. (2) follows from (1)
if we substitute $x=x_n$.\\

The first identity in (3) is clear from the definition of $e_k$. The
second one is obtained simply by inverting the first one (or an easy
induction on $k$ argument).
\end{pf}
\ \\

\PP{27.3}{Main theorem of symmetric polynomials}
%----------------------------------------------
Again, we keep the notations as above: $A$ is a unital commutative ring,
$R=A[x_1,\ldots,x_n]$, $S = R^{\Sym_n}$, and $e_k\in S$ ($1\leq k\leq n$).\\

\begin{thm}\hfill
\begin{enumerate}
\item $S$ is generated, as an algebra over $A$, by $\{e_1,\ldots,e_n\}$.\\

\item $\{e_1,\ldots,e_n\}$ are algebraically independent.\\

\item As an $S$--module, $R$ is free of rank $n!$. More precisely,
the following set of $n!$ monomials is an $S$--basis for $R$:
\[
\{x_1^{k_1}\cdots x_n^{k_n} : 0\leq k_j<j,\ \forall\ j=1,\ldots,n\}.
\]
\end{enumerate}
\end{thm}

\begin{pf}
Our proof of this theorem is going to be based on induction.
The base case of the induction is when $n=0$, $R=S=A$ and there is
nothing to show. Our induction hypothesis is that the theorem is
true for $n-1$, and we will prove it for $n$.\\

Let us prove (1) first. Let $p(x_1,\ldots,x_n)\in S$ be homogeneous
of degree $m$. We want to argue, by (a second) induction on $m$,
that $p$ can be written as a polynomial in $e_1,\ldots,e_n$, with
coefficients from $A$. If $m=0$, then $p\in A$ and there is nothing
to prove. Otherwise, consider:

\[
p' = p(x_1,\ldots,x_{n-1},0)\in A[x_1,\ldots,x_{n-1}]^{\Sym_{n-1}}.
\]

By induction hypothesis (on $n$) there exists a polynomial $P(y_1,\ldots, y_{n-1})
\in A[y_1,\ldots,y_{n-1}]$ such that $p' = P(e_1',\ldots,e_{n-1}')$. Here,
we are using the same notation as in Proposition 27.2 (2) above. Now
it is clear that $p(x_1,\ldots,x_n) - P(e_1,\ldots,e_{n-1})$ is divisible
by $x_n$. Since it is symmetric, it must also be divisible by
$x_1,x_2,\ldots,x_{n-1}$. That is,
\[
p(x_1,\ldots,x_n) - P(e_1,\ldots,e_{n-1}) = q(x_1,\ldots,x_n)\cdot (x_1\cdots x_n)
= q(x_1,\ldots,x_n)\cdot e_n,
\]
with $\deg(q)<\deg(p)$. This finishes the proof of (1).\\

Now we prove (2) and (3). For this, consider
\[
\wt{S} = R^{\Sym_{n-1}} = A[x_1,\ldots,x_n]^{\Sym_{n-1}}
= (A[x_n])[x_1,\ldots,x_{n-1}]^{\Sym_{n-1}}.
\]
That is, $\wt{S}$ is the ring of symmetric polynomials in $n-1$ variables
$x_1,\ldots,x_{n-1}$ with coefficients from $A[x_n]$. By induction hypothesis
(on $n$), $\wt{S}$ is generated (as an $A[x_n]$--algebra) by algebraically
independent elements $e_1',\ldots,e_{n-1}'$:
\[
\wt{S} = (A[x_n])[e_1',\ldots,e_{n-1}'].
\]

By Proposition 27.2 (3) above, the two sets $\{e_1',\ldots,e_{n-1}'\}$ and 
$\{e_1,\ldots,e_{n-1}\}$ are related by invertible linear (over $A[x_n]$)
transformations. Hence, we conclude that $e_1,\ldots,e_{n-1}$ are 
algebraically independent over $A[x_n]$ (in particular, over $A$), 
and can write:
\[
\wt{S} = A[x_n,e_1,\ldots,e_{n-1}] = (A[e_1,\ldots,e_{n-1}])[x_n].
\]

Let us pause and recollect what we know by now. We have shown that
$\{e_1,\ldots,e_{n-1},x_n\}$ are algebraically independent over $A$.
We also know that $S$ is generated by $\{e_1,\ldots,e_n\}$ as an
$A$--algebra. That is, if we write $C = A[e_1,\ldots,e_{n-1}]$,
then $S$ is the image of $\varphi: C[T] \to \wt{S}=C[x_n]$,
where:
\[
\varphi(T) = e_n = (-1)^{n+1}x_n^n + \sum_{j=1}^n (-1)^{n-j+1}
e_j x_n^{n-j},\ \text{ by Prop. 27.2 (2) above}.
\]

Note that $\varphi(T)$ is a degree $n$ polynomial in variable $x_n$,
whose leading coefficient $\pm 1$ is invertible. 
First of all, this implies that $\varphi$ is injective, since if
$p(T)\in C[T]$ is of degree $N\geq 1$ (in $T$ variable), then 
$\varphi(p(T))$ will have $(\pm 1)$ same leading coefficient as $p(T)$,
in degree $Nn$ in $x_n$ variable. This proves (2), that is,
$\{e_1,\ldots,e_n\}$ are algebraically independent over $A$.\\

Secondly, Euclidean division
algorithm can be performed to divide a given element of $C[x_n]$ by
$\varphi(T)$. Thus $\{1,x_n,\ldots,x_n^{n-1}\}$ is an $S$--basis
of $\wt{S}$. Combined with induction hypothesis, this proves (3).
\end{pf}
\ \\

\PP{27.4}{Remark}
%---------------
Note that our proof of (1) of Theorem 27.3 above gives an efficient
algorithm to express a given symmetric polynomial $p(x_1,\ldots,x_n)$
as a polynomial in $e_1,\ldots,e_n$. Namely:\\

\begin{itemize}
\item Set $x_n=0$ and express the resulting polynomial in $n-1$
variables, say $p'$, as a polynomial $P(e_1',\ldots,e_{n-1}')$.
\item Take the difference $p(x_1,\ldots,x_n)-P(e_1,\ldots,e_{n-1})$,
divide it by $e_n=x_1\cdots x_n$ to get another symmetric polynomial
$q(x_1,\ldots,x_n)$ of degree $\deg(p)-n$.
\item If $\deg(q)=0$, we are done. Otherwise, repeat the previous
two steps for $q$.\\
\end{itemize}

For instance, let $p=x_1^2(x_2+x_3)+x_2^2(x_3+x_1)+x_3^2(x_1+x_2)
\in \Z[x_1,x_2,x_3]^{\Sym_3}$. Then:
\[
p' = x_1^2x_2+x_1x_2^2 = e_1'e_2'.
\]
Now we compute:
\[
p-e_1e_2 = x_1^2(x_2+x_3)+x_2^2(x_3+x_1)+x_3^2(x_1+x_2)
-(x_1+x_2+x_3)(x_1x_2+x_2x_3+x_1x_3)\]
\[ 
= -3x_1x_2x_3 = -3e_3.
\]
So $p = e_1e_2-3e_3$.\\

\PP{27.5}{Application I}
%-----------------------
Let $K$ be any field, and let $T_1,\ldots,T_n$ be variables.\\

\begin{itemize}
\item Let $R=K[T_1,\ldots,T_n]$, $F(R) = K(T_1,\ldots,T_n)$ (field of fractions
of $R$). Note that we have a group homomorphism
$\Sym_n\to \Aut(F(R))$, the group of field
automorphisms of $F(R)$.\\

\item $S=R^{\Sym_n}=K[e_1,\ldots,e_n]$, where $e_k$'s are the elementary
symmetric polynomials in $T_1,\ldots,T_n$. Let $F(S)=K(e_1,\ldots,e_n)$ denote the
field of fractions of $S$.
\end{itemize}

\begin{cor}\hfill
\begin{enumerate}
\item $\ds F(S)=F(R)^{\Sym_n}$.
\item $F(R)$ is $n!$ dimensional $F(S)$--vector space. That is:
\[
\boxed{
[F(R):F(S)] = n!
}
\]
\end{enumerate}
\end{cor}

\begin{pf}
It is clear that $F(S)\subset F(R)^{\Sym_n}$. For the converse, assume
that $p/q\in F(R)$ is symmetric (where $p,q\in R=K[T_1,\ldots,T_n]$).

\[
\frac{p}{q} = \frac{p \prod_{\begin{subarray}{c}\sigma\in\Sym_n\\ \sigma\neq \Id
\end{subarray}} (\sigma\cdot q)} {\prod_{\sigma\in\Sym_n} (\sigma\cdot q)}.
\]
Now the denominator of the right--hand side is symmetric. Hence, so must
be the numerator since $p/q$ is symmetric. So $p/q\in F(S)$.\\

(2) follows from Theorem 27.3 (3) and an argument similar to the one above.
Consider the basis of $R$ as a rank $n!$ free $S$--module given
in Theorem 27.3 (3). For notational ease, we will write $m(\ul{k})$
for $x_1^{k_1}\cdots x_n^{k_n}$. The indexing set will be denoted by
$I$:
\[
I = \{\ul{k} = (k_1,\ldots,k_n) : 0\leq k_j<j,\ \forall\ j\},\qquad
B = \{m(\ul{k}) : \ul{k}\in I\} \subset R.
\]
We will show that $B$ spans $F(R)$ as $F(S)$--vector space, and is linearly
independent.

Let $p/q\in F(R)$. Multiplying and dividing this element by $\ds\prod_{\sigma}
(\sigma\cdot q)$, where the product is over all non--identity permutations,
we may assume that the denominator is symmetric. By Theorem 27.3 (3),
we can write:
\[
\frac{p}{q} = \frac{1}{q}\lp \sum_{\ul{k}\in I} c(\ul{k}) m(\ul{k})\rp
= \sum_{\ul{k}\in I} \frac{c(\ul{k})}{q} m(\ul{k})
\]
where $c(\ul{k})\in S$ for every $\ul{k}\in I$, hence $c(\ul{k})/q\in F(S)$. 
Therefore, $B$ spans $F(R)$. For linear independence, if we have
a linear dependence relation:
\[
\sum_{\ul{k}\in I} a(\ul{k}) m(\ul{k}) = 0,\ \text{ where }
a(\ul{k})\in F(S),
\]
then we can clear the denominator to assume that $a(\ul{k})\in S$. By
Theorem 27.3 (3), this implies that each $a(\ul{k})=0$.
\end{pf}
\ \\

\PP{27.6}{Application II. Discriminants}
%--------------------------------------
Let $K$ be a field, and assume that $p(x)\in K[x]$ is monic polynomial
of degree $n$. Let $L/K$ be an extension of $K$. Assume there exist
$r_1,\ldots,r_n\in L$ such that $\ds p(x) = (x-r_1)\cdots (x-r_n)$
in $L[x]$. As another application of Theorem 27.3, we have that
{\em every symmetric polynomial in} $r_1,\ldots,r_n$ {\em is an element
of} $K$, {\em which can be written as a polynomial in the coefficients
of} $p(x)$.\\

\begin{cor}
Let $P(x_1,\ldots,x_n)\in K[x_1,\ldots,x_n]^{\Sym_n}$.
Then, $P(r_1,\ldots,r_n)\in K$. Moreover, if 
$\ds p(x)=x^n+\sum_{j=1}^n a_j x^{n-j}$, where $a_1,\ldots,a_n\in K$,
then $P(r_1,\ldots,r_n)$ is a polynomial in $a_1,\ldots,a_n$.
\end{cor}

\begin{pf}
By Theorem 27.3, $P(x_1,\ldots,x_n) = Q[e_1,\ldots,e_n]$ for a
unique polynomial $Q$ in $K[y_1,\ldots,y_n]$. By
Proposition 27.2 (1), $e_j(r_1,\ldots,r_n) = (-1)^j a_j$.
This finishes the proof.
\end{pf}
\ \\

For instance, let $\Disc(p) = \prod_{i\neq j} (r_i-r_j) \in L$. Since
this is symmetric in the roots, we get that $\Disc(p)\in K$ is a polynomial
in the coefficients of $p$, called the {\em discriminant of } $p$.
By definition, $\Disc(p)=0$ if, and only if $p$ has {\em repeated roots}
in $L$.\\

As an example, if $p(x) = x^2+bx+c = (x-r)(x-s)$, then $r+s=-b$ and
$rs=c$. $\Disc(p)$ is computed as:
\[
(r-s)(s-r) = -(r-s)^2 = -((r+s)^2-4rs)=-(b^2-4c).
\]
\end{document}
