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\begin{document}

\chapter*{September 10:}

\begin{center}
{\Huge \bf Math 431 Class Lecture Notes}
\end{center}

\bigskip

\begin{itemize}
\item
Power sums and elementary symmetric functions
\item 
Cyclicity of finite subgroups of $F^*$
\end{itemize}

\section{Power sums and elementary symmetric functions}

The elementary symmetric functions of $n$ indeterminates are 
defined by
$$\prod_{i=1}^n (t-x_i) = \sum_{j=0}^n (-1)^j e_j t^{n-j}$$
and the power sums are defined by
$$ p_k := \sum_{i = 0}^n x_i^k$$
where $k$ is any nonnegative integer. 

A homework question asked for a proof of the identity
$$\sum_{j=0}^{k-1} (-1)^j p_{k-j} e_j + \quad (-1)^k k e_k = 0.$$
One approach is to find a relationship between the generating
functions of the $e$'s and $p$'s, since the only way that an identity
of the above shape can hold is if there is such a relationship.  

The generating
function for the $e$'s can be written
$$\sum_j (-1)^j e_j t^j = \prod_i (1 - t x_i).$$
The generating function for the $p$'s can be found by summing
a geometric series:
$$\sum p_k t^k = \sum_{i,k} x_i^k t^k = \sum_i \frac{x_i t}{1-x_it}.$$
With some work one finds a relationship between these generating
functions (roughly, the second is the logarithmic derivative of
the first) from which the identity follows.

Another approach is to first handle the case $k \ge n$.
Substitute $t = x_i$ in the defining
identities for the $e$'s and multiply by $x_i^{k-n}$ to get
$$\sum_{j=0}^{n} (-1)^j e_j x_i^{k-j} = 0.$$ 
Summing over~$i$ gives the desired identity (with a little careful
thought as to how to reconciel the two statements).

If $k < n$ then the following curious argument works.  Evaluate
the LHS of the identity (i.e., the function presumed to be 0).
In this symmetric function $g(x_1, \cdots , x_n)$
all monomials involve at most~$k$ different terms.
If this symmetric function is nonzero, then some monomial involving
at most the first $k$ variables is nonzero, i.e.,
$$g(x_1, \cdots , x_k, 0, \cdots , 0) \ne 0.$$
However, this reduces to the case $n = k$ of the identity, which
we already know to be true.

\section{Cyclicity of finite subgroups of $F^*$ }

\begin{theorem} If $G$ is a finite subgroup of the multiplicative
group $F^*$ of a field then $G$ is cyclic.
\end{theorem}

\begin{corollary} If $F$ is finite then $F^*$ is cyclic, i.e.,
``primitive roots exist.''
\end{corollary}

As we discovered in class, the proof is simplified by using the
following result.

\begin{lemma} If $x$ and $y$ are elements of an abelian group and
have relatively prime orders, then the order of $xy$ is the product
of the orders of $x$ and~$y$.
\end{lemma}

\begin{proof}
Suppose $x$ has order $m$, $y$ has order $n$, and $xy$ has order~$r$.  
Since the group is abelian
$$(xy)^{mn} = x^{mn} y^{mn} = 1$$
so the order of $r$ divides~$mn$.  
Also
$$1 = (xy)^{ra} = y^{ra}$$
and the $b$ divides $ra$.  Since $a$ and~$b$ are relatively prime,
$a$ divides~$r$.  Similarly, $b$ divides~$r$.  

Therefore $ab$ divides~$r$ and $r$ divides $ab$ so $r = ab$ as claimed.
\end{proof}

\begin{remark}
We have used two facts from elementary number theory: Suppose
that $u$ and $v$ are relatively prime; then if $u$ divides $vw$ then
$u$ divides~$w$; if both $u$ and $v$ divides~$w$ then $uv$ divides~$w$.
\end{remark}

Now we prove the theorem.  Let $G$ be a finite subgroup of the multiplicative
group~$F^*$ of a field~$F$.
Assume that the prime factorization of the order of~$G$ is
$$ \#G := n = \prod_{p_i^{a_i}}.$$
For each $i$, the polynomial $x^{n/p_i}-1$ can have at most $n/p_i$
roots.  Since $G$ has $n$ elements we can find an element
$u_i$ such that $u_i^{n/p_i}\neq\ 1$. 
The element $v_i := u_i^{n/p_i^{a_i}}$ has order $p_i^{a_i}$.
Indeed, it's clear that $v_i$ raised to the $p_i^{a_i}$ is equal 
to~1, so the order is a power of~$p_i$. But
$$v_i^{p_i^{a_i-1}} = u_i^{n/p_i} \ne 1$$
so the order is exactly equal to $p_i^{a_i}$.

By successive applications of the Lemma we see that the product of
the $v_i$ has order $n = \prod p_i^{a_i}$, and the group is cyclic
as claimed. 

\end{document}
