Field degrees and multiplicities for non-integral extensions

Bernd Ulrich
Clarence W. Wilkerson
Department of Mathematics, Purdue University, West Lafayette, IN 47907
Department of Mathematics, Purdue University, West Lafayette, IN 47907
ulrich@math.purdue.edu
cwilkers@purdue.edu

Let $k$ be a field and $S = k[t_1,\hdots,t_d]$ a polynomial ring with
variables $t_i$ of degree one. Consider a $k$-subalgebra $R$ generated
by $m$ homogeneous elements $\{x_1,\hdots,x_m\}$.  In general, if $x$ is
a homogeneous element in a graded object, we denote its degree by $|x|$.

{\bf Problem.} {\it Let $[S:R]$ denote the degree of the underlying
fraction field extension. If $S$ is algebraic over $R$, calculate
$[S:R]$ from the $\{|x_i|\}$ }.

First, one has a form of Bezout's Theorem:

\begin{thm}\label{BezoutsThm} If $S$ is integral over $R$, the following hold:
\begin{enumerate}
\item $[S:R]$ divides $\prod{|x_i|}$.
\item If $m=d$, then $[S:R] = \prod{|x_i|}$.
\end{enumerate}
\end{thm}

In this paper, we consider the case that $m = d$ and obtain a converse to
part (b) above:

\begin{thm}\label{MainTheorem} If $S$ is algebraic over $R$, $m=d$, and 
$[S:R] = \prod{|x_i|}$, then $S$ is integral over $R$ $($equivalently, $S$
is finitely generated as an $R$-module$)$.
\end{thm}

We also note that if $S$ is not integral over $R$, then $[S:R]$ need not
even divide $\prod{|x_i|}$.

Our proofs rely on reduction to the case of standard graded $k$-algebras. 

An interesting application of Theorem 1.2 is in the study of rings of
invariants of finite groups acting on a polynomial ring:

\begin{thm}\label{Invariants}
Let $V$ be a $d$-dimensional vector space over the field $k$, $V^\#$ its
$k$-dual, and $S = S[V^\#] = k[t_1,\hdots,t_d]$ the algebra of
polynomial functions on $V$. Let $W \subset GL(V)$ be a finite group.
There is an induced action on $S$. Then $S^W = R$ is a polynomial
algebra over $k$ if and only if there exist homogeneous elements $\{x_1,
\hdots,x_d\}$ of $R$ such that
\begin{enumerate}
\item $S$ is algebraic over $k[x_1,\hdots,x_d]$, and
\item $|W| = \prod{|x_i|}$.
\end{enumerate}
\end{thm}