## Why simple modules are often finite-dimensional II July 22, 2009

Posted by Akhil Mathew in algebra, representation theory.
Tags: , , , ,

I had a post a few days back on why simple representations of algebras over a field ${k}$ which are finitely generated over their centers are always finite-dimensional, where I covered some of the basic ideas, without actually finishing the proof; that is the purpose of this post.

So, let’s review the notation: ${k}$ is our ground field, which we no longer assume algebraically closed (thanks to a comment in the previous post), ${A}$ is a ${k}$-algebra, ${Z}$ its center. We assume ${Z}$ is a finitely generated ring over ${k}$, so in particular Noetherian: each ideal of ${Z}$ is finitely generated.

Theorem 1 (Dixmier, Quillen) If ${A}$ is a finite ${Z}$-module, then any simple ${A}$-module is a finite-dimensional ${k}$-vector space.

We know from the previous post that a simple representation of ${A}$ is just ${A/M}$ for ${M \subset A}$ a maximal left ideal; then ${M \cap Z \subset Z}$ is a left ${Z}$-ideal, hence a two-sided ${Z}$-ideal. Although ${M}$ was maximal, we don’t necessarily have ${M \cap Z}$ maximal in ${Z}$. So choose ${I \subset Z}$ to be maximal and containing ${M \cap Z}$.

Claim 1 ${I A \subset M}$.

First of all, ${IA}$ is a left ideal in ${A}$, because ${I}$ consists of central elements. Now, if ${IA \not\subset M}$, we would have by maximality

$\displaystyle M + IA = A$

or

$\displaystyle A/M \otimes_{Z} Z/I = A/(M+IA) = 0. \ \ \ \ \ (1)$

In other words, we are viewing ${A/M}$ as a ${Z}$-module.

Now we use:

Lemma 2 (Nakayama) Let ${R}$ be a commutative ring, ${J \subset R}$ be an ideal, and ${N}$ a finitely generated ${R}$-module such that

$\displaystyle N / JN = 0;$

then there exists ${x \in 1 + N}$ such that ${xN=0}$.

The proof uses the Cayley-Hamilton theorem, and is given here.  It’s essentially equivalent to other versions of Nakayama’s lemma.

So now, back to the claim. In (1), by Nakayama, we would get an element ${x \in 1 + I \subset Z}$ such that ${x(A/M) = 0}$, i.e. ${xA \in M}$, or ${x \in M}$. But then ${x \in M \cap Z = I}$, contradiction. We’ve thus proved our claim.

Now, onto the theorem itself: since ${IA \subset M}$, we can consider

$\displaystyle A/M$

as both a ${Z}$-module and a ${Z/I}$-module. As a ${Z}$-module it is finitely generated by assumption, so as a ${Z/I}$-module it is finitely generated.

But we can now invoke the following:

Theorem 3 (Generalized Nullstellensatz) Let ${k}$ be a field, ${R}$ a finitely generated commutative ring over ${k}$, and ${J \subset R}$ a maximal ideal. Then ${R/J}$ is a finite extension of ${k}$.

When ${k}$ is algebraically closed, any finite extension of ${k}$ is just ${k}$ itself, so this becomes a result from our prevoius post.

So, we see that ${A/M}$ is a finite-dimensional ${L := Z/I}$-vector space for ${L}$ a finite extension of ${k}$. Thus ${A/M}$ is a finite-dimensional ${k}$-vector space.