## Why simple modules are often finite-dimensional, I July 19, 2009

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

Today I want to talk (partially) about a general fact, that first came up as a side remark in the context of my project, and which Dustin Clausen, David Speyer, and I worked out a few days ago.  It was a useful bit of algebra for me to think about.

Theorem 1 Let ${A}$ be an associative algebra with identity over an algebraically closed field ${k}$; suppose the center ${Z \subset A}$ is a finitely generated ring over ${k}$, and ${A}$ is a finitely generated ${Z}$-module. Then: all simple ${A}$-modules are finite-dimensional ${k}$-vector spaces.

We’ll get to this after discussing a few other facts about rings, interesting in their own right.

Generalities

Recall an object ${E}$ of an abelian category is simple if any subobject is either zero or isomorphic to ${E}$. In the category of (left) ${R}$-modules for a ring ${R}$, this means the only proper submodule is zero.

We prove a general fact:

Theorem 2 Let ${R}$ be a ring (always with identity). Then any (nonzero) simple ${R}$-module is isomorphic to ${R/M}$ for ${M \subset R}$ a maximal left ideal.

Proof: A submodule of ${R/M}$ would be of the form ${L/M}$ for ${M \subset L}$ and ${L}$ a left ideal, by the isomorphism theorems. But ${M}$ is maximal. So we get one direction.

For the other, a simple ${R}$-module ${S}$ is generated by one element ${v \in S}$; just pick any nonzero ${v}$, and note that ${0 \neq Rv \subset S}$. So ${S \simeq R/I}$ for some left ideal ${I}$, the kernel of the surjection ${R \rightarrow S}$ given by ${x \rightarrow xv}$. If ${I}$ isn’t maximal, it’s contained in a maximal left ideal ${M}$, and we have ${0 \neq M/I \neq R/I}$, so ${R/I}$ isn’t simple. $\Box$

The Nullstellensatz

We want to consider the case ${A=Z}$ of the initial fact. So we have a finitely generated commutative ring ${A}$, and we will to show that its simple modules are one-dimensional. In other words, for any maximal ideal ${M}$, we have ${A/M \simeq k}$.

Theorem 3 Hypotheses as above, we have ${A/M \simeq k}$ for any maximal ideal ${M}$. In detail, if ${A}$ is a finitely generated commutative ring over the algebraically closed field ${k}$, then ${A/M \simeq k}$ for any maximal ideal ${M}$.

I will discuss a more concrete setting that may clarify it:

Since ${A}$ is commutative and finitely generated, we can write ${A = k[x_1, \dots, x_n]/I}$ for some maximal ideal ${I}$; let ${f: k[x_1, \dots, x_n] \rightarrow A}$ be the reduction map, and let ${N = f^{-1}(M)}$; then we have, by the isomorphism theorems

$\displaystyle A/M \simeq k[x_1, \dots, x_n]/(N+I);\ \ \ \ \ (1)$

this is a field, so ${N+I}$ is actually maximal. So we only need to consider the right hand side, i.e. find the maximal ideals in ${k[x_1, \dots, x_n]}$. The following two results will tells us what they are:

Theorem 4 (Hilbert’s Basis Theorem) The ring ${k[x_1, \dots, x_n]}$ is Noetherian, i.e. each ideal ${J \subset x_1, \dots, x_n}$ is finitely generated: ${J}$ can be written as ${J = (P_1, \dots, P_r)}$ for some polynomials ${P_1, \dots, P_r}$.

This might actually be a useful topic for a future post, but for now, I’ll simply quote it without proof.

Theorem 5 (The Weak Nullstellensatz) Let ${f_1, \dots, f_k}$ be polynomials in ${n}$ variables ${x_1, \dots, x_n}$, over the fixed algebraically closed field ${k}$. Suppose ${f_1, \dots, f_k}$ have no common zero. Then the ideal ${(f_1, \dots, f_k)=(1)}$, i.e. there are polynomials ${g_1, \dots, g_k}$ such that

$\displaystyle \sum g_i f_i = 1.$

So let’s see how Theorem 3 follows from these two results. Indeed, I claim that a maximal ideal ${J}$ of ${k[x_1, \dots, x_n]}$ is of the form ${(x_1-a_1, \dots, x_n-a_n)}$; then in (1) we will see ${A/M \simeq k}$. First of all, we have ${J = (f_1, \dots, f_k)}$ for some polynomials ${f_1, \dots, f_k}$ by the basis theorem. Next ${f_1, \dots, f_k}$ must have a common zero, otherwise ${J=(1)}$. Let the common zero be ${a_1, \dots, a_n}$. Then each ${f_j \subset (x_1-a_1, \dots, x_n-a_n)}$. This proves the claim.

Actually proving the Nullstellensatz could make another interesting algebraic post. For now, I’ll recommend David Speyer’s interpretation of it, and Terence Tao’s elementary proof.

In the next post, I’ll apply what we’ve discussed so far to prove our aims.