Why simple modules are often finite-dimensional, I July 19, 2009Posted by Akhil Mathew in algebra.
Tags: algebra, Hilbert basis theorem, Nullstellensatz, polynomials, simple modules
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 be an associative algebra with identity over an algebraically closed field ; suppose the center is a finitely generated ring over , and is a finitely generated -module. Then: all simple -modules are finite-dimensional -vector spaces.
We’ll get to this after discussing a few other facts about rings, interesting in their own right.
Recall an object of an abelian category is simple if any subobject is either zero or isomorphic to . In the category of (left) -modules for a ring , this means the only proper submodule is zero.
We prove a general fact:
Theorem 2 Let be a ring (always with identity). Then any (nonzero) simple -module is isomorphic to for a maximal left ideal.
Proof: A submodule of would be of the form for and a left ideal, by the isomorphism theorems. But is maximal. So we get one direction.
For the other, a simple -module is generated by one element ; just pick any nonzero , and note that . So for some left ideal , the kernel of the surjection given by . If isn’t maximal, it’s contained in a maximal left ideal , and we have , so isn’t simple.
We want to consider the case of the initial fact. So we have a finitely generated commutative ring , and we will to show that its simple modules are one-dimensional. In other words, for any maximal ideal , we have .
I will discuss a more concrete setting that may clarify it:
this is a field, so is actually maximal. So we only need to consider the right hand side, i.e. find the maximal ideals in . The following two results will tells us what they are:
Theorem 4 (Hilbert’s Basis Theorem) The ring is Noetherian, i.e. each ideal is finitely generated: can be written as for some polynomials .
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 be polynomials in variables , over the fixed algebraically closed field . Suppose have no common zero. Then the ideal , i.e. there are polynomials such that
So let’s see how Theorem 3 follows from these two results. Indeed, I claim that a maximal ideal of is of the form ; then in (1) we will see . First of all, we have for some polynomials by the basis theorem. Next must have a common zero, otherwise . Let the common zero be . Then each . This proves the claim.
In the next post, I’ll apply what we’ve discussed so far to prove our aims.