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 1Let 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.

** Generalities **

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 2Let 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.

** The Nullstellensatz **

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 .

Theorem 3Hypotheses as above, we have for any maximal ideal . In detail, if is a finitely generated commutative ring over the algebraically closed field , then for any maximal ideal .

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

Since is commutative and finitely generated, we can write for some maximal ideal ; let be the reduction map, and let ; then we have, by the isomorphism theorems

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 isNoetherian, 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.

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.

July 22, 2009 at 4:20 pm

Assuming that what I’ve thought of is a correct proof for the rest of Theorem 1, I don’t think algebraic closure is necessary: in Theorem 3, we just need that A/M is a finite extension of k, which is certainly true in general.

July 22, 2009 at 4:47 pm

Thanks for the comment! Yes, this should be true, using the extended Nullstellensatz (or some general algebraic fact); I believe someone else indicated it to me, in fact, so I should probably restate Theorem 1 in that form when I finish this loose end (ideally soon).

July 22, 2009 at 9:23 pm

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