I am now going to discuss Kempf’s proof of the theorem of Serre. Note that this is lifted verbatim of some notes I have been taking, so apologies if the style seems out of place as a result. I use the (highly nonstandard) notation for global sections of a sheaf for entirely logistical (and typo-errorgraphical) reasons. Since this is really better suited to a PDF, I’ll also post that.

(Note: You really should read the PDF, since some diagrams are missing from this post.)

Theorem 1 (Serre)Let be an affine scheme, a quasi-coherent sheaf. Then for .

We shall prove this result following Kempf. The idea is that has a very nice basis: namely, the family of all sets . These are themselves affine, and moreover the intersection of any two elements in this basis is still in this basis. For .

** 0.1. A lemma of Kempf **

First, we set up some notation, following Kempf. Given open and a sheaf on , define to be the sheaf on , where is the inclusion. This is equivalently the sheaf . There is a natural map , which of course induces maps on cohomology.

The elementary result that Kempf proves is:

Lemma 2 (Kempf)Let be a topological space and let be a basis for which is closed under finite intersections. Suppose is fixed.Suppose for a topological space is such that for all and . Suppose . Then there is a covering of by open sets such that the image of in is zero for each .

*Proof:* We will prove this result by induction on . First, suppose , and that the result is valid for . The base case will be handled next. We can embed in a flabby sheaf , and let be the cokernel. There is an exact sequence

and by the long exact sequence for cohomology (and since ), it follows that

is exact for every in the open basis . Now fix and consider the complex of sheaves

In general, we know that is exact, by looking at the stalks, but only that is left-exact. From this alone we get that (3) is exact except perhaps at the last step. But we also know that for any , we have that the sequence of sections

is exact in view of the definition of and exactness of (2). Consequently, since is a basis, we can pass to the direct limit to the stalks, and we see that (3) must be exact at the last step too.

But is also flabby and consequently has trivial cohomology. As a result, we find that for any , there is an isomorphism

Moreover, since is flabby and thus has trivial cohomology on , we get isomorphisms from the long exact sequence of (1)

:

This means that satisfies the conditions of the proposition with , and we have assumed inductively that the result is valid for . So maps to some ; this means there is an open cover of by various such that maps to zero in . This means that maps to zero in these by naturality. This completes the proof of the inductive step.

The base case remains, i.e. . Fix . We can still embed in a flabby sheaf and obtain an exact sequence as in (1). So we get an exact sequence:

However, exactness of (4) is now no longer valid, so we cannot conclude that (3) is exact. We do, however, have an exact sequence exact for some cokernel , and we can fit these into an exact commutative diagram

Let . Then lifts to some . We have a commutative diagram of exact sequences

So to say that is annihilated by the map to is the same as saying that the image of in lifts to something in . But if is small, surjectivity of implies that we can lift to something in . So we can cover by such sets in , completing the proof.

** 0.2. Proof of the vanishing theorem **

We now apply the lemma. *Proof:* Induction on .

Let be an affine scheme. Consider the basis of open sets ; this is obviously closed under intersection, as . Now if is the (quasi-coherent) sheaf on associated to an -module , then the pull-back to is the sheaf associated to . So the direct image to is just . In particular, these are quasi-coherent on .

We will now apply the previous lemma. Suppose that is fixed and for any quasi-coherent sheaf on and . Then, this is true for , so the previous lemma says that given , there is an open cover of such the cohomology class of in is zero. Now there is a map which is injective, since the generate the unit ideal; this induces a map of sheaves

where is also quasi-coherent. There is thus a long exact sequence, of which we write a piece:

Since is in the kernel of the second map, it is in the image of .

But, if , inductively we assumed was zero. This means . If , then we write out a bit more of the exact sequence:

As before, we find that is in the image of . But since is exact on quasi-coherent sheaves, it follows that the first map is surjective, and the map out of is zero. So again we find that . This proves Serre’s vanishing theorem.

August 11, 2010 at 2:00 pm

Thansk for pointing out this article by Kempf. I’ve started to read it as well.

In this post, some of the diagrams seems to be missing. For example, there should be something after “There is an exact sequence”; which is at the begining of the proof of Lemma 2.

August 12, 2010 at 12:47 am

Ack, you’re right. I really can’t fix it right now but will take a look at it shortly.

August 12, 2010 at 12:49 am

Akhil, how do you prepare these posts? (It seems you prepare nearly one per day which is pretty remarkable; I’d like to know how one approaches it.) I mean, do you just copy the theorems and proofs out of a textbook? Or is it something deeper than that? I’d like to start a blog like this one day on mathematics and I’d like to know if there is a way to efficiently prepare these mathematic sblogs.

August 12, 2010 at 12:51 am

I just read or otherwise learn material and post what I find interesting or would like to understand better, more or less. Sometimes I get into a scheme of repeated posting on the same topic, as well. It’s nothing deeper than that; I’d impute the frequency simply to the fact that it’s fun and that I had lots of free time before (something which seems to be less true now, potentially contributing to my recent silence…).