# Cohomology of Projective Stuff

Ogus’ approach to the cohomology of projective space is fairly similar to what’s done in Hartshorne. The only difference is Ogus avoids a few Noetherian hypothesis here and there. But just like Hartshorne, Ogus used Cech cohomology to prove the following (basis free) version of III.5.1

Thm1: Let be a projective R-module of rank . Set . Then

- is an iso.
- for all and .
- .
- is a perfect pairing.

Usually, when Noetherian hypotheses are removed (any qcoh things are involved), the subsequence proofs tend to rely on arguments utilizing the isomorphism . I’m not presenting the whole non noetherian proof here, but instead, here is a flavor of how claim #2 is proved.

Setup. Let be a line bundle on and . In this case, it makes sense to talk about as the open set where is locally a nonzero divisor. Then for qcoh, I can construct a directed system

which I denote . Then

Lemma: If is quasi compact and quasi separated (intersection of two quasi compacts is quasi compacts) then there is a map which is an isomorphism.

From this, Ogus proved if is separated then is an isomorphism. Recalling that higher cohomology of a qcoh sheaf vanishes on an affine scheme, I get

Cor: If is separated and is qcoh and is affine then for q>0.

Finally, the proof of #2: Use induction on n. The statement is vacuous for n = 1. Assume it holds for n-1. Choose a basis so that

As is a degree 1 global section, there are exact sequences

where G’ is a graded ring in one less variable, this gives rise to a map of sheaves

This gives rise to a long exact sequence

where . When q = 1 the first map is surjective because it is the surjection of the map of graded rings in the s.e.s above. Consequently, the last map is injective. For q>1 and q<n it follows that 0<q-1<n so by induction hypothesis says the second group is zero, so the last map again is injective. In summary, for all o<q<n,

is injective. This means all the maps are injective in the directed system , so each cohomology group is a subgroup of the direct limit, but is affine, so the corollary implies the direct limit is 0, so all the cohomology groups must be zero as well, which proves #2.

### Canonical Bundle

Say , and is qcoh on . There is projection . Its probably a simple exercise to show that

and that (using #3 of Thm1)

So define . Then Thm1 #3, #4 can be rewritten as

- is a perfect paring.

In fact,

Thm2: There is a canonical isomorphism .

proof (sketch): Recall from the discussion of PE in general that part of the data to is a universal line bundle which corresponds to a hyperplane . And recall the last thm in the diff3 post; it says

So in fact there is a s.e.s

Noting the first term has rank n, the middle rank n+1, then taking highest exterior powers gives

Q.E.D

### Other Stuff

Here are some fun facts. A projective morphism is proper. A quasi projective, proper morphism is in fact projective. At this point (about the end of february/early march) Ogus proved essentially the result III.5.2 in Hartshorne and then proceeded to talk about ample line bundles in order to later prove III.5.3.

I’ll end with this (unmotivated) definition. A scheme X/k is geometrically integral if for all K/k, the scheme is integral.

## About this entry

You’re currently reading “Cohomology of Projective Stuff,” an entry on Math Meandering

- Published:
- August 10, 2009 / 10:29 pm

- Category:
- alg. geo., Ogus Excerpts, Ogus/Hartshorne/Miranda

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