# Fun Comm. Alg. Result

The representability of was finished around the end of october. Then Ogus talked a little bit about some generalities of for a graded ring . For example, there is not an equivalence of categories between Graded modules and qcoh sheaves of modules on , but before that Ogus proved the following result.

Thm: a ring, an module and .

- is finitely generated iff has an affine open cover s.t. is finitely generated over .
- Same as 1 replacing ‘finitely generated’ with ‘finitely presented.’
- is finitely generated and projective iff is locally free of finite rank.

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### Proof of 1

One direction is clear: if is finitely generated, then is finitely generated over , for . For the converse, WLOG, replace the covering with a finite covering of distinguished affines: . That is we have

finitely generated (say by)

Multiplying by powers of we can assume . Then since we work with finitely many affines and finitely many generators, there is a finite set and a map

Such that the map become a surjection when localized at the , i.e. its locally surjective, and in particular surjective when localized at primes, so

is a surjective map of sheaves. By the exactness of on modules, we get is surjective. QED.

### Proof of 2

Again, the same direction as before is clear. For the converse the following lemma is useful

Lemma: Let be a finitely presented module. And and finite set of generators. Then is finitely generated.

pf: Say is a finite presentation for . As is free, and is surjective, this map lifts to factor through , also using the universal property of kernel I get the following commutative diagram:

The snake lemma says there is an exact sequence . In particular, and if finitely generated as there is a surjection . Therefore, the first and last terms of are finitely generated, hence so is the middle term. QED.

Now to prove 2 the other part of 2, say is locally finitely presented, note that by 1 is finitely generated, hence there is an exact sequence . It remains to show is finitely generated. Localizing the exact sequence I get (because localization is an exact functor)

So by 1, it suffices to show is finitely generated. And this follows from the lemma as is assumed to be finitely presented. QED.

### Proof of 3

Say first that is locally free of finite rank. The required statement is that if is exact then is exact. By lemma one in this post, it suffices to show is qcoh, and for this it suffices to show the natural map

is an isomorphism. This holds if is free (because localization commutes with direct sums). Using the 5-lemma and the previous observation, it can be shown that the required isomorphism holds also if is finitely presented. Now by assumption, is locally free, hence finitely presented, hence by 2, is finitely presented. So this direction is done.

For the converse, it is enough to prove that is locally free of finite rank, as will be finitely generated and also projective. So we can reduce this case to proving the following claim: if is a local ring and is finitely generated and projective, then is free

pf: for some finite we have an exact sequence . Now , that is, is a finite vector space. Subclaim: a basis for lifts to a generating set for (pf: let be arbitrary lifts of some basis of , let be the submodule they generate. Then so by Nakayama’s lemma says ). Now, as is projective, there is a lift , so the short exact sequence above splits, hence

which shows is finitely generated (since there is a surjection ), also so applying Nakayama again gives . QED.

The next big topic Ogus talked about was fibers a morphism and the valuation criterion for separatedness and properness. The treatment was not radically different (although it was more general) than what is found in Hartshorne, so I don’t plan on putting that stuff here. Next big topic: differentials….

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- Published:
- June 10, 2009 / 6:22 pm

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- alg. geo., Ogus Excerpts

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