Generic freeness I July 29, 2009Posted by Akhil Mathew in algebra, commutative algebra.
Tags: commutative algebra, devissage, generic freeness, localization, Noetherian rings, schemes
There is a useful fact in algebraic geometry that if you have a coherent sheaf over a Noetherian integral scheme, then it is locally free on some dense open subset. That is the content of today’s post, although I will use the language of commutative algebra than that of schemes (except at the end), to keep the presentation as elementary as possible. The goal is to get the generic freeness in a restricted case. Later, I’ll discuss the full “generic freeness” lemma of Grothendieck.
Noetherian Rings and Modules
All rings are assumed commutative in this post.
As I have already mentioned, a ring is Noetherian if each ideal of is finitely generated. Similarly, a module is Noetherian if every submodule is finitely generated. I will summarize the basic facts below briefly.
Proposition 1 A module is Noetherian iff every ascending chain
Proof: Indeed, we can write ; then is finitely generated, and each generator lies in some ; thus for some large , contains all the generators of and hence . The other implication is similar.
Proposition 2 Submodules and quotient modules of Noetherian modules are Noetherian.
Proof: In the quotient case, take inverse images in the initial module. In the submodule case, take images in the bigger initial module.
Proof: Given an ascending series , the ascending series and stabilize eventually. Now use:
Lemma 4 Given an exact sequence as above, if , and , then .
Indeed, given , we can write for some ; then , so , i.e. .
The following is the form in which Noetherian hypotheses are frequently used:
Proposition 5 Let be Noetherian; then any finitely generated -module is Noetherian.
Proof: By induction on the rank and Proposition 3, any finitely generated free -module is Noetherian; now any finitely generated -module is a quotient of such.
There are more interesting facts about Noetherian rings in general, which I will get to in another post. What we need here is the following filtration or dévissage lemma:
Proposition 6 (Dévissage) Let be a finitely generated module over a Noetherian ring . Then there is a filtration
such that each quotient for prime ideals .
Proof: The proof illustrates a useful technique that one often uses with Noetherian rings, an induction of sorts.
Consider the set defined as
Then has a maximal element . (Otherwise, we’d have an infinite strictly ascending collection of submodules.) There is thus such a filtration of . If we can find a submodule (coming from some ) with for some prime , then we would have by piecing together the filtration of with . But is bigger than , contradiction!
So we will be done if we show:
Lemma 7 Let be a module over a Noetherian ring . Then contains a submodule isomorphic to , for prime.
To prove this, we need to find an such that is prime. Choose such that is maximal; we may do this since is Noetherian. Then, for any , we have
I claim that is prime. So suppose and yet . Then . Thus , contradiction.
To state this result, I’ll need the notion of localization, which I don’t have time to define here yet; fortunately, Rigorous Trivialties has already covered it well.
The result is as follows:
Theorem 8 Let be a Noetherian integral domain, a finitely generated -module. Then there there exists with a free -module.
The result implies the corresponding fact over schemes:
Corollary 9 Let be a Noetherian integral scheme. Then if is a coherent sheaf on , there is an open dense such that the restriction is free.
Proof: Take an open affine ; since is irreducible, is dense. Say , where is a Noetherian domain. Then is the sheaf associated to a finitely generated -module . We can find with free over ; then one takes .
There is a more general formulation of this generic freeness lemma, which I’ll talk about next (as well as the proof, which goes by dévissage.)