Generic freeness II July 30, 2009Posted by Akhil Mathew in algebra, commutative algebra.
Tags: algebra, commutative algebra, generic freeness, localization
Today’s goal is to partially finish the proof of the generic freeness lemma; the more general case, with finitely generated algebras, will have to wait for a later time though.
Recall that our goal was the following:
Theorem 1 Let be a Noetherian integral domain, a finitely generated -module. Then there there exists with a free -module.
The argument proceeds using dévissage. By the last post, we can find a filtration
with isomorphic to for prime ideals . Now, consider the nonzero prime ideals that occur in the above filtration. Since is a domain, we have
and we may choose with . Then when we localize at , there is still a filtration
such that . (Essentially, this uses the fact that localization is an exact functor.) By the definition of localization and the choice of , this is zero when ; this is when .
So we have a filtration of by -modules:
such that each successive quotient is a free module of rank 1.
The proof will be completed by the following lemma:
Lemma 2 Suppose are free modules over a ring and is an -module. If there is an exact sequence
then is free.
Proof: The exact sequence splits. Indeed, we can lift a basis of to elements of by surjectivity; then define a map from that lifting, which gives a section of . Thus the sequence splits.
Now, by induction, we can show that the ‘s are free over . Hence is free.
[For a better proof, see the comments below. -AM, 8/17]