## Generic freeness II July 30, 2009

Posted by Akhil Mathew in algebra, commutative algebra.
Tags: , , ,

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 ${A}$ be a Noetherian integral domain, ${M}$ a finitely generated ${A}$-module. Then there there exists ${f \in A - \{0\}}$ with ${M_f}$ a free ${A_f}$-module.

The argument proceeds using dévissage. By the last post, we can find a filtration

$\displaystyle 0 = M_0 \subset M_1 \subset \dots \subset M_n = M,$

with ${M_{i+1}/M_i}$ isomorphic to ${A/\mathfrak{p}_i}$ for prime ideals ${\mathfrak{p}_i}$. Now, consider the nonzero prime ideals ${\mathfrak{p}'_j}$ that occur in the above filtration. Since ${A}$ is a domain, we have

$\displaystyle \prod \mathfrak{p}'_j \neq 0,$

and we may choose ${f \neq 0}$ with ${f \in \prod \mathfrak{p}_j'}$. Then when we localize at ${f}$, there is still a filtration

$\displaystyle 0 = (M_0)_f \subset (M_1)_f \subset \dots \subset (M_n)_f = M_f,$

such that ${(M_{i+1})_f/(M_i)_f = (A/\mathfrak{p}_i)_f}$. (Essentially, this uses the fact that localization is an exact functor.) By the definition of localization and the choice of ${f}$, this is zero when ${\mathfrak{p}_i \neq 0}$; this is ${A_f}$ when ${\mathfrak{p}_i=0}$.

So we have a filtration of ${M_f}$ by ${A_f}$-modules:

$\displaystyle 0 = N_0 \subset N_1 \subset \dots \subset N_m = M_f$

such that each successive quotient is a free module of rank 1.

The proof will be completed by the following lemma:

Lemma 2 Suppose ${F', F''}$ are free modules over a ring ${A}$ and ${M}$ is an ${A}$-module. If there is an exact sequence

$\displaystyle 0 \rightarrow F' \rightarrow M \rightarrow F'' \rightarrow 0,$

then ${M}$ is free.

Proof: The exact sequence splits. Indeed, we can lift a basis of ${F''}$ to elements of ${M}$ by surjectivity; then define a map ${F'' \rightarrow M}$ from that lifting, which gives a section of ${M \rightarrow F''}$. Thus the sequence splits. $\Box$

Now, by induction, we can show that the ${N_k}$‘s are free over ${A_f}$. Hence ${N_m = M_f}$ is free.

[For a better proof, see the comments below.  -AM, 8/17]

1. anonymous - August 17, 2009

Often times in algebraic geometry, to prove a statement “generically” it suffices to show it for the generic point. Then one is able to “spread out” from there.

Its clear that every coherent sheaf is free at the generic point because it becomes a module over a field, i.e. a vector space. Do you see a way of showing that the module will become free at some finite number of localizations without using your filtration lemma?

2. Akhil Mathew - August 17, 2009

Actually yes, because if you are working with a coherent sheaf $\mathcal{F}$ over a noetherian scheme $X$, then $F_x$ being free implies $F$ is free in some neighborhood of $x$. This follows because we can take finitely many sections $s_1, \dots, s_n$ in some neighborhood $U$ of $x$ which form a basis for $\mathcal{F}_x$. There is thus a homomorphism of sheaves $O_U^n \to \mathcal{F} |_U$ sending each coordiante to $s_i$. We can take the kernel $\mathcal{K}$ and the cokernel $\mathcal{C}$. Then $\mathcal{K}_x = \mathcal{C}_x = 0$. But the set of points where the stalk of a coherent sheaf is zero is open, since it is locally the complement of the supports of certain sections. Hence $\mathcal{K} = \mathcal{C} = 0$ in some neigborhood of $x$, so in that neighborhood $\mathcal{F}$ is free.

This is a better argument that what I posted above. Then again, it doesn’t (as far as I know) work for the general freeness lemma.

3. Anonymous - August 18, 2009

Wait — I thought if the stalk of *any* sheaf was zero then we could find a neighborhood where it was zero because the stalk is the direct limit. So … I think we can replace coherent with quasi-coherent, and this subsumes the general generic freeness lemma, no? (I’m not too clear on this stuff, its been a while, but I thought the general lemma was if X is a finite-type scheme over S and F is a coherent sheaf on X, we can find a neighborhood U in S so F is free on the pullback of U to X.)

4. Akhil Mathew - August 18, 2009

I’m pretty sure this is not the case, unless we know that $\mathcal{F}$ is locally finitely generated or something like that.

For instance consider the $\mathbb{Z}$-module
$M := \bigoplus_{n} \mathbb{Z}/n\mathbb{Z}$, and consider the associated quasi-coherent sheaf $\tilde{M}$ on $Spec \ \mathbb{Z}$. The stalk at the generic point $(0)$ is $M \otimes \mathbb{Q} = 0$, but the localization $M_f \neq 0$ for any nonzero $f \in \mathbb{Z}$, so $\tilde{M}$ is not zero on any open set.

Anonymous - August 18, 2009

What other “general” lemmas do you know of? Nakayama’s Lemma can be generalized as follows: let A->B be a local homomorphism of local rings and M be a finite B-module, then if m_a M = M, then M is trivial. But I’m not familiar with any other results in this vein. Ideas?

5. Akhil Mathew - August 18, 2009

There’s a result to the effect that if $S \subset A$ is a multiplicative subset of some Noetherian ring, and $M$ is a f.g. $A$-module with $S^{-1} M$ free over $S^{-1}A$, then there is $f \in S$ with $M_f$ free over $A_f$ (this e.g. implies Th. 1 above). The proof is basically the same as before: find a homomorphism $F \to M$ which becomes an isomorphism when tensoring with $S^{-1} A$, thus the kernels and cokernels vanish upon localization by $S$ (localization being exact), and then one uses the fact that the kernels and cokernels are finitely generated to find one element of $S$ annihilating them. We could also remove the hypothesis $A$ Noetherian if $M$ is finitely presented.

There are variants of Nakayama’s lemma that allow you to find generators of $M$ from $M/\mathfrak{m}M$. I might actually do a Nakayama post sometime, so I’ll give more details then. There is the standard fun corollary about f.g. projective modules over local rings being free. I kind of want to do that using $Tor$ though.

There is another lemma in Hartshorne that if you have local rings $A, B$ with a local homomorphism $A \to B$ which induces an isomorphism on the residue fields, makes $B$ into a f.g. $A$-module, and the map $m_A \to m_B/m_B^2$ is surjective, then $A \to B$ is surjective. The proof is basically Nakayama repeated several times, and I will probably mention it in the (future) Nakayama post.

Anonymous - August 19, 2009

Sorry; when I said “general”, I meant “Grothendieck-style” in the following sense. In classical A.G. you might have a result like: if M is a finite A-module satisfying P, then Q is true generically over A. I’m looking for results that look like: if M is a finite B-module where B is a f.g. algebra over A and M satisfies P, Q is true generically over A.

Examples are Grothendieck’s generic freeness and flatness lemmas, and the version of Nakayama I posted before. But I have no idea what sorts of lemmas can be generalized to this sort of statement. Do you have any idea?

Thanks!

6. Akhil Mathew - August 19, 2009

Well, first of all I think “generic freeness” and “generic flatness” actually refer to essentially the same argument, except that the conclusion of the former is of course stronger.
I’m not sure however how the extended version of Nakayama you posted above is “general” in the sense you gave- it doesn’t say anything holds generically, nor does it require any finite-type hypothesis.

Other results I know of that may count at least as “generic” are Chevalley’s theorem that fiber dimension is generic under suitable nice conditions (e.g. noetherian schemes, finite type) and even lower (or is it upper?) semicontinuous, and generic smoothness: with varieties over an algebraically closed field of characteristic zero $X,Y$ with $X$ nonsingular and a map $f: X \to Y$, there is an open subset $V \subset Y$ with $f^{-1}(V) \to V$ smooth. I may post about this kind of material in the future, since I’d like to understand it myself.

Anonymous - August 19, 2009

Ugh — I should really think before posting these comments. Under anonymity, there is no pressure to be careful or accurate or correct or anything

You’re right that I totally butchered the type of lemma that I was asking for. And that the form of Nakayama I gave didn’t at all fit into it. I guess what I meant is: there are a bunch of statements whose hypothesis includes: “M is a finite module over A”. Sometimes the hypothesis can be weakened: “M is a finite module over B, where B is an A-algebra satisfying [blah]“.

For Grothendieck’s generic freeness lemma, blah is finitely generated and for the version of Nakayama I mentioned, blah is nothing. I hear that this is a general phenomenon (and is useful in proving things in the relative framework of A.G.), but I know very few results of this form.

7. George - November 15, 2009

There is indeed categorical Nakayama Lemma as follows:

I should first introduce some notions:
C is an abelian category(with arbitrary intersection of subobjects)
Cpr is the proper subcategory generated by M of C such that for any nonzero M,any nonzero inclusion L—>M factors through a maximal proper subobject of M

Categorical Nakayama Lemma:
M is object of Cpr, and N—>M is an inclusion. The following conditions are equivalent:
1 N—>radM is inclusion, where radM=intersection of Kernel(M–>L), L goes through simple objects of Cpr.
2 if L—>M is inclusion such that N+L=M, then L=M.

Then the usual Nakayama Lemma is the corollary of this one.

Comments: Nakayama Lemma is crucial for proving noncommutative grassmannian is formally smooth.

Akhil Mathew - November 15, 2009

Sorry, I’m getting a bit confused here. Could you clarify what Cpr refers to in the case R-mod, for instance?