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Generic freeness I July 29, 2009

Posted by Akhil Mathew in algebra, commutative algebra.
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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 {A} 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 {M} is Noetherian iff every ascending chain

\displaystyle M_1 \subset M_2 \subset \dots \subset M

stabilizes.

Proof: Indeed, we can write {M' = \bigcup M_i \subset M}; then {M'} is finitely generated, and each generator lies in some {M_i}; thus for some large {N}, {M_N} contains all the generators of {M'} and hence {M_N = M'}. The other implication is similar. \Box

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.

Proposition 3 Given an exact sequence {0 \rightarrow M' \rightarrow M \rightarrow M'' \rightarrow 0}, if {M', M''} are Noetherian, then {M} is Noetherian.

Proof: Given an ascending series {M_i \subset M}, the ascending series {M' \cap M_i \subset M'} and {image(M_i) \subset M''} stabilize eventually. Now use:

Lemma 4 Given an exact sequence as above, if {N_0 \subset N \subset M}, and {N_0 \cap M' = N \cap M', image(N_0)=image(N)}, then {N_0 = N}.

Indeed, given {n \in N}, we can write {n \equiv n_0 \ (mod \ M')} for some {n_0 \in N_0}; then {n-n_0 \in N \cap M' = N_0 \cap M'}, so {n-n_0 \in N_0}, i.e. {n \in N_0}. \Box

The following is the form in which Noetherian hypotheses are frequently used:

Proposition 5 Let {A} be Noetherian; then any finitely generated {A}-module is Noetherian.

Proof: By induction on the rank and Proposition 3, any finitely generated free {A}-module is Noetherian; now any finitely generated {A}-module is a quotient of such. \Box

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 {M} be a finitely generated module over a Noetherian ring {A}. Then there is a filtration

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

such that each quotient {M_{i+1}/M_i \simeq A/\mathfrak{p}_i} for prime ideals {\mathfrak{p}_i}.

Proof: The proof illustrates a useful technique that one often uses with Noetherian rings, an induction of sorts.

Consider the set {S} defined as

\displaystyle S = \{ \mathrm{ \ submodules \ of \ } M \mathrm{ \ satisfying \ the \ proposition } \}.

Then {S} has a maximal element {N}. (Otherwise, we’d have an infinite strictly ascending collection of submodules.) There is thus such a filtration of {N}. If we can find a submodule {\bar{B} \subset M/N} (coming from some {B \supset N}) with {\bar{B} \simeq A/\mathfrak{p}} for some prime {\mathfrak{p}}, then we would have {B \in S} by piecing together the filtration of {N} with {B/N}. But {B} is bigger than {N}, contradiction!

So we will be done if we show:

Lemma 7 Let {C} be a module over a Noetherian ring {A}. Then {C} contains a submodule isomorphic to {A/\mathfrak{p}}, for {\mathfrak{p}} prime.

To prove this, we need to find an {x \in C} such that {ann(x) := \{ a \in A: ax = 0 \}} is prime. Choose {x \neq 0} such that {ann(x) \neq A} is maximal; we may do this since {A} is Noetherian. Then, for any {a \notin ann(x)}, we have

\displaystyle A \neq ann(ax) \supset ann(x), \quad \mathrm{so} \quad ann(ax) = ann(x).

I claim that {ann(x)} is prime. So suppose {c,d \notin ann(x)} and yet {cdx = 0}. Then {c(dx)=0}. Thus {c \in ann(dx)=ann(x)}, contradiction. \Box

Generic Freeness

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 {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 result implies the corresponding fact over schemes:

Corollary 9 Let {X} be a Noetherian integral scheme. Then if {\mathcal{F}} is a coherent sheaf on {X}, there is an open dense {U \subset X} such that the restriction {\mathcal{F}_U} is free.

Proof: Take an open affine {V \subset X}; since {X} is irreducible, {V} is dense. Say {V = \mathrm{Spec \ } A}, where {A} is a Noetherian domain. Then {\mathcal{F}_V} is the sheaf associated to a finitely generated {A}-module {M}. We can find {f \in A} with {M_f} free over {A_f}; then one takes {U = V_f = \mathrm{Spec \ } A_f}. \Box

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.)

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Comments»

1. hilbertthm90 - July 30, 2009

Proposition 6 is so useful (especially in the local case). I used it many times last quarter.

2. Akhil Mathew - July 30, 2009

Yes, though it took me a while to actually understand the significance, even though it was used quite a few times in commutative algebra books, and probably will be referred to again on this blog.

3. Generic freeness II « Delta Epsilons - August 13, 2009

[…] argument proceeds using dévissage. By the last post, we can find a […]

4. The Noetherian condition as compactness « Annoying Precision - November 28, 2009

[…] posts at Delta Epsilons here and here describe the basic properties of Noetherian rings well, including the proof of the […]


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