Download 0.1 A lemma of Kempf

yes no Was this document useful for you?
   Thank you for your participation!

* Your assessment is very important for improving the work of artificial intelligence, which forms the content of this project

Document related concepts

3-manifold wikipedia, lookup

Brouwer fixed-point theorem wikipedia, lookup

Covering space wikipedia, lookup

Grothendieck topology wikipedia, lookup

Group cohomology wikipedia, lookup

Étale cohomology wikipedia, lookup

Sheaf cohomology wikipedia, lookup

Sheaf (mathematics) wikipedia, lookup

Homological algebra wikipedia, lookup

Theorem 1 (Serre). Let X = SpecA be an affine scheme, F a quasi-coherent
sheaf. Then H i (X, F) = 0 for i ≥ 1.
We shall prove this result following [?]. The idea is that X has a very nice
basis: namely, the family of all sets D(f ), f ∈ A. These are themselves affine,
and moreover the intersection of any two elements in this basis is still in this
basis. For D(f g) = D(f ) ∩ D(g).
A lemma of Kempf
First, we set up some notation, following Kempf. Given V ⊂ X open and a
sheaf F on X, define V F to be the sheaf i∗ (i−1 F) on X, where i : V → X is
the inclusion. This is equivalently the sheaf U → F(V ∩ U ). There is a natural
map F → V F, which of course induces maps on cohomology.
The elementary result that Kempf proves is:
Lemma 1 (Kempf). Let X be a topological space and let A be a basis for X
which is closed under finite intersections. Suppose n ∈ N is fixed.
Suppose F ∈ Sh(X) for X a topological space is such that H i (U, F|U ) = 0
for all 0 < i < n and U ∈ A. Suppose α ∈ H n (X, F). Then there is a covering
of X by open sets V ∈ A such that the image of α in H n (X, V F) is zero for
each V .
Proof. We will prove this result by induction on n. First, suppose n > 1, and
that the result is valid for n − 1. The base case will be handled next. We can
embed F in a flabby sheaf G, and let H be the cokernel. There is an exact
0→F →G→H→0
and by the long exact sequence for cohomology (and since n > 1), it follows
0 → F(U ) → G(U ) → H(U ) → 0
is exact for every U in the open basis A. Now fix V ∈ A and consider the
complex of sheaves
0 → V F → V G → V H → 0.
In general, we know that i−1 is exact, by looking at the stalks, but only that
i∗ is left-exact. From this alone we get that (3) is exact except perhaps at the
last step. But we also know that for any U ∈ A, we have that the sequence of
0 → V F(U ) → V G(U ) → V H(U ) → 0
is exact in view of the definition of V and exactness of (2). Consequently, since
A is a basis, we can pass to the direct limit to the stalks, and we see that (3)
must be exact at the last step too.
But V G is also flabby and consequently has trivial cohomology. As a result,
we find that for any V ∈ A, there is an isomorphism
H n−1 (X, V H) ' H n (X, V F).
Moreover, since G is flabby and thus has trivial cohomology on X, we get isomorphisms from the long exact sequence of (1):
H n−1 (X, H) ' H n (X, F).
This means that H satisfies the conditions of the proposition with n − 1, and we
have assumed inductively that the result is valid for n − 1. So α maps to some
β ∈ H n−1 (X, H); this means there is an open cover of X by various V ∈ A such
that β maps to zero in H n−1 (X, V H). This means that α maps to zero in these
H n (X, V F) by naturality. This completes the proof of the inductive step.
The base case remains, i.e. n = 1. Fix α ∈ H 1 (X, F). We can still embed
F in a flabby sheaf and obtain an exact sequence as in (1). So we get an exact
0 → Gm(X, F) → Gm(X, G) → Gm(X, H) → H 1 (X, F) → 0.
However, exactness of (4) is now no longer valid, so we cannot conclude that (3)
is exact. We do, however, have an exact sequence 0 → V F → V G → K(V ) → 0
exact for some cokernel K(V ) , and we can fit these into an exact commutative
/ VF
/ VG
/ K(V )
Let α ∈ H 1 (X, F). Then α lifts to some β ∈ Gm(X, H). We have a commutative
diagram of exact sequences
/ H(X)
/ H 1 (X, F)
/ K(V ) (X)
/ H 1 (X, V F)
So to say that α is killed by the map to H 1 (X, V F) is the same as saying that the
image of β in K(V ) lifts to something in V G(X). But if V is small, surjectivity
of G → H implies that we can lift β to something in V G(X). So we can cover
X by such sets V in A, completing the proof.
Proof of the vanishing theorem
We now apply the lemma.
Proof of the vanishing theorem. Induction on n.
Let X = SpecA be an affine scheme. Consider the basis A of open sets
D(f ) = SpecAf ; this is obviously closed under intersection, as D(f g) = D(f ) ∩
f is the (quasi-coherent) sheaf on X associated to an A-module
D(g). Now if M
M , then the pull-back to D(f ) is the sheaf associated to M ⊗A Af = Mf . So
f to A is just M
gf . In particular, these are quasi-coherent
the direct image D(f ) M
on X.
We will now apply the previous lemma. Suppose that n is fixed and H i (X, F) =
0 for any quasi-coherent sheaf on X and 0 < i < n. Then, this is true for M
so the previous lemma says that given α ∈ H (X, M ), there is an open cover
f) is zero. Now
{D(fi )} of X such theLcohomology class of α in H n (X, D(fi ) M
there is a map M →
Mfi which is injective, since the fi generate the unit
ideal; this induces a map of sheaves
D(fi ) M → K → 0
where K is also quasi-coherent. There is thus a long exact sequence, of which
we write a piece:
f) → H n (X,
H n−1 (X, K) → H n (X, M
D(fi ) M ).
Since α is in the kernel of the second map, it is in the image of H n−1 (X, K).
But, if n > 1, inductively we assumed H n−1 (X, K) was zero. This means
α = 0. If n = 1, then we write out a bit more of the exact sequence:
D(fi ) M )
D(fi ) M ) → Gm(X, H) → H (X, F) → H (X,
As before, we find that α is in the image of Gm(X, H). But since Gm is exact
on quasi-coherent sheaves, it follows that the first map is surjective, and the
map out of Gm(X, H) is zero. So again we find that α = 0. This proves Serre’s
vanishing theorem.