I am now going to discuss Kempf’s proof of the theorem of Serre. Note that this is lifted verbatim of some notes I have been taking, so apologies if the style seems out of place as a result.  I use the (highly nonstandard) notation \mathcal{G}m for global sections of a sheaf for entirely logistical (and typo-errorgraphical) reasons.  Since this is really better suited to a PDF, I’ll also post that.

(Note: You really should read the PDF, since some diagrams are missing from this post.)

Theorem 1 (Serre) Let {X = \mathrm{Spec} A} be an affine scheme, {\mathcal{F}} a quasi-coherent sheaf. Then {H^i(X, \mathcal{F}) = 0} for {i \geq  1}.

We shall prove this result following Kempf. The idea is that {X} has a very nice basis: namely, the family of all sets {D(f), f \in A}. These are themselves affine, and moreover the intersection of any two elements in this basis is still in this basis. For {D(fg) = D(f) \cap  D(g)}.

0.1. A lemma of Kempf

First, we set up some notation, following Kempf. Given {V  \subset X} open and a sheaf {\mathcal{F}} on {X}, define {{}_V \mathcal{F}} to be the sheaf {i_*(i^{-1}\mathcal{F})} on {X}, where {i: V \rightarrow X} is the inclusion. This is equivalently the sheaf {U \rightarrow \mathcal{F}(V \cap  U)}. There is a natural map {\mathcal{F}  \rightarrow {}_V \mathcal{F}}, which of course induces maps on cohomology.

The elementary result that Kempf proves is:

Lemma 2 (Kempf) Let {X} be a topological space and let {\mathfrak{A}} be a basis for {X} which is closed under finite intersections. Suppose {n \in  \mathbb{N}} is fixed. Suppose {\mathcal{F} \in \mathrm{Sh}(X)} for {X} a topological space is such that {H^i(U, \mathcal{F}|_U) = 0} for all {0<i<n} and {U \in  \mathfrak{A}}. Suppose {\alpha \in H^n(X,  \mathcal{F})}. Then there is a covering of {X} by open sets {V \in  \mathfrak{A}} such that the image of {\alpha} in {H^n(X, {}_V  \mathcal{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 {\mathcal{F}} in a flabby sheaf {\mathcal{G}}, and let {\mathcal{H}} be the cokernel. There is an exact sequence

and by the long exact sequence for cohomology (and since {n>1}), it follows that

is exact for every {U} in the open basis {\mathfrak{A}}. Now fix {V \in  \mathfrak{A}} and consider the complex of sheaves

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 \in \mathfrak{A}}, we have that the sequence of sections

is exact in view of the definition of {{}_V} and exactness of (2). Consequently, since {\mathfrak{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 \mathcal{G}} is also flabby and consequently has trivial cohomology. As a result, we find that for any {V \in \mathfrak{A}}, there is an isomorphism

\displaystyle  H^{n-1}(X, {}_V \mathcal{H}) \simeq H^{n}(X, {}_V\mathcal{F}).

Moreover, since {\mathcal{G}} is flabby and thus has trivial cohomology on {X}, we get isomorphisms from the long exact sequence of (1)

:

\displaystyle  H^{n-1}(X, \mathcal{H}) \simeq H^{n}(X, \mathcal{F}).

This means that {\mathcal{H}} satisfies the conditions of the proposition with {n-1}, and we have assumed inductively that the result is valid for {n-1}. So {\alpha} maps to some {\beta \in H^{n-1}(X, \mathcal{H})}; this means there is an open cover of {X} by various {V  \in \mathfrak{A}} such that {\beta} maps to zero in {H^{n-1}(X, {}_V  \mathcal{H})}. This means that {\alpha} maps to zero in these {H^{n}(X,  {}_V\mathcal{F})} by naturality. This completes the proof of the inductive step.

The base case remains, i.e. {n=1}. Fix {\alpha \in H^1(X, \mathcal{F})}. We can still embed {\mathcal{F}} in a flabby sheaf and obtain an exact sequence as in (1). So we get an exact sequence:

\displaystyle  0 \rightarrow \mathcal{G}m(X, \mathcal{F}) \rightarrow  \mathcal{G}m(X, \mathcal{G}) \rightarrow \mathcal{G}m(X, \mathcal{H})  \rightarrow H^{1}(X, \mathcal{F}) \rightarrow 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 \rightarrow  {}_V\mathcal{F} \rightarrow {}_V\mathcal{G} \rightarrow  \mathcal{K}^{(V)} \rightarrow 0} exact for some cokernel {\mathcal{K}^{(V)}}, and we can fit these into an exact commutative diagram

Let {\alpha \in H^1(X, \mathcal{F})}. Then {\alpha} lifts to some {\beta \in \mathcal{G}m(X,  \mathcal{H})}. We have a commutative diagram of exact sequences

So to say that {\alpha} is annihilated by the map to {H^1(X, {}_V \mathcal{F})} is the same as saying that the image of {\beta} in {\mathcal{K}^{(V)}} lifts to something in {{}_V  \mathcal{G}(X)}. But if {V} is small, surjectivity of {\mathcal{G} \rightarrow  \mathcal{H}} implies that we can lift {\beta} to something in {{}_V  \mathcal{G}(X)}. So we can cover {X} by such sets {V} in {\mathfrak{A}}, completing the proof. \Box

0.2. Proof of the vanishing theorem

We now apply the lemma. Proof: Induction on {n}.

Let {X = \mathrm{Spec} A} be an affine scheme. Consider the basis {\mathfrak{A}} of open sets {D(f) = \mathrm{Spec} A_f}; this is obviously closed under intersection, as {D(fg) = D(f) \cap  D(g)}. Now if {\widetilde{M}} is the (quasi-coherent) sheaf on {X} associated to an {A}-module {M}, then the pull-back to {D(f)} is the sheaf associated to {M \otimes_A A_f = M_f}. So the direct image {{}_{D(f)} \widetilde{M}} to {A} is just {\widetilde{M_f}}. In particular, these are quasi-coherent on {X}.

We will now apply the previous lemma. Suppose that {n} is fixed and {H^i(X, \mathcal{F}) =  0} for any quasi-coherent sheaf on {X} and {0<i<n}. Then, this is true for {\widetilde{M}}, so the previous lemma says that given {\alpha \in H^n(X,  \widetilde{M})}, there is an open cover {\{D(f_i)\}} of {X} such the cohomology class of {\alpha} in {H^n(X,  {}_{D(f_i)}\widetilde{M})} is zero. Now there is a map {M \rightarrow \bigoplus M_{f_i}} which is injective, since the {f_i} generate the unit ideal; this induces a map of sheaves

\displaystyle  0 \rightarrow \widetilde{M} \rightarrow \bigoplus  {}_{D(f_i)} \widetilde{M} \rightarrow \mathcal{K} \rightarrow 0

where {\mathcal{K}} is also quasi-coherent. There is thus a long exact sequence, of which we write a piece:

\displaystyle  H^{n-1}(X, \mathcal{K}) \rightarrow H^{n}(X,  \widetilde{M}) \rightarrow H^n( X, \bigoplus {}_{D(f_i)} \widetilde{M}) .

Since {\alpha} is in the kernel of the second map, it is in the image of {H^{n-1}(X,  \mathcal{K})}.

But, if {n>1}, inductively we assumed {H^{n-1}(X, \mathcal{K})} was zero. This means {\alpha = 0}. If {n=1}, then we write out a bit more of the exact sequence:

\displaystyle  \mathcal{G}m(X, \bigoplus {}_{D(f_i)} \widetilde{M})  \rightarrow \mathcal{G}m(X, \mathcal{H}) \rightarrow H^1(X, \mathcal{F})  \rightarrow H^1(X, \bigoplus {}_{D(f_i)} \widetilde{M} )

As before, we find that {\alpha} is in the image of {\mathcal{G}m(X, \mathcal{H})}. But since {\mathcal{G}m} is exact on quasi-coherent sheaves, it follows that the first map is surjective, and the map out of {\mathcal{G}m(X, \mathcal{H})} is zero. So again we find that {\alpha = 0}. This proves Serre’s vanishing theorem. \Box

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