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Let $X$ and $Y \ $ be smooth varieties over a field or - depending on the answers - more general nice schemes (I don't know what one needs exactly as conditions). Let $p: X\times Y \rightarrow X$ be the projection morphism. It has constant fiber dimension $n:=dim(Y)$ over$X$, so is smooth of relative dimension n.

Is it then right that one has a vanishing of the higher direct images for any quasicoherent sheaf $\mathcal F$ on $X\times Y$:

$\mathcal R^i \mathcal F =0$ for all $i > n$ ?

Note that by Grothendieck's Vanishing Theorem one surely has it for $i>n+dim(X)$.

Also I really want Quasicoherence, not Coherence, so Semicontinuity Arguments don't seem to be at hand.

Furthermore, how far can one generalize this?

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It is true, and in a very general context. Note that the fibers are supported on a $n$-dimensional subset, hence the higher direct images vanish (recall that these are just the direct images of the direct image functor, which calculates $\Gamma(p^{-1}U,F)$.) –  Bonanza Dec 5 '11 at 16:48
    
@hilbert: But $p^{-1}(U)=U\times Y$. And this has dimension greater than n. So I don't see what exactly you mean. –  Cyril Dec 5 '11 at 19:08
    
Do you accept $Y$ to be projective ? –  user18119 Dec 6 '11 at 0:21
    
I would accept it proper. Does this help? –  Cyril Dec 6 '11 at 7:58
    
There is a fairly elementary proof for quasi-projective morphisms. For proper morphisms over a noetherian scheme, you could reduce to coherent sheaves considering a quasi-coherent sheaf as inductive limit of coherent sheaves. –  user18119 Dec 6 '11 at 9:28

1 Answer 1

up vote 1 down vote accepted

I am not sure what you call semicontinuity theorems. Your question is about vanishing of $R^i$'s.

Let $f : X\to S$ be a proper morphism to a locally noetherian scheme $S$ such that all its fibers have dimension $\le n$. Let $F$ be a quasi-coherent sheaf on $X$. Then $R^if_*F=0$ for all $i> n$.

Proof (admitting the result for coherent sheaves on $X$). The question is local on $S$, so we can suppose $S$ is affine. Then we have to show that $H^i(X, F)=0$ for $i>n$. Fix a finite affine covering $\mathcal U$ of $X$. We will compute $H^i(X,F)$ by Cech cohomology $H^i(\mathcal U, F)$. Let $$c\in \mathrm{ker}(C^{i}(\mathcal U, F)\to C^{i+1}(\mathcal U, F)).$$ By EGA I.9.4.9, $F$ is the inductive limit of its coherent submodules $(F_{\alpha})_{\alpha}$. As $c$ involves finitely many sections, there exists $\alpha$ such that $$c\in\mathrm{ker}(C^{i}(\mathcal U, F_\alpha)\to C^{i+1}(\mathcal U, F_\alpha))=\mathrm{Im}(C^{i-1}(\mathcal U, F_\alpha)\to C^{i}(\mathcal U, F_\alpha)).$$ Hence $$c \in \mathrm{Im}(C^{i-1}(\mathcal U, F)\to C^{i}(\mathcal U, F))$$ and $H^i(X,F)=H^i(\mathcal U, F)=0$.

Remark For a direct proof (without passing by coherent sheaves) in the case of projective morphisms to a locally noetherian scheme (I was wrong on quasi-projective morphisms), see Proposition 5.2.34 in "Algebraic geometry and arithmetic curves".

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You helped me a lot with this fine answer, thanks! –  Cyril Dec 8 '11 at 10:57

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