Let $\pi:X\rightarrow Y$ be a double cover where $X$ and $Y$ are projective smooth curves.

Is it true that $R^1\pi_*\mathcal O_X=0$ ? Why ?

Thanks in advance.

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    $\begingroup$ Yes. See [Hartshorne, Algebraic Geometry], Corollary 11.2 page 279. $\endgroup$ – Francesco Polizzi Jan 31 '15 at 12:19
  • $\begingroup$ Thanks @FrancescoPolizzi; in particular this is true for all finite covers. $\endgroup$ – Z.A.Z.Z Jan 31 '15 at 12:23
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    $\begingroup$ and for all coherent sheaves $\mathscr{F}$. $\endgroup$ – Francesco Polizzi Jan 31 '15 at 12:32

X and Y are proper and irreducible. As a double cover does not crush X to a point, it must be surjective. The fibres are then finite sets which implies that it is a finite morphism. Finite morphisms are affine. Affine morphisms do not have higher direct image (this is just the relative version of the theorem which says that affine schemes do not have higher cohomology with coefficients in quasi-coherent sheaves).

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  • $\begingroup$ +1 For your nice answer. I do not see why we need to invoke "proper+quasi finite = finite" (this is what seems to be implied from your sentence "The fibers are then finite sets which implies it is a finite morphism"). There is this result in Hartshorne II.6 which uses the finiteness of the integral closure to conclude that a non-constant map of curves is necessarily a finite surjective morphism. $\endgroup$ – Ben Lim Jan 31 '15 at 17:27
  • $\begingroup$ @BenLim It was just the chain of implications that came to mind (and makes sense for any X and Y of the same dimension). But you are quite right. $\endgroup$ – bananastack Jan 31 '15 at 22:32

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