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## The cohomology of the relative dualizing sheaf of a relative curve

Let $X\to S$ be a curve over $S=\mathrm{Spec} \ \mathbf{Z}$. Let $\omega$ be the relative dualizing sheaf of $X\to S$. Let $g$ be the genus of the generic fibre. Assume that $g\geq 2$.

I know that $\omega$ is ample. Even better, a theorem of Deligne and Mumford says that $\omega^{\otimes 3}$ is very ample.

So there is some integer $n$ depending on $X\to S$ such that $H^1(X,\omega^{\otimes m}) =0$ for all $m\geq n$.

Can we find such an $n$? Is it independent of $g$? Does $n=3$ work?

If not, can we find such an $n$ and bound it in terms of $g$?

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Yes, this is a simple consequence of "cohomology and base change". See, for example, math.ucdavis.edu/~osserman/math/… Theorem 1.2. The point is that one has vanishing for the generic fibre for $n>1$ (by duality) and since the fibres are all $1$-dimensional and the base is affine, the result follows from the Theorem. – ulrich May 28 2012 at 8:01
Thanks for your answer. The link doesn't work, unfortunately. What do you mean when you answer by yes? $n=3$ works? – Harry May 28 2012 at 8:05
Here's the link : math.ucdavis.edu/~osserman/math/… There was just a small typo in your hyperlink. – Harry May 28 2012 at 8:07
@ulrich. Let $n\geq 2$ and let $f:X\to S$ be a semi-stable curve. It's clear that the higher cohomology of $\omega^{\otimes n}$ vanishes on the generic fibre by duality. By loc. cit., this implies that $R^1 f_\ast \omega^{\otimes n} =0$. But why does this imply that $H^1(X,\omega^{\otimes n}) =0$? I don't see how $f_\ast$ and the global sections functor are related in this case, because the base is $\mathrm{Spec} \mathbf{Z}$. – Harry May 28 2012 at 9:23
@myself. I think I got it. The push-forward f_\ast coincides with taking global sections and then exact functor "tilde" which associates to a $\mathbf{Z}$-module $M$ the coherent sheaf $\widetilde{M}$ on $\mathrm{Spec} \mathbf{Z}$. – Harry May 28 2012 at 9:31
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