7
$\begingroup$

In p-adic properties of modular schemes and modular forms Katz formulates the following base change theorem as Theorem 1.7.1

Let $n\geq 3$ and $\overline{\mathcal{M}}_n$ be the compactified moduli scheme of elliptic curves with level-$n$-structure over $\mathbb{Z}[\frac1n]$. Let $K$ be any $\mathbb{Z}[\frac1n]$-module. Then the morphism $$K\otimes H^0(\overline{\mathcal{M}}_n; \omega^{\otimes k}) \to H^0(\overline{\mathcal{M}}_n; K\otimes \omega^{\otimes k})$$ is an isomorphism for $k\geq 2$.

The key step is to show that $H^1(\overline{\mathcal{M}}_n; \omega^{\otimes k})$ vanishes for $k\geq 2$. He claims that this follows from $\omega^{\otimes k}$ having degree strictly greater than $2g-2$ on every connected component of $\overline{\mathcal{M}}_n \otimes \mathbb{Z}[\frac1n, \zeta_n]$ (where $g$ is the genus of such a connected component) by Riemann-Roch.

I am familiar with this kind of Riemann-Roch argument for projective curves over a field, where it can be found in Hartshorne in nice cases and in more general cases in Liu's book. But Katz uses it in the case that the base is $\mathbb{Z}[\frac1n]$. What kind of Riemann-Roch argument works here?

$\endgroup$
3
  • 2
    $\begingroup$ You should probably read about the semicontinuity theorem and compatibility of cohomology and base change. This is Section 12 of Chapter III of Hartshorne's book. $\endgroup$ Commented Dec 22, 2015 at 14:25
  • 2
    $\begingroup$ Ah, I see. I check the vanishing in every fiber and then know that the rank of the first cohomology of $\omega^{\otimes k}$ is constant over all fibers - then I apply cohomology and base change to obtain that $R^1f_*\omega^{\otimes k}$ is locally free and hence also zero (for $f\colon \overline{\mathcal{M}}_n \to \mathrm{Spec}\mathbb{Z}[\frac1n]$). Thanks! $\endgroup$ Commented Dec 22, 2015 at 15:15
  • $\begingroup$ It would be useful to see this comment promoted to an answer I think $\endgroup$ Commented Apr 22, 2016 at 7:38

0

You must log in to answer this question.