3
$\begingroup$

Let $f \colon X \to S$ be a proper morphism form a seperated algebraic space $X$ to an affine noetherian scheme $S$.

Given a coherent sheaf $F$ on $X$, we know from Knutson's book, that the finiteness theorem holds: The $R^qf_*F$ are finite $\mathcal O_S$-modules. For complete $S$, we also have the theorem of formal functions.

Now, let us assume that $F$ is flat over $S$. If $X$ were a scheme, we would know that the function on $S$, given by the dimension of the cohomology of $F$ restricted to fibers, is upper semicontinous. Is the same also true for $X$ an algebraic space?

Moreover, if we knew that for some $i \ge 0$ the function $s \mapsto h^i(X_s, F_s)$ is constant, than for $X$ a scheme, it would follow that the base change map $R^i f_* (F) \otimes k(s) \to H^i(X_s, F_s)$ is an isomorphisms. What about algebraic spaces?

many thanks in advance.

Improve this question
$\endgroup$
1
  • 5
    $\begingroup$ For the final part of the question you have to assume $S$ is reduced (and can let $S$ be loc. noeth. alg. space). For both parts, look at proofs in Mumford's "Abelian Varieties" (upon which Hartshorne's proofs were based); I think those carry over verbatim, which is a reason there may be no reference. Usually proofs concerning cohomology for proper morphisms work with rather little (if any) change for algebraic spaces, perhaps after first making passage to affine base (as we may always do) and sometimes using etale covers in place of Zariski covers (as in Torsten's suggestion). $\endgroup$
    – BCnrd
    Commented Jul 31, 2010 at 18:09

1 Answer 1

Reset to default
7
$\begingroup$

This is definitely true and there ought to (but may not) exist a reference. One way to prove it is just to check that the usual proof for schemes extends. If we follow Hartshorne for instance the crucial part is Proposition III:12.2 which can be proven using an étale affine cover instead of an open affine one.

Improve this answer
$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .