Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Let $f:X\rightarrow S$ be a proper, flat, finite type morphism of noetherian schemes. $f$ is called cohomologically flat in degree 0, if formation of $f_*\mathcal{O}_X$ commutes with base change along any morphism $T\rightarrow S$.

Now let $f$ be a proper, flat, finite type morphism of noetherian schemes, and additionally require that $f$ has reduced geometric fibers. Then it follows from rather complicated results in EGAIII.2 (specifically 7.8.6), that $f$ is cohomologically flat in degree $0$.

My first question is: Is there are more elementary proof of this fact, possibly bypassing heavy cohomological machinery?

If in addition $f$ also has integral geometric fibers, then for every $T\rightarrow S$, the canonical map $$\mathcal{O}_T\rightarrow f_{T,*}\mathcal{O}_{X_T}$$ is an isomorphism, where the subscript $T$ denotes the objects base changed along $T$. For a proof, see e.g. Kleiman "The Picard Scheme", Solution to Ex. 3.11, which also uses heavy cohomological machinery.

My second question: Is there are more elementary proof of this fact?

share|improve this question
    
Well, the second fact follows immediately from the first. I am not aware of a proof that doesn't use the following fact: The complex $Rf_*\mathcal{O}_X$ is locally quasi-isomorphic to a perfect complex; that is, to a bounded complex of locally free modules of finite rank. Assuming this, the proof is quite elementary. I have some notes on precisely this on my webpage. –  Keerthi Madapusi Pera Apr 11 '11 at 21:46
    
Thanks, I'll check it out. –  Lars Apr 12 '11 at 12:40
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.