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When Grauert's Theorem is presented in Hartshorne, the statement goes as follows:

Let $f:X\rightarrow Y$ be a projective morphism of noetherian schemes, and $\mathcal F$ a coherent sheaf on $X$, flat over $Y$. Assume that $Y$ is integral and for some $i$, the function $h^i(y,\mathcal F)=\dim_{k(y)} H^i(X_y,\mathcal F_y)$ is constant on $Y$. Then $R^if_*(\mathcal F)$ is locally free on $Y$ and for every $y$ the natural map

$$R^if_*(\mathcal F)\otimes k(y)\rightarrow H^i(X_y,\mathcal F_y).$$

I was wondering if the assumption that $Y$ be integral can be removed. Certainly the proof in Hartshorne uses the integrality condition.

I know that the projective requirement can also be made more general in allowing proper morphisms. I have seen a statement to this effect (in allowing properness and any Noetherian scheme $Y$ as a base) in Nitsure's notes "Construction of Hilbert and Quot schemes" (Part 2 of "FGA:Explained"), and indeed part (3) of Theorem 5.10 there says precisely the statement above with properness replacing projective and without the requirement that the base be integral, but I was wondering if that was accurate. The reference given for the proof is the above result and proof in Hartshorne, which doesn't cover the general case.

In short, I would like to know whether this generalization (1) is indeed true, (2) if Hartshorne's proof can be easily modified, and (3) if not, is there a good reference (EGA is acceptable, but not preferred).

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See mathoverflow.net/questions/114238/… – Damian Rössler Dec 11 at 21:09
You can still hope that the following holds: suppose $f:X\to Y$ and $\mathcal{F}$ are like in your first sentence. Then if $R^i f_*(\mathcal{F})$ is locally free on $Y$, then $R^i f_*(\mathcal{F})$ commutes with base change. This I think is true and follows from the discussion of cohomology and base change in Mumford's "Abelian varieties". – Piotr Achinger Dec 12 at 0:50
@Piotr: What you suggest fails for $X = A \times A^t$, $Y = A^t$, $f$ the natural projection, and $F$ being the Poincare line bundle for an abelian variety $A$ of dimension $g$. Indeed, $R^i f_*(F) = 0$ for $i < g$, but the same is not true for the fibre over $0$. – anon Jan 13 at 6:06

2 Answers

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Did you look at Corollary 8.3.6.5 of Illusie's article in the "FGA:Explained" volume?

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There is a very clear exposition of the base change theory in Mumford's book on abelian varieties.

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It is very beautiful, but iirc requires a reduced base... – Daniel Litt Jan 9 at 22:21
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But uiam the dimension of cohomologies of fibers won't detect freeness of direct images if the base is not reduced. – ACL Jan 10 at 0:14

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