MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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).

share|cite|improve this question
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 '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 '13 at 6:06

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

share|cite|improve this answer
It is very beautiful, but iirc requires a reduced base... – Daniel Litt Jan 9 '13 at 22:21
But uiam the dimension of cohomologies of fibers won't detect freeness of direct images if the base is not reduced. – ACL Jan 10 '13 at 0:14

The standard proof (with only reduced hypotheses) is Theorem 28.1.5 in the June 11, 2013 version of the notes at

share|cite|improve this answer

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

share|cite|improve this answer

Your Answer


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

Not the answer you're looking for? Browse other questions tagged or ask your own question.