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Dear all,

Here is a problem I came across recently. Let $f:X\to Y$ be a proper morphism of Noetherian schemes and assume that $f_*\mathcal{O}_X\cong \mathcal{O}_Y$. Now let $S$ be a subscheme of $Y$, denote $T:=S\times_YX$ and let $f':T\to S$ be the projection.

Is it then true that $f'_*\mathcal{O}_T\cong \mathcal{O}_S$? If not, can you give an example where this does not work? In that case, are there may be general extra conditions on $X,Y$ and $f$ such that it does work?

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It's not true in general. Basically, you ask whether $f$ is cohomologically flat (in degree zero), see… For a specific counterexample and some discussion, see… Namely (quoting from Allen Knutson's answer to last question): let $f$ be a flat family of curves degenerating into a curve with an embedded point. – t3suji Apr 12 '11 at 12:42
An example is an elliptic fibration with wild fibres. See paper of Bombieri and Mumford on classification of surfaces in positive characteristic. – Torsten Ekedahl Apr 12 '11 at 13:05
@Ekedahl. Thank you, that is a nice example! Do you also happen to know an example in characteristic 0? – Bart van den Dries Apr 19 '11 at 12:13
@Bart: Not for elliptic fibrations, wild fibres only exist in positive characteristic. – Torsten Ekedahl May 3 '11 at 17:54

The property you want is called cohomological flatness. See this question. You can check out Hartshorne's chapter on semi-continuity and base change, but I would recommend Mumford's quick treatment in his book 'Lectures on curves on an algebraic surface' (this is just a great book in general). There are also notes by Nitin Nitsure in FGA Explained in his chapter on the Hilbert scheme. Of course, you can also always look at EGA III.

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Here is a very simple example where this doesn't work:

Let $Y$ be normal, $S\subset Y$ be non-normal and $f:X\to Y$ an embedded resolution of $S$. (To make it concrete, you may choose $Y=\mathbb P^2$, $S=Z(x^2-y^2)$ (or a nodal cubic) and $f:X\to Y$ the blow up of the singular point of $S$).

Then $T$ is the resolution of $S$ and since it is not-normal, $f'_*\mathscr O_T\not\simeq \mathscr O_S$, but $Y$ is normal, so $f_*\mathscr O_X\simeq \mathscr O_Y$.

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