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It is a basic fact that $H^n(X, F) = 0$ if $X$ is noetherian affine, $n > 0$, and $F$ a quasi-coherent sheaf.

If $Y \to X$ is a blow-up of a smooth variety in a smooth center, then then exceptional divisor is a projective bundle over the center, and so $H^n(Y, \mathcal{O}_Y) = H^n(X, \mathcal{O}_X)$. (right?)

I have not seen any examples of blow-ups of smooth varieties (with arbitrary center) whose fibres are not unions of projective spaces (sub-question: do such blow-ups exist?).

So I am lead to wonder: is it always true that $H^n(Y, \mathcal{O}_Y) = 0$ if $f: Y \to X$ is birational, $Y$ is integral, and $X$ is smooth and affine?

Edit: I mean't to include the hypothesis that $f$ is proper, and that $n$ is any integer $> 0$.

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    $\begingroup$ Am I missing something ? Let $F$ be arbitrary coherant sheaf. The $R^if_{*}(F)$ are quasi coherant so all their higher cohomology (on $X$) vanishes too. Hence the Leray SS degenerates and then your claim follows (for any $F$) since it is supported in dimension $= n-1$ and so standard base change shows $R^n f_{*}(F)$ vanishes $\endgroup$
    – meh
    Aug 12, 2013 at 16:12
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    $\begingroup$ @aginensky: I believe what you are saying is correct, but I think that is a little different from what the OP asks. In your comment, I believe that $n$ equals the dimension of $X$ (and $Y$). However, I do not see anywhere that the OP specifies that $n$ should be the dimension -- I think the OP wants the result for all $n>0$. $\endgroup$ Aug 12, 2013 at 17:50
  • $\begingroup$ @Jason- You are correct. If that is what he is asking, my argument won't do $\endgroup$
    – meh
    Aug 12, 2013 at 18:53
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    $\begingroup$ As stated this is not true: take $X=\mathbb{A}^2$, $Y = X\setminus (0, 0)$ and $f:Y\to X$ the inclusion. Then $H^1(Y, \mathcal{O}_Y) \neq 0$. Maybe assume $f$ proper? $\endgroup$ Aug 12, 2013 at 20:29

2 Answers 2

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Let me give a counter-example, where $f$ is moreover proper (see Piotr Achinger's comment), but $Y$ is not normal.

Let $X$ be the affine plane over a field $k$ and let $\tilde{Y}$ be the blow-up of the origin in $X$. We choose two $k$-points $P_1$ and $P_2$ on the exceptional divisor, and consider the variety $Y$ constructed by gluing $P_1$ and $P_2$ transversally. There is an induced proper and birational morphism $f:Y\to X$, and the contraction $c:\tilde{Y}\to Y$ is nothing else than the normalization of $Y$. Moreover, we have the following exact sequence on $Y$: $$0\to\mathcal{O}_Y\to c_*\mathcal{O}_{\tilde{Y}}\to\mathcal{O}_P\to 0,$$ where we denoted by $P$ the common image of $P_1$ and $P_2$ in $Y$, and where the last arrow is given by the difference of evaluations at $P_1$ and $P_2$.

We get a long exact sequence in cohomology (here we use that, the cohomology of $c_*\mathcal{O}_{\tilde{Y}}$ coincides with the cohomology of $\mathcal{O}_{\tilde{Y}}$ because $c$ is finite, and that $H^1(\tilde{Y},\mathcal{O}_\tilde{Y})=0$): $$0\to H^0(Y,\mathcal{O}_Y)\to H^0(\tilde{Y},\mathcal{O}_\tilde{Y})\to H^0(P,\mathcal{O}_P)\to H^1(Y,\mathcal{O}_Y)\to 0.$$ Every global function on $\tilde{Y}$ is constant on the exceptional divisor (as it is proper and integral), and hence descends to $Y$ showing that the first map is an isomorphism. As a consequence, $H^0(P,\mathcal{O}_P)\to H^1(Y,\mathcal{O}_Y)$ is an isomorphism, and $H^1(Y,\mathcal{O}_Y)$ is nonzero.

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As pointed out above in the comments by Piotr Achinger, I'm going to assume you meant the map to be proper.

For the first question then: sure, the fact that $R^i f_* O_Y = 0$ for $i > 0$ implies your first assertion that $H^n(Y, O_Y) = H^n(X, O_X)$ (here $f$ denotes the map $Y \to X$, you assumed $X$ was affine, but you don't need it for this). This condition is basically that the non-singular $X$ has rational singularities. In characteristic zero, this is a fairly straightforward application of Grauert-Riemenschnedier vanishing and holds as long as both $Y$ and $X$ are smooth (it doesn't matter whether you did blowups at smooth centers or not). Interestingly, we still don't know whether this vanishing holds for regular $Y$ and $X$ in mixed characteristic (although we do know this vanishing for regular $Y$ and $X$ in positive characteristic http://front.math.ucdavis.edu/0911.3599.

For the second question "fibres are not unions of projective spaces", you've already seen one non-normal example. Ulrich gives a nice example of a non-rational component in the comment below (originally I said something stupid here).

The last question, "is it always true that $H^n(Y,O_Y)=0$?", here's an example which this time is normal. It is Section III of Cutkosky's A new characterization of rational surface singularities Note the $R^2 h_* O_Z \neq 0$ in the papers notation is exactly the non-vanishing you were interested in.

Note: In the interest of full disclosure, this paper was pointed out to me by Hailong Dao in this answer on math overflow.

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    $\begingroup$ The irreducible components of the fibres need not be unirational: blow up a smooth point in a threefold and then a curve of genus at least one on the exceptional divisor. The fibre is rationally chain connected though, so each irreducible component is uniruled. $\endgroup$
    – naf
    Aug 13, 2013 at 5:21
  • $\begingroup$ Oh of course you are right. I was being dumb. Thanks. $\endgroup$ Aug 13, 2013 at 11:52

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