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I know the following result is well known to the experts, but what are the good references or proofs:

Let $X,Y$ be a smooth variety with projective, birational morphism $f: X \to Y$. Let $L$ be a line bundle on $Y$, then we have the property that the higher direct image $R^if_*f^* L = 0$ when $i > 0$.

This is equivalent to the statement that nonsingular variety is a rational singular variety.

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I think there are two issues here:

1) The "projection formula" tells you that $R^if_*(f^*L)$ is isomorphic to $R^if_*\mathcal{O}_X\otimes L$: see EGA chap. 0 , Prop. 12.2.3 (in EGA III.1).

2) Then the vanishing of $R^if_*\mathcal{O}_X$ for $i>0$. This is one of the many consequences of Hironaka's resolution of singularities: Resolution of singularities of an algebraic variety over a field of characteristic zero, I. Ann. of Math. (2) 79 (1964), 109–203 (the result you need is on p. 144-145).

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  • $\begingroup$ Thank you for your answer(and nice references)! But I was wondering if there is a simply way (like bare hand computation) without invoke the heavy machinery of resolution of singularity. $\endgroup$
    – Li Yutong
    Commented Apr 3, 2014 at 0:40
  • $\begingroup$ To convince you that there is no simple way, look at this talk by Grothendieck at the 1958 ICM: he considers your question as one of the major problems in cohomology ("problem B"). $\endgroup$
    – abx
    Commented Apr 3, 2014 at 5:57
  • $\begingroup$ Thank you for the answer, and it is also very interesting to read Grothendieck's talk when cohomology method was in its infant age. $\endgroup$
    – Li Yutong
    Commented Apr 4, 2014 at 13:26
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There is another way to view this (and prove the second part) that is probably worth explaining (and mentioning some characteristic $p > 0$ connections). As abx already pointed out, this problem reduces to computing $R^i f_* O_X$. You don't need resolution of singularities to prove that vanishing though.

Characteristic zero

In characteristic zero, by Grauert-Riemenschneider vanishing, we have that $R^i f_* \omega_X = 0$ for all $i > 0$ (even if $Y$ is singular). On the other hand, since the relative canonical divisor of a projective birational map between regular schemes $f : X \to Y$ is always effective (this is true whenever the relative canonical divisor makes sense, including mixed characteristic, see the work of Lipman on pseudo-rational singularities), we have that $f_* \omega_X = \omega_Y$.

In other words, $R f_* \omega_X \simeq \omega_Y$ in the derived category (they are quasi-isomorphic). Hence by Grothendieck duality for $f$ we see that $O_Y \simeq R f_* O_X$ in the derived category as well. This implies that $R^i f_* O_X = 0$ for all $i > 0$.

Characteristic $p > 0$

IF $Y$ is smooth, then the same results also holds in characteristic $p > 0$.
See

https://arxiv.org/abs/0911.3599

However, Grauert-Riemenschneider is known to fail in characteristic $p > 0$ for singular $Y$ so you really need $Y$ to at least have mild singularities. See for instance example 3.11 in

http://arxiv.org/pdf/1212.5105.pdf

It is reasonable to perhaps conjecture that if $Y$ has $F$-regular singularities then Grauert-Riemenschnedier vanishing is true and hence $R^i f_* O_X = 0$ for all $i > 0$ and all resolutions of singularities $f : X \to Y$.

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  • $\begingroup$ Thank you very much! I checked the Grauert-Riemenschneider vanishing. However, it use the fact ${\rm R}f_* \mathcal{O}_X = \mathcal{O}_X \in D(X)$ as the first step towards the proof. Moreover, it says that ${\rm R}f_* \mathcal{O}_X = \mathcal{O}_X$ follows from Leray spectral sequence. I was wondering if there was a direct proof using Leray spectral sequence? $\endgroup$
    – Li Yutong
    Commented Apr 3, 2014 at 0:45
  • $\begingroup$ The theorem of Grauert-Riemenschneider vanishing I mentioned is on page 45 of the following book mi.fu-berlin.de/users/esnault/books/esvibuch.pdf $\endgroup$
    – Li Yutong
    Commented Apr 3, 2014 at 0:47
  • $\begingroup$ There are proofs of Grauert-Riemenschneider vanishing in characteristic zero that certainly don't assume that $R f_* O_X = O_Y$. I think for instance you can see a proof on Page 186 of Lazarsfeld's book "Positivity in Algebraic Geometry II" $\endgroup$
    – Karl Schwede
    Commented Apr 3, 2014 at 18:31

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