The following question was asked on math.stackexchange.com with no reply for the past week or so. Let $f : X \to Y$ be a morphism of smooth (integral) varieties over $\Bbb{C}$ with generic fiber equal to $\Bbb{P}^1$.

Is it true that $R^if_\ast \mathcal{O}_X = 0$ for all $i > 0$?

By Cohomology and base change, we know that $R^if_\ast \mathcal{O}_X|_{\eta_Y} = 0$. However, because $f$ is not assumed proper, we do not know if $R^if_\ast \mathcal{O}_X$ is coherent. This leads me to suspect that the statement may be false, but I cannot come up with a counterexample.


No. Take $Y=\mathbb{A}^2_\mathbb{C}$ and $X=U\times \mathbb{P}^1$ where $j:U\hookrightarrow Y$ is the complement of the origin. One immediately finds $R^1 f_*\mathscr{O}_X=R^1 j_*\mathscr{O}_U$, which is nonzero.

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    $\begingroup$ If $Y$ is a DVR, take $X'=Y\times \mathbb{P}^1$, and $X=X'\smallsetminus\{z\}$ where $z$ is a point in the closed fiber. $\endgroup$ Jul 1 '15 at 10:05

Let me supplement Laurent Moret-Bailly's answer with a positive result.

First though, even if $f$ is proper and $R^if_* O_X$ vanishes generically, you cannot prove that $R^i f_* O_X=0$ just by invoking semicontinuity (as the comments at math.stackexchange.com were suggesting): semicontinuity says that fibers jump up at special points, so they could be nonzero somewhere.

On the other hand, for example in the case where $X$ and $Y$ are projective, $X$ is smooth, and $Y$ has rational singularities, this is (a special case of) Theorem 7.1 in Kollár, Higher direct images of dualizing sheaves I.


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