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Suppose $X\to Y $ is a morphism , under what conditions we have direct image sheaf $f_*(O_X)=O_Y$?

For example, suppose $\tilde{S}\to S$ is a blow up, do we have $f_*(O_{\tilde{S}})=O_S$?

Hartshorne III 11.3 says that $f_*(O_X)=O_Y$ implies the fibres are connected, is the convese true? That is: if the fibres of a morphism of are all connected, do we have $f_*(O_X)=O_Y$?

  • Last question: No chance. Take $Y$ to be the point and $X$ any connected scheme/variety. – Zhen Lin Apr 18 '14 at 11:03
  • You are right, but what about if $Y$ is a variety of dim > 0? –  Apr 18 '14 at 11:12
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    Well, then take $X = Y \times_k \mathbb{A}^1_k$, say. – Zhen Lin Apr 18 '14 at 11:15
  • You are right, thanks! Still a question on understanding this, fibration over a curve is "locally trival", but why can it have the $f_*(O_Y) = O_X$, which is quite different from the case above ? –  Apr 18 '14 at 11:29

1 Answers1

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Suppose $Y$ is noetherian for simplicity.

For a blowing-up morphism $f: X\to Y$, one has $f_*O_X=O_Y$ if $Y$ is integral and normal. Indeed, $f$ is birational and proper, hence $O_Y\to f_*O_X$ is finite by the direct image theorem, and $f_*O_X$ is contained in the function field of $Y$. So the normality of $Y$ implies that $O_Y=f_*O_X$.

If $Y$ is integral, but not normal, the normalization map $f: X\to Y$ is a blow-up morphism (if $Y$ is a quasi-projective variety), but $f_*O_X$ contains strictly $O_Y$.

A sufficient condition for $f_*O_X=O_Y$ is $f$ proper with geometrically integral fibers. Without properness, the example of Zhen Lin in the comments shows that $f$ can have geometrically integral fiber but still $f_*O_X\ne O_Y$.

  • Thanks for your answer! I still have a question, in the second paragraph, do you mean every normalization(of a quasi-proj variety) is a blow up? –  Apr 18 '14 at 14:27
  • Can you give a reference on the direct image theorem mentioned above? Thanks! –  Apr 18 '14 at 14:30
  • @mqx: yes, every projective birational morphism to an integral quasi-projective variety is a blowing-up. You can find this in Hartshorne in the subsection where blowing-ups are covered (iirc). The direct image theorem says that if $f$ is proper, then for any coherent sheaf $F$ on $X$, $f_F$ is a coherent sheaf on $Y$. This also holds for cohomological groups $R^if_F$. Again this can be found in Hartshorne III.11 or III.12 I think. –  Apr 18 '14 at 14:35
  • In the example mentioned above, if we replace the $\mathbb{A}^1$ by $\mathbb{P}^1$, $Y$ a projective variety, then we have the equation? But I cannot understand well because the two maps are the same on a open set. –  Apr 18 '14 at 14:36
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    Yes. Taking $f_*$ is same as taking global sections, and this of course leads to different results on $P^1$ and an open subset $A^1$. –  Apr 18 '14 at 14:37