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Assume we have a flat morphism $f: X\rightarrow Y$ of schemes such that $f_{*}O_X=O_Y$. Given a locally free sheaf $E$ of finite rank, the projection formula gives $f_{*}f^{*}E=f_{*}(O_X\otimes f^{*}E)=f_{*}O_X\otimes E=E$.

Using the flatness of $f$, can we replace locally free of finite rank by coherent or quasi coherent?

This is indicated in this answer, but I don't see how to use the flatness of $f$ to prove such a result.

Any advice on how to see this is appreciated. Maybe this can be translated in a commutative algebra question which is easier? References are also welcome.

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    $\begingroup$ Are you familiar with the derived version of the projection formula ? The second answer at your link gives you the reason $f_*f^*E=E$ if $f$ is flat. The proof of the derived version of the projection formula uses only standard techniques of homological algebra to reduce to the trivial case of a free module. If $E$ is not coherent but only quasi-coherent, you need an extra assumption on $f$ to ensure that $f_*$ commutes with infinite direct sums. $\endgroup$
    – Roland
    Commented Sep 26, 2015 at 12:20
  • $\begingroup$ @Roland : I remember something like: in the derived category we can replace the coherent sheaf by a complex of locally free sheaves of finite rank (a locally free resolution). Now since $f$ is flat we have $Lf^{*}=f^{*}$ and the projection formula in the MO-answer (there we should take $A=\mathcal{F}$) becomes the usual projection formula for locally free sheaves, hence it is also true for the resolved sheaf. Something like this? What condition on $f$ would help, maybe proper? $\endgroup$
    – Bernie
    Commented Sep 26, 2015 at 12:49
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    $\begingroup$ Yes that's it. You don't need $f$ proper, it is sufficient to assume either $E$ perfect or $f$ quasi-compact between noetherian schemes of finite Krull dimension. In the first case you can locally resolve by a bounded complex of free sheaves of finite dimension. In the second case, $f_*$ commutes with infinite direct sums hence it also works by passing to the limit. $\endgroup$
    – Roland
    Commented Sep 26, 2015 at 14:25

1 Answer 1

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For any (quasi-) coherent sheaf $F$ on $Y$and any $f:X\to Y$, you have a natural map $F\to f_*(f^*F)$ and so to check this is an isomorphism is local on $Y$. So, if $Y$ is locally Noetherian, we are reduced to the situation for a coherent sheaf $F$ with a presentation $\mathcal{O}_Y^m\to\mathcal{O}_Y^n\to F\to 0$ and now use $f_*\mathcal{O}_X=\mathcal{O}_Y$ to finish the proof.

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  • $\begingroup$ Thank you, but I am having troubles finishing the proof. Where do we need the flatness of f? To pullback the presentation so we get a presentation on $X$ of $f^{*}F$? And then apply $f_{*}$ and use $f_{*}O_X=O_Y$? Don't we get higher direct images then? $\endgroup$
    – Bernie
    Commented Sep 26, 2015 at 11:39
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    $\begingroup$ I am sorry, I was trying to say that this is all you really have. The argument gives $F\to f_*(f^*F)$ to be injective and nothing more. And in general, you can not do better. Consider $X=\mathbb{A}^2-\{0\}\to\mathbb{A}^2=Y$ as your $f$. Then $f_*\mathcal{O}_X=\mathcal{O}_Y$ and it is flat, but for a line $L$ through the origin, $f_*(f^*\mathcal{O}_L)$ is not even coherent. $\endgroup$
    – Mohan
    Commented Sep 26, 2015 at 17:53
  • $\begingroup$ I see. What if we assume that $f$ is proper, so that such things can't happen? $\endgroup$
    – Bernie
    Commented Sep 26, 2015 at 19:51

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