2
$\begingroup$

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.

$\endgroup$
  • 1
    $\begingroup$ 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 Sep 26 '15 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 Sep 26 '15 at 12:49
  • 1
    $\begingroup$ 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 Sep 26 '15 at 14:25
3
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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 Sep 26 '15 at 11:39
  • 1
    $\begingroup$ 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 Sep 26 '15 at 17:53
  • $\begingroup$ I see. What if we assume that $f$ is proper, so that such things can't happen? $\endgroup$ – Bernie Sep 26 '15 at 19:51

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.