6
$\begingroup$

Let $f:X\rightarrow Y$ be a birational morphism between smooth varieties and $F$ a torsion free sheaf. Is $f_{\ast} F$ torsion free? If not, are there conditions on either $F$, $f$, $X$ or $Y$ that could ensure the torsion freeness of $F$? For example if $f$ is surj. and $F$ is the canonical bundle on $X$ a theorem of Kollar ensure the torsion freeness of $f_*(\omega_X)$.

$\endgroup$
1
  • 5
    $\begingroup$ If $F$ is torsion free on $X$, then $F\to F\otimes K(X)$ is injective where $K(X)$ is the constant sheaf associated to the function field. The injectivity persists for $f_*F$ since $f_*$ is left exact. So yes, and you don't need anything as deep as Kollar's theorem. $\endgroup$ May 13, 2011 at 11:17

2 Answers 2

8
$\begingroup$

Actually much more is true. In fact there is the following result, whose proof can be found in

[Grothendieck - Dieudonné: EGA 1 (Elements de Géométrie Algebrique), Proposition 8.4.5 page 351].

Proposition. Let $X$, $Y$ be two integral schemes and $f \colon X \to Y$ be a dominant morphism. Then for any torsion-free $\mathcal{O}_X$-module $\mathcal{F}$, the push-forward $f_* \mathcal{F}$ is a torsion-free $\mathcal{O}_Y$-module.

Kollar's result is much deeper since he proves the torsion-freeness of the higher direct images $R^if_* \omega_X$.

$\endgroup$
2
  • $\begingroup$ Thank you very much!!! I believed it was true but I could not find the right reference!!! best wishes! $\endgroup$
    – Rurik
    May 13, 2011 at 11:54
  • $\begingroup$ Maybe we don't have the same edition (?), but in mine (french version) the result is Proposition 7.4.5 page 163 (still in EGA I). $\endgroup$
    – Henri
    Nov 10, 2016 at 19:54
6
$\begingroup$

As already mentioned by Donu and Francesco, you don't need such big guns as Kollár's theorem. Also, the Proposition Francesco cites might seem more serious than it is by virtue of being in EGA...

The point is this: For an open set $V\subseteq Y$, the module $f_*\mathscr F(V)$ is the same as the module $\mathscr F(f^{-1}V)$. Being torsion on an integral scheme is equivalent to having a non-empty support that is strictly smaller than the ambient scheme. If $f$ is surjective this already gives you what you want, and if it is only dominant you need to think a little more to see that the statement is true.

$\endgroup$
1
  • $\begingroup$ Thank you very much for your time! this was really helpfull. $\endgroup$
    – Rurik
    May 18, 2011 at 8:59

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.