4
$\begingroup$

The following is a part of proof of lemma 6.2 in the book.

$f:X \to Y$ a projective birational morphism of normal varieties

$D$: Weil divisor on $Y$, $E$: exceptional divisor of $f$

$\mathcal{O}_X(1)$: very ample

We get an injection

$a:\mathcal{O}_Y(mD)=f_*(\mathcal{O}_X(m))\subsetneq f_*(\mathcal{O}_X(m)(E))$ for $m\gg 0$.

Why is this a contradiction if $\mathcal{O}_Y(mD)$ is reflexive and $a$ is an isomorphism outside the codimension $2$ set $f(Ex(f))$?

$\endgroup$

1 Answer 1

3
$\begingroup$

Lemma Let $\alpha: \mathscr F\hookrightarrow \mathscr G$ be an injection of $\mathscr O_Y$-modules such that

  • $Y$ is normal,
  • $\alpha$ is an isomorphism in codimension $1$,
  • $\mathscr F$ is reflexive, and
  • $\mathscr G$ is torsion-free.

Then $\alpha$ is an isomorphism.

Proof: Since $\mathscr G$ is torsion-free, it embeds into its reflexive hull: $\mathscr G\hookrightarrow \mathscr G^{**}$. Since $Y$ is normal, a reflexive sheaf is determined by its restriction to any big open set (i.e., one with a codimension $2$ complement). This and the rest of the conditions imply that the composition $\mathscr F\hookrightarrow \mathscr G\hookrightarrow \mathscr G^{**}$ is an isomorphism, but then so is $\alpha$. $\square$

Note Actually, somewhat less than $Y$ being normal is enough...

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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