Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Let $X$ be a smooth degree $d$ $(d \ge 5)$ surface in $\mathbb{P}^3$. Let $D$ be an effective Cartier divisor (hence locally of complete intersection) on $X$ and $D_{red}$ the associated reduced scheme which is also a Cartier divisor on $X$. Then, we have a natural inclusion of ideal sheaves, $$0 \to \mathcal{O}_X(-D) \to \mathcal{O}_X(-D_{red}).$$

Taking the dual we have

$$\mathcal{O}_X(D_{red}) \to \mathcal{O}_X(D) \to 0.$$

Since, $\mathcal{O}_X(D_{red})$ and $\mathcal{O}_X(D)$ are locally free sheaves of rank $1$, this would imply that the kernel of the latter map is zero. This would imply that $\mathcal{O}_X(D_{red}) \cong \mathcal{O}_X(D)$. This result is very surprising. Is there a mistake in the proof or is there an explaination for this behaviour?

share|improve this question
There seems to be a problem with compilation of the math mode –  Naga Venkata Oct 8 '12 at 12:35
Use backtricks "`" around math asterisks. Regarding the mistake in your proof, recall that the derived functor of $Hom$ is $Ext$, so when you dualize at a certain point you have $Ext^1$ –  Francesco Polizzi Oct 8 '12 at 13:03
Regarding your question and Sandor's answer, note it has nothing to do with $X$ being a surface (of any degree) in $P^n$ –  aginensky Oct 9 '12 at 15:41

1 Answer 1

up vote 6 down vote accepted

The second short exact sequence is wrong. You should recognize this without knowing where the mistake is: $D_{\mathrm{red}}\leq D$, so $\mathscr{O}_X(D_{\mathrm{red}}) \subseteq \mathscr{O}_X(D)$ and so the map you have is surjective if and only if $D_{\mathrm{red}}= D$. In fact that map is always injective as you discovered... The underlying point is that $\mathscr Hom$ is left exact, but not right exact.

The right computation would be that the dual of $$0 \to \mathscr{O}_X(-D) \to \mathscr{O}_X(-D_{red})\to \mathscr F \to 0.$$ gives $$0 \to \mathscr Hom_X(\mathscr F, \mathscr O_X) \to \mathscr{O}_X(D) \to \mathscr{O}_X(D_{\mathrm{red}}) \to \mathscr Ext^1_X(\mathscr F, \mathscr O_X)\to \mathscr Ext^1_X(\mathscr O_X(-D), \mathscr O_X).$$

Now $\mathscr F$ is supported on $D$, so since $\mathscr O_X$ is torsion free $\mathscr Hom_X(\mathscr F, \mathscr O_X)=0$ and $\mathscr O_X(-D)$ is locally free, so $\mathscr Ext^1_X(\mathscr O_X(-D), \mathscr O_X)=0$ and hence you have a short exact sequence: $$0 \to \mathscr{O}_X(D_{\mathrm{red}}) \to \mathscr{O}_X(D) \to \mathscr Ext^1_X(\mathscr F, \mathscr O_X)\to 0.$$

share|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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