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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?

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There seems to be a problem with compilation of the math mode –  Naga Venkata Oct 8 '12 at 12:35
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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
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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
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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: $$0 \to \mathscr{O}_X(D_{\mathrm{red}}) \to \mathscr{O}_X(D) \to \mathscr Ext^1_X(\mathscr F, \mathscr O_X)\to 0.$$

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