linear system of non-reduced divisor and associated reduced divisors - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-21T17:53:54Z http://mathoverflow.net/feeds/question/109138 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/109138/linear-system-of-non-reduced-divisor-and-associated-reduced-divisors linear system of non-reduced divisor and associated reduced divisors Naga Venkata 2012-10-08T12:27:39Z 2012-10-09T06:34:16Z <p>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, <code>$$0 \to \mathcal{O}_X(-D) \to \mathcal{O}_X(-D_{red}).$$</code></p> <p>Taking the dual we have </p> <p><code>$$\mathcal{O}_X(D_{red}) \to \mathcal{O}_X(D) \to 0.$$</code></p> <p>Since, <code>$\mathcal{O}_X(D_{red})$</code> 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 <code>$\mathcal{O}_X(D_{red}) \cong \mathcal{O}_X(D)$</code>. This result is very surprising. Is there a mistake in the proof or is there an explaination for this behaviour?</p> http://mathoverflow.net/questions/109138/linear-system-of-non-reduced-divisor-and-associated-reduced-divisors/109203#109203 Answer by Sándor Kovács for linear system of non-reduced divisor and associated reduced divisors Sándor Kovács 2012-10-09T06:34:16Z 2012-10-09T06:34:16Z <p>The second short exact sequence is wrong. You should recognize this without knowing where the mistake is: <code>$D_{\mathrm{red}}\leq D$</code>, so <code>$\mathscr{O}_X(D_{\mathrm{red}}) \subseteq \mathscr{O}_X(D)$</code> 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.</p> <p>The right computation would be that the dual of <code>$$0 \to \mathscr{O}_X(-D) \to \mathscr{O}_X(-D_{red})\to \mathscr F \to 0.$$</code> gives <code>$$0 \to \mathscr Hom_X(\mathscr F, \mathscr O_X) \to \mathscr{O}_X(D) \to \mathscr{O}_X(D_{\mathrm{red}}) \to \mathscr Ext^1_X(\mathscr F, \mathscr O_X)\to \mathscr Ext^1_X(\mathscr O_X(-D), \mathscr O_X).$$</code></p> <p>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: <code>$$0 \to \mathscr{O}_X(D_{\mathrm{red}}) \to \mathscr{O}_X(D) \to \mathscr Ext^1_X(\mathscr F, \mathscr O_X)\to 0.$$</code></p>