O_X module with support Z \subset X vs O_S module? - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-19T16:36:11Z http://mathoverflow.net/feeds/question/16963 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/16963/o-x-module-with-support-z-subset-x-vs-o-s-module O_X module with support Z \subset X vs O_S module? John Doe 2010-03-03T10:17:48Z 2010-03-03T10:32:55Z <p>Given a $O_X$ module $\cal F$ whose support is a closed subscheme $Z \subset X$. Under what conditions can we say that $\cal F$ is an $O_S$ module ( how far off is $\cal F$ an $O_S$ module ? )</p> http://mathoverflow.net/questions/16963/o-x-module-with-support-z-subset-x-vs-o-s-module/16964#16964 Answer by Akhil Mathew for O_X module with support Z \subset X vs O_S module? Akhil Mathew 2010-03-03T10:22:08Z 2010-03-03T10:32:55Z <p>It has to be annihilated by the sheaf of ideals of $Z$. If you are working with a noetherian scheme and a coherent sheaf at least, we can at least filter $\mathcal{F}$ by subsheaves $\mathcal{I}^i \mathcal{F}$ (where $\mathcal{I}$ is the sheaf of ideals of $Z$) whose successive quotients are $O_S$-modules. Here $\mathcal{I}^i \mathcal{F} = 0$ for $i$ large by the assumption on support in $Y$. Indeed, $Y$ can be defined as $V(\mathcal{a})$ for some ideal $\mathcal{a}$. Then $M_{\mathfrak{p}} = 0$ if $\mathfrak{p} \not \supset \mathcal{a}$. Therefore, every associated prime must contain $\mathcal{a}$, so there is a filtration $0=M_0 \subset M_1 \subset \dots \subset M_k = M$ whose quotients are isomorphic to the form $A/\mathfrak{p}$ where $\mathfrak{p} \supset \mathcal{a}$ is a prime, which means $M$ is annihilated by $\mathcal{a}^k$.</p>