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Problem of the Week #291 - Jul 03, 2019

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Euge

MHB Global Moderator
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Jun 20, 2014
1,894
Here is this week's POTW:

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Suppose $(X, \mathscr{O}_X)$ is a locally ringed space. Show that for all $\mathscr{O}_X$-modules $\mathscr{F}$ and $\mathscr{G}$ of finite type, $\operatorname{Supp}(\mathscr{F}\otimes_{\mathscr{O}_X} \mathscr{G}) = \operatorname{Supp}(\mathscr{F})\cap \operatorname{Supp}(\mathscr{G})$.


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Remember to read the POTW submission guidelines to find out how to submit your answers!
 
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Euge

MHB Global Moderator
Staff member
Jun 20, 2014
1,894
No one answered this week's problem. You can read my solution below.


Let $\mathscr{F}$ and $\mathscr{G}$ be $\mathscr{O}_X$ modules of finite type. If $x\notin\operatorname{Supp}(\mathscr{F}) \cap \operatorname{Supp}(\mathscr{G})$, then either $\mathscr{F}_x = 0$ or $\mathscr{G}_x = 0$. It follows that $\mathscr{F}_x \otimes_{\mathscr{O}_{X,x}} \mathscr{G}_x = 0$, or $(\mathscr{F}\otimes_{\mathscr{O}_X} \mathscr{G})_x = 0$. Thus $x\notin \operatorname{Supp}(\mathscr{F}\otimes_X \mathscr{G})$. Conversely, if $x\notin \operatorname{Supp}(\mathscr{F}\otimes \mathscr{G})$, then $\mathscr{F}_x \otimes_{\mathscr{O}_{X,x}} \mathscr{G}_x = 0$. As $\mathscr{F}$ and $\mathscr{G}$ are of finite type, $\mathscr{F}_x$ and $\mathscr{G}_x$ are finitely generated modules over local ring $\mathscr{O}_{X,x}$. Nakayama's lemma implies $\mathscr{F}_x = 0$ or $\mathscr{G}_x = 0$, i.e., $x\notin \operatorname{Supp}(\mathscr{F}) \cap \operatorname{Supp}(\mathscr{G})$.
 
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