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In Commutative Algebra, we defined the Support of a Module $M$ $$ \operatorname{Supp}(M) = \{P \in \operatorname{Spec}(R) : M_P \neq \{0 \} \} = \{P \in \operatorname{Spec}(R): \exists m \in M: \operatorname{Ann}(m) \subseteq P \} $$ Today, I asked our tutor "Can you tell us why $\operatorname{Supp}(M)$ is interesting?" and he told me no, that he has never really seen it in action.

So I want to pose this question here. Why should $\operatorname{Supp}(M)$ be interesting?

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    $\begingroup$ For finitely generated modules the dimension of M is the supremum of the lengths of chains of primes in Supp(M). $\endgroup$
    – user26857
    Jun 24 '19 at 19:52
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Consider a sheaf $\mathcal{F}$ of $\mathcal{O}_X$-modules. Then the support of $\mathcal{F}$ is all the points of $X$ "where the stalk is non-zero", i.e. $$\mathrm{supp}(\mathcal{F}) = \{ x \in X \mid \mathcal{F}_x \neq 0 \}. $$ This indeed coincides with the definition you've given in your question. Note that one can generalise this notion to complexes of sheaves, i.e. $$ \mathrm{supp}(\mathcal{F}^\bullet) = \bigcup \mathrm{supp}(H^i(\mathcal{F}^\bullet)). $$ Then this notion is important in stating/proving many results from algebraic geometry. For example, if we know somthing like $x \notin \mathrm{supp}(\mathcal{F}^\bullet) $ then we can deduce that $\mathcal{F}^\bullet |_U$ is trivial, where $x \in U \subset X$ is an open neighbourhood of $x$.

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  • $\begingroup$ I do not yet know what a sheaf is but this answer seems really nice. I will definitely come back to it when I study Algebraic Geometry! Thank you! $\endgroup$
    – Qi Zhu
    Jun 25 '19 at 19:13
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    $\begingroup$ You're welcome! Algebraic geometry realises many concepts like this from commutative algebra in very nice geometric ways so this will be indeed good to come back to. Another nice example is the geometry behind flat modules. $\endgroup$
    – mathphys
    Jun 26 '19 at 12:16
  • $\begingroup$ Awesome. I am looking forward to that; thank you for the link! $\endgroup$
    – Qi Zhu
    Jun 26 '19 at 15:27
  • $\begingroup$ I can finally understand that this was the correct answer! (And I'm thinking I might have misinterpreted my tutor's answer, or I would at least be really surprised that he doesn't know such a basic notion.) $\endgroup$
    – Qi Zhu
    May 2 at 16:05

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