# About $\operatorname{Supp}_A M_{\mathfrak{p}}$

Let $M$ be a finitely generated $A$-module and $\mathfrak{p}\subset A$ a prime ideal. Is it true that $\operatorname{Supp}_A M_{\mathfrak{p}}$ is the closure of $\operatorname{Supp}_{A_{\mathfrak{p}}} M_{\mathfrak{p}} \subset \operatorname{Spec}A$ ?

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It is true if $A$ is Noetherian and $M=A$. It follows from the fact that $(A_{\mathfrak{p}})_{\mathfrak{q}}\neq0$ if and only if $\mathfrak{p}\cap\mathfrak{q}$ contains a prime. –  ashpool Sep 23 '11 at 16:43

I claim:

$$\mbox{Supp}_AM_{\mathfrak{p}}\subset\mbox{Cl}_{\mbox{ Spec} A}(\mbox{Supp}_{A_{\mathfrak{p}}}M_{\mathfrak{p}}),$$ eqaulity if $M$ f.g. and has only finitely many minimal associated primes inside $\mathfrak{p}$ (e.g., if $M$ is Noetherian).

Proof :

$\Rightarrow$ : Let $\mathfrak{q}\in\mbox{Supp}_AM_{\mathfrak{p}}$, so that $(M_{\mathfrak{p}})_{\mathfrak{q}}\neq0$. There is a $\mathbb{Z}$-module \iso\ $(M_{\mathfrak{p}})_{\mathfrak{q}}\simeq U^{-1} M$, where $U=(A\backslash\mathfrak{p})(A\backslash\mathfrak{q})$. Hence $\mbox{Ass}_{U^{-1} A}U^{-1} M\neq\emptyset.$ Since $\mbox{Ass}_{U^{-1} A}U^{-1} M\subset\mbox{Supp}_{U^{-1} A}U^{-1} M$, $\mbox{Supp}_{U^{-1} A}U^{-1} M\neq\emptyset$. Let $\mathfrak{r}\in\mbox{Supp}_{U^{-1} A}U^{-1} M$. $$\mbox{Supp}_{U^{-1} A}U^{-1} M=\mbox{Supp}_AM\cap\mbox{Spec} U^{-1} A,$$ so $\mathfrak{r}\cap U=\emptyset$, which implies $\mathfrak{r}\subset\mathfrak{p}\cap\mathfrak{q}$. Thus $\mathfrak{r}\subset\mathfrak{q}$ and $\mathfrak{r}\in\mbox{Supp}_{A_{\mathfrak{p}}}M_{\mathfrak{p}}$, so $\mathfrak{q}\in V(\cap_{\mathfrak{r}\in\mbox{ Supp}_{A_{\mathfrak{p}}}M_{\mathfrak{p}}}\mathfrak{r})=\overline{\mbox{Supp}_{A_{\mathfrak{p}}}M_{\mathfrak{p}}}$. Hence, $\mbox{Supp}_AM_{\mathfrak{p}}\subset\overline{\mbox{Supp}_{A_{\mathfrak{p}}}M_{\mathfrak{p}}}$.

$\Leftarrow$ : Suppose $\mathfrak{q}\in\overline{\mbox{Supp}_{A_{\mathfrak{p}}}M_{\mathfrak{p}}}$. $$\overline{\mbox{Supp}_{A_{\mathfrak{p}}}M_{\mathfrak{p}}}=V(\cap_{\mathfrak{r}\in\mbox{ Supp}_{A_{\mathfrak{p}}}M_{\mathfrak{p}}}\mathfrak{r})=V(\cap_{\mathfrak{r}\in\mathcal{M}}\mathfrak{r}),$$ where $\mathcal{M}$ is the set of minimal associated primes of $M$ inside $\mathfrak{p}$. Hence, if $|\mathcal{M}|<\infty$, then $\mathfrak{r}\subset\mathfrak{q}$ for some $\mathfrak{r}\in\mathcal{M}$, and since $$\mathfrak{r}\in\mbox{Ass}_AM\subset\mbox{Supp}_AM=V(\mbox{ann} M),$$ $\mbox{ann}_A M\subset\mathfrak{r}$. If $(M_{\mathfrak{p}})_{\mathfrak{q}}=0$ and if $M$ is f.g., then there are $s\not\in\mathfrak{p}$, $t\not\in\mathfrak{q}$ such that $st\in\mbox{ann}_AM$, hence $st\in\mathfrak{r}$. But $\mathfrak{r}\subset\mathfrak{p}\cap\mathfrak{q}$, so $st\not\in\mathfrak{r}$, a contradiction. Therefore, $(M_{\mathfrak{p}})_{\mathfrak{q}}\neq0$, and $\mathfrak{q}\in\mbox{Supp}_AM_{\mathfrak{p}}$. Hence $\overline{\mbox{Supp}_{A_{\mathfrak{p}}}M_{\mathfrak{p}}}\subset\mbox{Supp}_AM_{\mathfrak{p}}$.

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