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let $A$ a Noetherian ring and $M$ a finitely generated $A$-module. suppose that $t \in A$ is an $M$-regular element, i.e. multiplication map $t: M \to M. m \mapsto am$ is injective. recall the notation of support:

$\operatorname{Supp}(M):= \{\mathfrak{p} \in \operatorname{Spec}(R): M_{\mathfrak{p}} \neq 0 \}$

Q: why the equation $$\operatorname{Supp}(M/tM)= \operatorname{Supp}(M) \cap V(tA)$$ is true?

recall $V(tA):= \{\mathfrak{p} \subset A \text{ } \vert \text{ } tA \subset \mathfrak{p} \}$. the "$\subset$" part is easy: if $\mathfrak{p} \in \operatorname{Supp}(M/tM)$ prime with $(M/tM)_{\mathfrak{p}} \neq 0$ then obviously $M_{\mathfrak{p}} \neq 0$ and $tA \subset \mathfrak{p}$, since if otherwise $tA \not \subset \mathfrak{p}$ then a $ta \in A \backslash \mathfrak{p}$ would annulatate every $\bar{m} \in M/tM$ and thus $(M/tM)_{\mathfrak{p}} = 0$.

the "$\supset$" is harder. let $\mathfrak{p} \in \operatorname{Supp}(M/tM)$ with $tA \subset \mathfrak{p}$. possibly that can happen, that for every $m \in M$ we can find a $s_m \in \mathfrak{p}$ with $sm \in tM$,i.e. thus $(M/tM)_{\mathfrak{p}} = 0$. why this cannot happen?

alternative strategy might also be to show $\sqrt{Ann(M/tM)}= \sqrt{Ann(M)+ tA}$, since $\operatorname{Supp}(M) = V(Ann(M))$. trying this I also fail to show "$\supset$".

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    $\begingroup$ Indeed, $ann(M) + tA \subset ann (M/tM)$. Let $x \in ann(M) + tA$, then $x = y + tz$ for some $y \in ann(M)$ and $z \in A$. Then $x(M/tM) = y(M/tM) + t(M/tM) = 0$. $\endgroup$
    – Youngsu
    Oct 16 '19 at 18:20
  • $\begingroup$ that was the easy part. I can't show $\sqrt{ann(M) + tA} \supset ann (M/tM)$ $\endgroup$
    – user526728
    Oct 16 '19 at 20:43
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Let $\mathfrak{p}\in\text{Supp}(M)\cap V(tA)$. Now, suppose that $(M/tM)_\mathfrak{p}=0$. We have that $(M/tM)_\mathfrak{p}=M_\mathfrak{p}/tM_\mathfrak{p}$, and so $tM_\mathfrak{p}=M_\mathfrak{p}$.

Since $tA\subseteq\mathfrak{p}$ we have $\mathfrak{p}M_\mathfrak{p}=M_\mathfrak{p}$. Then certainly as $A_\mathfrak{p}$-modules we have $(\mathfrak{p}A_\mathfrak{p})M_\mathfrak{p}=M_\mathfrak{p}$. Since $M$ is a finitely generated $A$-module we have that $M_\mathfrak{p}$ is a finitely generated $A_\mathfrak{p}$-module.

Now, $\mathfrak{p}A_\mathfrak{p}$ is the unique maximal ideal of $A_\mathfrak{p}$, and so by Nakayama's Lemma we have $M_\mathfrak{p}=0$.

This is a contradiction, and so $(M/tM)_\mathfrak{p}\neq0$. Then $\mathfrak{p}\in\text{Supp}(M/tM)$ and so $\text{Supp}(M)\cap V(tA)\subseteq\text{Supp}(M/tM)$.

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Let $A$ be a commutative ring with unity, $I$ an $A$-ideal, and $M$ a finitely generated $A$-module. Then $\sqrt{\mathrm{Ann}(M/IM)}=\sqrt{\mathrm{Ann}(M)+I}$.

$\sqrt{\mathrm{Ann}(M/IM)}\supset\sqrt{\mathrm{Ann}(M)+I}$ by the comments above in the question.

For $\sqrt{\mathrm{Ann}(M/IM)}\subset\sqrt{\mathrm{Ann}(M)+I}$, on page 120 in Commutative algebra with a view toward algebraic geometry by David Eisenbud,

Theorem 4.3 (Cayley-Hamilton). Let $A$ be a commutative ring with unity, $I$ an $A$-ideal, and $M$ an $A$-module that can be generated by $n$ elements. Let $\varphi$ be an endomorphism of $M$. If $\varphi(M)\subset IM$, then there is a monic polynomial $p(x)=x^n+p_1x^{n-1}+\cdots+p_n$ with $p_j\in I^j$ for each $j$, such that $p(\varphi)=0$ as an endomorphism of $M$.

Now Let $a\in\sqrt{\mathrm{Ann}(M/IM)}\Rightarrow a^k\in\mathrm{Ann}(M/IM)\text{ for some positive integer $k$}\Rightarrow a^k(M/IM)=(a^kM)/IM=0\Rightarrow a^kM\subset IM$. Then define an endomorphism of $M$, $\varphi : M\rightarrow M$ by $m\rightarrow a^km$. Then since $\varphi(M)=a^kM\subset IM$, there is a monic polynomial, $p(x)=x^n+p_1x^{n-1}+\cdots+p_n$ with $p_j\in I^j$ for each $j$ such that $p(\varphi)=0$ as an endomorphism of $M$. So$((a^k)^n+p_1(a^k)^{n-1}+\cdots+p_n)M=0$ and $(a^k)^n+p_1(a^k)^{n-1}+\cdots+p_n\in\mathrm{Ann}(M)$.

Then $a^{kn}=(a^k)^n=\underbrace{(a^k)^n+(p_1(a^k)^{n-1}+\cdots+p_n)}_{\in\mathrm{Ann}(M)}-\underbrace{(p_1(a^k)^{n-1}+\cdots+p_n)}_{\in I\text{ since $p_j\in I^j$ for each $j$}}\in\mathrm{Ann}(M)+I\Rightarrow a\in\sqrt{\mathrm{Ann}(M)+I}.$

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