# Prove identity $\operatorname{Supp}(M/tM)= \operatorname{Supp}(M) \cap V(tA)$ between supports

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 "$$\supset$$".

• 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 $x(M/tM) = y(M/tM) + t(M/tM) = 0$. Oct 16 '19 at 18:20
• that was the easy part. I can't show $\sqrt{ann(M) + tA} \supset ann (M/tM)$
– user526728
Oct 16 '19 at 20:43

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 $$\text{Supp}(M)\cap V(tA)\subseteq\text{Supp}(M/tM)$$.

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