# If $M$ is a finitely generated module then $\sqrt{\text{ann}(M)}=\bigcap\text{supp}(M)$

Let $$R$$ be a unital commutative ring and let $$M$$ be a finitely generated $$R$$-module. Prove that $$\sqrt{\text{ann}(M)}=\bigcap \text {supp}(M)$$

Recall that:

• $$\text{ann}(M)=\{r\in R:\forall m\in m, rm=0\}$$
• $$\text{supp}(M)=\{p\in\text{spec}(R):M_p\ne 0\}$$
• $$\forall p\in\text{spec}(R), V(p)=\{q\in\text{spec}(R):p\subseteq q\}$$
• $$\forall A\subseteq R,\sqrt A=\{r\in R:\exists n, r^n\in A\}$$
• $$\operatorname{spec}(R)=\{I\subseteq R:I\text{ is a prime ideal}\}$$

According to a theorem, $$\text{supp}(M)=\bigcup_{p\in\text{ass}(M)}V(p)$$ But I don't really success to proceed.

• how is $\sqrt \ :\ \mbox{Set?} \to ?$ defined? Sep 20 '19 at 14:13
• I added the defenition. Sep 20 '19 at 14:20
• What is $\mbox{spec}(R)$ here? I know of "spectrum of a commutative ring" which consists of prime ideals of $R$. This is likely something else. Sep 20 '19 at 14:28
• I added the defenition. Sep 20 '19 at 14:29
• Ok and what is $M_p$, where $p$ is a (prime) ideal of $R$? Sep 20 '19 at 14:35

First, we show that if $$p \in supp(M)$$, then $$ann(M) \subset p$$.

Indeed, let $$x \in ann(M)$$ not in $$p$$: then multiplication by $$x$$ from $$M_p$$ to itself is an isomorphism, since $$x$$ is invertible in $$A_p$$. On the other hand, this function is zero since $$x$$ is in the annihilator of $$M$$, a contradiction.

Next, we show that if $$ann(M) \subset p$$, then $$M_p \neq 0$$.

For this, notice that if $$M_p=0$$, this means that for all $$m \in M$$, there is $$t \notin p$$ such that $$tm=0$$. Since $$M$$ is finitely generated, this implies the existence of $$t \notin p$$ such that $$tM=0$$, so $$t \in ann(M)$$ but $$t \notin p$$, a contradiction.

The rest is standard commutative algebra.

• I didn't understand why $0=M_p\Rightarrow \forall m\in M,\exists t\notin p:tm=0$. Indeed, if $0=M_p$ then $$0=M_p=\{{m\over t}:m\in M,t\notin p\}$$ Hence, $$\forall m\in M,\forall t\in R\setminus p, {m\over t}={0\over 1}\Rightarrow \\\forall m\in M,\forall t\in R\setminus p,\exists 0\ne r\in R, 0=r(m\cdot1-t\cdot0)=rm$$ So I conclude $$\forall m\in M,\exists 0\ne r\in R, rm=0$$ where do I miss? Sep 20 '19 at 18:45
• No. If $M_p=0$, for all $m \in M$, there is some $t \notin p$ such that $tm=0$. Let $m_1, \ldots, m_r$ be a generating subset of $M$, $t_k \notin p$ the associated elements. Then $\tau=\prod_{k=1}^r{t_k} \notin p$ is such that $\tau m_k=0$ for all $k$ so $\tau M=0$. Sep 20 '19 at 20:05
• Why does $B:=\bigcap\text{supp}(M)\subseteq \sqrt{\text{ann}(M)}=:A$? Suppose $b\in B$ then $\forall p\in \text{supp}(M), b\in p$. How does it imply $\exists n$ s.t. $p^n\in\text{ann}(M)$? We know that $\forall p\in\text{supp}(M), p$ contains $b$ *and $\text{ann}(M)$. But we are finish only if $p$ is cyclic, but we don't know that it is. Sep 21 '19 at 17:40
• There is the well-known result in commutative algebra, that for every ideal $I$ of a ring $R$, $\sqrt{I}$ is the intersection of all the prime ideals containing $I$. Sep 21 '19 at 17:41

$$\sqrt{\mathrm{ann}(M)}=\bigcap\limits_{p\in V(\mathrm{ann}(M))}^{}p$$ (Proposition 1.14 in Introduction to Commutative Algebra by M.F. Atiyah, I.G. MacDonald). But $$V(\mathrm{ann}(M))=\mathrm{supp}(M)$$ for a finitely generated $$R$$-module $$M$$(Theorem 1.5.5, https://faculty.math.illinois.edu/~r-ash/ComAlg/ComAlg1.pdf).

So, $$\sqrt{\mathrm{ann}(M)}=\bigcap\limits_{p\in\mathrm{Supp}(M)}^{}p$$.