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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.

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  • $\begingroup$ how is $\sqrt \ :\ \mbox{Set?} \to ?$ defined? $\endgroup$
    – AlvinL
    Sep 20 '19 at 14:13
  • $\begingroup$ I added the defenition. $\endgroup$
    – J. Doe
    Sep 20 '19 at 14:20
  • $\begingroup$ 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. $\endgroup$
    – AlvinL
    Sep 20 '19 at 14:28
  • $\begingroup$ I added the defenition. $\endgroup$
    – J. Doe
    Sep 20 '19 at 14:29
  • $\begingroup$ Ok and what is $M_p$, where $p$ is a (prime) ideal of $R$? $\endgroup$
    – AlvinL
    Sep 20 '19 at 14:35
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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.

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  • $\begingroup$ 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? $\endgroup$
    – J. Doe
    Sep 20 '19 at 18:45
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    $\begingroup$ 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 $\tau M=0$. $\endgroup$
    – Mindlack
    Sep 20 '19 at 20:05
  • $\begingroup$ 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. $\endgroup$
    – J. Doe
    Sep 21 '19 at 17:40
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    $\begingroup$ 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 $I$. $\endgroup$
    – Mindlack
    Sep 21 '19 at 17:41
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$\sqrt{\mathrm{ann}(M)}=\bigcap\limits_{p\in V(\mathrm{ann}(M))}^{}p$ (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, $\sqrt{\mathrm{ann}(M)}=\bigcap\limits_{p\in\mathrm{Supp}(M)}^{}p$.

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