# If $P \in \operatorname{Supp}(M)$ prove that $P$ contains a prime ideal $Q$ with $Q \in \operatorname{Ass}_R(M)$.

My problem is below,

Let $M$ be an $R$-module. The set of prime ideals $P$ of $R$ for which the localization $M_P$ is nonzero is called the support of $M$, denoted $\operatorname{Supp}(M)$. The set of prime ideals $Q$ of $R$ for which $Q$ is an annihilator for some element $m \in M$, denoted $\operatorname{Ass}_R(M)$.

Suppose that $R$ is Noetherian. If $P \in \operatorname{Supp}(M)$ prove that $P$ contains a prime ideal $Q$ with $Q \in \operatorname{Ass}_R(M)$.

My attempt,

Suppose there exists a prime ideal $P$ such that $M_P \neq 0$. Then there exists $x \in M$ such that $\operatorname{Ann}(x) \subset P$. Let $\mathcal{S} = \{ \operatorname{Ann}(x) : x \in M, \operatorname{Ann}(x) \subseteq P \}$. Take a maximal in $\mathcal{S}$ and I tried to prove that it is a prime ideal. But it fails until now.

Can anybody help me? Thank you.

This can be seen quickly if one uses certain results on the associated primes. For example:

answer 1: It is a fact that $\operatorname{Ass}(M) \subset \operatorname{Supp}(M)$ and the minimal elements of $\operatorname{Ass}(M)$ coincide with the minimal elements of $\operatorname{Supp}(M)$. So if $P \in \operatorname{Supp}(M)$ and if $P \not\in \operatorname{Ass}(M)$, then $P$ must contain a minimal element of $\operatorname{Supp}(M)$ which must be an element of $\operatorname{Ass}(M)$.

answer 2: Localize with respect to $P$ and obtain that $M_P$ is a non-zero finite module over the local ring $(R_P,PR_P)$. Now consider the set $\Phi = \left\{ann(\xi): 0 \neq \xi \in M_P \right\}$ and show that any maximal element of $\Phi$ must be prime. Then contract back $R \rightarrow R_P$ to show that the associated prime of $M_P$ is a localization of an associated prime of $M$ over $R$ contained in $P$.

But it can also be proved from entirely elementary considerations as well:

answer 3: Consider the set $\Phi=\left\{\operatorname{ann}(\xi): 0\neq \xi \in M_P\right\}$. This set of ideals is nonempty, since $M_P \neq 0$. Notice also that for every $\operatorname{ann}(\xi) \in \Phi$ we have $\operatorname{ann}(\xi) \subset P$. Let $I = \operatorname{ann}(\xi)$ be a maximal element of $\Phi$. Suppose $a,b$ are elements of $R$ such that $ab \in I$ and suppose that $a \not\in I$. There are two possibilities: i) $\operatorname{ann}(a\xi) \not\subset P$ and ii) $\operatorname{ann}(a \xi) \subset P$. If i) is true, then there exists an element $s \not\in P$ such that $sa\xi=0$. Now notice that $s \xi \neq 0$ and that $\operatorname{ann}(s\xi) \subset P$. Since $\operatorname{ann}(s \xi) \in \Phi$ and $\operatorname{ann}(\xi) \subset \operatorname{ann}(s \xi)$, the maximality of $\operatorname{ann}(\xi)$ in $\Phi$ gives $\operatorname{ann}(\xi) = \operatorname{ann}(s \xi)$. But then this implies that $a \in \operatorname{ann}(\xi)$, which is a contradiction on our hypothesis that $a \not\in \operatorname{ann}(\xi)$. Hence it must be the case that ii) is true, i.e. $\operatorname{ann}(a \xi) \subset P$. But then again the maximality of $\operatorname{ann}(\xi)$ gives $\operatorname{ann}(\xi) = \operatorname{ann}(a \xi)$, which implies that $b \in \operatorname{ann}(\xi)$ and so we have proved that $\operatorname{ann}(\xi)$ is prime.

• The alternative is the appropriate method for me. Thank you very much. – ljh8372 Dec 17 '14 at 3:39
• I have a question how I can know that $M_P$ is a non-zero finite module and the associated prime of $M_P$ is a localization of an associated prime of $M$ over $R$ contained in $P$. – ljh8372 Dec 17 '14 at 4:10
• That's a great question. In fact there is a theorem that is saying exactly that. Alternatively you can try and construct explicitly an element whose annihilator is the contraction of the associated prime of $M_P$ (and this is how the theorem is proved). But check out my third answer. I think that's what is you are looking for. – Manos Dec 17 '14 at 4:12
• I think that $ann(\xi)$ is a prime ideal in $R_P$ and a contraction of $ann(\xi)$ is a prime ideal in $R$. But I think that we need to show that the contraction of $ann(\xi)$ is an associated prime... for some element in $M$. – ljh8372 Dec 17 '14 at 4:29
• I am talking about answer 3. But I am confused between answer 2 and answer 3. – ljh8372 Dec 17 '14 at 4:40