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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.

user26857
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ljh8372
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1 Answers1

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.

Manos
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  • 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 04:10
  • 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 04:12
  • 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 04:29
  • I am talking about answer 3. But I am confused between answer 2 and answer 3. – ljh8372 Dec 17 '14 at 04:40
  • I suggest you forget about answer 2 for now. Answer 3 is self-contained, it does not make use of any results on associated primes. I have left some little easy details for you to prove. The big picture is that for a specific element $\xi \in M$ such that $ann(\xi) \subset P$, we show that $ann(\xi)$ is prime. Which is what we want. – Manos Dec 17 '14 at 04:42
  • I have a question in your answer 1 - why can you assure that any prime ideal in $\mathrm{Supp}(M)$ must contain a minimal element? Does Zorn's lemma holds in this set of prime ideals (ordered by reverse inclusion)? – Jerry.Li Apr 30 '20 at 16:50
  • @Manos There is no assumption on the $R$-module $M$. It may not be finitely generated. – Jerry.Li May 01 '20 at 06:56
  • @Manos I'm afraid there is no relationship between the set of minimal prime ideals and the set of minimal elements of $\mathrm{Supp}(M)$ - the support is just a subset of the spectrum, and may not be closed in general. – Jerry.Li May 01 '20 at 15:25
  • @Jerry.Li: You are right that answers 1 and 2 were using that $M$ finitely generated, so i deleted them. – Manos May 03 '20 at 04:33
  • @Jerry.Li And right about $Supp(M)$ if $M$ is not finitely generated. Thanks. – Manos May 03 '20 at 04:39