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Let $R$ be a commutative unitary ring, consider the exact sequence of $R$ - modules $$ 0\rightarrow N\rightarrow M\rightarrow L\rightarrow0. $$ We know that $$\text{Ass}(M)\subseteq \text{Ass}(N)\cup \text{Ass}(L).$$ My question is:

When (in what conditions on the ring $R$) do we have the equality $$ \text{Ass}(M)=\text{Ass}(N)\cup \text{Ass}(L)? $$ Can someone help me.

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I only know this happen when the the sequence is split exact. – tlquyen Nov 8 '12 at 8:45
What are $N,M,L$? You should make that clear in the question. – Cameron Buie Nov 8 '12 at 8:47
@tlquyen: you probably want to assume $R$ is noetherian. – user18119 Nov 8 '12 at 9:23
@ Cameron Buie: $M,N,L$ are arbitrary $R-$modules. – tlquyen Nov 8 '12 at 15:17
@ QiL: In case $R$ is neotherian, do we get the equality? – tlquyen Nov 8 '12 at 15:21
up vote 5 down vote accepted

Recall that a prime ideal $\mathfrak p$ of $R$ is an associated prime of a module $M$ if and only if $M$ contains a submodule isomorphic to $R/\mathfrak p$. Equivalently, there exists $x\in M$ such that $\mathfrak p=\mathrm{Ann}_R(x)$.

So we always have $\mathrm{Ass}(N)\subseteq \mathrm{Ass}(M)$. So the only problem is to see when $\mathrm{Ass}(L)\subseteq \mathrm{Ass}(M)$ holds.

If the Krull dimension $\dim R>0$ (equivalent to non-artinian if $R$ is noetherian), then there are distinct prime ideals $\mathfrak p\subset \mathfrak m$. Consider the canonical surjection $$M:=R/\mathfrak p\to L:=R/\mathfrak m.$$ Then $\{\mathfrak m \}=\mathrm{Ass}(L) \not\subset \{\mathfrak p\}=\mathrm{Ass}(M) $.

So the rings $R$ for which the desired equality holds for all exact sequences are dimension zero.

Edit 2. The above conclusion is frustrating because it doesn't give any positive information when $\dim R>0$. However:

Suppose $R$ is noetherian. Consider $\mathrm{Spec}(R)$ with its Zariski topology. Then $$\mathrm{Ass}(M)\subseteq \mathrm{Ass}(N)\cup \mathrm{Ass}(L)\subseteq \overline{\mathrm{Ass}(M)}.$$

Proof. Let $\mathfrak p\in \mathrm{Ass}(L)$. We have to show that $\mathfrak p$ contains a $\mathfrak q\in \mathrm{Ass}(M)$. We can replace $L$ by a suitable submodule (and $M$ by the preimage of this submodule) and suppose $L$ is isomorphic to $R/\mathfrak p$. Suppose that $\mathfrak p$ doesn't contain any associated prime of $M$. Then the localization $M_{\mathfrak p}$ (viewed as an $R$-module) has no associated prime (Matsumura: Commutative algebra, Lemma (7.C)). So $M_{\mathfrak p}=0$ (op. cit., Cor. 1, p. 50). Hence for any $x\in M$, there exists $a\in R\setminus \mathfrak p$ with $ax=0$. So $a\phi(x)=0$ and $\phi(x)=0$, $M=\mathrm{ker}(\phi)$. Impossible.

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Agree, $M:=\varphi^{-1}(L)\xrightarrow{\varphi}L$. – tlquyen Nov 9 '12 at 14:35
This means the equality holds in case $Ass(M)$ is closed set. Again, when does it happen? Now, if $M$ is finite module we have $Supp(M)$ is closed set. In case $R$ is noetherian and $M$ is finite length, we have $Ass(M)=Supp(M)$ (H.A. Nielsen: Elementary commutative algebra, Cor.9.3.14). So we get the desired equality. – tlquyen Nov 9 '12 at 14:53
I still think this holds in some case that is weaker. – tlquyen Nov 9 '12 at 15:00
This holds for example if the sequence is split (e.g. $L$ is projective) or if $\mathrm{Ass}(L)$ consists only in minimal prime ideals of $R$, or if $\mathrm{Ass}(M)$ is closed as you said (but this implies that the support of $M$ is a finite closed subset in the noetherian case). I don't know a general statement. – user18119 Nov 9 '12 at 20:31
I have just found an interested case. Let $R$ be a local ring and $M$ is a finite $R-$module. Then, $M$ has a simple submodule iff for each maximal submodule $U$ of $M$, we always have $$\mbox{Ass}(M)=\mbox{Ass}(U)\cup\mbox{Ass}(M/U).$$ – tlquyen Nov 10 '12 at 4:14

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