Suppose $M$ is an $R$ module and $N$ is a submodule of $M.$ I am trying to prove the following: $Ass(M) = Ass(N) \cup Ass(M/N)$. This seemed intuitively true but the proof proved to be difficult. If I have a prime ideal $P \in Ass(M)$ then $P = Ann(m)$ for some $m \in M.$ If $m \in N$ then $P \in Ass(N) \cup Ass(M/N).$ But suppose it isn't. We already have $P \subset Ann(m + N)$ so now I want to show that $Ann(m + N) \subset P.$ How do I go about doing this? If I have an element $x \in R$ such that $xm \in N$ I want to show that $xm = 0.$ If $R$ were a field this would be easy as I could simply multiply the inverse of $x$ and get that $m \in N$ which contradicts the assumption that $m \notin N.$ Should I continue in this approach?
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While the inclusions: $$\DeclareMathOperator{\Ass}{Ass}\Ass N\subset \Ass M\subset \Ass N\cup\Ass M/N$$ are always true, the reverse last inclusion is false in general: take a prime number $p$, and consider $M=\mathbf Z$, $N=p\mathbf Z$. Then $$\Ass(\mathbf Z)=\Ass(p\mathbf Z)=(0),\quad\text{but }\:\Ass(\mathbf Z/p\mathbf Z)=(p).$$
However, what is true is the following:
If $\Phi$ is a subset of $\Ass M$, the exists a submodule $N$ of $M$ such that $$ \Ass N =\Ass M -\Phi, \quad \Ass M/N=\Phi$$
(Bourbaki, Commutative Algebra, Ch. $4$: Associated Prime Ideals and Primary Decomposition, § $1$, n°1, Proposition 4).
Bernard
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oh yeah, you need it to split, right? – Tim kinsella Nov 30 '17 at 20:05
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Split $ $ what? – Bernard Nov 30 '17 at 20:22
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$N\rightarrow M \rightarrow M/N$ – Tim kinsella Nov 30 '17 at 20:29
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1Yes, in this this case it's true. More generally, $\operatorname{Ass}\displaystyle\Bigl(\bigoplus_{i\in I}M_i\Bigr)=\bigcup_{i\in I}\operatorname{Ass}M_i$. – Bernard Nov 30 '17 at 20:36
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Disclaimer: I'm not quite good enough to give a nice hint, or point you in the right direction. Here is the argument I am familiar with, but read on only if you want a spoiler
claim:$ Ass(M) \subset Ass(N) \cup Ass(M/N)$.
Let $P \in Ass(M)$. Note that $P$ is associated to $M$ if and only if there is an injection $f:R/P \hookrightarrow M$. Consider the image $\mathrm{Im}(f) \subset M$.
First assume that $\mathrm{Im}(f)$ is disjoint from $N$. Then $\pi:M \to M/N$ restricts to a monomorphism, which we consider as $\pi \circ f:R/P \to M/N$, which shows that $P \in Ass(M/N)$.
Now, assume that $\mathrm{Im}(f)$ intersects $N$ nontrivially, with $x \in \mathrm{Im}(f) \cap N$, but then $ann(x)=P$, since the image of $f$ is isomorphic to $R/P$, so $P$ is an associated prime $Ass(N)$, since it is the anhilator of some element $x \in N$.
Andres Mejia
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