Matsumura Commutative Ring Theory 6.9 on coprimary modules of finite length. This is the final exercise in the appendix to section 6 of Matsumura's Commutative Ring Theory, covering the ideas of secondary representations. To provide a brief, but not expansive detailing, we say that an $A$-module $M$ is secondary if for all $a \in A$, the endomorphism map $M\to M$ given by multiplication by $a$ is either surjective or nilpotent. In this case $\mathfrak{p} = \sqrt{\text{Ann}(M)}$ is a prime ideal, and we say that $M$ is $\mathfrak{p}$-secondary, and we note that any quotient of $M$ in this case is also $\mathfrak{p}$-secondary. We say that a module is coprimary if Ass$(M)$ is a singleton. We say an ideal $\mathfrak{p}$ is an attached ideal of $M$ if $M$ has a $\mathfrak{p}$-secondary quotient, and denote the set of attached ideals as Att$(M)$. The problem statement then is as follows:
Show that if $M$ is an $A$-module of finite length then $M$ is coprimary if and only if it is secondary. Show that such a module $M$ is a direct sum of secondary modules belonging to maximal ideals, and Ass$(M) = \text{Att}(M)$.
Via induction I was able to show that for all $n$, that a module of length $n$ is coprimary if and only if it is secondary, however I am not able to show that Ass$(M) = \text{Att}(M)$ even when noting that both are singleton sets. I have tried numerous things to attempt to somehow relate the two via submodules for which the two sets would have to be equal, if ann$(x)$ is an associated prime then we can look at $Ax$, the submodule generated by $x$, and if $Ax = M$ we are done, so we assume it is a non-zero submodule, and can look then at $M/Ax$. This let's us link Ass$(M)$ and Ass$(Ax)$, which in turn is linked to Att$(Ax)$, but I do not see how we could link this back up to Att$(M)$, and any other idea I have didn't really work. As for the final statement that it is a direct sum of secondary modules belonging to maximal ideals, in the $n = 1$ case it is simply equal to itself, and I had a suspicion that somehow $\sqrt{\text{Ann}(M)}$ would be maximal, but I no longer believe this to be the case.
I think it may be possible to resolve the issue of Ass$(M) = \text{Att}(M)$ by demonstrating that $M$ is the sum of length $1$ submodules, a previous theorem stated that an Artinian module is secondary if it is sum-irreducible, that is not the sum of two proper submodules, and so being length $1$ implies that it is secondary, which by our inductive hypothesis tells us that it is also coprimary, belonging to the associated prime of $M$. From here, the sum of $\mathfrak{p}$-secondary modules is also $\mathfrak{p}$-secondary, so since $M$ is the sum of $\mathfrak{p}$-secondary submodules it too is $\mathfrak{p}$-secondary, however $\{\mathfrak{p}\} = \text{Ass}(M)$ telling us that Ass$(M) = \text{Att}(M)$. We also note that our length $1$ submodules are isomorphic to $A/I$ for some $I$ since they are cyclic, and from this we can prove that $I$ is maximal, so this leads us close to the statement that $M$ is a sum of secondary submodules belonging to maximal ideals, all we would have to do is go from a normal sum to a direct sum somehow.
 A: Here's a proof that $\mathrm{Ass}(M) = \mathrm{Att}(M)$. First we show $\mathrm{Att}(M)= \{\sqrt{\mathrm{Ann}(M)}\}$. It's clear that any quotient of a secondary module is secondary, so we just need to show that if $N < M$ is a proper submodule, $\sqrt{\mathrm{Ann}(M/N)} = \sqrt{\mathrm{Ann}(M)}$. If $a \in \mathrm{Ann}(M/N)$, then $aM \subseteq N$, so multiplication by $a$ on $M$ isn't surjective, and thus is nilpotent. Thus $a \in \sqrt{\mathrm{Ann}(M)}$.
Thus what we really need to show is that $\mathrm{Ass}(M) = \{\sqrt{\mathrm{Ann}(M)}\}$. We may assume wlog that $\mathrm{Ann}(M) = 0$ by replacing $A$ with $A/\mathrm{Ann}(M)$. Any associated prime of $M$ contains the annihilator, so this doesn't really change $\mathrm{Ass}(M)$, and it's clear that $M$ is still secondary and artinian/noetherian since it has the same submodules over this new $A$. Then $M$ is a faithful artinian module over $A$, and so $A$ is artinian. Thus every prime ideal of $A$ is maximal. Since $M$ is secondary, $\sqrt{\mathrm{Ann}(M)} = \sqrt{0}$ is prime, and thus maximal, and thus the only prime ideal of $A$. Thus any associated prime of $M$ is equal to $\sqrt{\mathrm{Ann}(M)}$. Since $M$ is coprimary, it has an associated prime, and thus $\mathrm{Ass}(M) = \{\sqrt{\mathrm{Ann}(M)}\}$.
