Let A be a Noetherian ring with $1$
Let M be a finitely generated A-Module.
Show that $l_A$(M) $< \infty$ if and only if M is a direct sum of P-coprimary submodules where P runs through AP(M) – the set of associated primes of M
(Recall that a submodule N is p-coprimary if M/N is p-primary).
To Prove the forward implication, I tried the following:
Since M is of finite length, every associated prime is maximal. Let $m$ be a maximal ideal of A and $p \in$ AP(M).
Case 1 – If $m \neq p$, then $m$ is not an associated prime of M$_p$. Since AP(M) = Supp(M), we get that (M$_p$)$_m$ = 0 and so the map M$_m \rightarrow $(M$_p$)$_m$ is surjective.
Case 2 – If $m = p$, then the map M$_m \rightarrow $(M$_p$)$_m$ can be given by the identity map.
I was trying to then get that the induced map from M$_m$ to ($\bigoplus _{p \in AP(M)} M_p$)$_m$ is then an isomorphism and so then we would get that the map from M to ($\bigoplus _{p \in AP(M)} M_p$ is an isomorphism.
I'm not sure if this works or not since I don't seem to be using the Noetherian criterion. Furthermore, I don't know how to prove the reverse direction at all and any help would be appreciated!
Edit: Fixed an error in the question (Had a slightly different question before).
Best Answer
You are almost done the proof. $M_{m}=0$ if $m$ is not an associated prime of M. $M_{m}$ is isomorphic to $(M_{p})_{m}$ is $m=p$. So your map $$M_{m}\rightarrow\bigoplus _{p \in AP(M)} (M_p)_{m}$$ is zero to zero map is $m$ is not in $AP(M)$. And it is an isomorphicm $M_{m}\rightarrow (M_{m})_{m}$ if $m\in AS(M)$. In each case, it is an isomorphism.