Let $R$ be a Noetherian ring and let $M$ be a finitely generated $R$-module. Denote the set of all associated primes of $M$ by $Ass(M)$.
If $R= \oplus_{i=1}^{n} M_{i}$ where each $M_{i}$ is an $R$-module, then:
$Ass(R) = \cup_{i=1}^{n} M_{i}$. Call this (*)
I'm trying to show the above. Now there is a result in Reid's book (commutative algebra) that says that $Ass(M) \subseteq Ass(M/N) \cup Ass(N)$ where $N$ is a submodule of $M$.
So using the isomorphism $(M_{1} \oplus M_{2})/M_{1} \cong M_{2}$ one has:
$Ass(M_{1} \oplus M_{2}) \subseteq Ass(M_{1}) \cup Ass(M_{2})$
and then the result follows. So why (*) needs the asumption that $A$ is Noetherian and $M$ if finitely generated? What is a counterexample? Also, I'm curious, I assume the result is false if we consider arbitrary direct sum (i.e not necessarily finite), what is a counterexample?
Thank you