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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

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Perhaps we need $R$ Noetherian and $M$ finitely generated so that we can guarantee $Ass(M)$ is finite? – user6495 Apr 27 '11 at 17:27
up vote 1 down vote accepted

For * you don't need the assumption that $R$ is noetherian or $M$ is finitely generated. This is just a matter of convention. In fact $R$ is noetherian ensures that Ass$R$, $AssM$ are non empty and $M$ is finitely generated together with $R$ noetherian implies that Ass$M$ is finite. The statement is true for infinite direct sum also. Remember if $M_i$, $i\in I$ is an increasing chain of submodules of $M$ so that $M = \cup_{i \in I} M_i$, then $AssM = \cup_{i \in I} AssM_i$.

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thanks! – user6495 Apr 28 '11 at 13:58

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