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Theorem 6.8(ii) p.41 in Matsumura's Commutative Ring Theory, says that if $A$ is a Noetherian ring, $M$ a finite $A$-module and $N=N_1 \cap \cdots \cap N_s$ an irredundant primary decomposition of a proper submodule $N$ with $Ass(M/N_i) = \left\{P_i \right\}$, then $Ass(M/N) = \left\{P_1, \cdots, P_s \right\}$, where $Ass(\cdot)$ stands for "associated primes".

According to my understanding, the proof does not use the fact that $M$ is a finite $A$-module. I would like to confirm that the statement of theorem 6.8(ii) is valid even if $M$ is not finite over $A$ (however $A$ must still be Noetherian).

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  • $\begingroup$ It's probably the case that if you have an irr. primary decomposition the conclusion holds and the assumption that $M$ is finite ensures that every submodule has a primary decomposition. $\endgroup$
    – JSchlather
    Commented Jan 30, 2013 at 20:44
  • $\begingroup$ My findings corroborate your assertion. $\endgroup$
    – Manos
    Commented Jan 30, 2013 at 20:50

1 Answer 1

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$\DeclareMathOperator{\Ass}{Ass}$

Everything follows from two facts. The first is the "short exact sequence property" of associated primes. That is, given a short exact sequence of $A$-modules, the $\Ass$ of the left module is contained in the $\Ass$ of the middle, and the $\Ass$ of the middle is contained in the union of the $\Ass$'es of the two sides. The second is the fact that any nonzero $A$-module has an associated prime (which in turn requires that $A$ be Noetherian).

First, to simplify notation, we may mod out by $N$ to assume that $N=0$, so that what you really have is an irredundant primary decomposition of the zero module.

Next, there is a natural homomorphism $$M \hookrightarrow \bigoplus_{i=1}^s M/N_i$$ that is injective since its kernel is $\bigcap_{i=1}^s N_i = 0$, and hence $\Ass M \subseteq \Ass \left(\bigoplus_{i=1}^s M/N_i\right) = \bigcup_{i=1}^s \Ass(M/N_i) = \{P_1, \ldots, P_s\}$ (where the first equality follows from induction on the short exact sequence property).

For the reverse containment, by symmetry it is enough to show that $P_1 \in \Ass M$. Consider the following composition of natural homomorphisms: $$ N_2 \cap \cdots \cap N_s \hookrightarrow M \twoheadrightarrow M/N_1. $$ This composition is an injection. Hence, $\Ass (N_2 \cap \cdots \cap N_s) \subseteq \Ass(M/N_1) = \{P_1\}$. But $N_2 \cap \cdots \cap N_s \neq 0$ (by irredundancy) and since any nonzero $A$-module has an associated prime, $\Ass (N_2 \cap \cdots \cap N_s) = \{P_1\}$. But since $N_2 \cap \cdots \cap N_s$ is a submodule of $M$, it follows that $P_1 \in \Ass M$.

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  • $\begingroup$ Nice answer! I have a question: in the end of your first paragraph you say "any nonzero A-module has an associated prime". Is this true if and only $A$ is Noetherian? Second question on notation: what does this double arrow stand for in the map $M \rightarrow M/N_1$? $\endgroup$
    – Manos
    Commented Feb 8, 2013 at 15:58
  • $\begingroup$ 1. I don't think that "if and only if" holds. There may well be non-Noetherian rings over which any nonzero module has an associated prime. However, it is true that this property holds when the ring is Noetherian. 2. The double arrow means "surjection". $\endgroup$
    – neilme
    Commented Feb 9, 2013 at 8:29

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