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Let $N=\bigcap_{i=1}^nN_i$ be an irredundant primary decomposition of a submodule $N$ of the $R$-module $M,$ where $(N_i:M)$ is a $P_i$-primary ideal of $R$. (Irredundant means that all the $P_i$'s are different and $\bigcap_{i\neq j}N_j\nsubseteq N_i$, $1\leq i \leq n$).

We have that $(N:M)=\bigcap_{i=1}^n(N_i:M)$, is a primary decomposition of the ideal $(N:M)$ in $R$. I assumed (probably incorrect) that this decomposition is irredundant. All the $P_i$'s are distinct, but I have not been able to show that the conditions given above imply that $\bigcap_{i\neq j}(N_j:M)\nsubseteq (N_i:M)$, $1\leq i \leq n$.

(I am aware that this decomposition may be turned into an irredundant decomposition by simply remove superfluous terms). But doing this would remove some of the $P_i$'s.

To clarify, I hoped to prove that $\operatorname{Ass}(N:M)=\{P_1, P_2,\dots, P_n\}$. The above discussion only implies that $\operatorname{Ass}(N:M)\subseteq \{P_1, P_2,\dots, P_n\}$. (Using of course the uniqueness theorems for primary decomposition of ideals in $R$.)

Hence, I'm looking for help to prove that $\bigcap_{i\neq j}(N_j:M)\nsubseteq (N_i:M)$, $1\leq i \leq n$, in the context given above, OR as I suspect, verification that no such conclusion can be made. Thank you!

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up vote 3 down vote accepted

A very simple counterexample: $R=\mathbb Z$, $M=\mathbb Z\oplus \mathbb Z/p\mathbb Z$ ($p$ a prime number) and $N=0$. Then take $N_1=p\mathbb Z\oplus \{\hat 0\}$ and $N_2=\{0\}\oplus \mathbb Z/p\mathbb Z$. We have $(N_1:M)=p\mathbb Z$ and $(N:M_2)=(0)$.

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