I meet the following theorem:
Let $R$ be a Noetherian ring and $L$ be a $R$-submodule of $M$ s.t. $L\neq M$. Then there exists a family $\{ L(Q) \}_{Q\in\operatorname{Ass}_R(M/L)}$ of $R$-submodules of $M$ s.t.
1) $\operatorname{Ass}_R(M/L(Q))=\{Q\} $ for every $Q\in\operatorname{Ass}_R(M/L)$, and
2) $\bigcap_{Q\in\operatorname{Ass}_R(M/L)}L(Q)=L$.
So I'll sum up the proof:
Set $\mathcal F=\operatorname{Ass}_R(M/L)$. For each $Q\in\mathcal F$ we can find $L(Q)\supset L$ s.t. $\operatorname{Ass}_R(L(Q)/L)=\mathcal F\setminus \{Q\}$, so $\operatorname{Ass}_R(M/L(Q))=\{Q\}$. We set $N=\bigcap_{Q\in \mathcal F}L(Q) $. Then $L\subset N$. Also, $$\operatorname{Ass}_R(N/L) \subseteq \operatorname{Ass}_R(L(Q)/L)=\mathcal F\setminus \{Q\}$$ for all $Q$. So $\operatorname{Ass}_R(N/L)= \emptyset$, or $N=L$.
I can understand the proof, the daunting part is the Remark after that:
REMARK: If $\operatorname{Ass}_R(M/L)$ is finite, then the decomposition in (2) is irredundant.
So help me with this REMARK, I don't know why it happens. Thank you.