I read Prof. Goto's Notes in Commutative Algebra and I met a definition of primary module, like this:
"Suppose that $R$ is a Noetherian ring and $Q\in\operatorname{Spec}(R)$. Let $L\leq M$ be a $R$-submodule of $M$. Then we say that $L$ is a $Q$-primary submodule of $M$ if $\operatorname{Ass}_R(M/L)=\{Q\}$."
Also, the definition of $\operatorname{Ass}$ in these notes is:
"$\operatorname{Ass}_R M=$ $\{Q\in\operatorname{Spec}(R)\ \mid $ there exists an exact sequence $0\to R /Q \to M$ of $R$-modules$\}$"
I know the definition of primary module in Robert Ash's book, and I can show that under certain circumstances they are equivalent.
What I wanna ask here, is that whether the definition here is equivalent to the definition of primary ideals:
"An ideal $Q$ of $R$ is primary if $\forall$ $a,b\in Q$ s.t. $ab\in Q$, then $a\in Q$ or $b\in Q$"
In other words, the real question is:
Is it true that an ideal $Q$ is primary $\Leftrightarrow$ $\operatorname{Ass}_R(R/Q)=\{P\}$ for some $P\in\operatorname{Spec}(R)$?
I want to prove it's true with as few conditions as possible. But to make it simple, I assume $R$ is a Noetherian commutative ring and admit the primary decomposition of an arbitrary ideal. And this is how far I can go:
"($\Rightarrow$)
Suppose $Q$ is $P$-primary. We'll show that $\operatorname{Ass}_R(R/Q)=\{P\}$. Let $P_1\in \operatorname{Ass}_R(R/Q)$. Then $P_1=(Q:r_0)$ for some $r_0\notin Q$, now let $r\in P_1$, then $rr_0\in Q$ but $r_0\notin Q$, then $r\in \sqrt{Q}=P$, in other words $P_1\subset P$. We have $Q\subset (Q:r_0) \subset P$, then $Q=\sqrt{Q}\subset P_1=\sqrt{(Q:r_0)}\subset P$ or $Q=P$. So $\operatorname{Ass}_R(R/Q)\subseteq \{P\}$.
However, $\operatorname{Ass}_R(R/Q)\neq \emptyset$ since $R/Q\neq (0)$, so we have what we need.
($\Leftarrow$)
Suppose $Q=\bigcap Q_i$ is a primary decomposition of $Q$, in which $Q_i$ is $P_i$-primary. It is sufficient to prove that $\operatorname{Ass}_R(R/Q)=\{P_1,...,P_n\}$. We have an exact sequence $0\to R/Q\to \bigoplus R/Q_i$ so $\operatorname{Ass}_R(R/Q)\subseteq \{P_1,...,P_n\}$. The other inclusion, unfortunately, is not immediate, and I need help in this part.
Other approaches would be fine. And if it's ok, can we eliminate more conditions to prove it. Thank you.