A nonexample of a prime submodule?

Question

Let $M$ be an $A$-module. A submodule $P\subset M$ is called a prime submodule if it is proper and $am\in P$ implies $aM\subset P$ or $m\in P$. It is easy to see that if $P\subset M$ is a prime submodule, then the ideal $$(P:M):=\{a\in A \mid aM\subset P\}$$ is prime. I'm trying to find a counterexample of the converse; that is, a submodule $N\subset M$ which is not prime yet $(N:M)\subset A$ is a prime ideal.

Answer 1

Let $A=K[x,y]$ be the polynomial ring in two variables over a field $K$ and $M=A \oplus A$. We let $P = (x,y) \oplus (x) \subseteq M$, where $(x)$ and $(x,y)$ are the ideals generated by $x$ and $x,y$ respectively. Clearly $(P:M)=(x)$ is prime, but $P$ is not a prime submodule: Let $a=y$ and $m=(1,x) \in M$. We have $a \not\in (P:M)$ and $m \not\in P$, but $am \in P$.