Proving that $\mathfrak{p} \in \text{Ass}(M/L)$ and $\text{Ann}(L) \nsubseteq\mathfrak{p}$ implies $\mathfrak{p} \in \text{Ass}(M)$

Question

Let $L \subseteq M$ be $A$-modules and let $\mathfrak{p}$ be an associated prime of $M/L$ and assume that the annihilator of $L$ is not contained in $\mathfrak{p}$. How would one go about proving that $\mathfrak{p}$ is an associated prime of $M$?

Answer 1

Recall that $\mathfrak p$ being an associated prime of $M/L$ precisely means that $\mathfrak p$ is a prime ideal which is the annihilator of some element of $M/L$. If we think what this means in terms of the module $M$, it means that we can find an element $m \in M$ such that $\mathfrak p = \{a \in A \,| \, a m \in L\}.$

Now you want to replace $m$ by some other element $m' \in M$ with the property that $\mathfrak p = \{ a\in A \, | \, a m' = 0\},$ i.e. such that $\mathfrak p$ is the annihilator of $m'$. How do we find $m'$? Well, what we have to work with is the fact that $\mathfrak p$ does not contain the annihilator of $L$, so we may find $a \in A \setminus \mathfrak p$ such that $a L = 0.$

Do you see how to use $a$ and $m$ together to find the required $m'$?

P.S. From the wording of the question I wasn't sure if you wanted a hint or if you needed the complete solution, so to be on the safe side I'm giving a hint.