The Problem: Suppose $R$ is an integral domain, $M$ is an $R$-module. Suppose $N$ is a submodule of $M$. Let $\operatorname{Ann}(M)$ denote the annihilator of $M$ in $R$ (note that annihilators are ideals). Give an example where $\operatorname{Ann}(M/N) \, \operatorname{Ann}(N) \neq \operatorname{Ann}(M)$, where $\operatorname{Ann}(M/N) \, \operatorname{Ann}(N)$ is the set of all finite sums of elements of the form $ab$ with $a\in \operatorname{Ann}(M/N)$ and $b\in \operatorname{Ann}(N)$.
Note: I understand that $\operatorname{Ann}(M/N) \, \operatorname{Ann}(N) \subseteq \operatorname{Ann}(M)$, but what is a counterexample where the strict containment holds? Any help would be greatly appreciated.
It is not hard to check that $\operatorname{Ann}(M) \subseteq \operatorname{Ann}(M/N) \cap \operatorname{Ann}(N)$. Now we have a chain of two containments:
$$\operatorname{Ann}(M/N) \, \operatorname{Ann}(N) \subseteq \operatorname{Ann}(M) \subseteq \operatorname{Ann}(M/N) \cap \operatorname{Ann}(N).$$
We can find a counterexample if we can arrange for the intersection $\operatorname{Ann}(M/N) \cap \operatorname{Ann}(N)$ to strictly contain the product $\operatorname{Ann}(M/N) \, \operatorname{Ann}(N)$, and for the equality $\operatorname{Ann}(M) = \operatorname{Ann}(M/N) \cap \operatorname{Ann}(N)$. So let's find a ring $R$ and an ideal $I$ such that $I^2 \neq I$. Then if we can come up with an $R$-module $M$ and a submodule $N$ such that $\operatorname{Ann}(M) = \operatorname{Ann}(N) = \operatorname{Ann}(M/N) = I$, we will have it.
For instance, let $R=\mathbb{Z}$, let $F=\mathbb{Z}/p\mathbb{Z}$ be the integers mod $p$ for any prime $p$, let $N = F[x]$ be a polynomial ring in one variable over $F$, and let $M=F(x)$ be the field of rational functions in one variable over $F$. Since $M$ and $N$ are Abelian groups they are $\mathbb{Z}$-modules. The annihilator of both $M$ and $N$ is $p\mathbb{Z}$. Since $\operatorname{Ann}(M) \subseteq \operatorname{Ann}(M/N)$ and $\operatorname{Ann}(M)$ is maximal, the annihilator of $M/N$ is either equal to $\operatorname{Ann}(M)$ or equal to all of $\mathbb{Z}$. But since $M/N$ is not the zero module, its annihilator is not all of $\mathbb{Z}$. Thus $\operatorname{Ann}(M) = \operatorname{Ann}(N) = \operatorname{Ann}(M/N) = p\mathbb{Z}$. Finally observe that $\operatorname{Ann}(M/N) \, \operatorname{Ann}(N) = p^2\mathbb{Z}$ is a proper subset of $\operatorname{Ann}(M)$.