Let $R$ be a commutative ring with unity. Is it possible to characterize those $R$-modules $M$ such that for every positive integer $n$ and every submodule $N$ of $M^n$ , we have $Supp (M^n/N)=V(Ann (M^n /N))$ ?
I know that the above is true when $M$ is finitely generated, but I don't know whether the condition I have forces $M$ to be finitely generated or not, or can we give a nice characterization for all such $M$ (probably assuming some condition on $R$ ).
Note here that by $Supp (.)$ I mean the support of a module (https://en.wikipedia.org/wiki/Support_of_a_module) and $V(I)$ denotes the set of all prime ideals containing $I$.
This is just an explanation of my comment, which is too long for a comment. Let me do this for $n=2$, assuming the statement for $n=1$, rest is by an easy induction.
If $N\subset M^2$, let $N_1=M\cap N$, where $M$ is the first factor and let $N_2$ be the image of $N$ under the second projection. So, you have an exact sequence, $0\to M/N_1\to M^2/N\to M/N_2\to 0$. Since localization is exact, one has support of $M^2/N$ equal to the union of supports of $M/N_1, M/N_2$. By $n=1$ case, this is the union of $V(I_1), V(I_2)$ where$I_i$ is the annihilator of $M/N_i$. If $I$ is the annihilator of $M^2/N$, one has $I_1I_2\subset I\subset I_1\cap I_2$ and $(I_1\cap I_2)^2\subset I_1I_2$ and thus $V(I_1)\cup V(I_2)=V(I_1\cap I_2)=V(I)$, proving what you need.