Let $R$ be a commutative ring and $M, N$ modules over $R$ with $N$ finitely presented and $M$ finitely generated. Assume $\mathrm{Hom}_R(N, M) = 0$. Does there exist an element $r \in R$ which is regular on $M$ and annihilates $N$?
This is true in the Noetherian case, as if there is no such element then we can find some associated prime $\mathfrak{p}$ of $M$ with $\mathrm{Ann}(N) \subseteq \mathfrak{p}$ by prime avoidance, then find a nonzero map $N_\mathfrak{p} \twoheadrightarrow \kappa(\mathfrak{p}) \hookrightarrow M_{\mathfrak{p}}$ and conclude $\mathrm{Hom}_{R_\mathfrak{p}}(N_\mathfrak{p}, M_{\mathfrak{p}}) \cong \mathrm{Hom}_R(N, M)_{\mathfrak{p}}$ is nonzero.
I observed that in the general case, $\mathrm{Hom}(N, M) = 0$ if and only if $(0 :_M \mathrm{Fit}_0(N)) = 0$. You can see this by taking any presentation of $N$ and applying left exactness of $\mathrm{Hom}(-, M)$ to it, then applying a generalized version of McCoy's lemma that says (if $M \neq 0$) an $n$ by $m$ matrix $A$ defines an injection $M^m \to M^n$ if and only if $m \leq n$ and $(0 :_M D_m(A)) = 0$, where $D_k(A)$ is the ideal of $k\times k$ minors of $A$. Additionally since $\sqrt{\mathrm{Fit}_0(N)} = \mathrm{Ann}(N)$ and powers of regular elements are regular, if the lemma is true in general we must be able to find an element regular on $M$ which lies in $\mathrm{Fit}_0(N)$. The equation $(0 :_M \mathrm{Fit}_0(N)) = 0$ says that the elements of $\mathrm{Fit}_0(N)$ are "jointly regular" on $M$, but I don't see how we could get a single regular element from this.
