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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.

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The answer is no. If $M = R$ and $N = R/I$ for some ideal $I \subseteq R$, then $\operatorname{Hom}_R(N, M) = 0 :_R I$ is the annihilator of $I$. It thus suffices to give a finitely generated ideal $I \subseteq R$ consisting of zerodivisors, such that $0 :_R I = 0$. The ideal $J$ in this answer is such an example.

Addendum: in general, $0 :_M \operatorname{ann}(n) = 0$ for all $n \in N$ implies $\operatorname{Hom}_R(N, M) = 0$. The converse does not hold though, e.g. when $R = M = \mathbb{Z}, N = \mathbb{Q}$. If $N$ is finitely generated, say $N = Rx_1 + \ldots + Rx_p$, then the following are equivalent:

  1. $0 :_M \operatorname{ann}(n) = 0$ for all $n \in N$
  2. $0 :_M \operatorname{ann}(x_i) = 0$ for all $i = 1, \ldots, p$
  3. $0 :_M \operatorname{ann}(N) = 0$.
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