# Does Hom(N, M) = 0 imply the existence of an M-regular element in Ann(N) in the non-noetherian case? [duplicate]

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.

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. $$0 :_M \operatorname{ann}(N) = 0$$.