Annihilator of a finitely presented module

Question

Let $R$ be a commutative ring. Suppose $M$ is a finitely presented $R$-module i.e. there is a short exact sequence
$$
R^m \to R^n \to M \to 0.
$$
Is it true that the annihilator ${\rm Ann}_R(M)=\{x\in R \mid xm=0 \text{ for all } m \in M\}$ is a finitely generated ideal?

Answer 1

Generally no. For domains, this happens for all f.p. modules precisely when $R$ is coherent. In general I suspect it is a subtle property.

Consider the following three conditions:

$(i)$ Intersections of finitely generated $R$-submodules of free $R$-modules are finitely generated.

$(ii)$ Annihilators of finitely presented $R$-modules are finitely generated.

$(iii)$ Intersections of finitely generated ideals of $R$ are finitely generated.

We check that $(i) \implies (ii) \implies (iii)$.

To see that $(i) \implies (ii)$, consider $M$ finitely presented and $0 \rightarrow K \subseteq A^n \twoheadrightarrow M \rightarrow 0$. Let $F_i$ denote the free submodule of $A^n$ generated by the $i$th coefficient. By assumption, $F_i \cap K$ is finitely generated, and since it is a submodule of $F_i$, which is naturally identified with $R$, we see that $F_i \cap K$ is naturally identified with a finitely generated ideal $I_i$. Moreover, $(K :_R F_i) = I_i$, and therefore $\operatorname{Ann}(M) = (K :_R R^n) = \bigcap I_i$ is finitely generated.

To see $(ii) \implies (iii)$, let $I, J$ be two finitely generated ideals. Set $M = R/I \oplus R/J$. Clearly $M$ is finitely presented and $\operatorname{Ann}(M) = I \cap J$.

Additional notes: if $M$ is a coherent module, then its annihilator is clearly finitely presented. Indeed, each generator $m_i$ generates a cyclic finitely presented module (even a coherent module). Then the canonical map $R \rightarrow \bigoplus m_iR$ sending $1$ to $(m_1, \ldots, m_n)$ is a homomorphism from a finite module to a coherent module, and therefore its kernel is finitely generated.[Stacks 05CW]. But the kernel of this map is just the $R$-annihilator of $M$.

In particular, if $R$ is a coherent ring, then $(ii)$ holds.

Recall that a ring is coherent iff (1) annihilators of finitely generated ideals are finitely generated and (2) intersections of finitely generated ideals are finitely generated. In particular, a domain is coherent iff $(2)$ holds.

Furthermore, if $R$ is any ring in which $R$-elements have finitely generated annihilators, then $(ii)$ holds in $R$ iff $R$ is coherent. In particular, this is true for a domain.

In general, I'm not sure how coherence and $(ii)$ are related. Generalized valuation rings always satisfy the property, regardless of coherence. See [Stacks 0ASP]