If $A$ is a finitely generated $R$-module, is $\operatorname{Hom}_R(A,R)$ finitely generated?

Question

Let $R$ be an utterly arbitrary commutative, unital ring. Let $A$ be a finitely generated $R$-module.

Is $\operatorname{Hom}_R(A,R)$ finitely generated as an $R$-module?

Intuitively and based on the familiar cases, it seems to me it should indeed always be, but I'm not seeing the proof right now. If $A$ is free, it is generated (finitely) by the dual basis to a basis for $A$. But if $A$ is not free, there is no dual basis, so how do we get the result? (Or is it false?)

Thanks in advance!

Edit: This question seems to have been asked and answered before, in a more general setting, here. I find Hanno's answer useful in addition to the accepted answer at the linked question, however, since it includes an explanation of the motivation behind the proffered counterexample. I am not sure how to proceed given this.

Answer 1

No: For a local ring $(R,{\mathfrak m})$ with ${\mathfrak m}^2=0$ you have $\text{Hom}_R(R/{\mathfrak m},R)\cong\{x\in R\ |\ {\mathfrak m}x=0\}={\mathfrak m}$, so it suffices to choose $R$ such that ${\mathfrak m}$ is not finitely generated, e.g. $R := {\mathbb k}[x_1,x_2,\ldots]/(x_i^2, x_i x_j)$.