Let $(R,\mathfrak m)$ be a commutative Noetherian local ring and $M$ be an $R$-module such that $M \neq \mathfrak m M$ and $\mathrm{Hom}_R(M,R)\neq 0$. Then, is it true that $\mathrm{Hom}_R(M,R)\neq \mathfrak m \mathrm{Hom}_R(M,R)$ ?
If $M$ is finitely generated, this is of course just Nakayama's lemma, but what if $M$ is not finitely generated?
$\DeclareMathOperator\Hom{Hom}$Yes. Choose $f\neq 0$ in $\Hom_R(M,R)$. Choose $v\in M$ such that $f(v)\neq 0$. Let $v^\ast:\Hom_R(M,R)\to R$ be given by $v^\ast(g)= g(v)$. Since the image of $v^\ast $ is a non-zero ideal $I\subset R$, there is a non-zero module map $I\to R/\frak m$. The composite of this with the surjection $\Hom_R(M,R)\to I$ given by $v^\ast$ is a non-zero map $\Hom_R(M,R)\to R/\frak m$.
