Are coherent modules hopfian?

Question

As it is well-known, a noetherian module over an arbitrary ring is hopfian.

Are coherent modules also hopfian?

Answer 1

You can even find a ring $R$ such that $_RR$ is coherent but not Hopfian. (Keep in mind for rings being Hopfian amounts to being Dedekind finite.)

One such example is the endomorphism ring of an infinite dimensional vector space. It's left coherent because it's left semihereditary. It's well-known to be non-Dedekind finite.

Answer 2

Yes, if your base ring is commutative. More generally, if $R$ is commutative then any finitely generated $R$-module $M$ is hopfian (in particular, so is any coherent module).

Indeed, let $u:M\to M$ be a surjective $R$-linear map. Then $(P,x)\in R[X]\times M\mapsto P(u)(x)\in M$ endows $M$ with the structure of a finitely generated $R[X]$-module.

Since $u$ is surjective, it is easy to see that $R[X]X\cdot M=M$. Since $M$ is finitely generated, Nakayama's lemma implies that there is $Q\in R[X]$ such that $(1+XQ)\cdot M=0$, meaning that $x+(u\circ Q(u)) (x)=0$ for all $x\in M$. Now, if $x\in\ker(u)$, then $(u\circ Q(u))(x)=(Q(u)\circ u)(x)=0$, hence $x=0$ and $u$ is injective.