2

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

Are coherent modules also hopfian?

user26857
  • 53,190
Mariano Suárez-Álvarez
  • 139,300
  • That finitely generated modules over a commutative ring are Hopfian is so well-known (and well-covered on the site) I am wondering if you had not meant to stress the general case. Was it your intention? – rschwieb May 30 '22 at 20:43
  • 1
    By saying arbitrary ring and being quite explicit about coherence I thought I had stressed sufficiently much, to be honest! :-) – Mariano Suárez-Álvarez May 30 '22 at 22:07
  • nevertheless it attracted a commutative answer! Guess you can’t win. – rschwieb May 31 '22 at 00:06

2 Answers2

2

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.

rschwieb
  • 160,592
-1

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.

GreginGre
  • 16,641