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Fix a unital commutative ring $R$ and consider a left $R$-module $M$. $\newcommand{\End}{{\rm End}}$

(For the indirect application I have in mind, which would require another post, $\End_R(M)$ will never be simple.)

Suppose the following is true:

whenever $N$ is a left $R$-module and $\phi: \End_R(M)\to \End_R(N)$ is a surjective $R$-algebra homomorphism, then $\phi$ is injective.

(edit: thanks to Yves Cornulier for pointing out I should have said $N\neq 0$)

What can we say about $M$? In particular:

Q1. What examples are there of $M$ which satisfy this property, which are not finitely generated as $R$-modules?

The condition being imposed here is reminiscent both of Hopfian rings and Hopfian modules, but at present I don't see a way to reduce the condition I'm considering to either of these notions.

Q2. Is there an existing name for the condition I've just described above?

In my own rough notes I'm calling it "endo-Hopfian" but that seems unsatisfactory on several levels.

Update: In comments, Jeremy Rickard has pointed out that when $R$ is a field and $M$ is an $R$-vector space of countably-infinite dimension then $M$ satisfies the requirements of Q1. In some sense this is related to the indirect application which motivates this question. However, in that (functional-analytic) setting, the right analogy would somehow be a ring $R$ which does not have global dimension zero. (In particular, in my setting not all epis in $R$-Mod split.) I still want $\End_R(M)$ non-simple.

Let me try to turn Q1 into a more specific but speculative/wild question:

Q1b. If $M$ has the property described and is not finitely generated, and if $\End_R(M)$ is not simple, can we deduce anything about the (global) homological dimension of $R$? Or the homological dimension of $M$ in $R$-Mod?

It could be that the answer to Q1b is no for trivial reasons, in the sense that we can always cook up examples by extending a given $R$ in some way.

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  • $\begingroup$ You parenthetically mention the extra condition that $\text{End}_R(M)$ is not simple. Do you want answers to Q1 to satisfy this condition? If not, there are examples like $R=\mathbb{Z}$, $M=\mathbb{Q}$. $\endgroup$
    – Jeremy Rickard
    Commented Mar 27, 2018 at 8:01
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    $\begingroup$ An example where ${\rm End}_R(M)$ is not simple is $R$ a field and $M$ a countable dimensional vector space. $\endgroup$
    – Jeremy Rickard
    Commented Mar 27, 2018 at 8:03
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    $\begingroup$ By the way, “Hopfian” usually means that an epimorphism from the object to itself must be an isomorphism, which isn’t what you’re asking about. In particular, ${\rm End}_R(M)$ being Hopfian as an $R$-algebra doesn’t imply that $M$ satisfies your condition. $\endgroup$
    – Jeremy Rickard
    Commented Mar 27, 2018 at 8:37
  • $\begingroup$ @JeremyRickard Oops - you're quite right. Well, I've changed the title to something less misleading, and edited the text a little. And yes, ideally I want examples where End_R(M) is non-simple, just to get a flavour for what may or may not be going on in the functional-analytic problem that motivates this question $\endgroup$
    – Yemon Choi
    Commented Mar 27, 2018 at 23:33
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    $\begingroup$ For Q1b you could just replace $R$ with $R[x]/(x^2)$ in my example, with $x$ annihilating $M$. Then $M$ has infinite projective dimension. $\endgroup$
    – Jeremy Rickard
    Commented Mar 28, 2018 at 15:02

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