Can anyone provide an example of such a module? or show that no such module exists? For semisimple rings, we have coHopfian if and only if finitely generated. Perhaps the fact that QF rings (commutative) are finite products of local artinians (with simple socle) has something to lend to the situation.

1$\begingroup$ QFring = quasiFrobenius ring en.wikipedia.org/wiki/QuasiFrobenius_ring; coHopfian means: isomorphic to none of its proper submodules. $\endgroup$ – YCor Dec 5 '18 at 1:10

$\begingroup$ @YCor  Yes, that is correct. Vasconcelos showed that a commutative ring with identity has all finitely generated modules coHopfian if and only if it is 0dimensional. I am trying to show which 0dimensional Noetherian rings have the property that only the finitely generated modules are coHopfian. I have shown that this is true if the ring is semisimple (artinian). I'm trying to work up the ladder, so to speak. QF rings seemed a natural next choice since these rings have nice duality properties as well. The local case is 0dimensional Gorenstein, which are interesting in their own right. $\endgroup$ – Chris Leary Dec 5 '18 at 3:15

$\begingroup$ I believe I have an argument that covers the local case just mentioned, and I would like to know whether I can proceed with the general case, or discover an obstruction to it. $\endgroup$ – Chris Leary Dec 5 '18 at 3:18
Let $R$ be the fourdimensional algebra $k[x,y]/(x^2,y^2)$, where $k$ is an infinite field.
For each $\lambda\in k$, $M_\lambda=R/(x\lambda y)R$ is a twodimensional module with onedimensional radical. If $\lambda\neq\mu$, then every homomorphism $M_\lambda\to M_\mu$ has image contained in $\text{rad}(M_\mu)$ and kernel containing $\text{rad}(M_\lambda)$.
Since $k$ is infinite, $M=\bigoplus_{\lambda\in k}M_\lambda$ is not finitely generated, but I claim that it is coHopfian.
Suppose $\alpha$ is an injective endomorphism of $M$. In order that $\ker(\alpha)$ does not contain $\text{rad}(M_\lambda)$, the component $\alpha_{\lambda\lambda}:M_\lambda\to M_\lambda$ must be an isomorphism, for every $\lambda$. But the component $\alpha_{\mu\lambda}:M_\lambda\to M_\mu$ maps into $\text{rad}(M_\mu)$, so the map $M/\text{rad}(M)\to M/\text{rad}(M)$ induced by $\alpha$ is an isomorphism, and so $\alpha$ is surjective.

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