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Can anyone provide an example of such a module? or show that no such module exists? For semisimple rings, we have co-Hopfian 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.

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    $\begingroup$ QF-ring = quasi-Frobenius ring en.wikipedia.org/wiki/Quasi-Frobenius_ring; co-Hopfian 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 co-Hopfian if and only if it is 0-dimensional. I am trying to show which 0-dimensional Noetherian rings have the property that only the finitely generated modules are co-Hopfian. 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 0-dimensional 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
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Let $R$ be the four-dimensional 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 two-dimensional module with one-dimensional 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 co-Hopfian.

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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    $\begingroup$ Thanks, Jeremy, this helps quite a bit. $\endgroup$ – Chris Leary Dec 5 '18 at 17:52

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