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.

| cite | improve this question | | | | |
  • 1
    $\begingroup$ 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

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.

| cite | improve this answer | | | | |
  • 1
    $\begingroup$ Thanks, Jeremy, this helps quite a bit. $\endgroup$ – Chris Leary Dec 5 '18 at 17:52

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.