# example of a non-finitely generated co-Hopfian module over a commutative QF ring

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.

• QF-ring = quasi-Frobenius ring en.wikipedia.org/wiki/Quasi-Frobenius_ring; co-Hopfian means: isomorphic to none of its proper submodules. – YCor Dec 5 '18 at 1:10
• @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. – Chris Leary Dec 5 '18 at 3:15
• 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. – 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.