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.



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


