When does injective implies surjective for an $R$-module endomorphism? Let $R$ be a commutative ring.
It is known that if $M$ is a finitely generated $R$-module and if $f \in End_R(M)$, then surjectivity of $f$ implies injectivity of $f$ (Vasconcelos).
I am looking for sufficient conditions on $R$ or $M$ to have a converse to this statement.
For instance, if $R$ is a field or if $R$ is a finite commutative ring, this will be true (in the former case, because $M$ is a finite dimensional vector space over $R$, and in the latter case, because $|M| < \infty$).
If $M$ is a free module of finite rank, say $M \cong R^n$, then $f$ is just a matrix $A$. Injectivity is equivalent to the columns of $A$ to be linearly equivalent over $R$. By Exercise 5.23B in Exercises in Modules and Rings by T. Y. Lam, this is equivalent to $\det(A) \in R$ not being a zero divisor. If all such elements of $R$ are invertible, then $A$ and thus $f$ are invertible as well. By this question, all artinian rings $R$ work (and when $M$ is free of finite rank).
What are some other conditions?
 A: If $R$ is Artinian and $M$ is finitely generated, then any injective $R$-linear map $M \rightarrow M$ is surjective. Vasconcelos has also shown that all finitely generated $R$-modules have this property if and only if $R$ has Krull dimension $0.$
You can prove a nice theorem for finite rank free modules: $R^n$ has the property that all injective endomorphisms are surjective if and only if $R$ is a quoring (that is, all regular elements are invertible). Seehere. You have part of the argument already in your question.
Modules $M$ with the property you mention are called co-Hopfian. There is an extensive literature on them, and Lam has some exercises around them. It is worth noting that you can work with $M$ that are not finitely generatd as well. For intance, the abelian group $\oplus \mathbb{Z}(p),$ the sum over the positive primes with $\mathbb{Z}(p)$ the cyclic group of order $p,$ has this property. The property follows since there are no nonzero homomorphism between $\mathbb{Z}(p)$ and $\mathbb{Z}(q)$ if $p \neq q.$
