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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?

Alphonse
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1 Answers1

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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 $p \neq q.$

Chris Leary
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  • Ok, nice, thank you very much! Vasconcelos' result is in "Injective endomorphisms of finitely generated modules" e.g. https://www.researchgate.net/profile/Wolmer_Vasconcelos/publication/264961063_Injective_Endomorphisms_of_Finitely_Generated_Modules/links/56423f1408aebaaea1f8d19c.pdf – Alphonse Feb 20 '21 at 16:54
  • @Alphonse - Yes, that's right. I should have added that any Artinian $R$-module has the property. Interestingly, Artinian modules don't have to be finitely generated, for example, the Prufer $p$-group $\mathbb{Z}(p^{\infty}),$ the $p$-component of $\mathbb{Q}/\mathbb{Z}.$ – Chris Leary Feb 20 '21 at 17:43
  • Hi Chris, in the case that $R$ is Artinian, $M$ can actually be any module, cf. problem 1(b) in this HW solution. – Sam Wong Jan 08 '23 at 07:15
  • @SamWong - Hi, Sam. I think problem 1(b) specifies that the module is Artinian. A field is certainly an Artinian ring, but a countable dimension vector space over it is not co-Hopfian. The linear transformation that is a "shift operator,"e.g., the one sending the vector $(x_1, x_2, \ldots)$ to $(0, x_1, x_2, \ldots)$, is injective but not surjective. – Chris Leary Jan 09 '23 at 21:11