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Let $R$ be an Henselian discrete valuation ring with field of fractions $K$. Let $M$ be a torsion-free $R$-module of finite rank (i.e. $dim_K(M\otimes_RK)<+\infty$). Let $D$ be the maximal divisible $R$-submodule of $M$, then $M$ is said to be reduced if $D=0$. If I am not wrong if $M$ is reduced, of finite rank and torsion free, then $M$ is free when $R$ is complete. Is the same true if $R$ is only Henselian? What if $R$ is any discrete valuation ring?

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in fact when $R$ is complete, $M$ must even be finitely generated. – user91132 Feb 14 '12 at 8:36
Sorry, when I say that I want to prove that $M$ is free I mean "free and finitely generated". Is that what you mean? – Federigo Feb 14 '12 at 8:52
No. "Free" means a direct sum of (possibly infinitely many) copies of $R$. I meant that if your ring $R$ is complete, then not only is it true that any $M$ satisfying your conditions is free, but in fact it must also be finitely generated and free, i.e. a direct sum of finitely many copies of $R$. Given $M$ is free (this is not totally trivial), that $M$ is finitely generated follows trivially from the assumption that your $M$ has finite rank. Anyway, a-fortiori gave you exactly the counterexample you need below. – user91132 Feb 14 '12 at 9:10
Yes yes...that's what I meant!!! So I understood clearly from the beginning and that's what my question was asking even if not specified, thank you! The book in the library has been taken by someone else ... I have to wait ... – Federigo Feb 14 '12 at 9:44
up vote 5 down vote accepted

Theorem 19 in Kaplansky's Infinite Abelian Groups gives an example of a torsion-free, reduced, indecomposable rank 2 module for any incomplete discrete valuation ring.

In short, the construction is as follows: choose $\lambda\in\hat R\setminus R$. This induces a homomorphism $\tilde\lambda\colon K\to\hat R/R$. The short exact sequence $0\to R\to\hat R\to\hat R/R\to 0$ induces an injection $\mathrm{Hom}(K,\hat R/R)\to\mathrm{Ext}^1(K,R)$. The image of $\tilde\lambda$ under this injection corresponds to the desired rank 2 module $M$ sitting in a non-split extension $0\to R\to M\to K\to 0$. Any divisible element would induce a splitting, so $M$ is reduced. Since there is no surjection $R^2\to K$, the module $M$ cannot be free.

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Do you mean example of a torsion-free, reduced, indecomposable rank 2 module which is "not free"? Anyway... thank you, I'll check on this book. That's probably the reference I needed. – Federigo Feb 14 '12 at 8:37
Again, when I say that I want to prove that $M$ is free I mean "free and finitely generated". Sorry, but I think you understood. Ok, let me go to the library... – Federigo Feb 14 '12 at 8:46
Indeomposable of rank 2...yeah... I understand :-) – Federigo Feb 14 '12 at 10:06

Let $A$ be a local ring. Let $M$ be a finitely generated flat $A$-module. Then $M$ is free; see Theorem 1.2.16 in

That should answer your question.

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I am not assuming $M$ to be finitely generated, but only of finite rank. – Federigo Feb 14 '12 at 7:57
But your module is also torsion-free.. – Harized Feb 14 '12 at 8:25
I think the proof of Thm 1.2.16 can be altered slightly to make it work in your case. – Harized Feb 14 '12 at 8:27
When I say that I want to prove that $M$ is free I mean "free and finitely generated". Sorry. – Federigo Feb 14 '12 at 8:45

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