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? 



Is a reduced, torsion-free module of finite rank over an Henselian ring free?
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.


