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?
Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)
|
1
|
|||||||||||||||
|
|
5
|
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. |
|||||||||
|
You can accept an answer to one of your own questions by clicking the check mark next to it. This awards 15 reputation points to the person who answered and 2 reputation points to you.
|
0
|
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 http://ukcatalogue.oup.com/product/9780198502845.do That should answer your question. |
||||||||||||
|
