What is the structure of a finite rank torsion-free module over a DVR?

Question

Let $R$ be a DVR, $Q$ its quotient field, and suppose that $M$ is a finite rank torsion-free $R$-module, i.e. $M \otimes _R Q$ is a finite dimensional $Q$-vector space. In a posting on Physics Forum it was claimed that $M \cong R^m \oplus Q^n$ (presumably with $m+n= \mathrm{rank} M).$ Is this claim true? If it is, could someone please steer me toward a reference? I have been unable to locate such a result so far.

Answer 1

This is false. For instance, let $R$ be any DVR that is not complete and fix an element $x\in\hat{R}\setminus R$ (where $\hat{R}$ is the completion of $R$). Let $$M=\{(a,b)\in Q^2:ax+b\in\hat{R}\}.$$ Note that $R^2\subseteq M\subseteq Q^2$ so $M$ has rank $2$. Also no nonzero element of $M$ is infinitely divisible by the uniformizer since no nonzero element of $\hat{R}$ is (and $ax+b$ cannot be $0$ unless $a=b=0$ since $x\not\in R$). So, $M$ cannot have $Q$ as a summand. However, the first projection $M\to Q$ is surjective (for any $a\in Q$, there is $b\in Q$ such that $ax+b\in\hat{R}$ since we can find an element of $R$ which agrees with $x$ mod the denominator of $a$). In particular, this implies $M$ is not finitely generated, so cannot be isomorphic to $R^2$.

It is true if you additionally assume $R$ is complete. You can prove this by induction on the rank of $M$. Let $p$ be a uniformizer for $R$. If $M$ is $p$-divisible then it is a $Q$-vector space and we're done. Otherwise we can pick an element $x\in M$ which is not divisible by $p$. Then $M/Rx$ is still torsion-free, and has lower rank. So by induction, $M/Rx\cong R^m\oplus Q^n$ for some $m$ and $n$. To prove $M$ has the same form, it suffices to show the short exact sequence $$0\to Rx\to M\to M/Rx\to 0$$ splits, since $Rx\cong R$. This extension is classified by an element of $\operatorname{Ext}^1(R^m\oplus Q^n,R)\cong \operatorname{Ext}^1(Q,R)^n$ and so it suffices to show $\operatorname{Ext}^1(Q,R)=0$.

To prove this, consider the free resolution $$0\to R^{\oplus\mathbb{N}} \stackrel{f}\to R^{\oplus\mathbb{N}}\stackrel{g}\to Q\to 0$$ where $g$ sends the $i$th generator $e_i$ to $p^{-i}$ and the $f$ sends the $i$th generator to $e_i-pe_{i+1}$. Then $\operatorname{Ext}^1(Q,R)$ is the cokernel of the dual map $$f^*:R^\mathbb{N}\to R^\mathbb{N}$$ which is given by $$f^*(a_0,a_1,a_2,\dots)=(a_0-pa_1,a_1-pa_2,a_2-pa_3,\dots).$$ To show this is surjective, we need to know that for any $(b_0,b_1,b_2,\dots)\in R^\mathbb{N}$, we can solve for $a_0,a_1,a_2,\dots$ such that $$a_0=b_0+pa_1,\quad a_1=b_1+pa_2,\quad a_2=b_2+pa_3,\quad\dots$$ We can solve for $a_0$ by repeatedly substituting these equations, giving $$a_0=b_0+pb_1+p^2b_2+\dots$$ which gives a well-defined element of $R$ since $R$ is complete. Similarly we get $$a_1=b_1+pb_2+p^2b_3+\dots$$ and so on and it is easy to check that these $a_i$ do satisfy $a_i=b_i+pa_{i+1}$ for each $i$. Thus $f^*$ is surjective and $\operatorname{Ext}^1(Q,R)=0$.

(In fact, this calculation can be generalized to show that $\operatorname{Ext}^1(Q,R)\cong \hat{R}/R$ even if $R$ is not complete. The extension corresponding to an element of $\hat{R}$ (mod $R$) is exactly the example constructed above. So, in some sense, that is essentially the only example.)