Claim. Let $R$ be DVR and let $K$ be its fraction field. An $R$-submodule of $K^n$ is isomorphic to an $R$-module the form $K^m \oplus R^k$ for some non-negative integers $m$ and $k$ such that $m + k \le n$.
We will make use of
Lemma. Let $R$ be a DVR and let $K$ be its fraction field. Let $M \subseteq K^n$ be an $R$-submodule. Then the following are equivalent:
- $M$ contains a $K$-subspace of positive dimension.
- There is a surjective $R$-homomorphism from $M$ onto $K$.
- $M$ is not finitely generated.
Proof. Since a $K$-vector space is an injective $R$-module, hence a direct summand of any of its containing module, we infer that $(1)$ implies $(2)$. Since $K$ is not finitely generated as an $R$-module, we also infer that $(2)$ implies $(3)$. Assume no that $M$ is not finitely generated. Then some of its images under the projection maps $K^n \twoheadrightarrow K$ must be $K$ since otherwise $M$ would be embed into $R^n$. Therefore $(1)$ holds.
We are now in position to prove the claim.
Proof of the claim. Let $M$ be a submodule of $K^n$ and let $m \ge 0$ be the largest integer such that $M$ contains a $K$-subspace of $K^n$ of dimension $m$. Since $K^m$ is an injective $R$-module, it is a direct summand of $M$. Therefore we can write $M = K^m \oplus N$ for some $R$-submodule $N$ of $K^n$. As $N$ doesn't contain any $R$-submodule isomorphic to $K$, it is finitely generated by the above lemma. Because $N$ is torsion free, it is free of finite rank.