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Luc Guyot
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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 $m \le n$ and $k$.

Proof. 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 $K$, the image of any homomorphism from $N$ to $K$ is free of rank at most $1$. Thus $N$ embeds into $R^n$. In particular, $N$ is finitely generated and torsion free, hence it is free of finite rank.

Luc Guyot
  • 7.9k
  • 2
  • 30
  • 51