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If $M$ and $N$ are two isomorphic torsion-free $R$-modules of finite rank, then such modules have the same rank and the same $p\text{-rank}$, where $p\text{-rank}(M) = \dim_k(M/pM)$ and $k = R/p$. But these two invariants are not complete and most torsion-free $R$-modules of finite rank are not of the form $K^n \oplus R^m$. If $M$ embeds into $N$, $\text{rank}(M) = \text{N}$ and $p\text{-rank}(M) = p\text{-rank}(N)$, them $M$ is quasi-isomorphic to $N$ [2, Proposition 1.3][4, Lemma 1].($M$ and $N$ are quasi-isomorphic if $M$ embeds into $N$ and $M/N$ is a torsion module bounded by a power of $p$).

The focus of [4] is the class of purely indecomposable modules (pi-modules), i.e., torsion-free indecomposable modules $M$ of finite rank with $p\text{-rank}(M) = 1$, or equivalently indecomposable pure $R$-submodules of $R^{\ast}$ of finite rank. For instance, it is shown in [4, Proposition 1], that the set of isomorphism classes of pi-modules is a partially ordered with the ascending chain condition, but not the descending one, and $R$ is its smallest element. Theorem 1 of [4] shows that the class of pi-modules is closed under direct summands and exhibit numerical invariants that determine an isomorphism class.

If $M$ and $N$ are two isomorphic torsion-free $R$-modules of finite rank, then such modules have the same rank and the same $p\text{-rank}$, where $p\text{-rank}(M) = \dim_k(M/pM)$ and $k = R/p$. But these two invariants are not complete and most torsion-free $R$-modules of finite rank are not of the form $K^n \oplus R^m$. If $M$ embeds into $N$, $\text{rank}(M) = \text{rank}(N)$ and $p\text{-rank}(M) = p\text{-rank}(N)$, them $M$ is quasi-isomorphic to $N$ [2, Proposition 1.3][4, Lemma 1].($M$ and $N$ are quasi-isomorphic if $M$ embeds into $N$ and $M/N$ is a torsion module bounded by a power of $p$).

The focus of [4] is the class of purely indecomposable modules (pi-modules), i.e., torsion-free indecomposable modules $M$ of finite rank with $p\text{-rank}(M) = 1$, or equivalently indecomposable pure $R$-submodules of $R^{\ast}$ of finite rank. For instance, it is shown in [4, Proposition 1], that the set of isomorphism classes of pi-modules is a partially ordered set with the ascending chain condition, but not the descending one, and $R$ is its smallest element. Theorem 1 of [4] shows that the class of pi-modules is closed under direct summands and exhibit numerical invariants that determine an isomorphism class.

Claim 2. Let $R$ be a DVR with maximal ideal $p$ and let $K$ be its fraction field. Then $\text{Ext}^1(K, R) = K^{\ast}/K$ where $K^{\ast}$ is the $p$-adic completion of $K$.

The proof of Claim 2 relies essentially on the computations $$\text{Hom}_R(K, K/R) \simeq K^{\ast}, \quad \text{Hom}_R(K/R, K/R) \simeq R^{\ast}.$$ The latter can be proved for any Noetherian local ring $R$ if $K/R$ is replaced by the injective hull of the residual field of $R$; it is a building block of Matlis duality theory [3, Theorem 18.6.iv]. The former relies on the latter and holds for any DVR. Its computation is very similar to this MO computation by YCor.

Proof of Claim 2. If $R$ is any domain, then $K$ is its injective hull. If $R$ is any Dedekind domain, then $K/R$ is an injective $R$ module. (If $(R, p, k)$ is a DVR with maximal ideal $p$ and residue field $k$, then $K/R$ is moreover the injective hull of $k$.) Thus for any Dedekind domain $R$, the exact sequence $Q$ $$ 0 \rightarrow R \rightarrow K \rightarrow K/R \rightarrow 0$$ is an injective resolution of $R$. Therefore $$\text{Ext}_R^1(K, R) = H_{-1}(\text{Hom}_R(K, Q)) = \text{Hom}_R(K, K/R)/\text{Hom}_R(K, K)$$ where we have identified $\text{Hom}_R(K, K) \simeq K$ with its image in $\text{Hom}_R(K, K/R)$. From now on, the ring $R$ is assumed to be a DVR. Let $G = \text{Hom}_R(K, K/R)$ and let $G_n \subseteq G$ the $R$-submodule of $G$ consisting of the homomorphisms which vanishes on $p^nR$. Then $G_n \simeq \text{Hom}_R(K/p^nR, K/R) \simeq \text{Hom}_R(K/R, K/R) \simeq R^{\ast}$, the latter isomorphism being given by [3, Theorem 18.6.iv]. Since we have $G = \bigcup_{n \ge 0} G_n$ and $pG_{n + 1} = G_n$, we deduce that $G\simeq K^{\ast}$.

Claim 2. Let $R$ be a DVR with maximal ideal $p$ and let $K$ be its fraction field. Then $\text{Ext}^1(K, R) = R^{\ast}/R$ where $R^{\ast}$ is the $p$-adic completion of $R$.

The proof of Claim 2 relies essentially on the computations $$\text{Hom}_R(K, K/R) \simeq K^{\ast}, \quad \text{Hom}_R(K/R, K/R) \simeq R^{\ast}$$ where $K^{\ast}$ is the $p$-adic completion of $K$. The latter can be proved for any Noetherian local ring $R$ if $K/R$ is replaced by the injective hull of the residual field of $R$; it is a building block of Matlis duality theory [3, Theorem 18.6.iv]. The former relies on the latter and holds for any DVR. Its computation is very similar to this MO computation by YCor.

Proof of Claim 2. If $R$ is any domain, then $K$ is its injective hull. If $R$ is any Dedekind domain, then $K/R$ is an injective $R$ module. (If $(R, p, k)$ is a DVR with maximal ideal $p$ and residue field $k$, then $K/R$ is moreover the injective hull of $k$.) Thus for any Dedekind domain $R$, the exact sequence $Q$ $$ 0 \rightarrow R \rightarrow K \rightarrow K/R \rightarrow 0$$ is an injective resolution of $R$. Therefore $$\text{Ext}_R^1(K, R) = H_{-1}(\text{Hom}_R(K, Q)) = \text{Hom}_R(K, K/R)/\text{Hom}_R(K, K)$$ where we have identified $\text{Hom}_R(K, K) \simeq K$ with its image in $\text{Hom}_R(K, K/R)$. From now on, the ring $R$ is assumed to be a DVR. Let $G = \text{Hom}_R(K, K/R)$ and let $G_n \subseteq G$ the $R$-submodule of $G$ consisting of the homomorphisms which vanishes on $p^nR$. Then $G_n \simeq \text{Hom}_R(K/p^nR, K/R) \simeq \text{Hom}_R(K/R, K/R) \simeq R^{\ast}$, the latter isomorphism being given by [3, Theorem 18.6.iv]. Since we have $G = \bigcup_{n \ge 0} G_n$ and $pG_{n + 1} = G_n$, we deduce that $G\simeq K^{\ast}$. It is not difficult to show that $K^{\ast}/K \simeq R^{\ast}/R$. Indeed, the kernel of the map induced by the inclusion $R^{\ast} \rightarrow K^{\ast}/K$ is $R^{\ast} \cap K = R$ while $K^{\ast} = R^{\ast} + K$ follows from the density of $R$ in $R^{\ast}$.

The proof of Claim 2 relies essentially on the computations $$\text{Hom}_R(K, K/R) \simeq K^{\ast}, \quad \text{Hom}_R(K/R, K/R) \simeq R^{\ast}.$$ The latter can be proved for any Noetherian local ring $R$ ideal $K/R$ is replaces by the injective hull of the residual field of $R$; it is a building block of Matlis duality theory [3, Theorem 18.6.iv]. The former relies on the latter and holds for any DVR. Its computation is very similar to this MO computation by YCor.

The proof of Claim 2 relies essentially on the computations $$\text{Hom}_R(K, K/R) \simeq K^{\ast}, \quad \text{Hom}_R(K/R, K/R) \simeq R^{\ast}.$$ The latter can be proved for any Noetherian local ring $R$ if $K/R$ is replaced by the injective hull of the residual field of $R$; it is a building block of Matlis duality theory [3, Theorem 18.6.iv]. The former relies on the latter and holds for any DVR. Its computation is very similar to this MO computation by YCor.