**Proof.**
By Algebra, Lemma 10.157.4 we see that $R$ satisfies $(R_1)$ and $(S_2)$.

Assume (1). Then $M$ is torsion free by Lemma 15.23.2 and satisfies $(S_2)$ by Lemma 15.23.16. Thus (2) holds.

Assume (2). By definition $M' = \bigcap _{\text{height}(\mathfrak p) = 1} M_{\mathfrak p}$ is the kernel of the map

\[ M_ K \longrightarrow \bigoplus \nolimits _{\text{height}(\mathfrak p) = 1} M_ K/M_\mathfrak p \subset \prod \nolimits _{\text{height}(\mathfrak p) = 1} M_ K/M_\mathfrak p \]

Observe that our map indeed factors through the direct sum as indicated since given $a/b \in K$ there are at most finitely many height $1$ primes $\mathfrak p$ with $b \in \mathfrak p$. Let $\mathfrak p_0$ be a prime of height $1$. Then $(M_ K/M_\mathfrak p)_{\mathfrak p_0} = 0$ unless $\mathfrak p = \mathfrak p_0$ in which case we get $(M_ K/M_\mathfrak p)_{\mathfrak p_0} = M_ K/M_{\mathfrak p_0}$. Thus by exactness of localization and the fact that localization commutes with direct sums, we see that $M'_{\mathfrak p_0} = M_{\mathfrak p_0}$. Since $M$ has depth $\geq 2$ at primes of height $> 1$, we see that $M \to M'$ is an isomorphism by Lemma 15.23.14. Hence (3) holds.

Assume (3). Let $\mathfrak p$ be a prime of height $1$. Then $R_\mathfrak p$ is a discrete valuation ring by $(R_1)$. By Lemma 15.22.11 we see that $M_\mathfrak p$ is finite free, in particular reflexive. Hence the map $M \to M^{**}$ induces an isomorphism at all the primes $\mathfrak p$ of height $1$. Thus the condition $M = \bigcap _{\text{height}(\mathfrak p) = 1} M_{\mathfrak p}$ implies that $M = M^{**}$ and (1) holds.
$\square$

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