**Proof.**
Proof of (1). Let $M \subset F$ be a finite submodule of a free module $F$. Since $M$ is finite, we may assume $F$ is a finite free module (details omitted). Say $F = R^{\oplus n}$. We argue by induction on $n$. If $n = 1$, then $M$ is a finitely generated ideal, hence principal by our assumption that $R$ is Bézout. If $n > 1$, then we consider the image $I$ of $M$ under the projection $R^{\oplus n} \to R$ onto the last summand. If $I = (0)$, then $M \subset R^{\oplus n - 1}$ and we are done by induction. If $I \not= 0$, then $I = (f) \cong R$. Hence $M \cong R \oplus \mathop{\mathrm{Ker}}(M \to I)$ and we are done by induction as well.

Let $M$ be a finitely presented $R$-module. Since the localizations of $R$ are maximal ideals are valuation rings (Lemma 15.121.7) we may apply Lemma 15.121.4. Thus $M$ is a summand of a module of the form $R^{\oplus r} \oplus \bigoplus _{i = 1, \ldots , n} R/f_ iR$ with $f_ i \not= 0$. Since taking the torsion submodule is a functor we see that $M_{tors}$ is a summand of the module $\bigoplus _{i = 1, \ldots , n} R/f_ iR$ and $M/M_{tors}$ is a summand of $R^{\oplus r}$. By the first part of the proof we see that $M/M_{tors}$ is finite free. Hence $M \cong M_{tors} \oplus M/M_{tors}$ as desired.
$\square$

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