**Proof.**
Proof of (1). Since a PID is a Dedekind domain (Algebra, Lemma 10.120.15), it suffices to prove this for Dedekind domains. By Lemma 15.22.6 and Algebra, Lemma 10.39.18 it suffices to check the statement over $A_\mathfrak m$ for $\mathfrak m \subset A$ maximal. Since $A_\mathfrak m$ is a discrete valuation ring (Algebra, Lemma 10.120.17) we win by Lemma 15.22.10.

Proof of (2). Follows from Algebra, Lemma 10.78.2 and (1).

Proof of (3). Let $A$ be a PID and let $M$ be a finite torsion free module. By Lemma 15.22.7 we see that $M \subset A^{\oplus n}$ for some $n$. We argue that $M$ is free by induction on $n$. The case $n = 1$ expresses exactly the fact that $A$ is a PID. If $n > 1$ let $M' \subset R^{\oplus n - 1}$ be the image of the projection onto the last $n - 1$ summands of $R^{\oplus n}$. Then we obtain a short exact sequence $0 \to I \to M \to M' \to 0$ where $I$ is the intersection of $M$ with the first summand $R$ of $R^{\oplus n}$. By induction we see that $M$ is an extension of finite free $R$-modules, whence finite free.
$\square$

## Comments (2)

Comment #3540 by Laurent Moret-Bailly on

Comment #3672 by Johan on