**Proof.**
(For the parenthetical remark in the statement of the lemma, see Algebra, Lemma 10.120.15.) Proof of (1). 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 (4)

Comment #3540 by Laurent Moret-Bailly on

Comment #3672 by Johan on

Comment #6769 by Oliver Roendigs on

Comment #6931 by Johan on

There are also: