Bourbaki

Lemma 1

Tag 04TM, permanent. Cite it as [Bourbaki, Tag 04TM].

Commutative Algebra, chapter VI, § 3. Valuations

Let AA be a valuation ring. Every finitely generated torsion-free AA-module is free. Every finitely generated ideal a\mathfrak{a} of AA is principal. Every torsion-free AA-module is flat.

Let EE be a finitely generated torsion-free AA-module and let x1,…,xnx_1, \ldots, x_n be generators of EE which are minimal in number; we show that they are linearly independent. If ∑i=1naixi=0\sum_{i=1}^n a_i x_i = 0 (ai∈Aa_i \in A) is a non-trivial relation between the xix_i, one of the aia_i, say a1a_1, divides all the others since the set of principal ideals of AA is totally ordered by inclusion (§ 1, no. 2, Theorem 1); then a1≠0a_1 \neq 0 since the relation is non-trivial. As EE is torsion-free, we can divide by a1a_1, which amounts to assuming that a1=1a_1 = 1. But then x1x_1 is a linear combination of x2,…,xnx_2, \ldots, x_n, contrary to the minimal character of nn. Hence EE is free.

In particular every finitely generated ideal a\mathfrak{a} of AA is principal, all the elements of a system of generators of a\mathfrak{a} being multiples of one of them. Proposition 3 of Chapter I, § 2, no. 4 then shows that every torsion-free AA-module is flat.

Bourbaki puts the proof directly after the statement, so what is above is the statement and its proof together.

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