Bourbaki

Proposition 1

Tag 051P, permanent. Cite it as [Bourbaki, Tag 051P].

Commutative Algebra, chapter VII, § 3. Factorial domains

Let A be a Krull domain. If every divisorial ideal of A is invertible, then, for every maximal ideal m of A, AmA_m is factorial. The converse is true if it is also assumed that every divisorial ideal of A is finitely generated (in particular if A is Noetherian).

Suppose that every divisorial ideal of AA is invertible; as A,,A_{,,} is a Krull domain (§1§ 1, no. 4, Proposition 6), every divisorial ideal aa of A,,A_{,,} is the intersection of two principal fractional ideals (§1§ 1, no. 5, Corollary 2 to Proposition 9); hence a=bAma = bA_m, where bb is a divisorial ideal of AA (Chapter II, §2§ 2, no. 4); as bb is invertible by hypothesis, we deduce from Chapter II, §5§ 5, no. 6, Theorem 4 that aa is principal and hence A,,A_{,,} is a factorial domain (no. 1, Definition 1). Conversely, if all the A,,A_{,,} are factorial and cc is a finitely generated divisorial ideal of AA, cAmcA_m is a divisorial ideal of A,,A_{,,}, as follows from §1§ 1, no. 5, Corollary 2 to Proposition 9 and Chapter II, §2§ 2, no. 4; by hypothesis cAmcA_m is principal and hence it follows from Chapter 11, §5§ 5, no. 6, Theorem 4 that cc is invertible.

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

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