Corollary

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

Commutative Algebra, chapter VII, § 1. Krull domains

Let $A$ be a completely integrally closed domain and $a$ a divisorial fractional ideal of $A$ . Then, for every fractional ideal $b \neq 0$ of $A$ , $\operatorname{div}(a : b) = \operatorname{div} a - \operatorname{div} b$

By virtue of formula (1) of no. 1:

\operatorname{div}(a : b) = \operatorname{div}\left( \bigcap_{y \in b, y \neq 0} y^{-1}a \right) = \sup_{y \in b, y \neq 0} \operatorname{div}(y^{-1}a)

taking account of Proposition 2 and the fact that the fractional ideals $y^{-1}a$ are divisorial. But since $D(A)$ is an ordered group (Algebra, Chapter VI, § 1, no. 8):

\sup_{y \in b, y \neq 0} \operatorname{div}(y^{-1}a) = \sup_{y \in b, y \neq 0} (\operatorname{div} a - \operatorname{div}(y)) \\ = \operatorname{div} a - \inf_{y \in b, y \neq 0} \operatorname{div}(y) = \operatorname{div} a - \operatorname{div} b.

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

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Corollary 2, § 1

Corollary

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