Bourbaki

Definition 2

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

Commutative Algebra, chapter VII, § 1. Krull domains

Every element a of I(A) such that a=a~a = \tilde{a} is called a divisorial fractional ideal of A.

In other words a divisorial ideal is just a non-zero intersection of a non-empty family of fractional principal ideals. Every non-zero intersection of divisorial ideals is a divisorial ideal. If a is divisorial, so is ax for all x∈K∗x \in K^*, the mapping b↦bxb \mapsto bx being a bijection of the set of fractional principal ideals onto itself, For all a∈I(A)a \in I(A), 6 is the least divisorial ideal containing a and is equivalent to a modulo R. Moreover, if bb is a divisorial ideal equivalent to a modulo R, then a~=6=b\tilde{a} = 6 = b. Hence 5 is the unique divisorial ideal bb such that div a=a = div bb (in other words, the restriction of the mapping a↦a \mapsto div aa to the set of divisorial ideals is injective).

Let a and b be two fractional ideals of K. Recall (Chapter I, § 2, no. 10) that b:ab : a denotes the set of x∈Kx \in K such that xa⊂bxa \subset b; this is obviously an A-module; if b∈I(A)b \in I(A) and a∈I(A)a \in I(A), then b:a∈I(A)b : a \in I(A); for if dd is a non-zero element of A such that db⊂Adb \subset A and da⊂Ada \subset A and aa is a non-zero element of A∩aA \cap a, then da(b:a)⊂Ada(b : a) \subset A; on the other hand, if b≠0b \neq 0 belongs to bb, then bda⊂bbda \subset b, hence bd∈b:abd \in b : a and b:a≠0b : a \neq 0.

The definition of b:ab : a can also be written:

(1)b:a=⋂x∈a,x≠0bx−1. (1) \quad b : a = \bigcap_{x \in a, x \neq 0} b x^{-1}.

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

Cited by

Cites

Transcribed from the printed volume. Not translated.