Bourbaki

Exercise 12, § 4

Commutative Algebra, IX, p. 80

Let A be an integral domain, A‾\overline{A} its integral closure. Let a1,...,ara_1, ..., a_r be nonzero elements of A‾\overline{A}, b the fractional ideal ∑i=1rAai\sum_{i=1}^r Aa_i of A. a) There exists a fractional ideal aa of A such that aai⊂aaa_i \subset a for 1≤i≤r1 \leq i \leq r. One also has a~ai⊂a~\tilde{a}a_i \subset \tilde{a} for 1≤i≤r1 \leq i \leq r; we recall (VII, § 1, No. 1) that a~\tilde{a} is the intersection of the principal fractional ideals containing aa. b) Let yy be a nonzero element of b~\tilde{b}. Then

Ay−1⊃⋂i=1rAai−1anda~⊂a~y−1. A y^{-1} \supset \bigcap_{i=1}^r Aa_i^{-1} \quad \text{and} \quad \tilde{a} \subset \tilde{a}y^{-1}.
c) One has b~⊂A‾\tilde{b} \subset \overline{A}. d) Let zz be an element of A‾:A‾b\overline{A} : \overline{A}b. Then A‾\overline{A} contains b~z\tilde{b}z. e) One has b⊂A‾:(A‾:A‾b)b \subset \overline{A} : (\overline{A} : \overline{A}b). f) Let P\mathfrak{P} be a divisorial ideal of A‾\overline{A} and p=P∩Ap = \mathfrak{P} \cap A. Then pp is divisorial in A. (By e), one obtains p~⊂P\tilde{p} \subset \mathfrak{P}. But one also has p~⊂A\tilde{p} \subset A.

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