Exercise 30, § 1
Commutative Algebra, VII, 554
Bourbaki marks this one as harder than the rest.
An integral domain is called regularly integrally closed if, in the monoid D(A), every finitely generated divisor (Exercise 11) is a regular element. Every completely integrally closed domain is regularly integrally closed; every pseudo-Bezoutial domain (Exercise 2I) is regularly integrally closed. (a) If A is regularly integrally closed and are three finitely generated divisors, the relation implies . (b) Deduce from (a) that, for an integral domain A to be regularly integrally closed, it is necessary and sufficient that, for every fractional ideal a of A such that div(a) is a finitely generated divisor, a: ; in particular A is integrally closed (Exercise6 (b)). (Use the fact that A: (bc) = (A: b) : c for two fractional ideals b, c of A, taking .) Moreover, div(a) is then invertible in D(A).
(c) Let be a regularly integrally closed domain and its field of fractions; the monoid of finitely generated divisors of generates in a lattice-ordered group . For every polynomial , let denote the divisor of the fractional ideal of generated by the coefficients of ; for every rational function of , where are in and , is well defined on (independently of the expression of as a quotient of two polynomials); it is denoted by (cf. Exercise 12 (c)). If we then write , where is the fractional principal ideal of generated by , is an isomorphism of the ordered group (notation of § 3, no. 2) onto a subgroup of the product ordered group
Deduce that is a regularly integrally closed domain and that is isomorphic to (use (a) and (b) and Exercise 29 (b)).
(d) Let be an integrally closed domain, its field of fractions and the subring of the polynomial ring consisting of the polynomials whose constant term belongs to . Show that, if , is integrally closed but not regularly integrally closed (show that the divisor corresponds to a divisorial ideal such that ).
Solution
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