Exercise 19, § 2
Commutative Algebra, VII, 561
Bourbaki marks this one as harder than the rest.
An integral domain is called pseudo-Priiferian if the set of finitely generated divisors of (, Exercise 11) is a group. A pseudo-Priiferian domain is regularly integrally closed (, Exercise 29). A pseudo-Bezoutian domain (, Exercise 21) is pseudo-Priiferian; a Priiferian domain is pseudo-Priiferian; a Krull domain is pseudo-Priiferian.
(a) Show by examples that the converses of the last three assertions are not necessarily true.
(b) Let be an irrational number and the group ordered by taking the set of ordered pairs such that and as the set of positive elements; the ordered group is a complete lattice. Let be the pseudo-principal domain derived from by the procedure of , Exercise 22 and let be the subring the intersection of and the algebra over of the subgroup of . Show that is a completely integrally closed domain but is not pseudo-Priiferian (note that, if is the field of fractions of and the group of invertible elements of , the ordered group is isomorphic to the group , ordered by the ordering induced by that on ).
(c) If the domain is pseudo-Pruferian, show that the polynomial domain is pseudo-Pruferian (cf. § 1, Exercise 30 (c)).
(d) Let be a pseudo-Pruferian domain and its field of fractions. Show that the integral closure of in an algebraic extension of is a pseudo-Pruferian domain (method of Exercise 16, using § 1, Exercise 29 (a)).
(e) Deduce from Exercise 30 (d) of § 1 an example of an increasing sequence of factorial domains with the same field of fractions whose union is not a regularly integrally closed domain (nor a fortiori a pseudo-Pruferian domain) (in the example quoted, take to be a discrete valuation ring). \*
Solution
No solution has been written yet.