Bourbaki

Exercise 19, § 2

Commutative Algebra, VII, 561

Bourbaki marks this one as harder than the rest.

An integral domain AA is called pseudo-Priiferian if the set of finitely generated divisors of D(A)D(A) (§1§ 1, Exercise 11) is a group. A pseudo-Priiferian domain is regularly integrally closed (§1§ 1, Exercise 29). A pseudo-Bezoutian domain (§1§ 1, 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 θ\theta be an irrational number >0> 0 and Γ\Gamma the group R2\mathbf{R}^2 ordered by taking the set of ordered pairs (a,β)(a, \beta) such that a≥0a \geq 0 and β≥θa\beta \geq \theta a as the set of positive elements; the ordered group Γ\Gamma is a complete lattice. Let BB be the pseudo-principal domain derived from Γ\Gamma by the procedure of §1§ 1, Exercise 22 and let AA be the subring the intersection of BB and the algebra over kk of the subgroup Q2\mathbf{Q}^2 of R2\mathbf{R}^2. Show that AA is a completely integrally closed domain but is not pseudo-Priiferian (note that, if KK is the field of fractions of AA and UU the group of invertible elements of AA, the ordered group K∗/UK^*/U is isomorphic to the group Γ∩Q2\Gamma \cap \mathbf{Q}^2, ordered by the ordering induced by that on Γ\Gamma).

(c) If the domain A\mathbf{A} is pseudo-Pruferian, show that the polynomial domain A[X]\mathbf{A}[X] is pseudo-Pruferian (cf. § 1, Exercise 30 (c)).

(d) Let A\mathbf{A} be a pseudo-Pruferian domain and K\mathbf{K} its field of fractions. Show that the integral closure B\mathbf{B} of A\mathbf{A} in an algebraic extension L\mathbf{L} of K\mathbf{K} 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 (Ai)(\mathbf{A}_i) 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 B\mathbf{B} to be a discrete valuation ring). \*

Solution

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Transcribed from the printed volume. Not translated.