Bourbaki

Exercise 22, § 1

Commutative Algebra, VII, 551

Bourbaki marks this one as harder than the rest.

(a) Let Γ′\Gamma' be a lattice-ordered additive group and AA the algebra of Γ\Gamma over a field kk; AA is an integral domain (Algebra, Chapter II, § 11, no. 4, Proposition 8). Every element x≠0x \neq 0 in AA may be written uniquely as

x=∑i=1nαievi, x = \sum_{i=1}^n \alpha_i e^{v_i},
where the viv_i are distinct elements of Γ\Gamma, the evie^{v_i} the corresponding elements of the canonical basis of AA over kk and the aia_i elements ≠0\neq 0 of kk. Write ϕ(x)=inf⁡(v1,…,vn)\phi(x) = \inf(v_1, \ldots, v_n) in Γ\Gamma; show that ϕ(x+y)≥inf⁡(ϕ(x),ϕ(y))\phi(x+y) \geq \inf(\phi(x), \phi(y)) if x,yx, y and x+yx+y are ≠0\neq 0 in AA and ϕ(xy)=ϕ(x)+ϕ(y)\phi(xy) = \phi(x) + \phi(y) if xy≠0xy \neq 0 in AA. (To show the second assertion, establish first the following lemma: given a finite family (ξj)(\xi_j) of elements of Γ\Gamma, in order that inf⁡(ξj)=0\inf(\xi_j) = 0, it is necessary and sufficient that for all η>0\eta > 0 in Γ′\Gamma' there exist an index jj and an element ζ∈Γ\zeta \in \Gamma such that 0<ζ≤η0 < \zeta \leq \eta and inf⁡(ξj,ζ)=0\inf(\xi_j, \zeta) = 0. Apply this lemma reducing it to the case where ϕ(x)=ϕ(y)=0\phi(x) = \phi(y) = 0.)

(b) Deduce from (a) that, if KK is the field of fractions of AA, ϕ\phi can be extended to a homomorphism from K∗K^* to Γ\Gamma (also denoted by ϕ\phi) such that

ϕ(x+y)≥inf⁡(ϕ(x),ϕ(y)) \phi(x+y) \geq \inf(\phi(x), \phi(y))
if x,yx, y and x+yx+y are ≠0\neq 0 in KK. Deduce that, if BB is the set of x∈Kx \in K such that x=0x = 0 or ϕ(x)≥0\phi(x) \geq 0, BB is a domain whose field of fractions is KK and such that, if UU is the group of invertible elements of BB, K∗/UK^*/U is isomorphic to Γ\Gamma (and in particular BB is a pseudo-Bezoutian domain). Derive examples of a pseudo-Bezoutian domain which is not completely integrally closed, a completely integrally closed pseudo-Bezoutian domain which is not pseudo-principal (cf.

Algebra, Chapter VI, § 1, Exercise 31) \* and a pseudo-principal domain which is not factorial. \*

(c) Derive from (b) an example of an integral domain which is the union of a right directed family of factorial subdomains and is completely integrally closed but not pseudo-Bezoutian. (Let θ\theta be an irrational number in [0,1][0, 1] and for every integer jj let qjq_j be the greatest integer such that qj/2j<0q_j / 2^j < 0. For all jj define on the product group Q×Q\mathbf{Q} \times \mathbf{Q} a lattice-ordered group structure by taking the set (Gj)+(\mathscr{G}_j)_+ of elements ≥0\geq 0 of this group to be the ordered pairs (ξ,η)(\xi, \eta) such that ξ≥0\xi \geq 0 and 0≤η≤(qj2−j)ξ0 \leq \eta \leq (q_j 2^{-j}) \xi. *)

Solution

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