Bourbaki

Theorem 2

Tag 06XU, permanent. Cite it as [Bourbaki, Tag 06XU].

Algebra, chapter VI, § 1. Ordered groups. Divisibility

— Let GG be a filtered group. Then the following properties are equivalent : a) GG is isomorphic to an ordered group of the form Z(I)\mathbf{Z}^{(I)}. b) GG is lattice ordered and satisfies the following condition : (MIN) Every nonempty set of positive elements of GG has a minimal element. c) GG satisfies condition (MIN) and every irreducible element of GG has property (P). d) GG is generated by its irreducible elements, and every irreducible element of GG has property (P).

Let us show first that a) implies b). The group Z(I)\mathbf{Z}^{(I)} is lattice ordered, as the direct sum of totally ordered groups. On the other hand let EE be a nonempty set of positive elements of Z(I)\mathbf{Z}^{(I)} and let x=∑nιeιx = \sum n_\iota e_\iota be an element of EE (where (eι)(e_\iota) denotes the natural basis of Z(I)\mathbf{Z}^{(I)}) ; there are a finite number ∏(nι+1)\prod (n_\iota + 1) of elements yy of Z(I)\mathbf{Z}^{(I)} such that 0≤y≤x0 \leq y \leq x, so the set FF of elements of EE which are less than or equal to xx is aa fortiori finite ; since it is nonempty, it contains a minimal element (Set Theory, III, p. 170, Cor. 2), which is clearly a minimal element of EE.

It is clear that b) implies c), by Prop. 14. Let us show that c) implies d). Since GG is filtered, it is enough (VI, p. 4, Prop. 4) to check that the set FF of positive elements of GG which are sums of irreducible elements is equal to G+−{0}G_+ - \{0\}. If this were not true, it would follow from (MIN) that the complement of FF in G+−{0}G_+ - \{0\} would have a minimal element aa; by definition aa is not irreducible, so is the sum of two strictly positive elements xx and yy; since x<ax < a and y<ay < a, these elements belong to FF, and so are sums of irreducible elements, and it follows that so is aa, which is a contradiction. Finally, d) implies a) by Prop. 15.

We will apply Th. 2 to the theory of divisibility in principal ideal domains (VII, p. 4) and in unique factorisation domains (AC, VII, § 3), as well as to the study of ideals in a Dedekind ring (AC, VII, § 2).

Bourbaki puts the proof directly after the statement, so what is above is the statement and its proof together.

Cited by

Cites

  • AC, VII, § 2 not yet on this site: AC VII
  • AC, VII, § 3 not yet on this site: AC VII
  • Prop. 14
  • Prop. 15
Transcribed from the printed edition, 2003, Springer. Not translated.