Theorem 2
Tag 06XU, permanent. Cite it as [Bourbaki, Tag 06XU].
Algebra, chapter VI, § 1. Ordered groups. Divisibility
— Let be a filtered group. Then the following properties are equivalent : a) is isomorphic to an ordered group of the form . b) is lattice ordered and satisfies the following condition : (MIN) Every nonempty set of positive elements of has a minimal element. c) satisfies condition (MIN) and every irreducible element of has property (P). d) is generated by its irreducible elements, and every irreducible element of has property (P).
Let us show first that a) implies b). The group is lattice ordered, as the direct sum of totally ordered groups. On the other hand let be a nonempty set of positive elements of and let be an element of (where denotes the natural basis of ) ; there are a finite number of elements of such that , so the set of elements of which are less than or equal to is 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 .
It is clear that b) implies c), by Prop. 14. Let us show that c) implies d). Since is filtered, it is enough (VI, p. 4, Prop. 4) to check that the set of positive elements of which are sums of irreducible elements is equal to . If this were not true, it would follow from (MIN) that the complement of in would have a minimal element ; by definition is not irreducible, so is the sum of two strictly positive elements and ; since and , these elements belong to , and so are sums of irreducible elements, and it follows that so is , 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.