Proposition 15

— In an ordered group G, let $(p_i)_{i \in I}$ be a family of pairwise distinct positive elements of G satisfying condition (P) (and hence irreducible). Then the map

is an isomorphism of the ordered group $\mathbf{Z}^{(I)}$, the direct sum of the ordered groups $Z$ (VI, p. 7) onto the ordered subgroup of $G$ generated by the $p_\iota$.

It is enough to show that the relation $\sum_{\iota \in I} n_\iota p_\iota \geq 0$ is equivalent to $n_\iota \geq 0$ for all $\iota$, for in particular the relation $\sum_{\iota \in I} n_\iota p_\iota = 0$ will imply $n_\iota = 0$ for all $\iota$, hence this will show that the family $(p_\iota)$ is linearly independent. Now let $I'$ (resp. $I''$) be the finite subset of $I$ consisting of those $\iota$ such that $n_\iota > 0$ (resp. $n_\iota < 0$) ; we have

In particular, for $\lambda \in I''$, this implies that $p_\lambda \leq \sum_{\iota \in I'} n_\iota p_\iota$, and it follows by induction from property (P) that we must have $p_\lambda \leq p_\iota$ for some $\iota \in I'$; since $p_\iota$ is irreducible, this would imply $p_\lambda = p_\iota$, which is absurd. Hence $I'$ is empty, which proves the proposition.