Proposition 14

— For an element $x > 0$ of an ordered group G to be irreducible it is sufficient that it satisfy the following property :

(P) The relations $x \leq y + z,\ y \geq 0,\ z \geq 0$ imply $x \leq y$ or $x \leq z$. This condition is necessary when G is lattice ordered.

If G is lattice ordered and x is irreducible, we have just seen that y is either greater than x or coprime to x; in the latter case Cor. 2 of VI, p. 15 shows that z is greater than x. Conversely, suppose the condition is satisfied: if $0 \leq y \leq x$ then it follows, by putting $x = y + z\ (z \geq 0)$, that either $x \leq y$ or $x \leq z$; in the first case we have $x = y$; in the second we have $x \leq x - y$, so $y \leq 0$ and so $y = 0$; this shows that x is indeed irreducible.