Theorem 1

For an extension $E$ of $K$ to admit the structure of an ordered extension of $K$, the following condition is necessary and sufficient: (OE) The relation $p_1 x_1^2 + \ldots + p_n x_n^2 = 0$ implies

for any finite sequence $(x_i, p_i)$ of pairs of elements $x$, of $E$ and positive elements $p_i$ of $K$.

Condition (OE) is clearly equivalent to: (OE') The element – 1 is not a sum of elements of the form $px^2$ ($x \in E, p \in K, p \geq 0$).

Condition (OE) is necessary: if E is an ordered extension of K then the elements $p_i x_i^2$ are positive in E, so zero if their sum is zero. On the other hand $p_i x_i^2 = 0$ is equivalent to $p_i x_i = 0$.

Conversely, suppose condition (OE) is satisfied, then we will define an ordering on E by constructing a subset P of E which satisfies conditions (AP_I), (AP_{II}), (AP_{III}) and (AP_{IV}), and which contains the set $K_+$ of positive elements of K. Such a subset P will certainly make E an ordered extension of K, for we will have $K \cap P = K_+$; indeed, if P were to contain an element $-a < 0$ of K, then a would belong to $P \cap (-P)$, contradicting (APIII).

To define P, let us consider the set $\mathscr{M}$ of subsets of E which satisfy (AP_I), (AP_{II}) and (AP_{III}), and which contain the union of $K_+$ and the set C of squares of elements of E. This set $\mathscr{M}$ is nonempty, for it contains the set $P_0$ of elements of the form $\sum p_i x_i^2$ (that $P_0$ satisfies (AP_{III}) follows immediately from (OE)).

Moreover $\mathscr{M}$ is inductive (Set Theory, III, p. 154, Def. 3). Thus there exists, by Th. 2 of Set Theory, III, p. 154, a maximal element in $\mathscr{M}$, which it remains for us to prove satisfies (AP_{IV}); now this follows from the following lemma: