Theorem 2

— Every ordered field K admits an ordered algebraic extension which is a maximal ordered field.

One can show that this ordered extension is unique up to K-isomorphism (VI, p. 40, Ex. 15).

Let $\Omega$ be an algebraic closure of K, and let $\mathfrak{N}$ be the set of pairs $(A, w)$, where $A$ is a sub-K-extension of $\Omega$, and $w$ is an ordering on $A$ making $A$ an ordered extension of $K$. Order $\mathfrak{N}$ by the relation « L is an ordered extension of M » between M and $L$. Equipped with this ordering, $\mathfrak{N}$ is an inductive ordered set: indeed if $(L_i)$ is a totally ordered family of elements of $\mathfrak{N}$, then the field $L = \bigcup L_i$, ordered by taking $L_1 = \bigcup (L_i)_1$, is an upper bound for the $L_i$. Then $\mathfrak{N}$ has a maximal element, by Set Theory, III, p. 154, Th. 2, which completes the proof.