Theorem 3 (Euler-Lagrange)

— Let K be an ordered field. Then the following three properties are equivalent: a) The field $K(i)$ is algebraically closed (where i denotes a square root of -1). b) The ordered field K is maximal. c) Every positive element of K is a square, and every polynomial of odd degree in $K[X]$ has a root in K.

It is clear that a) implies b): indeed K has only two algebraic extensions up to isomorphism, the field K itself and $K(i)$, which cannot be ordered since -1 is a square.

The fact that b) implies c) is nothing other than Prop. 6 of VI, p. 25.

It remains for us to prove that c) implies a). That will follow from the next two propositions.