Artin–Schreier theorem

Remarks

The proof uses Galois theory on the finite extension $C/F$. One first shows that $C/F$ is separable, so that $C/F$ is a finite Galois extension. In characteristic $p$ such an extension of degree greater than $1$ cannot exist, by an argument with the roots of $x^p - x - a$. In characteristic $0$, one shows that $i \notin F$, that $C = F(i)$, and that $-1$ is not a sum of squares in $F$, by studying cyclic subextensions of prime degree. Example: $F = \R$ and $C = \C$, where $[\C : \R] = 2$, as the fundamental theorem of algebra says. Without the finiteness hypothesis nothing of this kind holds: for $F = \Q$, the degree $[\overline{\Q} : \Q]$ is infinite.

Going deeper

For a student

In terms of Galois groups: a field whose absolute Galois group is finite and nontrivial has absolute Galois group of order $2$, and it is real closed. So the only nontrivial finite subgroups of absolute Galois groups have order $2$, and they come from orderings of the field. The theorem is part of the theory of real closed fields developed by Artin and Schreier. That theory also shows that a field can be ordered if and only if it is formally real.