The Artin-schreier Theorem

The algebraic closure of R is C, which is a finite extension. Are there other fields which are not algebraically closed but have an algebraic closure which is a finite extension? Yes. An example is the field of real algebraic numbers. Since complex conjugation is a field automorphism fixing Q, and the real and imaginary parts of a complex number can be computed using field operations and complex conjugation, a complex number a + bi is algebraic over Q if and only if a and b are algebraic over Q (meaning a and b are real algebraic numbers), so the algebraic closure of the real algebraic numbers is obtained by adjoining i. This example is not too different from R, in an algebraic sense. Is there an example which does not look like the reals, such as a field of positive characteristic or a field whose algebraic closure is a finite extension of degree greater than 2? Amazingly, it turns out that if a field F is not algebraically closed but its algebraic closure C is a finite extension, then in a sense which will be made precise below, F looks like the real numbers. For example, F must have characteristic 0 and C = F (i). This is a theorem of Artin and Schreier. Proofs of the Artin-Schreier theorem can be found in [5, Theorem 11.14] and [6, Corollary 9.3, Chapter VI], although in both cases the theorem is proved only after the development of some general theory which is useful for more than just the ArtinSchreier theorem. In [7] there is an elementary proof (based on the original one, as all known proofs are) under the hypothesis that F has characteristic 0; we essentially reproduce much of that proof below, but we add on some extra details to prove in an elementary manner that F must have characteristic 0. The prerequisites are a knowledge of basic field theory and Galois theory of finite extensions, including Kummer extensions and Artin-Schreier extensions. Applications of the Artin–Schreier theorem to the Galois theory of infinite extensions will be mentioned after the proof.