TheoremProved

Fundamental theorem of Galois theory

Statement

For a finite Galois extension $K/F$ with Galois group $G=\operatorname{Gal}(K/F)$ , there is an inclusion-reversing bijection between the subgroups of $G$ and the intermediate fields $F\subseteq E\subseteq K$ , given by $H \mapsto K^H$ (the fixed field of $H$ ) and $E \mapsto \operatorname{Gal}(K/E)$ ; under this correspondence $E/F$ is a normal extension iff $H$ is a normal subgroup of $G$ , and then $\operatorname{Gal}(E/F)\cong G/H$ .

Why is it true?

Instead of studying an unwieldy field extension directly, Galois theory trades it for its (usually much smaller, finite) symmetry group $G$ . The correspondence says this trade loses nothing: every intermediate field sits between $F$ and $K$ in exactly the same pattern that a subgroup sits inside $G$ , just flipped upside down — bigger fields correspond to smaller groups. Questions about towers of fields (like 'can this be built with square roots?') become questions about the structure of a finite group, which are usually far easier to answer.

Proof sketch

Given a subgroup $H\le G$ , $K^H$ is shown to be a field with $F\subseteq K^H\subseteq K$ ; linear independence of characters (Artin's lemma) shows $[K:K^H]=|H|$ , and separately $[K:F]=|G|$ , which pins down $[K^H:F]=[G:H]$ . Injectivity and surjectivity of $H\mapsto K^H$ follow from Artin's theorem that $\operatorname{Gal}(K/K^H)=H$ for every subgroup $H$ ; the normality statement follows because conjugating $H$ by $g\in G$ corresponds to applying $g$ to the fixed field $K^H$ , so $H$ normal in $G$ is equivalent to $K^H$ being stable under all of $G$ , i.e. $K^H/F$ Galois.

Proved by

proved 1832Évariste Galois

Topics that use this theorem

Galois theory

Related theorems

Abel–Ruffini theorem Lagrange's theorem (group theory)

Step-by-step proofs

No step-by-step proof yet for this theorem.

References

Ian Stewart (2015). Galois Theory · DOI:10.1201/b18187