Let \(K/F\) be a finite Galois extension and let \(E\) be an intermediate field (\(F \subseteq E \subseteq K\)). (1) Prove that \(K/E\) is always Galois with \(\operatorname{Gal}(K/E) \le \operatorname{Gal}(K/F)\). (2) Prove that \(E/F\) is Galois if and only if \(\operatorname{Gal}(K/E) \trianglelefteq \operatorname{Gal}(K/F)\). (3) Prove that when \(E/F\) is Galois, \(\operatorname{Gal}(E/F) \cong \operatorname{Gal}(K/F) / \operatorname{Gal}(K/E)\).
Solution
Write \[ G=\operatorname{Gal}(K/F),\qquad H=\operatorname{Gal}(K/E). \]
Because \(K/F\) is finite Galois, every element of \(K\) is separable over \(F\), hence also over \(E\). Moreover, if \(K\) is the splitting field over \(F\) of a separable polynomial \(f\in F[x]\), then \[ K=F(\text{roots of }f)=E(\text{roots of }f), \] so \(K\) is also the splitting field of \(f\) over \(E\). Thus \(K/E\) is Galois. Every automorphism fixing \(E\) fixes \(F\), so \(H\le G\).
Under the Galois correspondence, the fixed field of \(\sigma H\sigma^{-1}\) is \(\sigma(E)\). Therefore \[ H\trianglelefteq G \iff \sigma(E)=E\quad\text{for every }\sigma\in G. \] Since \(E/F\) is separable, the latter condition is equivalent to \(E/F\) being normal, hence Galois. This proves \[ E/F\text{ is Galois}\iff H\trianglelefteq G. \]
Assume now that \(E/F\) is Galois. Restriction defines \[ \rho:G\longrightarrow \operatorname{Gal}(E/F),\qquad \rho(\sigma)=\sigma|_E. \] Its kernel is exactly \(H\). Every \(F\)-automorphism of \(E\) extends to an \(F\)-automorphism of the normal extension \(K/F\), so \(\rho\) is surjective. Hence the first isomorphism theorem gives \[ \operatorname{Gal}(E/F)\cong G/H =\operatorname{Gal}(K/F)/\operatorname{Gal}(K/E). \]