Show that if \(K/E/F\) with \(K/F\) Galois then \(K/E\) is always Galois with \(g(K/E) \leq g(K/F)\).
Show additionally \(E/F\) is Galois \(\iff g(K/E) \normal g(K/F)\).
Show that in this case, \(g(E/F) = g(K/F) / g(K/E)\).
Solution
Let \[ G=\Gal(K/F),\qquad H=\Gal(K/E). \] Because \(K/F\) is finite Galois, \(K\) is the splitting field over \(F\) of a separable polynomial. The same polynomial lies in \(E[x]\) and still splits in \(K\), so \(K/E\) is Galois. Also every \(E\)-automorphism of \(K\) fixes \(F\), hence \[ H\le G. \]
Under the fundamental Galois correspondence, the fixed field of \(H\) is \(E\). For \(\sigma\in G\), \[ \Fix(\sigma H\sigma^{-1})=\sigma(E). \] Therefore \[ H\trianglelefteq G \iff \sigma(E)=E\ \text{for all }\sigma\in G. \] Since \(E/F\) is automatically separable, the right-hand condition is equivalent to \(E/F\) being normal, hence Galois. Thus \[ E/F\text{ is Galois}\iff H\trianglelefteq G. \]
When these equivalent conditions hold, restriction gives a homomorphism \[ \rho:G\to\Gal(E/F),\qquad \sigma\mapsto \sigma|_E. \] Its kernel is \(H\). Every \(F\)-automorphism of \(E\) extends to an \(F\)-automorphism of the normal extension \(K/F\), so \(\rho\) is surjective. By the first isomorphism theorem, \[ \boxed{\Gal(E/F)\cong G/H} =\frac{\Gal(K/F)}{\Gal(K/E)}. \]