Suppose $K$ is a Galois extension of $F$. Consider $E/F$ a finite extension such that $K\cap E=F$. Show that $[KE:K]=[E:F]$.

Question

I found the following problem:

Suppose $K$ is a Galois extension of $F$. Consider $E/F$ a finite extension such that $K\cap E=F$. Show that $[KE:K]=[E:F]$.

Can someone give me a hint? I remember that $[KE:K]\leq[E:F]$ and equality happens when $[K,F],[E,F]$ are relatively prime. But I'm not seeing what forces a Galois extensions to make $[K,F],[E,F]$ relatively prime.

Answer 1

$[EK:K]=\frac{[EK:E][E:F]}{[K:F]}$. So we need to show $[EK:E]=[K:F]$. For this, show that $EK/E$ is Galois and that we have a natural isomorphism $Gal(EK/E) \rightarrow Gal(K/F)$.