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

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.

$$[EK:K]=\frac{[EK:E][E:F]}{[K:F]}$$. So we need to show $$[EK:E]=[K:F]$$. For this, show that $$EK/E$$ is Galois and that we have a natural isomorphism $$Gal(EK/E) \rightarrow Gal(K/F)$$.
• How do you show it is surjective? Usually I say that $K=F(a)$, since $K/F$ is Galois the conjugates thus the coefficients of $a$'s $E$-minimal polynomial are in $K$, if $E\cap K=F$ then $a$'s $E$-minimal polynomial is $a$'s $F$-minimal polynomial ie. $[EF:E]=[K:F]$. Apr 22 at 20:07
• @reuns: let $H$ be the image of the morphism. By definition $K^H \subset K$ and clearly $K^H \subset E$, so that $K^H \subset K \cap E=F$, thus $H=Gal(K/K^H)=Gal(K/F)$. Apr 22 at 22:37