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 $Gal(EK/E) \rightarrow Gal(K/F)$.

  • $\begingroup$ 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. $[EF:E]=[K:F]$. $\endgroup$
    – reuns
    Apr 22 at 20:07
  • $\begingroup$ @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 $H=Gal(K/K^H)=Gal(K/F)$. $\endgroup$
    – Mindlack
    Apr 22 at 22:37

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