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Let $E/\mathbb{Q}$ be a finite field extension. Let $F,K \subset E$ subfields which contains $\mathbb{Q}$. Let M the smallest subfield of E which contains $F$ and $K$. If $K/\mathbb{Q}$ is a Galois extension then $M/F$ and $K/(K\cap F)$ are Galois extensions and \begin{equation} r : Gal(M/F) → Gal(K/(K ∩ F))\\ σ → σ|_K \end{equation} is a well defined homomorphism.

Can you help me to prove it?

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    $\begingroup$ So what's the question? $\endgroup$
    – Alex
    Commented Jan 13, 2017 at 21:41
  • $\begingroup$ @Alex how to prove it, sorry. I'm going to edit it. $\endgroup$ Commented Jan 13, 2017 at 21:43
  • $\begingroup$ @Watson I just want a hint, but I'll edit with my efforts ^^ $\endgroup$ Commented Jan 13, 2017 at 21:50

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$K/(K\cap F)$ is a Galois extension since $K\cap F$ is an extension of $\Bbb Q$. Then you essentially copy the proof of the diamond isomorphism theorem from Dummit and Foote to show that the homomorphism $Aut(M/F) → Gal(K/(K ∩ F)), σ \mapsto σ|_K $ is well-defined, and an isomorphism. By considering the degrees of the extensions, you conclude that $M=FK$ is a Galois extension of $F$.

Since you just want a hint, this should be enough; let me know if you get stuck somewhere.

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  • $\begingroup$ I've just finished but in my language. Thank you. $\endgroup$ Commented Jan 13, 2017 at 22:08
  • $\begingroup$ Si tu quieres, yo puedo mirar a tu respuesta. Porque la parte " show that the homomorphism ... is well-defined" requiere muchas detalles. $\endgroup$
    – Alex
    Commented Jan 13, 2017 at 22:12
  • $\begingroup$ No, I'm fine with it. Thank you, but I need to check other dem. It's easier, but I need to be sure. $\endgroup$ Commented Jan 13, 2017 at 22:54

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