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?

  • 2
    $\begingroup$ So what's the question? $\endgroup$ – Alex Jan 13 '17 at 21:41
  • $\begingroup$ @Alex how to prove it, sorry. I'm going to edit it. $\endgroup$ – Rafael Gonzalez Lopez Jan 13 '17 at 21:43
  • $\begingroup$ @Watson I just want a hint, but I'll edit with my efforts ^^ $\endgroup$ – Rafael Gonzalez Lopez Jan 13 '17 at 21:50

$K/(K\cap F)$ 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.

  • $\begingroup$ I've just finished but in my language. Thank you. $\endgroup$ – Rafael Gonzalez Lopez Jan 13 '17 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 Jan 13 '17 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. $\endgroup$ – Rafael Gonzalez Lopez Jan 13 '17 at 22:54

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.