I was wondering, when the restriction map $Gal(L/F) \rightarrow Gal(K/F)$ is a surjection? I found a good answer to this question here. In point 3), Starfall explains why with the hypothesis of the theorem, the automorphism extends, and I understand that in order to extend the automorphism we need that $L/F$ is Galois. But why we need $K/F$ to be Galois? I also posted a comment there but maybe I won't get a response.

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    $\begingroup$ If $K/F$ is a normal extension then $\sigma \mapsto \sigma|_K$ in an homomorphism $Gal(L/F) \to Gal(K/F)$ and it is surjective if $L/F$ is normal (thus with kernel $Gal(L/K)$). If $K/F$ is not normal then you need pick first the subgroup $\{ \sigma \in Gal(L/F), \sigma(K) = K\} $ before applying $\sigma \mapsto \sigma|_K$. $\endgroup$ – reuns Dec 1 '18 at 23:26
  • $\begingroup$ Okay, but I am a little confused. I am reading a book called introduction to abstract algebra by Keith Nicholson, and there is this theorem that says : Let $E/F$ be a Galois extension and let $G=gal(E:F)$. If $K$ is a stable intermediate field, then $K'=gal(E:K)$ is normal in $G$ and $G/K' \cong \{ \lambda \in gal(K:F) \mid \lambda \text{ extends to an automorphism of E} \}$. But since $E/F$ is Galois, don't we even have $G/K' \cong gal(K:F)$ by the arguments you gave? $\endgroup$ – roi_saumon Dec 2 '18 at 13:07
  • $\begingroup$ Yes, and by the argument given by starfall that any $\sigma\in Aut(K/F)$ extends to automorphisms of the Galois (normal) extensions $L/F$ above $K/F$. For the intuition : in the usual separable setting take a primitive element $K = F(a)$ and look at $L$ the splitting field of $a$'s minimal polynomial, then at $L(b)$. $\endgroup$ – reuns Dec 2 '18 at 22:12

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