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Let $K/F$ be a finite Galois extension and $a\in K$. Let's say $m_a \in F[x]$ is a minimal polynomial of $a$ and $L$ is a splitting field of $m_a$ over $F$. As $F(a) /F$ is separable, $m_a$ is also separable so we can write as $$ m_a = (x-\alpha_1) (x-\alpha_2) \cdots (x-\alpha_n) $$ where $\alpha_1 = a$ and distinct $\alpha_i \in L$.

Now define $\tau_i: F(a) \to F(\alpha_i)$ by $\tau_i(a) = \alpha_i$ and $\tau_i \vert _F = 1_F$. I wonder if there always exists $\sigma_i \in \operatorname{Gal}(L/F)$ such that $\sigma_i \vert _{F(a)} = \tau_i$. Considering $L/F$ as a vector space, then I know that there is a bijective linear transformation of $L$ whose restriction to $F(a)$ is $\tau_i$. It's done by just expanding each basis of $F(a)$ and $F(\alpha_i)$ to $L$'s basis and take each basis elements to the corresponding basis.

But the existence of the bijective linear transformation does not necessarily guarantee the existence of a field isomorphism.

Counterexample) Let $\mathbb{C} / \mathbb{R}$ be a vector space and think a map $\mathbb{C} \to \mathbb{C}$ such that $a+bi \mapsto a+2bi$ where $a,b\in \mathbb{R}$. It's a bijective linear operator but not a field homomorphism.

This approach cannot show the existence of an isomorphism of $L$ extending $\tau_i$ exists in this way, but I believe such isomorphism exists even if $L/F$ is not separable. (Actually I'm not sure.) Can one give me some hints?

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Since $m_a$ is irreducible, we have that $\operatorname{Gal}(L/F)$ acts transitively on the roots of $m_a$. In particular there is some $\sigma_i$ with $\sigma_i(a)=\alpha_i$.

Of course we have $\sigma_{i|F(a)} = \tau_i$ since both maps coincide on $a$ and fix $F$, so they coincide on whole of $F(a)$.

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  • $\begingroup$ Why does Galois group act transitively? $\endgroup$ – user258265 Nov 12 '15 at 16:40
  • $\begingroup$ What is your background? Have u heard of extending field homomorphisms to algebraic extensions? $\endgroup$ – MooS Nov 12 '15 at 16:55
  • $\begingroup$ Ah now I get how it works. Thanks! $\endgroup$ – user258265 Nov 12 '15 at 17:34
  • $\begingroup$ Actually I think it is a circular argument. Proving that Galois group act transitively on the set of roots requires the answer of what I questioned. But now I proved it myself. $\endgroup$ – user258265 Nov 14 '15 at 9:04
  • $\begingroup$ Yes, thats why I asked for your background $\endgroup$ – MooS Nov 14 '15 at 9:06

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