Is true that, although $\Bbb{Q}(\sqrt{2})$ and $\Bbb{Q}(\sqrt{3})$ are not isomorphic, their Galois groups over $\Bbb{Q}$ are isomorphic (since both of then has order 2). But, there is a way to "refine" this extension: Let $A,B$ be intermediate fields of a finite Galois extension $L/k$, with a field homomorphism $\alpha:A \to B$ over $k$.
Is true that, there is a group homomorphism $\varphi_{\alpha}: \operatorname{Gal}(L/B) \to \operatorname{Gal}(L/A)$ and $\beta \in \operatorname{Gal}(L/k)$ where $\varphi_{\alpha}(\sigma)=\beta \ \circ \sigma \ \circ \beta^{-1}$, for all $\sigma \in \operatorname{Gal}(L/B)$?
I mean, it looks fine, but I can't prove it with a formal way. In Lang's Algebra, there is something that looks like it, on Chapter VI, before the theorem 1.10. The problem is, he supposes that there is the homomorphism and I want to prove that. Any help would be appreciated! Thanks in advance!
Best Answer
If $\alpha$ is a field homomorphism over $k$ (that is, $\alpha|_{k}=\mathrm{id}_k$), then vieweing it as an embedding $A\hookrightarrow \overline{k}$; we know that we can extend it to an embedding $\gamma\colon L\hookrightarrow \overline{k}$, since $L$ is algebraic over $A$. And because $L$ is Galois over $k$, the image of this embedding is contained in $L$. That is, we can extend $\alpha$ to an automorphism of $L$ over $k$, so $\gamma\in\mathrm{Gal}(L/k)$ and $\gamma|_A = \alpha$.
I claim that conjugation by $\beta=\gamma^{-1}$ yields the morphism you want. This is just a matter of verifying if $\sigma\in\mathrm{Gal}(L/B)$, then $\gamma^{-1}\circ\sigma\circ\gamma\in \mathrm{Gal}(L/A)$. And for that, we just need to verify that it fixes the elements of $A$ pointwise.
Indeed, if $a\in A$, then $\gamma(a)=\alpha(a)\in B$, so $\sigma(\gamma(a))=\gamma(a)$; therefore, $\gamma^{-1}\sigma\gamma(a) = \gamma^{-1}(\gamma(a)) = a$. Thus, $\gamma^{-1}\circ\sigma\circ\gamma \in \mathrm{Gal}(L/B)$, as desired.