$\operatorname{Gal}(A/B)=\operatorname{Gal}(AC/BC)$

Question

Let $A/B$ be a Galois extension, and $C/B$ another extension. Under what conditions on $C$ do we have:

$$\operatorname{Gal}(A/B)=\operatorname{Gal}(AC/BC)?$$

Thanks in advance for any help or reference.

Answer 1

I assume that all your extensions are algebraic and everything happens in a fixed algebraic closure of $B$.

Given a (finite) Galois extension $A/B$ and a field extension $C/B$, then $AC/C$ and $A/A\cap C$ are (finite) Galois. Then the restriction map $$ {\rm Gal}(AC/C)\longrightarrow {\rm Gal}(A/A\cap C),\quad \sigma\longmapsto \sigma\big|_A $$ is a well-defined group isomorphism. For well-definedness, notice that if $\sigma\in {\rm Gal}(AC/C)$, then $\sigma\big|_A$ fixes $A\cap C$ pointwise and hence lies in ${\rm Gal}(A/A\cap C)$. It is also clear that restriction is a group homomorphism.
Injectivity: Given $\sigma\in {\rm Gal}(AC/C)$ with $\sigma\big|_A = {\rm id}_A$, then $\sigma$ fixes both $A$ and $C$ pointwise, and hence $AC$, i. e. $\sigma = {\rm id}_{AC}$.
Surjectivity: Let $H$ be the image of ${\rm Gal}(AC/C)$ in ${\rm Gal}(A/A\cap C)$. Then we have $A^H = A\cap C$: $A\cap C\subseteq A^H$ is clear. Conversely, if $a\in A^H$, then $\sigma(a) = a$ for all $\sigma\in {\rm Gal}(AC/C)$ and hence $a\in (AC)^{{\rm Gal}(AC/C)} = C$, i. e. $a\in A\cap C$. By the Galois correspondence, it follows that $H = {\rm Gal}(A/A\cap C)$.

Therefore, we have ${\rm Gal}(AC/C) = {\rm Gal}(A/B)$ if and only if $A\cap C = B$.

This also holds for infinite Galois extensions (with extended proofs).