# Question in proof of $\mathrm{Gal}(K_1K_2/F) \cong H \leq \mathrm{Gal}(K_1/F) \times \mathrm{Gal}(K_2/F)$

In section 14.4 of the text by Dummit and Foote, there is the following proposition:

Proposition. Let $K_1$ and $K_2$ be Galois over a field $F$. Then:

(1) $K_1 \cap K_2$ is Galois over $F$.

(2) $K_1K_2$ is Galois over $F$, with \begin{equation*} \text{Gal}(K_1K_2/F) \cong \{(\sigma, \tau): \sigma|_{K_1 \cap K_2} = \tau|_{K_1 \cap K_2} \} \leq \text{Gal}(K_1/F) \times \text{Gal}(K_2/F) \end{equation*}

Proving (1) is not hard, and it is also straightforward to prove that $K_1K_2$ is Galois over $F$. The next few lines of the proof proceed as follows:

• The map $\phi: \text{Gal}(K_1K_2/F) \to \text{Gal}(K_1/F) \times \text{Gal}(K_2/F)$ defined by \begin{equation*} \sigma \mapsto (\sigma|_{K_1}, \sigma|_{K_2}) \end{equation*} is a homomorphism

• If $\sigma$ is in the kernel of this map, then $\sigma$ is trivial on $K_1$ and $K_2$, and so also trivial on $K_1K_2$. Thus the map is injective.

• The image of this map is indeed in the subgroup $H$, since: \begin{equation*} (\sigma|_{K_1})|_{K_1 \cap K_2} = (\sigma|_{K_2})|_{K_1 \cap K_2} \end{equation*}

• The order of $H$ can be computed by observing that for every $\sigma \in \text{Gal}(K_1/F)$, there are $|\text{Gal}(K_2/K_1 \cap K_2)|$ elements $\tau \in \text{Gal}(K_2/F)$ whose restrictions to $K_1 \cap K_2$ are $\sigma|_{K_1 \cap K_2}$.

It is this last point that I am lost on.

Where does the number $|\text{Gal}(K_2/K_1 \cap K_2)|$ come from?

Since (from a previous corollary) \begin{equation*} \left| \text{Gal}(K_2/K_1 \cap K_2) \right|= \frac{|\text{Gal}(K_2/F)|}{|\text{Gal}(K_1 \cap K_2/F)|} \end{equation*}

it seems like there are "initially" $|\text{Gal}(K_2/F)|$ choices for $\tau$, but then there is some double counting going on, and so we need to divide by $|\text{Gal}(K_1 \cap K_2/F)|$ to account for this?

$H$ is the subgroup of $Gal(K_1/F) \times Gal(K_2/F)$ defined by $$H= \{(\sigma, \tau): \sigma|_{K_1 \cap K_2} = \tau|_{K_1 \cap K_2} \}$$ So using that all those extensions are Galois $$|H| = \sum_{\sigma \in Gal(K_1/F)} \# \{ \tau \in Gal(K_2/F),\sigma|_{K_1 \cap K_2} = \tau|_{K_1 \cap K_2}\}= \sum_{\sigma \in Gal(K_1/F)} |\text{Gal}(K_2/K_1 \cap K_2)|$$
$$= |\text{Gal}(K_1/F)| \, |\text{Gal}(K_2/K_1 \cap K_2)| =|\text{Gal}(K_1/F)| \,\frac{|\text{Gal}(K_2/F)|}{|\text{Gal}(K_1 \cap K_2/F)|}$$
Finally write the inverse map of $\phi$ so that $\text{Gal}(K_1K_2/F) \cong H$
• I'm lost on the equality: $\#\{\tau\in \text{Gal}(K_2/F): \sigma \text{ and } \tau \text{ agree on } K_1 \cap K_2\} = |\text{Gal}(K_2/K_1 \cap K_2)|$ – Sam Y. Apr 19 '17 at 21:08
• @SamY. come on... Split $\tau$ in two parts : one acting on $K_1 \cap K_2/F$ and the other on $K_2/K_1 \cap K_2$. Since everything is Galois, the two parts are independent. – reuns Apr 19 '17 at 22:41