Let L/K, M/K be finite Galois extensions contained in some common field extension $\mathbb{K}$ of $K$, so we can speak about $L \cap M$ and the composite $LM$ , which is defined to be the smallest subfield of $\mathbb{K}$ containing $L$ and $M$.
Prove that:
$$ Gal(LM/K) \cong Gal(L/K) \times _{Gal(L\cap M/K)} Gal(M/K)
:= \{(\sigma,\tau) \in Gal(L/K) \times Gal(M/K) : \sigma|_{L\cap M} = \tau |_{L\cap M} \}.$$
I recognize that this looks very similar to cofibre products, and feels like an equivalent statement to the Seifert-Van Kampen theorem.
I've got that $LM, L\cap M$ are both Galois over K, thus $Gal(LM/L\cap M)$ is normal and so are the intermediate fields generated by the sub-fields $L$ and $M$. I have also proved that $Gal(LM/L) \cap Gal(LM/M) = e$. I feel like I have to find a way to embed the group on the right hand side in $Gal(LM/M)$, but I can't figure out a way to do that. What am I missing?
Best Answer
The reason this looks similar to Van Kampen's theorem is because the Galois group of the compositum is a pullback and the fundamental group computed in Van Kampen's theorem is a pushout.
Since $L$ and $M$ are Galois over $K$, any automorphism of $LM/K$ will stabilize the subfields $L$ and $M$. This means that you have a map $\mathrm{Gal}(LM/K) \to \mathrm{Gal}(L/K)$ and $\mathrm{Gal}(LM/K) \to \mathrm{Gal}(M/K)$. Consequently, you have a map $\mathrm{Gal}(LM/K) \to \mathrm{Gal}(L/K) \times \mathrm{Gal}(M/K)$.
Show that the kernel of $\mathrm{Gal}(LM/K) \to \mathrm{Gal}(L/K)$ is $\mathrm{Gal}(LM/L)$, and similarly for the other map. You have shown that the intersection of the kernels is trivial, so the map to the product is injective. Furthermore, it is clear that the image is contained in the subgroup $\mathrm{Gal}(L/K) \times_{\mathrm{Gal}(L \cap M/K)} \mathrm{Gal}(M/K)$. Can you conclude that this map is an isomorphism by order considerations?