proof of Galois group of the compositum of two Galois extensions

Proof.

Consider the diagram

(1): Let $p(x)\in K[x]$ with a root $\alpha \in E\cap F$. Then since $E$ (resp. $F$) is Galois over $K$, all the roots of $p$ lie in $E$ (resp. $F$) and thus in $E\cap F$. The result follows.

(2): We first show that $EF$ is Galois over $K$. Choose separable polynomials $p(x),q(x)\in K[x]$ so that $E$ (resp. $F$) is a splitting field for $p$ (resp. $q$). Then $EF$ is a splitting field for the squarefree part of $pq$, which is separable since it is squarefree and since $p(x),q(x)$ are separable.

Now, define

$$\theta :\mathrm{Gal}(EF/K)\to \mathrm{Gal}(E/K)\times \mathrm{Gal}(F/K):\sigma \mapsto ({\sigma |}_E,{\sigma |}_F)$$

This map is a group homomorphism; its kernel is precisely those elements that leave both $E$ and $F$ fixed. Any such element must thus leave $EF$ fixed, so that $\theta$ is injective. The image obviously lies in

$$H=\{(\sigma ,\tau ):{\sigma |}_{E\cap F}={\tau |}_{E\cap F}\}$$

by construction: ${({\sigma |}_E)|}_{E\cap F}={\sigma |}_{E\cap F}={({\sigma |}_F)|}_{E\cap F}$. We will show that $H$ is precisely the image of $\theta$ by showing that the order of $H$ is the same as the index of the field extension $[EF:K]$.

For each $\sigma \in \mathrm{Gal}(E/K)$, there are precisely $\left|\mathrm{Gal}(F/E\cap F)\right|$ elements of $\mathrm{Gal}(F/K)$ whose restrictions to $E\cap F$ are ${\sigma |}_{E\cap F}$. Thus directly from the definition of $H$,

$$\left|H\right|=\left|\mathrm{Gal}(E/K)\right|\cdot \left|\mathrm{Gal}(F/E\cap F)\right|=\left|\mathrm{Gal}(E/K)\right|\cdot \frac{\left|\mathrm{Gal}(F/K)\right|}{\left|\mathrm{Gal}((E\cap F)/K)\right|}$$

By the corollary to the theorem regarding the compositum of a Galois extension and another extension (http://planetmath.org/CorollaryToTheCompositumOfAGaloisExtensionAndAnotherExtensionIsGalois), we have

$$[EF:K]=[EF:F][F:K]=[E:E\cap F][F:K]=\frac{[E:K][F:K]}{[E\cap F:K]}$$

so that

$$\left|H\right|=[EF:K]$$

∎