# proof of Galois group of the compositum of two Galois extensions

###### Proof.

Consider the diagram

 $\xymatrix@R1pc@C1pc{&\ar@{-}[ld]\ar@{-}[rd]EF\\ \ar@{-}[rd]E&&\ar@{-}[ld]F\\ &\ar@{-}[d]E\cap F\\ &K}$

(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

 $\theta:\operatorname{Gal}(EF/K)\to\operatorname{Gal}(E/K)\times\operatorname{% Gal}(F/K):\sigma\mapsto(\sigma|_{E},\sigma|_{F})$

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

 $\left\lvert H\right\rvert=\left\lvert\operatorname{Gal}(E/K)\right\rvert\cdot% \left\lvert\operatorname{Gal}(F/E\cap F)\right\rvert=\left\lvert\operatorname{% Gal}(E/K)\right\rvert\cdot\frac{\left\lvert\operatorname{Gal}(F/K)\right\rvert% }{\left\lvert\operatorname{Gal}((E\cap F)/K)\right\rvert}$

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

 $\left\lvert H\right\rvert=[EF:K]$

## References

• 1 Dummit, D., Foote, R.M., Abstract Algebra, Third Edition, Wiley, 2004.
Title proof of Galois group of the compositum of two Galois extensions ProofOfGaloisGroupOfTheCompositumOfTwoGaloisExtensions 2013-03-22 18:42:01 2013-03-22 18:42:01 rm50 (10146) rm50 (10146) 6 rm50 (10146) Proof msc 11R32 msc 12F99