proof of Galois group of the compositum of two Galois extensions
ProofOfGaloisGroupOfTheCompositumOfTwoGaloisExtensions
2009-01-05 19:09:48
2009-01-06 20:28:10
Proof
Galois group of the compositum of two Galois extensions
0
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\begin{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
\[\theta: \Gal(EF/K)\to \Gal(E/K)\times \Gal(F/K): \sigma\mapsto (\sigma|_E,\sigma|_F)\]
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 $\sigma\in \Gal(E/K)$, there are precisely $\Order{\Gal(F/E\cap F)}$ elements of $\Gal(F/K)$ whose restrictions to $E\cap F$ are $\sigma|_{E\cap F}$. Thus directly from the definition of $H$,
\[
\Order{H} = \Order{\Gal(E/K)}\cdot\Order{\Gal(F/E\cap F)}
= \Order{\Gal(E/K)}\cdot\frac{\Order{\Gal(F/K)}}{\Order{\Gal((E\cap F)/K)}}
\]
By the \PMlinkname{corollary to the theorem regarding the compositum of a Galois extension and another extension}{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
\[
\Order{H} = [EF:K]
\]
\end{proof}
\begin{thebibliography}{10}
\bibitem{bib:df}
Dummit,~D.,~Foote,~R.M., \emph{Abstract Algebra, Third Edition}, Wiley, 2004.
\end{thebibliography}