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<record version="2" id="6794">
 <title>Galois group of the compositum of two Galois extensions</title>
 <name>GaloisGroupOfTheCompositumOfTwoGaloisExtensions</name>
 <created>2005-02-22 10:30:07</created>
 <modified>2005-03-10 15:39:17</modified>
 <type>Theorem</type>
<parent id="1511">composite field</parent>
 <creator id="2414" name="alozano"/>
 <author id="2414" name="alozano"/>
 <classification>
	<category scheme="msc" code="12F99"/>
	<category scheme="msc" code="11R32"/>
 </classification>
 <related>
	<object name="CompositumOfAGaloisExtensionAndAnotherExtensionIsGalois"/>
	<object name="GaloisExtension"/>
 </related>
 <keywords>
	<term>compositum</term>
	<term>Galois group</term>
 </keywords>
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 <content>\begin{thm}
Let $E$ and $F$ be Galois extensions of a field $K$. Then:
\begin{enumerate}
\item The intersection $E\cap F$ is Galois over $K$.\\
\item The compositum $EF$ is Galois over $K$. Moreover, the Galois group $\Gal(EF/K)$ is isomorphic to the subgroup $H$ of the direct product $G=\Gal(E/K)\times \Gal(F/K)$ given by:
$$H=\{ (\sigma, \psi) : \sigma|_{E\cap F}=\psi|_{E\cap F} \}$$
i. e. $H$ consists of pairs of elements of $G$ whose restrictions to $E\cap F$ are equal.
\end{enumerate}
\end{thm}

\begin{cor}
Let $E$ and $F$ be Galois extensions of a field $K$ such that $E\cap F=K$. Then $EF$ is Galois over $K$ and the Galois group is isomorphic to the direct product:
$$\Gal(EF/K)\cong \Gal(E/K) \times \Gal(F/K).$$
\end{cor}</content>
</record>
