# proof that the compositum of a Galois extension and another extension is Galois

###### Proof.

The diagram of the situation of the theorem is:

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

To see that $EF/F$ is Galois, note that since $E/K$ is Galois, $E$ is a splitting field  of a set of polynomials    over $K$; clearly $EF$ is a splitting field of the same set of polynomials over $F$. Also, if $f\in K[x]$ is separable  over $K$, then also $f$ is separable over $F$. Thus $EF$ is normal and separable over $F$, so is Galois. $E$ is obviously Galois over $E\cap F$ since $E\cap F\supset K$.

 $r:H=\operatorname{Gal}(EF/F)\to\operatorname{Gal}(E/K):\sigma\mapsto\sigma|_{E}$

$r$ is clearly a group homomorphism  , and since $E$ is normal over $K$, $r$ is well-defined.

Claim $r$ is injective  . For suppose $\sigma\in\operatorname{Gal}(EF/F)$ and $\sigma|_{E}$ is the identity     . Then $\sigma$ is fixed on $F$ (since it is in $\operatorname{Gal}(EF/F)$ and on $E$ (since its restriction to $E$ is the identity), so is fixed on $EF$ and thus is itself the identity.

Now, the image of $r$ is a subgroup   of $\operatorname{Gal}(E/K)$ with fixed field $L$, and thus the image of $r$ is $\operatorname{Gal}(E/L)$. Claim $E\cap F=L$. $\subset$ is obvious: any element $x\in E\cap F$ is fixed by $\sigma|_{E}$ for each $\sigma\in H$ since $\sigma$ fixes $F$. Thus $E\cap F\subset L$. To see the reverse inclusion, choose $x\in L$; then $x$ is fixed by each $r(\sigma)$ for $\sigma\in H$. But $x\in L\subset E$, so that (as an element of $E$), $x$ is fixed by each $\sigma\in H$. Thus $x\in F$ so that $x\in E\cap F$.

## References

• 1
Title proof that the compositum of a Galois extension  and another extension    is Galois ProofThatTheCompositumOfAGaloisExtensionAndAnotherExtensionIsGalois 2013-03-22 18:41:58 2013-03-22 18:41:58 rm50 (10146) rm50 (10146) 6 rm50 (10146) Proof msc 11R32 msc 12F99