proof of Galois group of the compositum of two Galois extensions
Proof.
Consider the diagram
$$\text{xymatrix}\mathrm{@}R1pc\mathrm{@}C1pc\mathrm{\&}\text{ar}\mathrm{@}-[ld]\text{ar}\mathrm{@}-[rd]EF\text{ar}\mathrm{@}-[rd]E\mathrm{\&}\mathrm{\&}\text{ar}\mathrm{@}-[ld]F\mathrm{\&}\text{ar}\mathrm{@}-[d]E\cap F\mathrm{\&}K$$ |
(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
$$\theta :\mathrm{Gal}(EF/K)\to \mathrm{Gal}(E/K)\times \mathrm{Gal}(F/K):\sigma \mapsto ({\sigma |}_{E},{\sigma |}_{F})$$ |
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
$$\left|H\right|=[EF:K]$$ |
∎
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 |
---|---|
Canonical name | ProofOfGaloisGroupOfTheCompositumOfTwoGaloisExtensions |
Date of creation | 2013-03-22 18:42:01 |
Last modified on | 2013-03-22 18:42:01 |
Owner | rm50 (10146) |
Last modified by | rm50 (10146) |
Numerical id | 6 |
Author | rm50 (10146) |
Entry type | Proof |
Classification | msc 11R32 |
Classification | msc 12F99 |