# Proof Verification/Strategy: Galois Group of the splitting field of a polynomial is solvable.

Hi I just wanted to pose a question and see if my approach is valid. I have a more rigorous way of explaining it but it's a little lengthy and I wanted to be as brief as I could. I'm working to prove that the Galois group Gal($\mathbb{E}$/$\mathbb{F}$) is solvable where $\mathbb{E}$ is the splitting field of $$f=x^{n}-\alpha$$ where $f \in \mathbb{F}[x]$. Basically, The Fundamental Theorem of Galois gives me that Gal($\mathbb{E}$/$\mathbb{B}$) is normal to Gal($\mathbb{E}$/$\mathbb{F}$) and that Gal($\mathbb{B}$/$\mathbb{F}$) $\cong$ Gal($\mathbb{E}$/$\mathbb{F}$)/Gal($\mathbb{E}$/$\mathbb{B}$)

($\mathbb{B}$ is the splitting field of $x^{n}-1$ over $\mathbb{F}$).

So I have to prove Gal($\mathbb{E}$/$\mathbb{B}$) and Gal($\mathbb{B}$/$\mathbb{F}$) are solvable. I'm trying to show Gal($\mathbb{E}$/$\mathbb{B}$) is solvable and I wanted to be sure that what I'm saying is valid. Basically every $\phi$ in Gal($\mathbb{E}$/$\mathbb{B}$) only permutes $\beta$ (where $\beta$ is the n$^{th}$ root of $\alpha$). It must be the case that $\phi_{k}(\beta)=\beta\zeta \in$ Gal( $\mathbb{E}$/$\mathbb{B}$) and this is a generator for this group. So it is a cyclic group. So every subgroup is cyclic and every quotient group is cyclic. So every factor in the composition series is cyclic and thus the group is solvable.

I'm slightly suspicious of whether or not that particular $\phi_{k}$ is definitely in the Galois group. My reasoning is that, since it's coming from Gal($\mathbb{E}$/$\mathbb{F}$) and it fixes $\zeta$ it's essentially just $\{ \phi_{i} \in Gal(\mathbb{E}/\mathbb{F}): \phi_{i}(\beta)=\beta\zeta^{i},\phi_{i}(\zeta)=\zeta, 1 \leq i \leq n\}$ and $\phi(\beta)=\beta\zeta$ is definitely in there.

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What is $\Bbb B$? – Greg Martin Jun 18 '14 at 6:23
the splitting field of $x^{n}-1$ over $\mathbb{F}$ – Thomas Hughes Jun 18 '14 at 7:02

$\newcommand{\Gal}{\text{Gal}}$$G = \Gal(\mathbb{E}/\mathbb{B})$ is abelian. Fix one root $\gamma$ of $f$. Then the roots of $f$ are of the form $\gamma z$, where $z \in \mathbb{B}$ is a root of $x^n -1$. Thus an element of $G$ is determined by its action on $\gamma$. So if $g, h \in G$, and $g$ sends $\gamma$ to $\gamma z$, and $h$ sends $\gamma$ to $\gamma w$, then $gh$ and $hg$ send $\gamma$ to $\gamma z w$, hence they coincide.
Thank you so much! I actually went ahead and finished the proof and found that both Galois groups Gal($\mathbb{E}/\mathbb{B}$) and Gal($\mathbb{B}/\mathbb{F}$) are abelian and solvable and, thus, the Galois group Gal($\mathbb{E}/\mathbb{F}$) is solvable! Thanks again! – Thomas Hughes Jun 20 '14 at 16:46