Exercise 9.1.11 (Solvability of Galois group)

Question

Use Lemmas 9.1.6 and 9.1.8 to show that $\mathrm{Gal}_{\mathbb{Q}}(t^n - a)$ is solvable for all $a \in \mathbb{Q}$.

Hassaan Naeem · Accepted Answer

extbf{Solution:}
A group is solvable if it can be constructed from abelian groups through extensions.
Hence we have $\{1\} = G_0 \unlhd G_1 \unlhd ... \unlhd G_k = G$ such that $G_{i-1}$ is normal in $G_i$ and $G_{i} / G_{i-1}$
is an abelian group, for $i = 1, 2, ..., k$.
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We have that $Gal_{\mathbb{Q}}(t^n-a) = Gal(SF_{\mathbb{Q}}(t^n - a): \mathbb{Q})$.
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If $a=0$ then $Gal_{\mathbb{Q}}(t^n) = Gal(SF_{\mathbb{Q}}(t^n): \mathbb{Q}) = Gal(\mathbb{Q}: \mathbb{Q}) = \{1\}$ which is trivially solvable. 
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For $a 
eq 0$, we need to show that $Gal_{\mathbb{Q}}(t^n-a)$ is abelian. If we choose a positive rational root $\xi$ of $t^n - a$ and let $\omega = e^{2\pi i /n}$, then
all the roots of $t^n - a$ are $\xi, \omega\xi, ..., \omega^{n-1}\xi$. Hence $SF_{\mathbb{Q}}(t^n - a)$ contains $(\omega^i\xi)/\xi \ \forall i$, therefore $t^n-1$ splits in $SF_{\mathbb{Q}}(t^n - a)$.
Hence we have $\mathbb{Q} \subseteq SF_{\mathbb{Q}}(t^n-1) \subseteq SF_{\mathbb{Q}}(t^n-a)$.
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By 	extbf{Lemma 9.1.6} we have that $Gal_{\mathbb{Q}}(t^n-1)$ is abelian.  Let $K = SF_{\mathbb{Q}}(t^n-1)$, then by 	extbf{Lemma 9.1.8} $Gal_K(t^n-a) = Gal(SF_K(t^n-a):K)$ is abelian.
Since all the extensions are splitting fields, we also know they are all normal.
Hence we have $\{1\} \unlhd Gal(K:\mathbb{Q}) \unlhd Gal(SF_{K}(t^n-a): K)$.
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We know that $Gal(K:\mathbb{Q})/\{ 1 \} = Gal(K:\mathbb{Q})$ which is abelian.
We also know that $Gal(SF_{K}(t^n-a):K)/Gal(K:\mathbb{Q})$ is abelian since the quotient group of an abelian group is also abelian.
Therefore $Gal_{\mathbb{Q}}(t^n - a)$  is solvable.

Richard Ganaye · Answer

ewcommand{\Q}{\mathbb{Q}} ewcommand{\Z}{\mathbb{Z}} ewcommand{\N}{\mathbb{N}} ewcommand{\R}{\mathbb{R}} ewcommand{\C}{\mathbb{C}} ewcommand{\F}{\mathbb{F}} ewcommand{\U}{\mathbb{U}} ewcommand{ e}{\,\mathrm{Re}\,} ewcommand{\im}{\,\mathrm{Im}\,} ewcommand{\ord}{\mathrm{ord}} ewcommand{\Gal}{\mathrm{Gal}} \begin{proof} For $a= 0$, $\Gal(K/\Q)$ is trivial, so we can assume that $a e 0$. Write $M = \mathrm{SF}_{\mathbb{Q}}(t^n - a)$ and $K = \mathrm{SF}_{\mathbb{Q}}(t^n - 1)$. If $\omega = e^{2i\pi/n}$, then $t^n - 1 = (t-1)(t- \omega)\cdots(t-\omega^{n-1})$, so $K =\Q(1,\omega,\cdots,\omega^{n-1}) = \Q(\omega)$. Let $\xi = \sqrt[n]{a}$ be the positive real root of $t^n -a$. Then $t^n -a = (t-\xi)(t- \omega \xi)\cdots(t- \omega^{n-1} \xi)$, and $M = \Q(\xi, \omega \xi,\ldots, \omega^{n-1} \xi) = \Q(\omega, \xi)$. This shows that $K = \Q(\omega) \subset \Q(\omega, \xi) = M$. Consider these extensions and their Galois groups: \medskip \begin{center} \begin{tikzpicture} ode(1) at (4,6) {$1$} ; ode (H1) at (4,4) {$\Gal(M :K)$} ; ode (H2) at (4,2) {$G = \Gal(M:\Q) = \Gal_\Q(t^n-a)$} ; \draw [<-] (H2) edge (H1) ; \draw [<-] (H1) edge (1) ; ode [<-](M) at (0,6) {$M = \Q(\omega, \xi) = \mathrm{SF}_\Q(t^n-a)$} ; ode (FH1) at (0,4) {$K = \Q(\omega) = \mathrm{SF}_\Q(t^n-1)$} ; ode (FH2) at (0,2) {$\Q$} ; \draw [->](FH2) edge (FH1) ; \draw [->](FH1) edge (M) ; \end{tikzpicture} \end{center} Then $K$ is a normal extension of $\Q$, since it is the splitting field of $t^n -1$ over $\Q$. Therefore $\Gal(M:K) \lhd \Gal(M:\Q)$, and, by the Fundamental Theorem, \begin{align} \Gal(M:\Q)/\Gal(M:K) \simeq \Gal(K/\Q). \end{align} By Lemma 9.1.6, $ \Gal(K/\Q) = \Gal_\Q(t^n-1)$ is abelian, and by Lemma 9.1.8, since $t^n - 1 = (t-1)(t- \omega)\cdots(t-\omega^{n-1})$ splits in $K$, $\Gal(M : K) = \Gal_K(t^n-a)$ is abelian. Using $$\{1_M\} \lhd \Gal(M:K) \lhd \Gal(M:\Q),$$ where $\Gal(M:K)/\{1_M\} \simeq \Gal(M:K)$ is abelian, and $\Gal(M:\Q)/\Gal(M:K) \simeq \Gal(K/\Q)$ is abelian, we conclude that $\Gal(K/\Q) = \Gal_\Q(t^n-a)$ is solvable. \end{proof}

Exercise 9.1.11 (Solvability of Galois group)

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Exercise 9.1.11 (Solvability of Galois group)

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