Question

Suppose $\mathit{\zeta }\mathbf{,}{\mathit{\zeta }}^{\mathbf{2}}\mathbf{,}\mathbf{.}\mathbf{.....}{\mathit{\zeta }}^{\mathbf{n}}\mathbf{=}\mathbf{1}$ are $\mathit{n}$distinct root of ${\mathit{x}}^{\mathbf{n}}\mathbf{-}\mathbf{1}$some extension field of $\mathit{Q}$. Prove that $\mathit{G}\mathit{a}{\mathit{l}}_{\mathbf{Q}}\mathit{Q}\mathbf{\left(}\mathbf{\zeta }\mathbf{\right)}$is abelian.

Short answer

${\mathbf{Gal}}_{\mathbf{Q}}\mathbf{Q}\mathbf{\left(}\mathbf{\zeta }\mathbf{\right)}$is abelian.

Step-by-step solution

Definition of the abelian group.

An abelian group is commutative group and satisfy the condition for all $\mathit{a}\mathbf{,}\mathit{b}$ in field extension is $\mathit{a}\mathbf{\cdot }\mathit{b}\mathbf{=}\mathit{b}\mathbf{\cdot }\mathit{a}$.

Step-2: Showing that GalQQ(ζ) is abelian.

Let $\mathbf{\zeta }\mathbf{,}{\mathbf{\zeta }}^{\mathbf{2}}\mathbf{,}\mathbf{.}\mathbf{.....}{\mathbf{\zeta }}^{\mathbf{n}}\mathbf{=}\mathbf{1}$are $\mathit{n}$ distinct root of the minimal polynomial ${\mathit{x}}^{\mathbf{n}}\mathbf{-}\mathbf{1}$over the extension field of $\mathit{Q}$ such that $\mathit{G}\mathit{a}{\mathit{l}}_{\mathbf{Q}}\mathit{Q}\mathbf{\left(}\mathbf{\zeta }\mathbf{\right)}$ is abelian and the condition of abelian for two element $\mathit{a}\mathbf{,}\mathit{b}$are $\mathit{a}\mathbf{\cdot }\mathit{b}\mathbf{=}\mathit{b}\mathbf{\cdot }\mathit{a}$.

Therefore the minimal polynomial ${\mathit{x}}^{\mathbf{n}}\mathbf{-}\mathbf{1}$ has $\mathit{n}$ roots ${\mathit{\zeta }}^{\mathbf{n}}\mathbf{=}\mathbf{1}$ it can be represent as a $\mathbf{\{}\mathbf{1}\mathbf{,}\mathbf{\zeta }\mathbf{,}{\mathbf{\zeta }}^{\mathbf{2}}\mathbf{,}\mathbf{......}{\mathbf{\zeta }}^{\mathbf{n}\mathbf{-}\mathbf{1}}\mathbf{\}}$. Now the Galois group of the set of root of polynomial ${\mathbf{Gal}}_{\mathbf{Q}}\mathbf{Q}\mathbf{\left(}\mathbf{\zeta }\mathbf{\right)}\mathbf{,}{\mathbf{\varphi }}_{\mathbf{i}}\mathbf{\left(}\mathbf{\zeta }\mathbf{\right)}\mathbf{=}{\mathbf{\zeta }}^{\mathbf{i}}$ and the abelian expression is $\mathbf{\{}{\mathbf{Gal}}_{\mathbf{Q}}\mathbf{Q}\mathbf{\left(}\mathbf{\zeta }\mathbf{\right)}\mathbf{,}\mathbf{o}\mathbf{\}}$ for any two point $\mathit{i}\mathbf{,}\mathit{j}$.

$\mathbf{\{}{\mathbf{\varphi }}_{\mathbf{i}}\mathbf{\circ }{\mathbf{\varphi }}_{\mathbf{j}}\mathbf{\}}\mathbf{\left(}\mathbf{\zeta }\mathbf{\right)}\mathbf{=}\mathbf{\{}{\mathbf{\varphi }}_{\mathbf{j}}\mathbf{\circ }{\mathbf{\varphi }}_{\mathbf{i}}\mathbf{\}}\mathbf{\left(}\mathbf{\zeta }\mathbf{\right)}$

Use left hand side of the expression

$\begin{array}{c}\{{\mathrm{\varphi }}_{\mathrm{i}}\circ {\mathrm{\varphi }}_{\mathrm{j}}\}\left(\mathrm{\zeta }\right)={\mathrm{\varphi }}_{\mathrm{i}}\left({\mathrm{\varphi }}_{\mathrm{j}}\left(\mathrm{\zeta }\right)\right)\\ ={\mathrm{\varphi }}_{\mathrm{i}}\left({\mathrm{\zeta }}^{\mathrm{j}}\right)\\ ={\left({\mathrm{\varphi }}_{\mathrm{i}}\left(\mathrm{\zeta }\right)\right)}^{\mathrm{j}}\\ ={\mathrm{\zeta }}^{\mathrm{ij}}\end{array}$

Further solve the

$\begin{array}{c}\{{\varphi }_i\circ {\varphi }_j\}\left(\zeta \right)={\zeta }^{ij}\\ ={\left({\zeta }^i\right)}^j\\ ={\left({\varphi }_i\left(\zeta \right)\right)}^j\\ ={\varphi }_j\left({\varphi }_i\left(\zeta \right)\right)\end{array}$

Now use the abelian property.

$\mathbf{\{}{\mathbf{\varphi }}_{\mathbf{i}}\mathbf{\circ }{\mathbf{\varphi }}_{\mathbf{j}}\mathbf{\}}\mathbf{\left(}\mathbf{\zeta }\mathbf{\right)}\mathbf{=}\mathbf{(}{\mathbf{\varphi }}_{\mathbf{j}}\mathbf{\circ }{\mathbf{\varphi }}_{\mathbf{i}}\mathbf{)}\mathbf{\left(}\mathbf{\zeta }\mathbf{\right)}$

So left hand side is equal to right hand side. Thus Galois group of the given root of the set ${\mathbf{Gal}}_{\mathbf{Q}}\mathbf{Q}\mathbf{\left(}\mathbf{\zeta }\mathbf{\right)}$is abelian.