Exercise 12.1.13

Question

\def\imp{\Rightarrow}
\def\gcro{\mbox{[\hspace{-.15em}[}}% intervalles d'entiers 
\def\dcro{\mbox{]\hspace{-.15em}]}}

ewcommand{\be} {\begin{enumerate}}

ewcommand{\ee} {\end{enumerate}}

ewcommand{\deb}{\begin{eqnarray*}}

ewcommand{\fin}{\end{eqnarray*}}

ewcommand{\ssi} {si et seulement si }

ewcommand{\D}{\mathrm{d}}

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}}

ewcommand{\legendre}[2]{\genfrac{(}{)}{}{}{#1}{#2}}

Let $\zeta$ be a primitive $n$th root of unity, and let ${\alpha = x_1+\zeta x_2+\cdots + \zeta^{n-1}x_n}$. Prove that $H(\alpha^n) = \langle (1\,2\ldots n)angle \subset S_n$.

Richard Ganaye · Accepted Answer

\def\imp{\Rightarrow}
\def\gcro{\mbox{[\hspace{-.15em}[}}% intervalles d'entiers 
\def\dcro{\mbox{]\hspace{-.15em}]}}

ewcommand{\be} {\begin{enumerate}}

ewcommand{\ee} {\end{enumerate}}

ewcommand{\deb}{\begin{eqnarray*}}

ewcommand{\fin}{\end{eqnarray*}}

ewcommand{\ssi} {si et seulement si }

ewcommand{\D}{\mathrm{d}}

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}}

ewcommand{\legendre}[2]{\genfrac{(}{)}{}{}{#1}{#2}}

\begin{proof}
$(1\,2\ldots n)\cdot \alpha =  x_2+\zeta x_3+\cdots + \zeta^{n-1}x_1 = \zeta^{-1} \alpha$, therefore $(1\,2\ldots n)\cdot \alpha^n = (\zeta^{-1} \alpha)^n = \alpha^n$, so 
$$ \langle (1\,2\ldots n)angle \subset H(\alpha^n).$$

Conversely, suppose that $\sigma \in H(\alpha^n)$.  Then $\sigma \cdot \alpha^n = \alpha^n$, so
$$(x_{\sigma(1)}+ \zeta x_{\sigma(2)} + \cdots +\zeta^{n-1} x_{\sigma(n)})^n = (x_1+\zeta x_2+\cdots + \zeta^{n-1}x_n)^n.$$
Therefore, there exists a $n$th root of unity $\xi$ such that
$$x_{\sigma(1)}+ \zeta x_{\sigma(2)} + \cdots + \zeta^{n-1} x_{\sigma(n)} = \xi (x_1+\zeta x_2+\cdots + \zeta^{n-1}x_n).$$

Then
\begin{align*}
\xi \sum_{i=1}^n \zeta^{i-1} x_{i} &= \sum_{j=1}^n \zeta^{j-1} x_{\sigma(j)}\
&=\sum_{i=1}^n \zeta^{\sigma^{-1}(i)-1} x_i, \qquad (i = \sigma(j))
\end{align*}
Therefore, for all $i=1,\ldots,n$,
$$\xi\, \zeta^{i-1} = \zeta^{\sigma^{-1}(i)-1}$$
For $i=1$, we obtain $\xi =\zeta^{\sigma^{-1}(1)-1}$, thus $\zeta^{\sigma^{-1}(1)-1+ i-1} = \zeta^{\sigma^{-1}(i)-1}$.

Since $\zeta$ is a primitive $n$th root of unity,
$$\sigma^{-1}(1)+ i-1 \equiv \sigma^{-1}(i) \pmod n \qquad (1\leq i \leq n).$$
If $k = \sigma^{-1}(1) -1$, then 
$$\sigma^{-1}(i) \equiv i+k \pmod n,$$
therefore $\sigma^{-1} = (1\,2\ldots n)^k, \sigma = (1\,2\ldots n)^{n-k}$ are in the subgroup  $\langle (1\,2\ldots n)angle$.

$$H(\alpha^n) = \langle (1\,2\ldots n)angle.$$
\end{proof}

Exercise 12.1.13

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Exercise 12.1.13

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