Exercise 9.1.3

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 $\zeta_n = e^{2\pi i/n} \in \C$. Prove that $\zeta_n^i$ for $0\leq i <n$ and $\gcd(i,n) = 1$ are the primitive $n$th roots of unity in $\C$.

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}
Let $\U_n$ be the group of $n$-th roots of unity in $\C$. Then $\U_n = \langle \zeta_n angle$, where $\zeta_n = e^{2i\pi/n}$. Write $o(x)$ the order of an element $x\in G$. Then $o(x)= \vert \langle x angle \vert$.

Recall that if $d>0$, $o(x)= d \iff (\forall k \in \Z,\  x^k = e \iff d \mid k)$.

For all $i \in \Z$,
$$o(\zeta_n^i)=\frac{n}{n\wedge i}.$$
Indeed, for all $k\in\Z$,
$$(\zeta_n ^i)^k = 1 \iff \zeta_n^{ik} = 1 \iff n \mid ik \iff \frac{n}{n\wedge i} \mid \frac{i}{n\wedge i} k \iff  \frac{n}{n\wedge i} \mid k$$
(since $\frac{n}{n\wedge i} \wedge \frac{i}{n\wedge i} = 1$).
So $o(\zeta_n^i)=\frac{n}{n\wedge i}$.

By definition, $\zeta$ is a primitive $n$th root of unity if and only if $\zeta$ is a generator of $\U_n$, if and only if $o(\zeta)= n$, so

$$\U_n = \langle \zeta_n^i angle \iff o(\zeta_n^i)= n \iff \frac{n}{n\wedge i} = n \iff n \wedge i = 1.$$
\end{proof}

Exercise 9.1.3

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Exercise 9.1.3

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