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\begin{document}

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\section*{root of unity}

\hypertarget{contents}{}\section*{{Contents}}\label{contents}

\noindent\hyperlink{idea}{Idea}\dotfill \pageref*{idea} \linebreak
\noindent\hyperlink{properties}{Properties}\dotfill \pageref*{properties} \linebreak
\noindent\hyperlink{over_a_field}{Over a field}\dotfill \pageref*{over_a_field} \linebreak
\noindent\hyperlink{inside_the_multiplicative_group}{Inside the multiplicative group}\dotfill \pageref*{inside_the_multiplicative_group} \linebreak
\noindent\hyperlink{related_concepts}{Related concepts}\dotfill \pageref*{related_concepts} \linebreak
\noindent\hyperlink{references}{References}\dotfill \pageref*{references} \linebreak


\hypertarget{idea}{}\subsection*{{Idea}}\label{idea}

An $n$th \emph{root of unity} in a [[ring]] $R$ is an element $x$ such that $x^n = 1$ in $R$, hence is a [[root]] of the [[equation]] $x^n-1 = 0$.

\hypertarget{properties}{}\subsection*{{Properties}}\label{properties}

\hypertarget{over_a_field}{}\subsubsection*{{Over a field}}\label{over_a_field}

In a [[field]] $k$, a [[torsion subgroup|torsion]] element of the multiplicative group $k^\ast$ is a [[root of unity]] by definition. Moreover we have the following useful result.

\begin{theorem}
\label{}\hypertarget{}{}
Let $G$ be a [[finite group|finite]] [[subgroup]] of the [[multiplicative group]] $k^\ast$ of a field $k$. Then $G$ is [[cyclic group|cyclic]].

\end{theorem}
\begin{proof}
Let $e$ be the \emph{[[exponent of a group|exponent]]} of $G$, i.e., the smallest $n \gt 0$ such that $g^n = 1$ for all $g \in G$, and let $m = order(G)$. Then each element of $G$ is a root of $x^e - 1$, so that $\prod_{g \in G} (x - g)$ divides $x^e - 1$, i.e., $m \leq e$. But of course $g^m = 1$ for all $g \in G$, so $e \leq m$, and thus $e = m$.

This is enough to force $G$ to be cyclic. Indeed, consider the prime factorization $e = p_1^{r_1} p_2^{r_2} \ldots p_k^{r_k}$. Since $e$ is the least common multiple of the orders of elements, the exponent $r_i$ is the maximum multiplicity of $p_i$ occurring in orders of elements; any element realizing this maximum will have order divisible by $p_i^{r_i}$, and some power $y_i$ of that element will have order exactly $p_i^{r_i}$. Then $y = \prod_i y_i$ will have order $e = m$ by the following lemma and induction, so that powers of $y$ exhaust all $m$ elements of $G$, i.e., $y$ generates $G$ as desired.

\end{proof}
\begin{lemma}
\label{}\hypertarget{}{}
If $m, n$ are relatively prime and $x$ has order $m$ and $y$ has order $n$ in an abelian group, then $x y$ has order $m n$.

\end{lemma}
\begin{proof}
Suppose $(x y)^k = x^k y^k = 1$. For some $a, b$ we have $a m - b n = 1$, and so $1 = x^{k a m} y^{k a m} = y^{k a m} = y^k y^{k b n} = y^k$. It follows that $n$ divides $k$. Similarly $m$ divides $k$, so $m n = lcm(m, n)$ divides $k$, as desired.

\end{proof}
Clearly there is at most one subgroup $G$ of a given order $n$ in $k^\ast$, which will be the set of $n^{th}$ [[roots of unity]]. If $G$ is a finite subgroup of order $n$ in $k^\ast$, then a generator of $G$ is called a \textbf{primitive} $n^{th}$ [[root of unity]] in $k$.

\begin{cor}
\label{}\hypertarget{}{}
Every [[finite field]] has a [[cyclic group|cyclic]] [[multiplicative group]].

\end{cor}
\hypertarget{inside_the_multiplicative_group}{}\subsubsection*{{Inside the multiplicative group}}\label{inside_the_multiplicative_group}

Given a base [[commutative ring]] $R$, such that the [[affine line]] is

\begin{displaymath}
\mathbb{A}^1 = Spec( R[t] )
\end{displaymath}
and the [[multiplicative group]] is

\begin{displaymath}
\mathbb{G}_m = Spec(R[t,t^{-1}])
\end{displaymath}
then the group of $n$th roots of units is

\begin{displaymath}
\mu_n = Spec( R[t]/(t^n - 1) ) 
  \,.
\end{displaymath}
(For review see e.g. \hyperlink{Watts}{Watts, def. 2.3}, \hyperlink{Sutherland}{Sutherland, example 6.7}).

As such this is part of the [[Kummer sequence]]

\begin{displaymath}
\mu_n \longrightarrow \mathbb{G}_m \stackrel{(-)^n}{\longrightarrow} \mathbb{G}_m
\end{displaymath}
\hypertarget{related_concepts}{}\subsection*{{Related concepts}}\label{related_concepts}

\begin{itemize}%
\item [[Tate twist]]

\end{itemize}
\hypertarget{references}{}\subsection*{{References}}\label{references}

\begin{itemize}%
\item Jordan Watts, \emph{The Kummer sequence} (\href{http://www.math.toronto.edu/jwatts/papers/galois.pdf}{pdf})


\item Tom Sutherland, \emph{\'E{}tale cohomology} (\href{http://www-dimat.unipv.it/sutherland/etale.pdf}{pdf})



\end{itemize}
[[!redirects roots of unity]]

[[!redirects primitive root of unity]] [[!redirects primitive roots of unity]]



\end{document}