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\begin{document}

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\section*{root}

\begin{quote}%
This entry is about the notion of \emph{root} in [[algebra]]. For the notion in [[representation theory]] see at \emph{[[root (in representation theory)]]}.


\end{quote}
\hypertarget{context}{}\subsubsection*{{Context}}\label{context}

\hypertarget{algebra}{}\paragraph*{{Algebra}}\label{algebra}

[[!include higher algebra - contents]]

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

\noindent\hyperlink{definitions}{Definitions}\dotfill \pageref*{definitions} \linebreak
\noindent\hyperlink{rootsunity}{Roots of unity in fields}\dotfill \pageref*{rootsunity} \linebreak
\noindent\hyperlink{related_concepts}{Related concepts}\dotfill \pageref*{related_concepts} \linebreak


\hypertarget{definitions}{}\subsection*{{Definitions}}\label{definitions}

A \textbf{[[square root]]} of an element $a$ in a [[monoid]] $M$ is a solution to the equation $x^2 = a$ within $M$. If $M$ is the multiplicative monoid of non-zero elements of a (commutative) [[field]] or [[integral domain]] $K$ (or a submonoid of this, such as the [[group of units]]), then there are exactly two square roots, denoted $\pm \sqrt{a}$, if there are any; the element $0$ has only one square root, $\pm \sqrt{0} = 0$. But if $K$ is even a non-integral domain or a non-commutative [[skew field]], then there may be more; in the skew field $\mathbb{H}$ of [[quaternions]], there are [[continuum|continuumly]] many square roots of $-1$.

More generally, for $n \in \mathbb{N}$ a solution to the [[equation]] $x^n = a$ is called an \textbf{$n$th root} of $a$. Specifically for $a = 1$ and working in the [[field]] of [[complex number]]s, one speaks of an \textbf{$n$th [[root of unity]]}. This terminology can be applied to other fields as well; for example, the field of \href{http://ncatlab.org/nlab/show/p-adic+number}{7-adic numbers} contains non-trivial cube (or $3^{rd}$) [[roots of unity]].

More generally, for any [[polynomial]] $P(x)$ of $x$ with coefficients in a field $K$, a solution to $P(x) = 0$ in $K$ is called a \textbf{root} of $P$. When $P(x) \in K[x]$ has no solution in $k$, one can speak of a [[splitting field]] obtained by ``adjoining roots'' of $P$ to $K$, meaning that one considers roots in an [[extension field]] $i: K \hookrightarrow E$ of the corresponding polynomial $Q = (i \otimes_K K[x])(P) \in E[x]$, i.e., applying the evident composite map

\begin{displaymath}
K[x] \cong K \otimes_K K[x] \stackrel{i \otimes_K 1}{\to} E \otimes_K K[x] \cong E[x]
\end{displaymath}
to $P$ to get $Q$, and passing the smallest [[intermediate subfield]] between $K$ and $E$ that contains the designated roots of $Q$ (often writing $P$ for $Q$ by abuse of language).

More generally still, one may refer to roots even of non-polynomial [[functions]] $f$ defined on a field, for example of [[meromorphic functions]] $f \colon \mathbb{C} \to \mathbb{C}$, although it is much more usual to speak of \emph{zeroes} of $f$ instead of roots of $f$ (e.g., zeroes of the [[Riemann zeta function]]); see [[zero set]] and [[intermediate value theorem]].

\hypertarget{rootsunity}{}\subsection*{{Roots of unity in fields}}\label{rootsunity}

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 $x^e - 1$, so $m \leq e$ by comparing [[polynomial|degrees]]. 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, write $e = p_1^{r_1} p_2^{r_2} \ldots p_k^{r_k}$. 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{related_concepts}{}\subsection*{{Related concepts}}\label{related_concepts}

\begin{itemize}%
\item \textbf{root}


\item [[square root]]


\item [[Newton's method]]


\item [[Chern root]]



\end{itemize}
[[!redirects root]] [[!redirects roots]]



\end{document}