symmetric monoidal (∞,1)-category of spectra
transfinite arithmetic, cardinal arithmetic, ordinal arithmetic
prime field, p-adic integer, p-adic rational number, p-adic complex number
arithmetic geometry, function field analogy
An $n$th 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 $x^n-1 = 0$.
In a field $k$, a torsion element of the multiplicative group $k^\ast$ is a root of unity by definition. Moreover we have the following useful result.
Let $G$ be a finite subgroup of the multiplicative group $k^\ast$ of a field $k$. Then $G$ is cyclic.
Let $e$ be the 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.
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 $m n$.
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.
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 primitive $n^{th}$ root of unity in $k$.
Every finite field has a cyclic multiplicative group.
Given a base commutative ring $R$, such that the affine line is
and the multiplicative group is
then the group of $n$th roots of units is
(For review see e.g. Watts, def. 2.3, Sutherland, example 6.7).
As such this is part of the Kummer sequence
Last revised on December 18, 2020 at 15:04:20. See the history of this page for a list of all contributions to it.