nLab root of unity

Idea

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 $x^n-1 = 0$.

Properties

Over a field

Proposition

(sum of roots of unity)
For $k \in \mathbb{N}_{\gt 0}$, let $\zeta$ be a $k$th root of unity in a field, $\zeta^k = 1$. Then

(1)$$\tfrac{1}{k} \sum_{n=0}^{k-1} \zeta^n \;=\; \left\{ \begin{array}{lcl} 1 &\vert& \zeta = 1 \\ 0 &\vert& otherwise \end{array} \right. \,.$$

Proof

The case $\zeta = 1$ is immediate. For the case $\zeta \neq 1$ observe that

$$(1 - \zeta) \sum_{n=0}^{k-1} \zeta^n \;=\; 1 - \zeta^{k} \;=\; 0 \,.$$

This may be understood as expressing the discrete Fourier transform of the Kronecker delta. For more subtle variants of the expression (1) see at Gauss sum.

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:

Theorem

Let $G$ be a finite subgroup of the multiplicative group $k^\ast$ of a field $k$. Then $G$ is cyclic.

Proof

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.

Lemma

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 $m n$.

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.

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 $k$.

Corollary

Every finite field has a cyclic multiplicative group.

Inside the multiplicative group

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

$$\mathbb{A}^1 = Spec( R[t] )$$

and the multiplicative group is

$$\mathbb{G}_m = Spec(R[t,t^{-1}])$$

then the group of $n$th roots of units is

$$\mu_n = Spec( R[t]/(t^n - 1) ) \,.$$

(For review see e.g. Watts, def. 2.3, Sutherland, example 6.7).

As such this is part of the Kummer sequence

$$\mu_n \longrightarrow \mathbb{G}_m \stackrel{(-)^n}{\longrightarrow} \mathbb{G}_m$$

Related concepts

• Gauss sum

• Q/Z

• Tate twist

References

• Jordan Watts, The Kummer sequence (pdf, archived)

• Tom Sutherland, Étale cohomology (pdf)