root of unity



An nnth root of unity in a ring RR is an element xx such that x n=1x^n = 1 in RR, hence is a root of the equation x n1=0x^n-1 = 0.


Over a field

In a field kk, a torsion element of the multiplicative group k *k^\ast is a root of unity by definition. Moreover we have the following useful result.


Let GG be a finite subgroup of the multiplicative group k *k^\ast of a field kk. Then GG is cyclic.


Let ee be the exponent of GG, i.e., the smallest n>0n \gt 0 such that g n=1g^n = 1 for all gGg \in G, and let m=order(G)m = order(G). Then each element of GG is a root of x e1x^e - 1, so that gG(xg)\prod_{g \in G} (x - g) divides x e1x^e - 1, i.e., mem \leq e. But of course g m=1g^m = 1 for all gGg \in G, so eme \leq m, and thus e=me = m.

This is enough to force GG to be cyclic. Indeed, consider the prime factorization e=p 1 r 1p 2 r 2p k r ke = p_1^{r_1} p_2^{r_2} \ldots p_k^{r_k}. Since ee is the least common multiple of the orders of elements, the exponent r ir_i is the maximum multiplicity of p ip_i occurring in orders of elements; any element realizing this maximum will have order divisible by p i r ip_i^{r_i}, and some power y iy_i of that element will have order exactly p i r ip_i^{r_i}. Then y= iy iy = \prod_i y_i will have order e=me = m by the following lemma and induction, so that powers of yy exhaust all mm elements of GG, i.e., yy generates GG as desired.


If m,nm, n are relatively prime and xx has order mm and yy has order nn in an abelian group, then xyx y has order mnm n.


Suppose (xy) k=x ky k=1(x y)^k = x^k y^k = 1. For some a,ba, b we have ambn=1a m - b n = 1, and so 1=x kamy kam=y kam=y ky kbn=y k1 = x^{k a m} y^{k a m} = y^{k a m} = y^k y^{k b n} = y^k. It follows that nn divides kk. Similarly mm divides kk, so mn=lcm(m,n)m n = lcm(m, n) divides kk, as desired.

Clearly there is at most one subgroup GG of a given order nn in k *k^\ast, which will be the set of n thn^{th} roots of unity. If GG is a finite subgroup of order nn in k *k^\ast, then a generator of GG is called a primitive n thn^{th} root of unity in kk.


Every finite field has a cyclic multiplicative group.

Inside the multiplicative group

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

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

and the multiplicative group is

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

then the group of nnth roots of units is

μ n=Spec(R[t]/(t n1)). \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

μ n𝔾 m() n𝔾 m \mu_n \longrightarrow \mathbb{G}_m \stackrel{(-)^n}{\longrightarrow} \mathbb{G}_m


  • Jordan Watts, The Kummer sequence (pdf)

  • Tom Sutherland, Étale cohomology (pdf)

Last revised on August 27, 2018 at 13:05:23. See the history of this page for a list of all contributions to it.