Root of unity modulo n

In mathematics, namely ring theory, a k-th root of unity modulo n for positive integers k, n ≥ 2, is a root of unity in the ring of integers modulo n, that is, a solution x to the equation (or congruence) $x^{k}\equiv 1{\pmod {n}}$. If k is the smallest such exponent for x, then x is called a primitive k-th root of unity modulo n. See modular arithmetic for notation and terminology.

Do not confuse this with a primitive root modulo n, which is a generator of the group of units of the ring of integers modulo n. The primitive roots modulo n are the primitive $\varphi (n)$-roots of unity modulo n, where $\varphi$ is Euler's totient function.

Roots of unity

Properties

• If x is a k-th root of unity, then x is a unit (invertible) whose inverse is $x^{k-1}$. That is, x and n are coprime.
• If x is a unit, then it is a (primitive) k-th root of unity modulo n, where k is the multiplicative order of x modulo n.
• If x is a k-th root of unity and $x-1$ is not a zero divisor, then $\sum _{j=0}^{k-1}x^{j}\equiv 0{\pmod {n}}$, because

$(x-1)\cdot \sum _{j=0}^{k-1}x^{j}\equiv x^{k}-1\equiv 0{\pmod {n}}.$

Number of k-th roots

For the lack of a widely accepted symbol, we denote the number of k-th roots of unity modulo n by $f(n,k)$. It satisfies a number of properties:

• $f(n,1)=1$ for $n\geq 2$
• $f(n,\lambda (n))=\varphi (n)$ where λ denotes the Carmichael function and $\varphi$ denotes Euler's totient function
• $n\mapsto f(n,k)$ is a multiplicative function
• $k\mid l\implies f(n,k)\mid f(n,l)$ where the bar denotes divisibility
• $f(n,\mathrm {lcm} (a,b))=\mathrm {lcm} (f(n,a),f(n,b))$ where $\mathrm {lcm}$ denotes the least common multiple
• For prime $p$, $\forall i\in \mathbb {N} \ \exists j\in \mathbb {N} \ f(n,p^{i})=p^{j}$. The precise mapping from $i$ to $j$ is not known. If it were known, then together with the previous law it would yield a way to evaluate $f$ quickly.

Primitive roots of unity

Properties

• The maximum possible radix exponent for primitive roots modulo $n$ is $\lambda (n)$, where λ denotes the Carmichael function.
• A radix exponent for a primitive root of unity is a divisor of $\lambda (n)$.
• Every divisor $k$ of $\lambda (n)$ yields a primitive $k$-th root of unity. You can obtain one by choosing a $\lambda (n)$-th primitive root of unity (that must exist by definition of λ), named $x$ and compute the power $x^{\lambda (n)/k}$.
• If x is a primitive k-th root of unity and also a (not necessarily primitive) ℓ-th root of unity, then k is a divisor of ℓ. This is true, because Bézout's identity yields an integer linear combination of k and ℓ equal to $\mathrm {gcd} (k,\ell )$. Since k is minimal, it must be $k=\mathrm {gcd} (k,\ell )$ and $\mathrm {gcd} (k,\ell )$ is a divisor of ℓ.

Number of primitive k-th roots

For the lack of a widely accepted symbol, we denote the number of primitive k-th roots of unity modulo n by $g(n,k)$. It satisfies the following properties:

• $g(n,k)={\begin{cases}>0&{\text{if }}k\mid \lambda (n),\\0&{\text{otherwise}}.\end{cases}}$
• Consequently the function $k\mapsto g(n,k)$ has $d(\lambda (n))$ values different from zero, where $d$ computes the number of divisors.
• $g(n,1)=1$
• $g(4,2)=1$
• $g(2^{n},2)=3$ for $n\geq 3$, since -1 is always a square root of 1.
• $g(2^{n},2^{k})=2^{k}$ for $k\in [2,n-1)$
• $g(n,2)=1$ for $n\geq 3$ and $n$ in (sequence A033948 in the OEIS)
• $\sum _{k\in \mathbb {N} }g(n,k)=f(n,\lambda (n))=\varphi (n)$ with $\varphi$ being Euler's totient function
• The connection between $f$ and $g$ can be written in an elegant way using a Dirichlet convolution:

$f=\mathbf {1} *g$, i.e. $f(n,k)=\sum _{d\mid k}g(n,d)$

You can compute values of $g$ recursively from $f$ using this formula, which is equivalent to the Möbius inversion formula.

Testing whether x is a primitive k-th root of unity modulo n

By fast exponentiation you can check that $x^{k}\equiv 1{\pmod {n}}$. If this is true, x is a k-th root of unity modulo n but not necessarily primitive. If it is not a primitive root, then there would be some divisor ℓ of k, with $x^{\ell }\equiv 1{\pmod {n}}$. In order to exclude this possibility, one has only to check for a few ℓ's equal k divided by a prime. That is, what needs to be checked is:

$\forall p{\text{ prime dividing}}\ k,\quad x^{k/p}\not \equiv 1{\pmod {n}}.$

Finding a primitive k-th root of unity modulo n

Among the primitive k-th roots of unity, the primitive $\lambda (n)$-th roots are most frequent. It is thus recommended to try some integers for being a primitive $\lambda (n)$-th root, what will succeed quickly. For a primitive $\lambda (n)$-th root x, the number $x^{\lambda (n)/k}$ is a primitive $k$-th root of unity. If k does not divide $\lambda (n)$, then there will be no k-th roots of unity, at all.

Finding multiple primitive k-th roots modulo n

Once you have a primitive k-th root of unity x, every power $x^{l}$ is a $k$-th root of unity, but not necessarily a primitive one. The power $x^{l}$ is a primitive $k$-th root of unity if and only if $k$ and $l$ are coprime. The proof is as follows: If $x^{l}$ is not primitive, then there exists a divisor $m$ of $k$ with $(x^{l})^{m}\equiv 1{\pmod {n}}$, and since $k$ and $l$ are coprime, there exists an inverse $l^{-1}$ of $l$ modulo $k$. This yields $1\equiv ((x^{l})^{m})^{l^{-l}}\equiv x^{m}{\pmod {n}}$, which means that $x$ is not a primitive $k$-th root of unity because there is the smaller exponent $m$.

That is, by exponentiating x one can obtain $\varphi (k)$ different primitive k-th roots of unity, but these may not be all such roots. However, finding all of them is not so easy.

Finding an n with a primitive k-th root of unity modulo n

You may want to know, in what integer residue class rings you have a primitive k-th root of unity. You need it for instance if you want to compute a Discrete Fourier Transform (more precisely a Number theoretic transform) of a $k$-dimensional integer vector. In order to perform the inverse transform, you also need to divide by $k$, that is, k shall also be a unit modulo $n$.

A simple way to find such an n is to check for primitive k-th roots with respect to the moduli in the arithmetic progression $k+1,2k+1,3k+1,\dots$. All of these moduli are coprime to k and thus k is a unit. According to Dirichlet's theorem on arithmetic progressions there are infinitely many primes in the progression, and for a prime $p$ it holds $\lambda (p)=p-1$. Thus if $mk+1$ is prime then $\lambda (mk+1)=mk$ and thus you have primitive k-th roots of unity. But the test for primes is too strong, you may find other appropriate moduli.

Finding an n with multiple primitive roots of unity modulo n

If you want to have a modulus $n$ such that there are primitive $k_{1}$-th, $k_{2}$-th, ..., $k_{m}$-th roots of unity modulo $n$, the following theorem reduces the problem to a simpler one:

For given $n$ there are primitive $k_{1}$-th, ..., $k_{m}$-th roots of unity modulo $n$ if and only if there is a primitive $\mathrm {lcm} (k_{1},...,k_{m})$-th root of unity modulo n.

Proof

Backward direction: If there is a primitive $\mathrm {lcm} (k_{1},...,k_{m})$-th root of unity modulo $n$ called $x$, then $x^{\mathrm {lcm} (k_{1},\dots ,k_{m})/k_{l}}$ is a $k_{l}$-th root of unity modulo $n$.

Forward direction: If there are primitive $k_{1}$-th, ..., $k_{m}$-th roots of unity modulo $n$, then all exponents $k_{1},\dots ,k_{m}$ are divisors of $\lambda (n)$. This implies $\mathrm {lcm} (k_{1},\dots ,k_{m})\mid \lambda (n)$ and this in turn means there is a primitive $\mathrm {lcm} (k_{1},...,k_{m})$-th root of unity modulo $n$.