# examples of cyclotomic polynomials

In this entry we calculate a number of cyclotomic polynomials  , $\Phi_{d}(x)\in\mathbb{Q}[x]$, for various $d\geq 1$. The interested reader can find specific examples at the bottom of the entry. We will concentrate first on the theory details which allow us to calculate these polynomials.

## 1 The theory behind the examples

The following simple lemma is also useful when calculating cyclotomic polynomials:

###### Lemma 1.

Let $n,d\geq 2$ be positive integers. If $d$ divides $n$ then $\Phi_{d}(x)$ divides $q_{n}(x)=\frac{x^{n}-1}{x-1}$.

###### Proof.

Let $\zeta_{d}$ be a primitive $d$th root of unity  and let $\Phi_{d}(x)$ be the $d$th cyclotomic polynomial (which is the minimal polynomial of $\zeta_{d}$ over $\mathbb{Q}$). If $d$ divides $n$ then there is a natural number  $m\geq 1$ such that $n=dm$. Thus

 $\zeta_{d}^{n}=(\zeta_{d}^{d})^{m}=1^{m}=1$

and so, $\zeta_{d}$ is also an $n$th root of unity and, in particular, it is a root of $q_{n}(x)$. By the properties of minimal polynomials, $\Phi_{d}(x)$ must divide $q_{n}(x)$. ∎

###### Lemma 2.

The polynomial $\Phi_{d}(x)$ is of degree $\varphi(d)$, where $\varphi$ is Euler’s phi function.

###### Proof.

This is an immediate consequence of the definition: for any positive integer $d$, we define $\Phi_{n}(x)$, the $d$th cyclotomic polynomial, by

 $\Phi_{d}(x)=\prod_{\lx@stackrel{{j=1,}}{{\scriptscriptstyle{(j,d)% =1}}}}^{d}(x-{\zeta_{d}}^{j})$

where $\zeta_{d}=e^{\frac{2\pi i}{d}}$, i.e. $\zeta_{d}$ is an $d$th root of unity. Therefore, the degree is $\varphi(d)$. ∎

We begin with the $p$th cyclotomic polynomials for a prime $p\geq 2$.

###### Proposition 1.

The polynomial

 $q_{p}(x)=\frac{x^{p}-1}{x-1}=x^{p-1}+x^{p-2}+\cdots+x^{2}+x+1$

is irreducible over $\mathbb{Q}$.

###### Proof.

In order to show that $q(x)=q_{p}(x)$ is irreducible, we perform a change of variables $x\mapsto x+1$, and define $q^{\prime}(x)=q(x+1)$. Clearly, $q^{\prime}(x)$ is irreducible over $\mathbb{Q}$ if and only if $q(x)$ is irreducible. Also:

 $\displaystyle q^{\prime}(x)$ $\displaystyle=$ $\displaystyle q(x+1)=\frac{(x+1)^{p}-1}{(x+1)-1}$ $\displaystyle=$ $\displaystyle\frac{x^{p}+{p\choose 1}x^{p-1}+{p\choose 2}x^{p-2}+\cdots+{p% \choose p-2}x^{2}+{p\choose p-1}x}{x}$ $\displaystyle=$ $\displaystyle x^{p-1}+{p\choose 1}x^{p-2}+{p\choose 2}x^{p-3}+\cdots+{p\choose p% -2}x+{p\choose p-1}$

Since all the binomial coefficients  ${p\choose n}=\frac{p!}{(p-n)!n!}$, for $n=1,\ldots,p-1$, are integers divisible by $p$, and ${p\choose p-1}=p$ is not divisible by $p^{2}$, we can use Eisenstein’s criterion to conclude that $q^{\prime}(x)$ is irreducible over $\mathbb{Q}$. Thus $q(x)$ is irreducible as well, as desired. ∎

As a corollary, we obtain:

###### Theorem 1.

Let $p\geq 2$ be a prime. Then the $p$th cyclotomic polynomial is given by

 $\Phi_{p}(x)=\frac{x^{p}-1}{x-1}=x^{p-1}+x^{p-2}+\cdots+x+1.$
###### Proof.

By the lemma, the polynomial $\Phi_{p}(x)\in\mathbb{Q}[x]$ divides $q(x)=\frac{x^{p}-1}{x-1}$ and, by the proposition  above, $q(x)$ is irreducible. Hence $\Phi_{p}(x)=q(x)$ as claimed. ∎

The following proposition will be very useful as well:

###### Proposition 2.

Let $n$ be a positive integer.  Then the binomial  $x^{n}\!-\!1$  has as many prime factors  (http://planetmath.org/PrimeFactorsOfXn1) with integer coefficients as the integer $n$ has positive divisors   , both numbers thus being $\tau(n)$ (http://planetmath.org/TauFunction).

###### Proof.

A proof can be found in this entry (http://planetmath.org/FactorsOfNAndXn1). ∎

## 2 The examples

A generous list of examples can be found in this entry (http://planetmath.org/PrimeFactorsOfXn1). The examples of $\Phi_{d}(x)$ can be calculated by recursively factoring the polynomials $x^{n}-1$, for $n\geq 1$, using (a) the fact that $\Phi_{p}(x)=(x^{p}-1)/(x-1)$ for primes $p\geq 2$ and (b) the polynomial $\Phi_{d}(x)$ is a divisor of $x^{n}-1$ if and only if $n$ is a multiple  of $d$ (and $\Phi_{d}(x)$ appears with multiplicity  one as a factor, because $x^{n}-1$ does not have repeated roots). Hence, we can calculate:

 $x-1,\ x^{2}-1=(x-1)(x+1),\ x^{3}-1=(x-1)(x^{2}+x+1)$

and deduce

 $\Phi_{1}(x)=x-1,\ \Phi_{2}(x)=x+1,\ \Phi_{3}(x)=x^{2}+x+1.$

Before factoring $x^{4}-1$, note that we know that $(x-1)$ divides it, $\Phi_{2}(x)$ divides it and $x^{4}-1$ has as many divisors as $\tau(4)=3$. Therefore $\Phi_{4}(x)=(x^{4}-1)/((x-1)(x+1))=x^{2}+1$.

The polynomial $\Phi_{5}(x)$ is $x^{4}+x^{3}+x^{2}+x+1$ (by the Theorem). In order to calculate $\Phi_{6}(x)$ we factor $x^{6}-1$. Once again, note that $6$ has $4$ positive divisors, and we already know the following divisors: $x-1$, $\Phi_{2}(x)$, $\Phi_{3}(x)$. Hence:

 $\Phi_{6}(x)=\frac{x^{6}-1}{(x-1)(x+1)(x^{2}+x+1)}=x^{2}-x+1.$

Notice that we knew a priori (by a Lemma above) that the degree of $\Phi_{6}(x)$ is in fact $\varphi(6)=2$. Similarly, suppose we want to calculate $\Phi_{12}$. This is a polynomial of degree $\varphi(12)=4$, and divides $x^{12}-1$. On the other hand, $x^{12}-1$ has $\tau(12)=6$ irreducible factors and we already know the factors corresponding to $n=1,2,3,4,6$. Thus:

 $\Phi_{12}(x)=\frac{x^{12}-1}{(x-1)(x+1)(x^{2}+x+1)(x^{2}+1)(x^{2}-x+1)}=x^{4}-% x^{2}+1.$

Incidentally, we can find an explicit root of $\Phi_{12}(x)$ in terms of radicals. The roots are simply given by:

 $x^{2}=\frac{1\pm\sqrt{-3}}{2},\quad x=\pm\sqrt{\frac{1\pm\sqrt{-3}}{2}}.$
Title examples of cyclotomic polynomials ExamplesOfCyclotomicPolynomials 2013-03-22 17:20:03 2013-03-22 17:20:03 alozano (2414) alozano (2414) 10 alozano (2414) Example msc 11R60 msc 11R18 msc 11C08 calculating cyclotomic polynomials PrimeFactorsOfXn1 RootOfUnity