# examples of cyclotomic polynomials

In this entry we calculate a number of cyclotomic polynomials^{}, ${\mathrm{\Phi}}_{d}(x)\in \mathbb{Q}[x]$, for various $d\ge 1$. 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 $n\mathrm{,}d\mathrm{\ge}\mathrm{2}$ be positive integers. If $d$ divides $n$ then ${\mathrm{\Phi}}_{d}\mathit{}\mathrm{(}x\mathrm{)}$ divides ${q}_{n}\mathit{}\mathrm{(}x\mathrm{)}\mathrm{=}\frac{{x}^{n}\mathrm{-}\mathrm{1}}{x\mathrm{-}\mathrm{1}}$.

###### Proof.

Let ${\zeta}_{d}$ be a primitive $d$th root of unity^{} and let ${\mathrm{\Phi}}_{d}(x)$ 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^{} $m\ge 1$ 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, ${\mathrm{\Phi}}_{d}(x)$ must divide ${q}_{n}(x)$. ∎

###### Lemma 2.

The polynomial ${\mathrm{\Phi}}_{d}\mathit{}\mathrm{(}x\mathrm{)}$ is of degree $\phi \mathit{}\mathrm{(}d\mathrm{)}$, where $\phi $ is Euler’s phi function.

###### Proof.

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

$${\mathrm{\Phi}}_{d}(x)=\prod _{\stackrel{j=1,}{(j,d)=1}}^{d}(x-\zeta _{d}{}^{j})$$ |

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

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

###### Proposition 1.

The polynomial

$${q}_{p}(x)=\frac{{x}^{p}-1}{x-1}={x}^{p-1}+{x}^{p-2}+\mathrm{\cdots}+{x}^{2}+x+1$$ |

is irreducible over $\mathrm{Q}$.

###### 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:

${q}^{\prime}(x)$ | $=$ | $q(x+1)={\displaystyle \frac{{(x+1)}^{p}-1}{(x+1)-1}}$ | ||

$=$ | $\frac{{x}^{p}+\left(\genfrac{}{}{0pt}{}{p}{1}\right){x}^{p-1}+\left(\genfrac{}{}{0pt}{}{p}{2}\right){x}^{p-2}+\mathrm{\cdots}+\left(\genfrac{}{}{0pt}{}{p}{p-2}\right){x}^{2}+\left(\genfrac{}{}{0pt}{}{p}{p-1}\right)x}{x}$ | |||

$=$ | ${x}^{p-1}+\left({\displaystyle \genfrac{}{}{0pt}{}{p}{1}}\right){x}^{p-2}+\left({\displaystyle \genfrac{}{}{0pt}{}{p}{2}}\right){x}^{p-3}+\mathrm{\cdots}+\left({\displaystyle \genfrac{}{}{0pt}{}{p}{p-2}}\right)x+\left({\displaystyle \genfrac{}{}{0pt}{}{p}{p-1}}\right)$ |

Since all the binomial coefficients^{} $\left(\genfrac{}{}{0pt}{}{p}{n}\right)=\frac{p!}{(p-n)!n!}$, for $n=1,\mathrm{\dots},p-1$, are integers divisible by $p$, and $\left(\genfrac{}{}{0pt}{}{p}{p-1}\right)=p$ 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 $p\mathrm{\ge}\mathrm{2}$ be a prime. Then the $p$th cyclotomic polynomial is given by

$${\mathrm{\Phi}}_{p}(x)=\frac{{x}^{p}-1}{x-1}={x}^{p-1}+{x}^{p-2}+\mathrm{\cdots}+x+1.$$ |

###### Proof.

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

The following proposition will be very useful as well:

###### Proposition 2.

Let $n$ be a positive integer. Then the binomial ${x}^{n}\mathrm{-}\mathrm{1}$ has as many prime factors^{} (http://planetmath.org/PrimeFactorsOfXn1) with integer coefficients as the integer $n$ has positive divisors^{}, both numbers thus being $\tau \mathit{}\mathrm{(}n\mathrm{)}$ (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 ${\mathrm{\Phi}}_{d}(x)$ can be calculated by recursively factoring the polynomials ${x}^{n}-1$, for $n\ge 1$, using (a) the fact that ${\mathrm{\Phi}}_{p}(x)=({x}^{p}-1)/(x-1)$ for primes $p\ge 2$ and (b) the polynomial ${\mathrm{\Phi}}_{d}(x)$ is a divisor of ${x}^{n}-1$ if and only if $n$ is a multiple^{} of $d$ (and ${\mathrm{\Phi}}_{d}(x)$ 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

$${\mathrm{\Phi}}_{1}(x)=x-1,{\mathrm{\Phi}}_{2}(x)=x+1,{\mathrm{\Phi}}_{3}(x)={x}^{2}+x+1.$$ |

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

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

$${\mathrm{\Phi}}_{6}(x)=\frac{{x}^{6}-1}{(x-1)(x+1)({x}^{2}+x+1)}={x}^{2}-x+1.$$ |

Notice that we knew a priori (by a Lemma above) that the degree of ${\mathrm{\Phi}}_{6}(x)$ is in fact $\phi (6)=2$. Similarly, suppose we want to calculate ${\mathrm{\Phi}}_{12}$. This is a polynomial of degree $\phi (12)=4$, 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:

$${\mathrm{\Phi}}_{12}(x)=\frac{{x}^{12}-1}{(x-1)(x+1)({x}^{2}+x+1)({x}^{2}+1)({x}^{2}-x+1)}={x}^{4}-{x}^{2}+1.$$ |

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

$${x}^{2}=\frac{1\pm \sqrt{-3}}{2},x=\pm \sqrt{\frac{1\pm \sqrt{-3}}{2}}.$$ |

Title | examples of cyclotomic polynomials |
---|---|

Canonical name | ExamplesOfCyclotomicPolynomials |

Date of creation | 2013-03-22 17:20:03 |

Last modified on | 2013-03-22 17:20:03 |

Owner | alozano (2414) |

Last modified by | alozano (2414) |

Numerical id | 10 |

Author | alozano (2414) |

Entry type | Example |

Classification | msc 11R60 |

Classification | msc 11R18 |

Classification | msc 11C08 |

Synonym | calculating cyclotomic polynomials |

Related topic | PrimeFactorsOfXn1 |

Related topic | RootOfUnity |