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 cyclotomicpolynomial, 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}}$

$=$${\displaystyle \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:

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