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examples of cyclotomic polynomials

(Example)

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.

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 divides then $\Phi_d(x)$ divides $q_n(x)=\frac{x^n-1}{x-1}$ .

Proof. Let $\zeta_d$ be a primitive

th root of unity and let $\Phi_d(x)$ be the

th cyclotomic polynomial (which is the minimal polynomial of $\zeta_d$ over $\mathbb{Q}$ ). If

divides

then there is a natural number $m\geq 1$ such that

. Thus

$\displaystyle \zeta_d^n=(\zeta_d^d)^m=1^m=1$

and so, $\zeta_d$ is also an

th root of unity and, in particular, it is a root of

. By the properties of minimal polynomials, $\Phi_d(x)$ must divide

. $\qedsymbol$

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

, we define $\Phi_n(x)$ , the

th cyclotomic polynomial, by

$\displaystyle \Phi_d(x)=\prod_{\stackrel{j=1,}{\scriptscriptstyle{(j,d)=1}}}^d(x-{\zeta_d}^j)$

where $\zeta_d=e^{\frac{2\pi i}{d}}$ , i.e. $\zeta_d$ is an

th root of unity. Therefore, the degree is $\varphi(d)$ . $\qedsymbol$

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

Proposition 1 The polynomial

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

is irreducible over $\mathbb{Q}$ .

Proof. In order to show that

is irreducible, we perform a change of variables $x\mapsto x+1$ , and define

. Clearly,

is irreducible over $\mathbb{Q}$ if and only if

is irreducible. Also:

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

Since all the binomial coefficients ${p\choose n} = \frac{p!}{(p-n)!n!}$ , for $n=1,\ldots,p-1$ , are integers divisible by

, and ${p\choose p-1}=p$ is not divisible by

, we can use Eisenstein's criterion to conclude that

is irreducible over $\mathbb{Q}$ . Thus

is irreducible as well, as desired. $\qedsymbol$

As a corollary, we obtain:

Theorem 1 Let $p\geq 2$ be a prime. Then the th cyclotomic polynomial is given by

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

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,

is irreducible. Hence $\Phi_p(x)=q(x)$ as claimed. $\qedsymbol$

The following proposition will be very useful as well:

Proposition 2 Let be a positive integer. Then the binomial $x^n\!-\!1$ has as many prime factors with integer coefficients as the integer has positive divisors, both numbers thus being $\tau(n)$ .

Proof. A proof can be found in this entry. $\qedsymbol$

The examples

A generous list of examples can be found in this entry. The examples of $\Phi_d(x)$ can be calculated by recursively factoring the polynomials , 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 if and only if is a multiple of (and $\Phi_d(x)$ appears with multiplicity one as a factor, because does not have repeated roots). Hence, we can calculate:

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

and deduce

$\displaystyle \Phi_1(x)=x-1,\ \Phi_2(x)=x+1,\ \Phi_3(x)=x^2+x+1.$

Before factoring

, note that we know that

divides it, $\Phi_2(x)$ divides it and

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 (by the Theorem). In order to calculate $\Phi_6(x)$ we factor . Once again, note that has positive divisors, and we already know the following divisors: , $\Phi_2(x)$ , $\Phi_3(x)$ . Hence:

$\displaystyle \Phi_6(x)=\frac{x^6-1}{(x-1)(x+1)(x^2+x+1)}=x^2-x+1.$

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

. Thus:

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

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

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

"examples of cyclotomic polynomials" is owned by alozano.

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