examples of cyclotomic polynomials

examples of cyclotomic polynomials

In this entry we calculate a number of

cyclotomic polynomials

,

Φ

d

⁢

(

x

)

∈

ℚ

⁢

[

x

]

, for various

d

≥

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

≥

2

be

positive integers

. If

d

divides

n

then

Φ

d

⁢

(

x

)

divides

q

n

⁢

(

x

)

=

x

n

-

1

x

-

1

.

Proof.

Let

ζ

d

be a

primitive

d

th

root of unity

and let

Φ

d

⁢

(

x

)

be the

d

th cyclotomic polynomial (which is the

minimal polynomial

of

ζ

d

over

ℚ

). If

d

divides

n

then there is a

natural number

m

≥

1

such that

n

=

d

⁢

m

. Thus

ζ

d

n

=

(

ζ

d

d

)

m

=

1

m

=

1

and so,

ζ

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,

Φ

d

⁢

(

x

)

must divide

q

n

⁢

(

x

)

.
∎

Lemma 2

.

The polynomial

Φ

d

⁢

(

x

)

is of degree

φ

⁢

(

d

)

, where

φ

is Euler’s phi function.

Proof.

This is an immediate

consequence

of the

definition

: for any positive integer

d

, we define

Φ

n

⁢

(

x

)

, the

d

th

cyclotomic
polynomial

, by

Φ

d

⁢

(

x

)

=

∏

(

j

,

d

)

=

1

j

=

1

,

d

(

x

-

ζ

d

j

)

where

ζ

d

=

e

2

⁢

π

⁢

i

d

, i.e.

ζ

d

is an

d

th root of unity. Therefore, the degree is

φ

⁢

(

d

)

.
∎

We begin with the

p

th cyclotomic polynomials for a prime

p

≥

2

.

Proposition 1

.

The polynomial

q

p

⁢

(

x

)

=

x

p

-

1

x

-

1

=

x

p

-

1

+

x

p

-

2

+

⋯

+

x

2

+

x

+

1

is

irreducible

over

Q

.

Proof.

In order to show that

q

⁢

(

x

)

=

q

p

⁢

(

x

)

is irreducible, we perform a change of variables

x

↦

x

+

1

, and define

q

′

⁢

(

x

)

=

q

⁢

(

x

+

1

)

. Clearly,

q

′

⁢

(

x

)

is irreducible over

ℚ

if and only if

q

⁢

(

x

)

is irreducible. Also:

q

′

⁢

(

x

)

=

q

⁢

(

x

+

1

)

=

(

x

+

1

)

p

-

1

(

x

+

1

)

-

1

=

x

p

+

(

p

1

)

⁢

x

p

-

1

+

(

p

2

)

⁢

x

p

-

2

+

⋯

+

(

p

p

-

2

)

⁢

x

2

+

(

p

p

-

1

)

⁢

x

x

=

x

p

-

1

+

(

p

1

)

⁢

x

p

-

2

+

(

p

2

)

⁢

x

p

-

3

+

⋯

+

(

p

p

-

2

)

⁢

x

+

(

p

p

-

1

)

Since all the

binomial coefficients

(

p

n

)

=

p

!

(

p

-

n

)

!

⁢

n

!

, for

n

=

1

,

…

,

p

-

1

, are

integers

divisible

by

p

, and

(

p

p

-

1

)

=

p

is not divisible by

p

2

, we can use Eisenstein’s criterion to conclude that

q

′

⁢

(

x

)

is irreducible over

ℚ

. Thus

q

⁢

(

x

)

is irreducible as well, as desired.
∎

As a

corollary

, we obtain:

Theorem 1

.

Let

p

≥

2

be a prime. Then the

p

th cyclotomic polynomial is given by

Φ

p

⁢

(

x

)

=

x

p

-

1

x

-

1

=

x

p

-

1

+

x

p

-

2

+

⋯

+

x

+

1

.

Proof.

By the lemma, the polynomial

Φ

p

⁢

(

x

)

∈

ℚ

⁢

[

x

]

divides

q

⁢

(

x

)

=

x

p

-

1

x

-

1

and, by the

proposition

above,

q

⁢

(

x

)

is irreducible. Hence

Φ

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

τ

⁢

(

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

Φ

d

⁢

(

x

)

can be calculated by recursively

factoring

the polynomials

x

n

-

1

, for

n

≥

1

, using (a) the fact that

Φ

p

⁢

(

x

)

=

(

x

p

-

1

)

/

(

x

-

1

)

for primes

p

≥

2

and (b) the polynomial

Φ

d

⁢

(

x

)

is a divisor of

x

n

-

1

if and only if

n

is a

multiple

of

d

(and

Φ

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

Φ

1

⁢

(

x

)

=

x

-

1

,

Φ

2

⁢

(

x

)

=

x

+

1

,

Φ

3

⁢

(

x

)

=

x

2

+

x

+

1

.

Before factoring

x

4

-

1

, note that we know that

(

x

-

1

)

divides it,

Φ

2

⁢

(

x

)

divides it and

x

4

-

1

has as many divisors as

τ

⁢

(

4

)

=

3

. Therefore

Φ

4

⁢

(

x

)

=

(

x

4

-

1

)

/

(

(

x

-

1

)

⁢

(

x

+

1

)

)

=

x

2

+

1

.

The polynomial

Φ

5

⁢

(

x

)

is

x

4

+

x

3

+

x

2

+

x

+

1

(by the

Theorem

). In order to calculate

Φ

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

,

Φ

2

⁢

(

x

)

,

Φ

3

⁢

(

x

)

. Hence:

Φ

6

⁢

(

x

)

=

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

Φ

6

⁢

(

x

)

is in fact

φ

⁢

(

6

)

=

2

. Similarly, suppose we want to calculate

Φ

12

. This is a polynomial of degree

φ

⁢

(

12

)

=

4

, and divides

x

12

-

1

. On the other hand,

x

12

-

1

has

τ

⁢

(

12

)

=

6

irreducible factors and we already know the factors corresponding to

n

=

1

,

2

,

3

,

4

,

6

. Thus:

Φ

12

⁢

(

x

)

=

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

Φ

12

⁢

(

x

)

in terms of

radicals

. The roots are simply given by:

x

2

=

1

±

-

3

2

,

x

=

±

1

±

-

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

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