proof that the cyclotomic polynomial is irreducible

proof that the cyclotomic polynomial is irreducible

We first prove that

Φ

n

⁢

(

x

)

∈

ℤ

⁢

[

x

]

. The

field extension

ℚ

⁢

(

ζ

n

)

of

ℚ

is the

splitting field

of the

polynomial

x

n

-

1

∈

ℚ

⁢

[

x

]

, since it splits this polynomial and is generated as an

algebra

by a single root of the polynomial. Since splitting fields are normal, the

extension

ℚ

⁢

(

ζ

n

)

/

ℚ

is a

Galois extension

. Any element of the

Galois group

, being a

field automorphism

, must map

ζ

n

to another

root of unity

of exact order

n

. Therefore, since the Galois group of

ℚ

⁢

(

ζ

n

)

/

ℚ

permutes the roots of

Φ

n

⁢

(

x

)

, it must fix the coefficients of

Φ

n

⁢

(

x

)

, so by

Galois theory

these coefficients are in

ℚ

. Moreover, since the coefficients are

algebraic integers

, they must be in

ℤ

as well.

Let

f

⁢

(

x

)

be the

minimal polynomial

of

ζ

n

in

ℚ

⁢

[

x

]

. Then

f

⁢

(

x

)

has

integer

coefficients as well, since

ζ

n

is an algebraic integer. We will prove

f

⁢

(

x

)

=

Φ

n

⁢

(

x

)

by showing that every root of

Φ

n

⁢

(

x

)

is a root of

f

⁢

(

x

)

. We do so via the following claim:

Claim:

For any prime

p

not dividing

n

, and any

primitive

n

th

root of unity

ζ

∈

ℂ

, if

f

⁢

(

ζ

)

=

0

then

f

⁢

(

ζ

p

)

=

0

.

This claim does the job, since we know

f

⁢

(

ζ

n

)

=

0

, and any other primitive

n

th

root of unity can be obtained from

ζ

n

by successively raising

ζ

n

by prime powers

p

not dividing

n

a finite number of times

1

1

Actually, if one applies Dirichlet’s

theorem

on primes in arithmetic progressions here, it turns out that one prime is enough, but we do not need such a sharp result here.

.

To prove this claim, consider the factorization

x

n

-

1

=

f

⁢

(

x

)

⁢

g

⁢

(

x

)

for some polynomial

g

⁢

(

x

)

∈

ℤ

⁢

[

x

]

. Writing

𝒪

for the

ring of integers

of

ℚ

⁢

(

ζ

n

)

, we treat the factorization as taking place in

𝒪

⁢

[

x

]

and proceed to mod out both sides of the factorization by any

prime ideal

𝔭

of

𝒪

lying over

(

p

)

. Note that the polynomial

x

n

-

1

has no repeated roots mod

𝔭

, since its

derivative

n

⁢

x

n

-

1

is relatively prime to

x

n

-

1

mod

𝔭

. Therefore, if

f

⁢

(

ζ

)

=

0

mod

𝔭

, then

g

⁢

(

ζ

)

≠

0

mod

𝔭

, and applying the

p

th

power

Frobenius map

to both sides yields

g

⁢

(

ζ

p

)

≠

0

mod

𝔭

. This means that

g

⁢

(

ζ

p

)

cannot be 0 in

ℂ

, because it doesn’t even equal

0

mod

𝔭

. However,

ζ

p

is a root of

x

n

-

1

, so if it is not a root of

g

, it must be a root of

f

, and so we have

f

⁢

(

ζ

p

)

=

0

, as desired.

Title

proof that the

cyclotomic polynomial

is

irreducible

Canonical name

ProofThatTheCyclotomicPolynomialIsIrreducible

Date of creation

2013-03-22 12:38:04

Last modified on

2013-03-22 12:38:04

Owner

djao (24)

Last modified by

djao (24)

Numerical id

9

Author

djao (24)

Entry type

Proof

Classification

msc 12E05

Classification

msc 11R60

Classification

msc 11R18

Classification

msc 11C08

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