2.8 The Cyclotomic Polynomials

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Let n be a positive integer, and let _ be the complex number _ = e2_i/n. Set

_n(x) =

Y

d

(x − _d),

where 1 _ d _ n, and gcd(d, n) = 1. We call _n(x), the n-th cyclotomic

polynomial. A little thought reveals that

xn − 1 =

Y

d|n

_d(x);

in particular, we have (using induction) that _n(x) 2 Z[x].

Proposition 2.8.1 _n(x) is irreducible in Z[x].

Proposition 2.8.2 Let F be a field, and let f(x) = xn − 1 2 F[x]. If G is

the Galois group of f(x), then G is isomorphic to a subgroup of the group

Un of units in the ring Z/(n) of integers modulo n.

Corollary 2.8.2.1 Same hypotheses as above, except that F = Q. Then

G is also the Galois group of _n(x), and is isomorphic to Un.

As an application of the above simple result, we have the following result.

Proposition 2.8.3 Let A be any abelian group. Then there exists a finite

Galois extension K _ Q such that Gal(K/Q) _= A. In fact, there exists an

integer n such that Q _ K _ Q(_), where _ = e2_i/n.

We can easily outline the proof here, we will. The main ingredient is the

following special case of the so-called Dirichlet Theorem on Primes in an

Arithmetic Progression, namely, if n is any integer, then there are infinitely

many primes p such that p _ 1 mod n. Assuming this result, the proof

proceeds as follows. If A is an abelian group, then A can be decomposed as

a product of cyclic groups:

A _= Zn1 × Zn2 × · · · × Znk .

Choose distinct primes p1, p2, . . . pk such that pi _ 1 mod ni, i = 1, 2, . . . , k.

Now set n = p1p2 · · · pk, _ = e2_i/n. Then

Gal (Q(_)/Q) _= Un

_= Zp1−1 × Zp2−1 × · · · × Zpk−1.

2.8. THE CYCLOTOMIC POLYNOMIALS 67

Choose generators _1, _2, . . . , _k in each of the factors and let

H = h_(p1−1)/n1

1 , _(p2−1)/n2

2 , . . . , _(pk−1)/nk

k i; setting K = invH(Q(_)) we get

Gal(K/Q) _= Gal Q(_)/H _= A.

Exercises 2.7

1. Compute _n(x), 1 _ n _ 20.

2. Suppose that p is prime and that n _ 1. Show that

_pn(x) =

n _n(xp) if p|n,

_n(xp)/_n(x) if p/ n.

3. If n is a positive odd integer, show that _2n(x) = _n(−x).

4. (For those that know M¨obius inversion). Show that

_n(x) =

Y

d|n

(xd − 1)μ(n/d).

5. Let _ = e2_i/5. Show that Q(_ + _−1) = Q(

p

5)

6. Let n be a positive integer and let _ = e2_i/n, a primitive n-th root

of unity. Let n = 2r1pr2

2 · · · prk

k is the prime factorization of n, and

compute the structure of Gal(Q(_ +_−1)/Q). (Use Exercises 1 and 2

of Section 1.5 of Chapter 1.)

7. Let n be a positive integer. Show that

(a) If 8|n, then Q(cos 2_/n) = Q(sin 2_/n);

(b) If 4|n, 8/ n, but n 6= 4, then Q(sin 2_/n) _ Q(cos 2_/n), and

[Q(cos 2_/n) : Q(sin 2_/n)] = 2;

(c) If 4/ n, but n 6= 1, 2 then Q(cos 2_/n) _ Q(sin 2_/n), and

[Q(sin 2_/n) : Q(cos 2_/n)] = 2.

68 CHAPTER 2. FIELD AND GALOIS THEORY

8. Show that if n 6= 1, 2, then the degree of the minimal polynomial of

cos 2_/n is _(n)/2. Using Exercise 7, compute the degree of the minimal

polynomial of sin 2_/n. (The minimal polynomial of cos 2_/n

can be computed in principle in terms of the so-called Chebychev

polynomials.)1

9. Let n _ 3, and set _n−2 = e2_i/2n + e−2_i/2n

2 R. Show that _1 =

p

2, _2 =

p

2 +

p

2, . . . , _n =

r

2 +

q

2 +

p

· · · +

p

2 (n times). (Show

that _2n

= 2 + _n−1, n _ 2.)

10. Notation as above. Show that Q _ Q(_n) is a Galois extension whose

Galois group has order 2n, n = 1, 2, . . .. (This will require Exercise 2,

of Section 1.5.)

11. Let m, n be relatively prime positive integers, and set _ = e2_i/n. Show

that _m(x) is irreducible in Q(_)[x].

1See W. Watkins and J. Zeitlin, The minimal polynomial of cos 2_/2, Amer. Math.

Monthly 100, (1993), no. 5, 474-474, MR 94b:12001.

2.9. SOLVABILITY BY RADICALS 69

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