5.3. Cyclotomic extensions.

Back

A primitive nth root of 1 in F is an element of order n in F×. Such an element can exist

only if F has characteristic 0 or characteristic p not dividing n.

Proposition 5.7. Let F be a field of characteristic 0 or characteristic p not dividing n.

Let E be the splitting field of Xn − 1.

(a) There exists a primitive nth root of 1 in E.

(b) If ζ is a primitive nth root of 1 in E, then E = F[ζ].

(c) The field E is Galois over F, and the map

Gal(E/F) → (Z/nZ)

×

sending σ to [i] if σζ = ζi is injective.

Proof. (a) The roots of Xn − 1 are distinct, because its derivative nXn−1 has only zero

as a root (we use here the condition on the characteristic), and so E contains n distinct nth

roots of 1. The nth roots of one form a finite subgroup of E×, and so (see Exercise 3) they

form a cyclic group. Any generator will have order n, and hence will be a primitive nth root

of 1.

(b) The roots of Xn − 1 are the powers of ζ, and F[ζ] contains them all.

(c) If ζ is one primitive nth root of 1, then the remaining primitive nth roots of 1 are the

elements ζi with i relatively prime to n. Since σζ is again a primitive nth root of 1 for any

automorphism σ of E, it equals ζi for some i relatively prime to n, and the map σ → i

mod n is injective because ζ generates E over F. It obviously is a homomorphism (and is

independent of the choice of ζ).

The map σ → i : Gal(F[ζ]/F )→ (Z/nZ)× need not be surjective. For example, if F = C,

then its image is {1}, and if F = R, it is {Ѓ}1} (n _= 2)—because F[ζ] = C, Gal(C/R) is

generated by complex conjugation ι, and ιζ = ￣ζ = ζn−1. On the other hand, when n = p is

prime, we saw in (1.31) that [Q[ζ] : Q] = p−1, and so the map is surjective. We shall prove

that the map is surjective for all n when F = Q.

The polynomial Xn−1 has some obvious factors in Q[X], namely, the polynomials Xd−1

for any d|n. The quotient of Xn−1 by all these factors for d < n is called the nth cyclotomic

polynomial Φn. Thus

Φn =

_

(X − ζ) (product over the primitive nth roots of 1).

It has degree ϕ(n), the order of (Z/nZ)×. Since every nth root of 1 is a primitive dth root of

1 for exactly one d dividing n, we see that

Xn − 1 =

_

d|n

Φd(X).

For example, Φ1(X) = X − 1, Φ2(X) = X + 1, Φ3(X) = X2 + X + 1, and

Φ6(X) =

X6 − 1

(X − 1)(X + 1)(X2 + X + 1)

= X2 − X + 1.

This gives an easy inductive method of computing the cyclotomic polynomials. Alternatively

ask Maple by typing: with(numtheory); cyclotomic(n,X);. Because Xn − 1 has

coefficients in Z and is monic, any monic factor of it has coefficients in Z (see (1.6)). In

particular, the cyclotomic polynomials lie in Z[X].

40 J.S. MILNE

Lemma 5.8. Let F be a field of characteristic 0 or p not dividing n, and let ζ be a primitive

nth root of 1 in some extension field. The following are equivalent:

(a) the nth cyclotomic polynomial Φn is irreducible;

(b) the degree [F[ζ] : F] = ϕ(n);

(c) the homomorphism

Gal(F[ζ]/F )→ (Z/nZ)

×

is an isomorphism.

Proof. Because ζ is a root of Φn, the minimum polynomial of ζ divides Φn. It is equal to

it if and only if [F[ζ] : F] = ϕ(n), which is true if and only if the injection Gal(F[ζ]/F ) _→

(Z/nZ)× is onto.

Theorem 5.9. The nth cyclotomic polynomial Φn is irreducible in Q[X].

Proof. Let f(X) be a monic irreducible factor of Φn in Q[X]. Its roots will be primitive

nth roots of 1, and we have to show they include all primitive nth roots of 1. For this it

suffices to show that

ζ a root of f(X) =⇒ ζi a root of f(X) for all i such that gcd(i, n) = 1.

Such an i is a product of primes not dividing n, and so it suffices to show that

ζ a root of f(X) =⇒ ζp a root of f(X) for all primes p  n.

Write

Φn(X) = f(X)g(X).

Again (1.6) implies that f(X) and g(X) lie in Z[X]. Suppose ζ is a root of f, but that for

some prime p not dividing n, ζp is not a root of f. Then ζp is a root g(X), which implies

that ζ is a root of g(Xp). Since f(X) and g(Xp) have a common root, their greatest common

divisor (in Q[X]) is _= 1 (see 3.1). Write h(X) → ￣h(X) for the map Z[X] → Fp[X], and note

that

gcd(f(X), g(Xp)) _= 1 =⇒ gcd( ￣ f(X), ￣g(Xp)) _= 1.

But ￣g(Xp) = ￣g(X)p (use the mod p binomial theorem and that ap = a for all a ∈ Fp), and

so gcd( ￣ f(X), ￣g(X)p) _= 1, which implies that ￣ f(X) and ￣g(X) have a common factor. Hence

Xn − 1 (regarded as an element of Fp[X]) has multiple roots, but we saw in the proof of 5.7

that it doesn’t. Contradiction.

Remark 5.10. This proof is very old—in essence it goes back to Dedekind in 1857—but

its general scheme has recently become very popular: take a statement in characteristic zero,

reduce modulo p (where the statement may no longer be true), and exploit the existence

of the Frobenius automorphism a → ap to obtain a proof of the original statement. For

example, commutative algebraists use this method to prove results about commutative rings,

and there are theorems about complex manifolds5 that have only been proved by reducing

things to characteristic p.

There are some beautiful and mysterious relations between what happens in characteristic

0 and in characteristic p. For example, let f(X1, ...,Xn) ∈ Z[X1, ...,Xn]. We can

(i) look at the solutions of f = 0 in C, and so get a topological space;

(ii) reduce mod p, and look at the solutions of ￣ f = 0 in Fpn.

5This is from my old notes—I no longer remember what I was thinking of.

FIELDS AND GALOIS THEORY 41

The Weil conjectures (Weil 1949;p roved by Grothendieck and Deligne 1973) assert that the

Betti numbers of the space in (i) control the cardinalities of the sets in (ii).

Theorem 5.11. The regular n-gon is constructible if and only if n = 2kp1 ・ ・ ・ ps where

the pi are distinct Fermat primes.

Proof. The regular n-gon is constructible if and only if cos 2π

n (or ζ = e2πi/n) is constructible.

We know that Q[ζ] is Galois over Q, and so (according to 1.27 and 3.22) ζ is

constructible if and only if [Q[ζ] : Q] is a power of 2. But (see Groups 3.10)

ϕ(n) =

_

p|n

(p − 1)pn(p)−1, n=

_

pn(p),

and this is a power of 2 if and only if n has the required form.

Remark 5.12. The final section of Gauss’s, Disquisitiones Arithmeticae (1801) is titled

“Equations defining sections of a Circle”. In it Gauss proves that the nth roots of 1 form

a cyclic group, that Xn − 1 is solvable (this was before the theory of abelian groups had

been developed, and before Galois), and that the regular n-gon is constructible when n is as

in the Theorem. He also claimed to have proved the converse statement6. This leads some

people to credit him with the above proof of the irreducibility of Φn, but in the absence of

further evidence, I’m sticking with Dedekind.

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