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A cyclotomic extension Q(ζn) of the rationals is formed by adjoining a primitive nth

root of unity ζn. In this chapter, we will find an integral basis and calculate the field

discriminant.

**7.1 Some Preliminary Calculations**

**7.1.1 The Cyclotomic Polynomial**

Recall that the cyclotomic polynomial Φn(X) is defined as the product of the terms X−ζ,

where ζ ranges over all primitive nth roots of unity in C. Now an nth root of unity is

a primitive dth root of unity for some divisor d of n, so Xn − 1 is the product of all

cyclotomic polynomials Φd(X) with d a divisor of n. In particular, let n = pr be a prime

power. Since a divisor of pr is either pr or a divisor of pr−1, we have

Φpr (X) = Xpr − 1

Xpr−1 − 1

= tp − 1

t − 1

= 1+t + · · · + tp−1

where t = Xpr−1. If X = 1 then t = 1, and it follows that Φpr (1) = p.

Until otherwise specified, we assume that n is a prime power pr.

**7.1.2 Lemma**

Let ζ and ζ be primitive (pr)th roots of unity. Then u = (1−ζ )/(1−ζ) is a unit in Z[ζ],

hence in the ring of algebraic integers.

Proof. Since ζ is primitive, ζ = ζs for some s (not a multiple of p). It follows that

u = (1−ζs)/(1−ζ) = 1+ζ+· · ·+ζs−1 ∈ Z[ζ]. By symmetry, (1−ζ)/(1−ζ ) ∈ Z[ζ ] = Z[ζ],

and the result follows. ♣

**7.1.3 Lemma**

Let π = 1− ζ and e = ϕ(pr) = pr−1(p − 1), where ϕ is the Euler phi function. Then the

principal ideals (p) and (π)e coincide.

unit in Z[ζ]. The result follows. ♣

We can now give a short proof of a basic result, but remember that we are operating

under the restriction that n = pr.

**7.1.4 Proposition**

The degree of the extension Q(ζ)/Q equals the degree of the cyclotomic polynomial,

namely ϕ(pr). Therefore the cyclotomic polynomial is irreducible over Q.

Proof. By (7.1.3), p has at least e = ϕ(pr) prime factors (not necessarily distinct) in the

ring of algebraic integers of Q(ζ). By the ram-rel identity (4.1.6), e ≤ [Q(ζ) : Q]. But

[Q(ζ) : Q] cannot exceed the degree of a polynomial having ζ as a root, so [Q(ζ) : Q] ≤ e.

If ζ were a root of an irreducible factor of Φpr , then the degree of the cyclotomic extension

would be less than ϕ(pr), contradicting what we have just proved. ♣

**7.1.5 Lemma**

Let B be the ring of algebraic integers of Q(ζ). Then (π) is a prime ideal (equivalently,

π is a prime element) of B. The relative degree f of (π) over (p) is 1, hence the injection

Z/(p) → B/(π) is an isomorphism.

Proof. If (π) were not prime, (p) would have more than ϕ(pr) prime ideal factors, which

is impossible, in view of the ram-rel identity. This identity also gives f = 1. ♣

We will need to do several discriminant computations, and to prepare for this, we do

some calculations of norms. The symbol N with no subscript will mean the norm in the

extension Q(ζ)/Q.

**7.1.6 Proposition**

N(1 − ζ) = ±p, and more generally, N(1 − ζps) = ±pps

, 0 ≤ s < r.

Proof. The minimal polynomial of 1−ζ is Φpr (1−X), which has constant term Φpr (1−0) =

p by (7.1.1). This proves the first assertion. If 0 < s < r, then ζps is a primitive (pr−s)th

root of unity, so by the above calculation with r replaced by r − s,

N1(1 − ζps) = ±p

where N1 is the norm in the extension Q(ζps )/Q. By transitivity of norms [see (2.1.7)]

applied to the chain Q(ζ),Q(ζps ),Q, and the formula in (2.1.3) for the norm of an element

of the base field, we get

N(1 − ζps) = N1((1 − ζps )b)

where b = [Q(ζ) : Q(ζps )] = ϕ(pr)/ϕ(pr−s) = ps. Thus N(1 − ζps) = ±pb, and the result

follows. ♣

In (7.1.6), the sign is (−1)ϕ(n); see (2.1.3).

**7.1.7 Proposition**

Let D be the discriminant of the basis 1, ζ, . . . , ζϕ(pr)−1. Then D = ±pc, where c =

pr−1(pr − r − 1).

Proof. By (2.3.6), D = ±N(Φ

pr (ζ)). Differentiate the equation

(Xpr−1 − 1)Φpr (X) = Xpr − 1

to get

(Xpr−1 − 1)Φ

pr (X) + pr−1Xpr−1−1Φpr (X) = prXpr−1.

Setting X = ζ and noting that ζ is a root of Φpr , we have

(ζpr−1 − 1)Φ

pr (ζ) + 0 = prζpr−1.

Thus

Φ

pr (ζ) = prζpr−1

ζpr−1 − 1.

The norm of the denominator has been computed in (7.1.6). The norm of ζ is ±1, as

ζ is a root of unity. The norm of pr is prϕ(pr) = prpr−1(p−1). By (2.1.3), the norm is

multiplicative, so the norm of Φ

pr (ζ) is ±pc, where

c = r(p − 1)pr−1 − pr−1 = pr−1(pr − r − 1). ♣

**7.1.8Remarks**

In (4.2.5), we related the norm of an ideal I to the field discriminant d and the discriminant

D(z) of a basis z for I. It is important to notice that the same argument works if I is

replaced by any free Z-module J of rank n. Thus if B is the ring of algebraic integers,

then

D(z) = |B/J|2d.

Applying this result with z = {1, ζ, . . . , ζϕ(pr)−1} and J = Z[ζ], we find that

D = |B/Z[ζ]|2d.

Thus if we can show that the powers of ζ form an integral basis, so that Z[ζ] = B, then

in view of (7.1.7), we are able to calculate the field discriminant up to sign. Also, by the

exercises in Section 4.2, the only ramified prime is p.

Let π = 1− ζ as in (7.1.3), and recall the isomorphism Z/(p) → B/(π) of (7.1.5).

**7.1.9 Lemma**

For every positive integer m, we have Z[ζ] + pmB = B.

Proof. We first prove the identity with p replaced by π. If b ∈ B, then b+(π) = t+(π) for

some integer t, hence b−t ∈ (π). Thus Z[ζ]+πB = B, and consequently πZ[ζ]+π2B = πB.

Now iterate: If b ∈ B, then b = b1 + b2, b1 ∈ Z[ζ], b2 ∈ πB. Then b2 = b3 + b4, b3 ∈

πZ[ζ] ⊆ Z[ζ], b4 ∈ π2B. Observe that b = (b1 + b3) + b4, so Z[ζ] + π2B = B. Continue

in this fashion to obtain the desired result. Now by (7.1.3), πϕ(pr) is p times a unit, so if

m = ϕ(pr), we can replace πmB by pB, so that Z[ζ] + pB = B. But we can iterate this

equation exactly as above, and the result follows. ♣

**7.1.10 Theorem**

The set {1, ζ, . . . , ζϕ(pr)−1} is an integral basis for the ring of algebraic integers of Q(ζpr ).

Proof. By (7.1.7) and (7.1.8), |B/Z[ζ]| is a power of p, so pm(B/Z[ζ]) = 0 for sufficiently

large m. Therefore pmB ⊆ Z[ζ], hence by (7.1.9), Z[ζ] = B. ♣

Problems For Section 7.1

This problem set will indicate how to find the sign of the discriminant of the basis

1, α, . . . , αn−1 of L = Q(α), where the minimal polynomial f of α has degree n.

1. Let c1, . . . , cr1 be the real conjugates of α, that is, the real roots of f, and let

cr1+1, cr1+1, . . . , cr1+r2 , cr1+r2 be the complex (=non-real) conjugates. Show that the

sign of the discriminant is the sign of

r2

i=1

(cr1+i − cr1+i)2.

2. Show that the sign of the discriminant is (−1)r2 , where 2r2 is the number of complex

embeddings.

3. Apply the results to α = ζ, where ζ is a primitive (pr)th root of unity. (Note that a

nontrivial cyclotomic extension has no real embeddings.)

**7.2 An Integral Basis of a Cyclotomic Field**

In the previous section, we found that the powers of ζ form an integral basis when ζ is a

power of a prime. We will extend the result to all cyclotomic extensions.

**7.2.1 Notation and Remarks**

Let K and L be number fields of respective degrees m and n over Q, and let KL be

the composite of K and L. Then KL consists of all finite sums aibi with ai ∈ K

and bi ∈ L. This is because the composite can be formed by adjoining basis elements of

K/Q and L/Q one at a time, thus allowing an induction argument. Let R, S, T be the

algebraic integers of K,L,KL respectively. Define RS as the set of all finite sums aibi

with ai ∈ R, bi ∈ S. Then RS ⊆ T, but equality does not hold in general. For example,

Lemma

Assume that [KL : Q] = mn. Let σ be an embedding of K in C and τ an embedding of

L in C. Then there is an embedding of KL in C that restricts to σ on K and to τ on L.

Proof. The embedding σ has [KL : K] = n distinct extensions to embeddings of KL in

C, and if two of them agree on L, then they agree on KL (because they coincide with

σ on K). This contradicts the fact that the extensions are distinct. Thus we have n

embeddings of KL in C with distinct restrictions to L. But there are only n embeddings

of L in C, so one of them must be τ , and the result follows. ♣

**7.2.3 Lemma**

Again assume [KL : Q] = mn. Let a1, . . . , am and b1, . . . , bn be integral bases for R and

S respectively. If α ∈ T, then

α =

m

i=1

n

j=1

cij

r

aibj, cij ∈ Z, r ∈ Z

with r having no factor (except ±1) in common with all the cij .

Proof. The assumption that [KL : Q] = mn implies that the aibj form a basis for KL/Q.

[See the process of constructing KL discussed in (7.2.1).] In fact the aibj form an integral

basis for RS. (This is because RS consists of all finite sums viwi, vi ∈ R, wi ∈ S.

Each vi is a linear combination of the ak with integer coefficients, and so on.) It follows

that α is a linear combination of the aibj with rational coefficients. Form a common

denominator and eliminate common factors to obtain the desired result. ♣

**7.2.4 Proposition**

We are still assuming that [KL : Q] = mn. If d is the greatest common divisor of the

discriminant of R and the discriminant of S, then T ⊆ 1

dRS. Thus if d = 1, then T = RS.

Proof. It suffices to show that in (7.2.3), r divides d. To see this, write

cij

r

= cij(d/r)

d

.

In turn, it suffices to show that r divides the discriminant of R. Then by symmetry, r

will also divide the discriminant of S, and therefore divide d.

Let σ be an embedding of K in C. By (7.2.2), σ extends to an embedding (also called

σ) of KL in C such that σ is the identity on L. By (7.2.3), if α ∈ T we have

σ(α) =

i,j

cij

r

σ(ai)bj .

If we set

xi =

n

j=1

cij

r

bj ,

we have the system of linear equations

m

i=1

σ(ai)xi = σ(α)

where there is one equation for each of the m embeddings σ from K to C. Solving for xi

by Cramer’s rule, we get xi = γi/δ, where δ is the determinant formed from the σ(ai) and

γi is the determinant obtained by replacing the ith column of δ with the σ(α). Note that

by (2.3.3), δ2 is the discriminant of R, call it e. Since all the σ(ai) and σ(α) are algebraic

integers, so are δ and all the γi. Now

xi = γi

δ

= γiδ

δ2 = γiδ

e

so exi = γiδ is an algebraic integer. By definition of xi,

exi =

n

j=1

ecij

r

bj ,

an algebraic integer in RS. But e is a Z-linear combination of the ai, and the aibj are

an integral basis for RS, so ecij/r is an integer. Thus r divides every ecij . By (7.2.3), r

has no factor (except the trivial ±1) in common with every cij . Consequently, r divides

e, the discriminant of R. ♣

We need one more preliminary result.

**7.2.5 Lemma**

Let ζ be a primitive nth root of unity, and denote the discriminant of {1, ζ, . . . , ζϕ(n)−1}

by disc(ζ). Then disc(ζ) divides nϕ(n).

Proof. Let f (= Φn, the nth cyclotomic polynomial) be the minimal polynomial of ζ

over Q. Since ζ is a root of Xn − 1, we have Xn − 1 = f(X)g(X) for some g ∈ Q[X].

But f ∈ Z[X] (because ζ is an algebraic integer), and f, hence g, is monic, so g ∈ Z[X].

Differentiate both sides of the equation to get nXn−1 = f(X)g (X)+f (X)g(X). Setting

X = ζ, which is a root of f, we have nζn−1 = f (ζ)g(ζ). But ζn−1 = ζn/ζ = 1/ζ, so

n = ζf

(ζ)g(ζ).

Now [Q(ζ) : Q] = ϕ(n), so taking the norm of each side yields

nϕ(n) = N(f

(ζ))N(ζg(ζ)).

But by (2.3.6), N(f (ζ)) = ±disc (ζ), and N(ζg(ζ)) ∈ Z by (2.2.2). The desired result

follows. ♣

**7.2.6 Theorem**

If ζ is a primitive nth root of unity, then the ring of algebraic integers of Q(ζ) is Z[ζ]. in

other words, the powers of ζ form an integral basis.

Proof. We have proved this when ζ is a prime power, so let n = m1m2 where the mi are

relatively prime and greater than 1. Now

ζm1 = (ei2π/n)m1 = ei2πm1/n = ei2π/m2 = ζ2,

a primitive (m2)th root of unity, and similarly ζm2 = ζ1, a primitive (m1)th root of unity.

Thus Q(ζ1) and Q(ζ2) are contained in Q(ζ). On the other hand, since m1 and m2 are

relatively prime, there are integers r, s such that rm2 + sm1 = 1. Thus

ζ = ζrm2+sm1 = ζr

1 ζs

2 .

It follows that Q(ζ) = Q(ζ1)Q(ζ2), and we can apply (7.2.4). In that proposition, we

take K = Q(ζ1), L = Q(ζ2),KL = Q(ζ),R = Z[ζ1], S = Z[ζ2] (induction hypothesis),

T = RS. The hypothesis on the degree [KL : Q] is satisfied because ϕ(n) = ϕ(m1)ϕ(m2).

By (7.2.5), disc(ζ1) divides a power of m1 and disc(ζ2) divides a power of m2. Thus the

greatest common divisor of disc(R) and disc(S) is 1, and again the hypothesis of (7.2.4)

is satisfied. The conclusion is that the ring T of algebraic integers of KL coincides with

RS. But the above argument that Q(ζ) = Q(ζ1)Q(ζ2) may be repeated verbatim with Q

replaced by Z. We conclude that Z[ζ] = Z[ζ1]Z[ζ2] = RS = T. ♣

**7.2.7 The Discriminant of a General Cyclotomic Extension**

The field discriminant of Q(ζ), where ζ is a primitive nth root of unity, is given by

(−1)ϕ(n)/2nϕ(n)

p|n pϕ(n)/(p−1) .

A direct verification, with the aid of (7.1.7) and Problem 3 of Section 7.1, shows that

the formula is correct when n = pr. The general case is handled by induction, but the

computation is very messy.

In the next chapter, we will study factorization of primes in Galois extensions. The

results will apply, in particular, to cyclotomic extensions.

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