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If you face any problem to loading this page, click here for PDF file of this chapter. 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.
If you face any problem to loading this page, click here for PDF file of this chapter.
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