Roots of unity and the four periods
Compute h(Kp)
Minkowski bound
Factor-base ideals
Relations
h(Kp)
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Lesson 12 · Class numbers and L-functions · 65–80 minutes
The Quartic Field Kp
Add up the pth roots of unity over each coset Cj of Lesson 07 and you get four complex numbers, the Gaussian periods. They generate a field Kp of degree 4, the only one inside the pth roots of unity, and for p ∈ P* it is imaginary. In this lesson you build Kp from its periods, see how small primes split in it, and then compute its class number directly: find enough elements of small norm, factor them into prime ideals, and read the class number off a determinant. The answer lines up with the quartic value of Lesson 08: k2 + t2 = 2h*(p).
Where this lesson is headed
ηj = ∑n ∈ Cj ζn, Kp = ℚ(η0), h*(p) = h(Kp)/h(ℚ(√p)), N(β) = k2 + t2 = 2h*(p)
In words: the four periods are the conjugates of a degree-4 algebraic integer, and they form a basis of the integers of Kp. Its class group is generated by prime ideals of small norm (Minkowski), so a finite search for relations computes h(Kp). Divide by the class number of the real subfield ℚ(√p) from Lesson 11 to get the relative class number h*(p). For every prime computed here, h*(p) = N(β)/2. Lesson 16 proves this in general.
You will use
The cosets Cj = 2jH, −Cj = Cj+2, and the quartic Euler criterion (Lesson 07); β = k + it (Lesson 08); h(ℚ(√p)) by cycles of forms (Lesson 11). Three standard facts are stated and used: the periods form an integral basis (discriminant p3); Minkowski's bound; and ∑n (n/p)ζn = √p for p ≡ 1 (mod 4).
p: a prime in P*; ζ = cos(2π/p) + i sin(2π/p)
ηj: the Gaussian periods, j = 0, 1, 2, 3
Kp = ℚ(η0), with ring of integers O = ℤη0 + ℤη1 + ℤη2 + ℤη3
N(α): the norm, product of the 4 conjugates of α
h(Kp): class number; h*(p) = h(Kp)/h(ℚ(√p))
Mp = (4!/44)(4/π)2p3/2: the Minkowski bound
01. Four periods and a class number
Start with p = 13. The plane shows the twelve 13th roots of unity colored by coset, and the four periods ηj, the sums over each coset. η2 is the mirror image of η0, so K13 is not inside the real numbers. Then press Compute to run the class-group calculation, and compare with N(β).
Guided experiments
Do these in order. "Set this up" chooses p.
What the computation actually does
Every ideal class of Kp contains a prime ideal of norm at most Mp, so those primes generate the class group. Primes of residue degree 2 or 4, and the prime above p, turn out to be principal, so only the degree-1 primes need tracking: four above each q with χ4(q) = 1.
Each element α = a0η0 + ⋯ + a3η3 with small coefficients whose norm factors over those q gives a relation: the product of the prime ideals dividing it is principal. The class group is the free group on the factor base modulo all such relations, so its order is the determinant of the relation lattice. More relations can only shrink that determinant; it settles on h(Kp).
02. Building Kp and counting its classes
Each step has the general statement on the left and your p from Stage 01 on the right.
1
The Gaussian periods
ηj = ∑n ∈ Cj ζn. Since the Cj partition 1, …, p − 1 and 1 + ζ + ⋯ + ζp−1 = 0,
η0 + η1 + η2 + η3 = −1
Complex conjugation sends ζn to ζ−n, and −Cj = Cj+2, so η̄j = ηj+2: the periods are not real.
2
A cyclic field of degree 4
The substitution σ: ζ ↦ ζ2 respects all algebraic relations among powers of ζ, and maps Cj to 2Cj = Cj+1, so σ(ηj) = ηj+1. The polynomial (x − η0)(x − η1)(x − η2)(x − η3) is unchanged by σ, and its coefficients turn out to be integers. Kp = ℚ(η0) has degree 4, and σ generates its symmetries: a cyclic quartic field, with σ2 = complex conjugation.
3
The real quadratic subfield
η0 + η2 is the sum of ζn over the squares, fixed by σ2. With the Gauss sum ∑(n/p)ζn = √p (stated; Lesson 14 proves |·|2 = p):
η0 + η2 = (−1 + √p)/2, η1 + η3 = (−1 − √p)/2
So Kp contains ℚ(√p), the real field of Lesson 11, and Kp is a quadratic extension of it.
4
Arithmetic with the periods
A product ηaηb = ∑x∈Ca, y∈Cb ζx+y is again an integer combination of periods (the counts of x + y in each coset are the "cyclotomic numbers"). So O = ℤη0 + ⋯ + ℤη3 is closed under multiplication. Its discriminant det(Tr(ηaηb)) equals p3, the discriminant of Kp, so O is the full ring of integers (stated). Norms of α ∈ O are integers: N(α) = |α|2|σα|2.
5
How a prime q splits
For q ≠ p, the residue degree f of the primes above q is the order of χ4(q) as a root of unity, and there are 4/f of them:
χ4(q) = 1 (q ∈ H): f = 1, four primes of norm q; they correspond to the four ring maps O → 𝔽q. χ4(q) = −1: f = 2, two primes of norm q2, each coming from a prime of ℚ(√p). χ4(q) = ±i: f = 4, q stays prime. q = p: totally ramified, (p) = 𝔭4.
6
Which primes can be non-principal
Inert primes are generated by q itself. The ramified prime 𝔭 is generated by an element of norm p, namely ∏h∈H(1 − ζh). A degree-2 prime is the extension of a prime of ℚ(√p), which is principal when h(ℚ(√p)) = 1 (Lesson 11). So for these p, only the degree-1 primes of norm q ≤ Mp matter, and by Minkowski they generate the whole class group.
7
Relations and a determinant
Search α with coefficients in [−B, B]. When |N(α)| factors as pe × (primes q with χ4(q) ≠ 1) × (distinct primes q ∈ H), the ideal (α) is a product of known principal pieces and exactly one degree-1 prime above each such q, found by testing which ring map O → 𝔽q kills α. Record its exponent vector. Together with (q) = 𝔮1𝔮2𝔮3𝔮4, these vectors span a lattice L inside the lattice Λ of all principal relations, and
h(Kp) = [ℤn : Λ] divides det L
det L only drops as relations are added; it stabilizes at h(Kp). When det L = 1, this proves h = 1.
03. What is proved, and what is observed
This lesson mixes proofs, stated facts and computation. Here is the ledger, then the checks.
Proved here. η0 + ⋯ + η3 = −1; η̄j = ηj+2, so Kp is imaginary; σ permutes the periods cyclically; products of periods are integer combinations of periods. For each listed p, the relation lattice has determinant equal to the number shown, which is a multiple of h(Kp); for p ≤ 61 it is 1, which proves h(Kp) = 1.
Stated (standard). The periods form an integral basis; Minkowski's bound; the splitting rule; ∑(n/p)ζn = √p; h(ℚ(√p)) = number of cycles.
Observed. For every p computed, the determinant stabilizes at (k2 + t2)/2 × h(ℚ(√p)), i.e. k2 + t2 = 2h*(p). Lesson 16 proves this identity for all p ∈ P*.
Stated (standard). The periods form an integral basis; Minkowski's bound; the splitting rule; ∑(n/p)ζn = √p; h(ℚ(√p)) = number of cycles.
Observed. For every p computed, the determinant stabilizes at (k2 + t2)/2 × h(ℚ(√p)), i.e. k2 + t2 = 2h*(p). Lesson 16 proves this identity for all p ∈ P*.
Summary of the argument
Structure. The periods sum to −1, are permuted cyclically by σ: ζ ↦ ζ2, and satisfy η̄j = ηj+2 because −1 ∈ C2. Products ηaηb are integer combinations of periods by counting x + y over the cosets.
Reduction to degree-1 primes. Given the stated facts, each class contains a prime of norm ≤ Mp; inert primes, the ramified prime and (when h(ℚ(√p)) = 1) the degree-2 primes are principal. So the degree-1 primes above q ∈ H, q ≤ Mp, generate the class group, and Cl(Kp) ≅ ℤn/Λ.
Determinant. Every relation found lies in Λ, so the found lattice L ⊆ Λ and det L = h·[Λ : L]. In particular det L = 1 forces h = 1. For the other primes, det L is an upper bound for h that has stabilized; an independent run with a different search box gives the same value. ∎
Computer check
For all twelve primes: checks disc(O) = det(Tr(ηaηb)) = p3; the period polynomial has integer coefficients; η0 + η2 = (−1 + √p)/2; runs the class-group search twice with different search boxes and checks both determinants agree; and compares with (k2 + t2)/2.
Not run yet.
Exit problem
Use p = 149.