Lesson 16 · Capstone · 70–90 minutes

The Class Number of K_p

Lesson 12 counted the ideal classes of K_p by brute force and found h*(p) = N(β)/2 for every prime it could reach. This lesson proves it for every p ∈ P*. The zeta function of K_p splits into four Dirichlet L-functions, one for each character that is trivial on H. The class number formula relates its residue at s = 1 to h(K_p). Dividing by the same formula for ℚ(√p) leaves exactly |L(1, χ₄)|², and Lesson 15 turned that into N(β). Every lesson of the course feeds into the proof.

The capstone theorem

ζ_{K_p}(s) = ζ(s)·L(s, χ₄)·L(s, (·/p))·L(s, χ̄₄), h*(p) = (p/2π²)|L(1, χ₄)|² = (k² + t²)/2

In words: for every prime p ≡ 5 (mod 8), p > 5, the relative class number of the quartic field K_p is half the norm of the quartic value β = B_1,χ₄. Because k and t are odd, h*(p) ≡ 1 (mod 4); in particular h*(p) is odd and never 0.

You will use

The splitting of primes in K_p, its discriminant p³ and the class number of ℚ(√p) (Lessons 11 and 12); Euler products (Lesson 13); |L(1, χ₄)|² = π²N(β)/p (Lesson 15); k, t odd (Lesson 08). Stated and used: unique factorization of ideals, the analytic class number formula, and Hasse's unit index Q = 1 for imaginary cyclic quartic fields.

p: a prime in P*; K = K_p, F = ℚ(√p) ⊂ K

ζ_K(s) = ∑_𝔞 N(𝔞)^−s over nonzero ideals: the Dedekind zeta function

w: number of roots of unity; d: discriminant; R: regulator; r₁, r₂: numbers of real and complex-pair embeddings

ε: the fundamental unit of F; h*(p) = h(K)/h(F)

01. One zeta function, four L-functions

Start with p = 13. Pick a prime q: the panel shows how q splits in K₁₃, the Euler factor it contributes to ζ_K, and the product of the four L-factors at q. They always match. Below, the same identity for the whole functions at s = 2, then the table of h*(p) for every prime of P* below 1000.

Matching Euler factors at q

q · a prime

character	χ(q)	factor (1 − χ(q)x)⁻¹

The identity at s = 2

h*(p) against p

Guided experiments

Do these in order. "Set this up" chooses p and q.

Why four L-functions?

K_p is the fixed field of H inside ℚ(ζ_p). The characters mod p that are trivial on H are exactly the four with χ⁴ = χ₀: χ₀, χ₄, χ₄² = (·/p) and χ₄³ = χ̄₄. Each prime q splits in K according to χ₄(q), and the four characters, evaluated at q, reproduce exactly the local factor.

For F = ℚ(√p), the fixed field of the squares, only χ₀ and (·/p) appear. Dividing the two zeta functions cancels ζ(s) and L(s, (·/p)) and leaves the two odd characters, the ones Lesson 15 evaluates in closed form.

ζ_K/ζ_F = L(s, χ₄)L(s, χ̄₄). Stage 02 takes residues at s = 1 and reaches N(β).

02. From residues to β

Each step has the general argument on the left and your p from Stage 01 on the right.

The Dedekind zeta function

ζ_K(s) = ∑_𝔞 N(𝔞)^−s converges for s > 1. Since nonzero ideals of O factor uniquely into prime ideals (stated) and norms multiply,

ζ_K(s) = ∏_𝔭 (1 − N(𝔭)^−s)⁻¹

exactly as the Euler product of Lesson 13 came from unique factorization of integers. Grouping the 𝔭 by the rational prime q below them gives one local factor per q.

The local factors match, prime by prime

Write x = q^−s and let ω = χ₄(q). The four characters take the values 1, ω, ω², ω³ at q, so their L-factors multiply to ∏_j(1 − ω^jx)⁻¹. By Lesson 12's splitting rule:

ω = 1: four primes of norm q: (1 − x)⁻⁴ = ∏(1 − x)⁻¹ ✓

ω = −1: two primes of norm q²: (1 − x²)⁻² = [(1 − x)(1 + x)]⁻² ✓

ω = ±i: one prime of norm q⁴: (1 − x⁴)⁻¹ = [(1 − x)(1 − ix)(1 + x)(1 + ix)]⁻¹ ✓

q = p: one prime of norm p: (1 − x)⁻¹, and only χ₀'s factor, which is ζ's, is nontrivial ✓

ζ_K(s) = ζ(s)L(s, χ₄)L(s, (·/p))L(s, χ̄₄)

The same for F, and the quotient

In F = ℚ(√p), q splits when (q/p) = 1 and stays prime when (q/p) = −1, so the same comparison gives ζ_F(s) = ζ(s)L(s, (·/p)). Dividing:

ζ_K(s)/ζ_F(s) = L(s, χ₄)·L(s, χ̄₄) = |L(s, χ₄)|² (s real)

The right side is continuous at s = 1 (Lesson 13), so the ratio of the residues of ζ_K and ζ_F at s = 1 is |L(1, χ₄)|².

The class number formula (stated)

For a number field of degree n, near s = 1 (Dirichlet, Dedekind):

(s − 1)ζ_K(s) → 2^r₁(2π)^r₂ h R / (w√|d|)

For K: r₁ = 0, r₂ = 2, d = p³. For F: r₁ = 2, r₂ = 0, w = 2, d = p, R_F = log ε. As a check, for F this says L(1, (·/p)) = 2h(F) log ε/√p, the even case left open in Lesson 15; the panel confirms it numerically.

Roots of unity and units of K

w_K = 2. K ⊂ ℚ(ζ_p), whose roots of unity are ±ζ_p^k; ζ_p ∉ K because ℚ(ζ_p) has degree p − 1 > 4. So only ±1.

R_K = 2R_F. K and F both have unit rank 1. By Hasse's theorem for imaginary cyclic quartic fields (stated), the units of K are ±εⁿ, the same as F's (unit index Q = 1). The regulator of K uses 2 log|σ(ε)| at a complex place, so R_K = 2 log ε = 2R_F.

Take the ratio

Divide the residue for K by the residue for F:

Res ζ_K = (2π)²·h(K)·2 log ε/(2p^3/2) = 4π²h(K) log ε/p^3/2

Res ζ_F = 2²·h(F)·log ε/(2√p) = 2h(F) log ε/√p

Res ζ_K/Res ζ_F = 2π²h*(p)/p

and by step 3 this ratio is |L(1, χ₄)|². So

h*(p) = (p/(2π²))·|L(1, χ₄)|²

The unknowns log ε and the factors of 2 cancel; what survives is a single L-value.

Insert Lesson 15

Lesson 15 proved |L(1, χ₄)|² = π²N(β)/p. Substituting:

h*(p) = (p/(2π²))·(π²N(β)/p) = N(β)/2 = (k² + t²)/2

The observation of Lesson 12 is now a theorem, for every p ∈ P*, including those like 229 where the direct computation was out of reach.

A corollary: h*(p) ≡ 1 (mod 4)

k and t are odd (Lesson 08), so k² ≡ t² ≡ 1 (mod 8), N(β) ≡ 2 (mod 8), and

h*(p) = N(β)/2 ≡ 1 (mod 4)

In particular h*(p) is odd, and since h* ≥ 1 the identity also gives β ≠ 0 a second time.

Stage 03 assembles the proof and maps it onto the course.

03. The whole course in one proof

State the theorem, trace each ingredient to the lesson that supplied it, then run the final checks.

Theorem. Let p be a prime with p ≡ 5 (mod 8), p > 5, let χ₄ be the quartic character mod p with χ₄(2) = i, and let β = B_1,χ₄ = k + it. Let K_p be the quartic subfield of ℚ(ζ_p) and F = ℚ(√p). Then

h(K_p)/h(F) = (k² + t²)/2,

and this number is ≡ 1 (mod 4).

ingredient	where it came from
Cosets C_j = 2^jH, the pin χ₄(2) = i	L02, L07
β = k + it with k, t odd	L08
Characters, orthogonality, parity	L06
Norms in ℤ[i], N(β)	L04
ℚ(√p), its class number and fundamental unit	L11
K_p, its periods, discriminant p³, splitting of primes	L12
Euler products, L(1, χ) ≠ 0 and continuity at s = 1	L13
\|τ(χ)\|² = p and the Fourier expansion	L14
\|L(1, χ₄)\|² = π²N(β)/p	L15
Unique factorization of ideals; class number formula; unit index Q = 1	stated (standard theorems)

1. Why does ζ_K have an Euler product? Ideals factor uniquely into prime ideals, and the norm is multiplicative. Because elements of K factor uniquely. Because K is cyclic.

2. For q with χ₄(q) = −1, why do the local factors match? Because q does not divide p. Two primes of norm q² give (1 − x²)⁻², and the characters take values 1, −1, 1, −1 at q. Because both sides equal 1.

3. What is ζ_K/ζ_F? ζ(s). L(s, (·/p)). L(s, χ₄)L(s, χ̄₄), the two odd characters.

4. Why is w_K = 2? The only roots of unity in ℚ(ζ_p) are ±ζ_p^k, and ζ_p is not in the degree-4 field K. Because K is real. Because i ∈ K.

5. What cancels when the two residues are divided? The class numbers. The regulator log ε (since R_K = 2R_F) and powers of 2. The L-values.

6. How does Lesson 15 finish the proof? By computing log ε. By showing h(F) = 1. |L(1, χ₄)|² = π²N(β)/p turns (p/2π²)|L|² into N(β)/2.

7. Why is h*(p) ≡ 1 (mod 4)? k and t are odd, so k² + t² ≡ 2 (mod 8). Because p ≡ 1 (mod 4). Because h(F) = 1.

Final computer check

For all 42 primes of P* below 1000: the local factors match for every prime q < 2000; ζ_K(2) as a product over prime ideals equals ζ(2)L(2, χ₄)L(2, (·/p))L(2, χ̄₄); p|L(1, χ₄)|²/(2π²) equals N(β)/2 and is ≡ 1 (mod 4); L(1, (·/p)) = 2h(F) log ε/√p. Then it reruns Lesson 12's direct class-group computation for its twelve primes and compares.

Not run yet.

Exit problem

Use p = 101, where β = 3 − i.

End of the course · From the order of an element mod p to the class number of a quartic field: h*(p) = (k² + t²)/2.

COURSE COMPLETEWhere to go from here

The same pattern, a class number equal to a product of B₁'s of odd characters, holds for every imaginary abelian field (the analytic class number formula for CM fields). The Growing Clock and this course together give you the tools to read it. Back to the start: Lesson 01.