Norms x2 + p y2
Reduced forms of discriminant −4p
| a | b | c | b² − 4ac | reduced? | kind |
|---|
Lesson 11 · Class numbers and L-functions · 60–75 minutes
Quadratic Fields and Forms
In ℤ[i] factorization is unique (Lesson 05 relied on it). Adjoin √−13 instead of √−1 and it fails: 14 = 2 × 7 = (1 + √−13)(1 − √−13), with all four factors irreducible. The class number measures how badly unique factorization fails. Gauss found a concrete way to compute it: count the reduced binary quadratic forms of a given discriminant. You will reduce forms, count them, prove the class number of ℚ(√−p) is even for every p ≡ 1 (mod 4), and meet the real field ℚ(√p), whose class number Lesson 12 needs.
Where this lesson is headed
h(−4p) = #{(a, b, c) : b2 − 4ac = −4p, |b| ≤ a ≤ c, b ≥ 0 if |b| = a or a = c}, h(−4p) even
In words: every positive definite form of discriminant −4p is equivalent to exactly one reduced form, so the number of classes is a finite count. For p ≡ 1 (mod 4) exactly two reduced forms are their own mirror images, (1, 0, p) and (2, 2, (p + 1)/2), and all others come in pairs, so h(−4p) is even and never 1. The same count also equals −B1 of the character n ↦ χ−4(n)(n/p), which Lesson 15 derives from Dirichlet's class number formula.
You will use
Norms in ℤ[i] and unique factorization there (Lesson 05). The Legendre symbol and χ−4 (Lesson 03, Lesson 06). B1,χ (Lesson 08). Two facts are stated here and used, not proved: forms of discriminant D correspond to ideal classes of the quadratic field, and h = 1 exactly when factorization is unique.
p: a prime, p ≡ 1 (mod 4)
ℤ[√−p] = {x + y√−p}; N(x + y√−p) = x2 + py2
form (a, b, c): ax2 + bxy + cy2; discriminant D = b2 − 4ac
equivalent forms: related by an integer change of variables of determinant 1
reduced (D < 0): |b| ≤ a ≤ c, with b ≥ 0 if |b| = a or a = c
h(D): number of classes of primitive forms of discriminant D (positive definite if D < 0)
01. When factorization fails
Start with p = 13. The norm table lists x2 + 13y2: the numbers of the form N(x + y√−13). Neither 2 nor 7 appears, so nothing in ℤ[√−13] has norm 2 or 7. That is what makes 14 = 2 × 7 = (1 + √−13)(1 − √−13) two genuinely different factorizations. Then count reduced forms, and reduce a form of your own.
Guided experiments
Do these in order. "Set this up" chooses p.
The real field ℚ(√p)
For discriminant +p the forms are indefinite, and reduction no longer lands on a single form: reduced forms fall into cycles, and the number of cycles is the class number of ℚ(√p). Lesson 12 needs it.
| cycle | reduced forms (a, b, c) in order |
|---|
02. Reduce, count, and pair up
Each step has the general argument on the left and your p from Stage 01 on the right.
1
Norms in ℤ[√−p]
As in ℤ[i], N(x + y√−p) = (x + y√−p)(x − y√−p) = x2 + py2, and N(αβ) = N(α)N(β). The units are ±1 (norm 1 forces y = 0, x = ±1). An element is irreducible if it has no factorization into two non-units.
2
Two factorizations of p + 1
p + 1 = 2 · (p + 1)/2 = (1 + √−p)(1 − √−p). A norm below p must be a perfect square (y = 0), and 2 is not one, so no element has norm 2. Hence 2 (norm 4) is irreducible, and so is 1 ± √−p (norm p + 1, whose only splittings would need a factor of norm 2 or of norm (p + 1)/2; the latter is impossible too when (p + 1)/2 is not a square). Now 2 divides the product (1 + √−p)(1 − √−p) but neither factor, since (1 ± √−p)/2 is not in ℤ[√−p]. An irreducible that divides a product without dividing a factor: factorization is not unique.
3
Forms and equivalence
The norm x2 + py2 is the form (1, 0, p). In general f = ax2 + bxy + cy2 has discriminant D = b2 − 4ac. A substitution x = αx′ + βy′, y = γx′ + δy′ with αδ − βγ = 1 gives a form with the same discriminant and the same set of values; such forms are equivalent. Two basic moves generate everything:
T: (a, b, c) ↦ (a, b + 2a, a + b + c) (x ↦ x + y)
S: (a, b, c) ↦ (c, −b, a) (x ↦ −y, y ↦ x)
4
Every form reduces
For D < 0 and a > 0: use Tk to bring b into (−a, a]; if then a > c, apply S. Each S strictly lowers a, so the process stops, at a form with |b| ≤ a ≤ c (and the sign conventions fixed by one last S or T). From 4ac − b2 = |D| and b2 ≤ a2 ≤ ac:
3a2 ≤ |D| ⇒ a ≤ √(|D|/3)
so there are only finitely many reduced forms. (Uniqueness, that two different reduced forms are never equivalent, is the classical theorem of Lagrange and Gauss: a is the smallest nonzero value of the form, and c the next; it is stated here.)
5
The class number
h(D) is the number of reduced primitive forms. For D = −4p every form is primitive: a common factor g would need g2 | 4p, so g = 2, leaving a form of discriminant −p ≡ 3 (mod 4), which is impossible since discriminants are 0 or 1 mod 4. The ring of integers of ℚ(√−p) is ℤ[√−p], and (stated) its ideal classes correspond to these form classes, so h(−4p) = h(ℚ(√−p)). In particular factorization in ℤ[√−p] is unique exactly when h = 1.
6
h(−4p) is even
If (a, b, c) is reduced with 0 < |b| < a < c, then (a, −b, c) is a different reduced form: these come in pairs. The unpaired ones are the ambiguous forms, with b = 0, b = a or a = c. For D = −4p (b even):
b = 0: ac = p, a ≤ c: only (1, 0, p).
b = a = 2m: m(2c − m) = p forces m = 1, c = (p + 1)/2: the form (2, 2, (p + 1)/2).
a = c, b = 2k: (a − k)(a + k) = p forces b = p − 1 > a: not reduced.
Exactly two ambiguous forms, so h(−4p) = 2 + (pairs) is even, and at least 2.
7
A character sum for the same number
Let χ(n) = χ−4(n)·(n/p) for odd n, and 0 for even n: a real, odd character mod 4p. Then (derived in Lesson 15 from Dirichlet's class number formula)
h(−4p) = −B1,χ = −(1/(4p)) ∑n=14p χ(n)·n
The same shape as β = B1,χ4: class numbers live in B1 of odd characters. That is the thread that leads to h*(p) = N(β)/2 in Lesson 12.
8
The real field ℚ(√p)
For p ≡ 1 (mod 4) the ring of integers of ℚ(√p) is ℤ[(1 + √p)/2], of discriminant p. Forms of discriminant p are indefinite. Call (a, b, c) reduced if 0 < b < √p and √p − b < 2|a| < √p + b. There are finitely many, and the operation ρ(a, b, c) = (c, b′, (b′2 − p)/(4c)), with b′ ≡ −b (mod 2c) chosen in (√p − 2|c|, √p), permutes them in cycles. (Stated, after Gauss:) the number of cycles is the class number h(ℚ(√p)). (Strictly, cycles count the narrow class number; for p ≡ 1 (mod 4) prime the fundamental unit has norm −1, so the two agree.) For p ≡ 1 (mod 4) prime it is often 1; the first exception is p = 229, with h = 3.
03. The class number of ℚ(√−p) is even
Assemble the argument, then test it by computer.
Theorem. Let p ≡ 1 (mod 4) be prime. Every positive definite form of discriminant −4p is equivalent to a reduced form (a, b, c) with a ≤ √(4p/3). The reduced forms of discriminant −4p are all primitive; exactly two of them, (1, 0, p) and (2, 2, (p + 1)/2), are ambiguous; so h(−4p) is even and at least 2. In ℤ[√−p], if 2 and (p + 1)/2 are not squares, p + 1 has two essentially different factorizations into irreducibles.
Complete proof
Non-unique factorization. Norms below p are squares, so if 2 and (p + 1)/2 are not squares, no element has norm 2 or (p + 1)/2. Then 2 and 1 ± √−p are irreducible (a proper factor would have norm 2 or (p + 1)/2), and 2 ∤ 1 + √−p since (1 + √−p)/2 ∉ ℤ[√−p]. So the two factorizations of p + 1 differ.
Reduction. T and S have determinant 1 and preserve D. Normalizing b with T and swapping with S when c < a strictly lowers a; at the end |b| ≤ a ≤ c, and 3a² ≤ |D|.
Evenness. All forms of discriminant −4p are primitive. The map (a, b, c) ↦ (a, −b, c) is an involution on reduced forms whose fixed points are the ambiguous ones. Solving b = 0, b = a and a = c with b² − 4ac = −4p leaves exactly (1, 0, p) and (2, 2, (p + 1)/2) (the latter needs (p + 1)/2 odd, true for p ≡ 1 mod 4, for primitivity, and 2 ≤ (p + 1)/2). Hence h(−4p) ≡ 2 (mod 2), i.e. even, and h(−4p) ≥ 2. ∎
Computer check
For every prime p ≡ 1 (mod 4) below 1000: counts the reduced forms of discriminant −4p, checks the count is even with exactly two ambiguous forms, checks the reduction algorithm on 200 random forms, and compares h(−4p) with −(1/4p)∑χ(n)n.
Not run yet.
Exit problem
Use p = 37.