Bound and determine a quadratic class group

Try

In O=ℤ[√−5], we already know that I=(2,1+√−5) represents a nonidentity invertible ideal class with square equal to the identity class. That proves at least two classes. How can we prove there are no others without guessing ideal generators?

We will find a small-index integral ideal in every class. The new ingredient is geometric: sufficiently large planar regions must contain a difference of two points that is a nonzero lattice vector. This gives a short element of an ideal, and the proved multiplicativity of ideal norms converts its size into a bound on a representative’s index. We use the preceding result that every nonzero ideal of this full quadratic integer ring is invertible; the earlier nonmaximal-order counterexample is not covered by that theorem.

Predict

A planar lattice Λ has the form Bℤ² for an invertible real 2 by 2 matrix B. Its covolume is d=|det B|, the area of a fundamental parallelogram. Choose the half-open one

D=❴B(u,v):0≤u<1, 0≤v<1❵.

The translates D+λ, λ∈Λ, partition the plane exactly: write each coordinate of B⁻¹x as its integer floor plus a remainder in [0,1). The half-open convention prevents double counting at shared edges. Predict why a bounded measurable set S of area greater than d must have two distinct points differing by a nonzero lattice vector. The points need not themselves be lattice points.

We use ordinary planar Lebesgue area here: measurable indicators have integrals equal to area, translations preserve area, and finite sums integrate term by term. These basic area facts are the analytic inputs. No theorem about convex bodies containing lattice points is assumed.

Experiment

Embed a+b√−5 as (a,b√5) in ℝ². The following picture uses equal scales on the two axes. Identify the nonzero lattice points inside the disk of radius √3 before reading the caption. Compare the disk area 3π with four times the lattice covolume 4√5.

For a general ideal, its basis need not be rectangular. Predict which quantity survives a change of integer lattice basis: a drawn side length, or the absolute determinant? The determinant is the invariant covolume used in the proof.

Understand

Prove the area pigeonhole lemma. For bounded measurable S define, on D, the integer multiplicity

m(x)=Σ over λ∈Λ of 1_S(x+λ).

Only finitely many terms can be nonzero anywhere on D. Indeed, if (D+λ) meets S, then λ lies in the bounded set S−D. Applying the invertible map B⁻¹ puts these λ in a bounded set of integer pairs, which is finite. The translates partition the plane, so translation invariance and finite additivity give

∫ over D of m(x) dx = Σ area(S∩(D+λ)) = area(S).

If area(S)>d, m cannot be at most 1 everywhere; otherwise its integral would be at most area(D)=d. Thus some x∈D and distinct λ,μ satisfy x+λ,x+μ∈S. These are distinct points of S with difference λ−μ∈Λ nonzero. This is the bounded measurable form of Blichfeldt’s argument, including the finite-overlap justification.

Deduce a convex-body bound. Let C be a bounded measurable convex set centrally symmetric about 0, with area(C)>4d. Its half-scale S=C/2 has area greater than d. The lemma gives distinct x,y∈S with x−y a nonzero lattice vector. Since 2x,2y∈C, central symmetry gives −2y∈C, and convexity puts their midpoint x−y in C. Hence C contains a nonzero lattice point. Central symmetry and convexity each have a specific role; area alone would not suffice for this conclusion.

For a disk let r₀²=4d/π. Every closed disk of radius r>r₀ satisfies the strict-area hypothesis. To obtain the non-strict bound, take radii r₀+1/n. All selected nonzero lattice points lie in the fixed disk of radius r₀+1. There are finitely many lattice points in a bounded disk, by the same inverse-coordinate argument as above. One selected point occurs for infinitely many n; its norm is at most every corresponding radius, hence at most r₀. We have proved the existence of 0≠α∈Λ with |α|²≤4d/π. We did not apply a strict inequality theorem directly at equality.

Turn geometry into ideal representatives. In the complex embedding of O=ℤ[√−5], the basis 1,√−5 has covolume √5. If an integral ideal A has integer basis matrix B in this basis, its covolume is √5|det B|=√5 N(A), by the ideal lattice index theorem. The disk bound supplies 0≠α∈A with

|α|² ≤ (4√5/π) N(A).

Define J=(α)A⁻¹. It is integral because α∈A implies αA⁻¹⊆AA⁻¹=O. Also AJ=(α). Ideal norm multiplicativity and the principal norm formula give

N(J)=N((α))/N(A)=|α|²/N(A)≤4√5/π.

The use of |α|² here is specific to this imaginary quadratic embedding: it equals the positive field norm αᾱ. To bound the constant rigorously, √5<9/4 since 5<81/16, and π>3, so 4√5<9<3π. Thus N(J)<3. As a positive integer, it is 1 or 2; no decimal rounding decides this step.

Why does this reach every class? Any invertible fractional ideal F has a nonzero denominator d∈O with dF integral. Scaling by d changes its class by a principal ideal, so every class has an integral representative A. The construction above gives [J]=[A]⁻¹. Inversion is a bijection on the group of invertible ideal classes. Equivalently, to target a specified class C, begin with an integral representative of C⁻¹. Therefore every class, not merely some classes, has an integral representative of norm at most 2.

Exhaust the small representatives. Norm 1 means J=O. If N(J)=2, the quotient ring has two elements and nonzero identity, hence is F₂. Any unital quotient map O→F₂ sends √−5 to a root of t²+5=t²+1 in F₂; its only root is 1. Thus its kernel is uniquely I=(2,1+√−5): a+b√−5 maps to a+b modulo 2. Every index-two ideal is this kernel. There are consequently at most two classes, represented by O and I.

The earlier inverse-ideal certificate proves I²=2O and I is nonprincipal. Recall the obstruction: a principal ideal of index 2 would have generator a+b√−5 with a²+5b²=2, impossible over integers. Thus [I] is nonidentity and has order two. Combining this lower bound with the exhaustive representative bound proves that the entire class group is C₂. The exhaustion is the new step that upgrades an example of an order-two class to a class-number computation.

Apply

Repeat the argument for the Gaussian integer ring ℤ[i]. Determine its lattice covolume, the representative norm bound, and the resulting class group. Explain why this proof uses ideal inversion and norm multiplicativity even though a separate Euclidean proof of principality is available.

Check the changed imaginary quadratic field

The basis 1,i has covolume 1. For every integral ideal A the covolume is N(A). The same disk argument gives α∈A nonzero with |α|²≤(4/π)N(A), and J=(α)A⁻¹ is integral with N(J)≤4/π<2 because π>2. Clearing denominators and ranging over inverse classes again covers every class. The only positive integer below 2 is 1, so J=ℤ[i] and every class is principal. Its class group is trivial. This particular proof depends on the established ideal inverse and norm theorems; it does not silently replace them with a Euclidean algorithm. Different quadratic covolumes produce different bounds, so this calculation is not a claim that all imaginary quadratic class groups are trivial or have order two.

Continue

Audit the chain of dependencies in your own solution: half-open cell and area count; convex symmetry; endpoint passage using lattice finiteness; ideal covolume; invertibility and norm multiplicativity; inverse-class coverage; exhaustive small-index ideals. Each step serves a different purpose. The result computes a whole class group for this one ring; it does not establish a general class-number formula or replace the proofs needed for other rings.