Try
Put s=√−5 and O=ℤ[s]. An element a+bs is represented by the integer coordinate pair (a,b). Consider the lattice generated over ℤ by 2 and 1+s. Its elements are
2m+n(1+s)=(2m+n)+ns, m,n∈ℤ.
Which coordinate pairs occur? Solving n=b and 2m=a−b shows that membership is exactly a−b even. Thus 3+s belongs, but 2+s does not. There are two additive cosets, represented by 0 and 1.
This calculation does not yet prove the lattice is an ideal. An ideal must also remain inside itself after multiplication by every element of O. We will check that condition, then explain why every nonzero ideal in a quadratic integer ring has a finite quotient. That quotient gives an exact numerical invariant even when no single element generates the ideal.
Predict
Use the proved full quadratic integer ring and integral basis. Write K for a quadratic number field and O=ℤ[ω] for its full integer ring, with basis 1,ω over ℤ. There are integers t,u such that ω²=tω−u. The polynomial X²−tX+u is irreducible over ℚ, and the other root is ω̄=t−ω.
For α=a+bω, multiplication by α is an integer-linear map on O. Predict its matrix by multiplying α first by 1 and then by ω. We use column coordinates throughout: the two resulting coordinate vectors are the two columns, not the rows.
A nonzero integral ideal I means an ideal contained in O with I≠0; the whole ring is allowed. Its ideal norm will mean the index N(I)=[O:I], the number of additive cosets. The element norm N(α)=αᾱ is a different construction and can be negative in a real quadratic field. We will prove the precise relation for principal ideals, with an absolute value.
Experiment
Return to O=ℤ[s], s²=−5. For a lattice with an integer basis, multiplying the basis vectors by s is enough to test ideal closure. Indeed every ring element is a+bs, and integer combinations are already allowed.
| Integer generators | Membership condition on a+bs | Products by s | Ideal? |
|---|
| 2, 1+s | a−b even | 2s=2(1+s)−2; s(1+s)=(1+s)−3·2 | Yes |
| 3, 1+s | a−b divisible by 3 | 3s=3(1+s)−3; s(1+s)=(1+s)−2·3 | Yes |
| 3, 1−s | a+b divisible by 3 | 3s=3−3(1−s); s(1−s)=2·3−(1−s) | Yes |
| 2, s | a even | s·s=−5 fails membership | No |
Read the expressions such as −3·2 as integer multiples of the displayed generator 2. The first three rows prove the indicated ℤ-spans are ideals and therefore equal the O-ideals generated by the same lists: an ideal containing the generators contains their ℤ-span, while the closed ℤ-span already is such an ideal. The last row is a full-rank lattice of index two but is not an ideal. Rank and index alone do not supply multiplicative closure.
For each valid row, predict the quotient field by reducing the membership condition modulo 2 or 3. Check not only addition but also the equation s²=−5.
Check the quotient maps
For I=(2,1+s), the map a+bs↦a+b modulo 2 respects s²=−5 because 1²=−5 in ℤ/2ℤ. Its kernel is the first lattice and it is onto, so O/I≅F₂. For J=(3,1+s), use a−b modulo 3, sending s to −1. For L=(3,1−s), use a+b modulo 3, sending s to 1. Both satisfy the quadratic relation, have the claimed kernels and are onto F₃. Thus their indices are 2,3,3 and all three ideals are maximal. These maps are ring certificates, stronger than merely counting lattice cells.
Understand
Multiplication has a norm determinant
For α=a+bω, the two products are α·1=a+bω and α·ω=−bu+(a+bt)ω. Therefore its multiplication matrix is
Mα = [[a,−bu],[b,a+bt]].
Its determinant is a²+abt+b²u. Direct conjugation gives the same value:
αᾱ=(a+bω)(a+b(t−ω))=a²+abt+b²u.
This proves the formula without choosing complex coordinates or assuming the determinant equals a norm by definition. If α≠0, its conjugate is nonzero as well, since conjugation is a field automorphism. Thus the determinant is a nonzero integer. Multiplication matrices satisfy Mαβ=MαMβ, giving multiplicativity of element norms; this is not yet a theorem about indices of arbitrary ideal products.
Every nonzero ideal has rank two and finite index
Choose 0≠α∈I. Then αO⊆I⊆O. Under the integral basis, αO is the column image Mαℤ². Since det Mα≠0, its two columns are independent over ℚ and hence over ℤ. The subgroup theorem for ℤ² gives a free integer basis for I of rank at most two; the contained independent columns force its rank to be exactly two.
For finite index there is also a direct containment certificate. The integer adjugate identity Mα adj(Mα)=(det Mα)I₂ shows that (det Mα)ℤ²⊆Mαℤ². Thus I contains a nonzero integer multiple of both basis directions. Only finitely many residue pairs remain, so O/I is finite. This proves existence of a finite index without assuming that I is principal or invoking ideal factorization.
For the exact index of a principal ideal, apply the already proved Smith classification. Unimodular row and column changes reduce the full-rank matrix Mα to positive diagonal entries d₁,d₂. The additive quotient is ℤ/d₁ℤ⊕ℤ/d₂ℤ, with d₁d₂=|det Mα| elements. Consequently
N(αO)=|N(α)| for α≠0.
Likewise any integer basis matrix B for I gives N(I)=|det B|. One must first know that its columns are a complete ℤ-basis for I. A list of O-generators is not automatically such a basis; closure and equality with the proposed lattice need proof. The zero ideal is excluded: its quotient is the infinite additive group ℤ² and its rank is zero. The whole ring has norm 1.
Why every nonzero prime ideal is maximal
A prime ideal P is proper by definition. Since P≠0, the quotient O/P is a finite nonzero integral domain. For any nonzero class c, multiplication by c is injective by cancellation in a domain. On a finite set it is therefore surjective, so some class maps to 1. Every nonzero class has an inverse, making O/P a field. Hence P is maximal.
Both hypotheses matter. The zero ideal is prime because O is a domain, but is not maximal: O is not a field, for example 2 has no inverse in its integer basis. And a finite quotient alone need not be a domain. A composite index can yield zero divisors; it does not prove primality.
Changed integral basis and a complete membership certificate
Take K=ℚ(√13) and ω=(1+√13)/2. The preceding integrality theorem gives the full ring O=ℤ[ω], and ω²=ω+3. Thus t=1,u=−3. For α=3+ω,
Mα=[[3,3],[1,4]], det Mα=9.
The inverse matrix over ℚ gives coordinates ((4a−3b)/9,(−a+3b)/9) for a+bω relative to the columns. Both are integers exactly when a−3b is divisible by 9: substituting a=3b+9k makes the first numerator 9b+36k as well. Hence the principal ideal is precisely this lattice.
The map a+bω↦a−3b modulo 9 respects ω²−ω−3=0 because (−3)²−(−3)−3=9. It is onto and has the computed kernel, so O/(3+ω)≅ℤ/9ℤ as rings. Its nine elements agree with the norm determinant, but the nonzero class 3 squares to zero. Thus this ideal is not prime. This also illustrates why an additive index and a field conclusion are different claims.
Apply
In the same ring ℤ[(1+√13)/2], change α to 2+ω. Compute its multiplication matrix, exact index and a quotient map. Then test the proposed lattice generated by 4 and 1+s in ℤ[√−5]. Is its index a legitimate ideal norm?
Check the changed ideal and the failed lattice
For α=2+ω the matrix is [[2,3],[1,3]], with determinant 3. Its column lattice has membership a−2b divisible by 3. The map a+bω↦a+b modulo 3 has the same kernel, respects ω²−ω−3 and is onto F₃. Thus the principal ideal is maximal and has norm 3.
The lattice generated by 4 and 1+s has membership a−b divisible by 4, hence additive index four. But s(1+s)=−5+s has a−b=−6, not divisible by 4. It is not an ideal. Writing (4,1+s) as an O-ideal instead would generate additional integer vectors; its index cannot be read from those two columns without first enlarging to a closed lattice.
Continue
Reuse integer presentations for column relations and subgroup freeness, and quadratic norms and factorization for the nonprincipal ideal I above 2. Its norm is 2 even though it has no element generator.
Continue to fractional ideals and invertible classes. The finite-index and principal determinant results here do not yet prove that every nonzero ideal is invertible, that arbitrary ideal norms multiply, or that prime ideal factorization is unique. Those statements require their own proofs.
