Prove ideal norms multiply

Try

In O=ℤ[s], s²=−5, the ideal I=(2,1+s) has index 2 and satisfies I²=(2). Its square has index 4. Does a quotient with four elements have to be a field? No: O/(2) contains a nonzero element ε represented by 1+s, and ε²=0.

This lesson proves why ideal indices multiply, then uses exact quotient maps to distinguish fields, products and rings with nilpotent elements. An element is nilpotent if a positive power is zero; a nonzero nilpotent cannot occur in a field. Cardinality alone does not tell us which ring structure is present.

The Dedekind ideal-factorization theorem now supplies a new hypothesis unavailable in the earlier example lessons: every nonzero integral ideal of the full quadratic integer ring factors into invertible nonzero prime ideals. We use that proved theorem, not an analogy with factorization of elements.

Predict

For a nonzero integral ideal A⊆O, write N(A)=[O:A], its finite additive index. The preceding ideal-lattice lesson proves finiteness and N(αO)=|N(α)| for nonzero elements α. The unit ideal O has norm 1. The zero ideal is excluded throughout: its quotient is infinite.

Suppose P is a nonzero prime ideal and A is invertible. Predict the number of cosets in A/PA. It will be N(P), independently of A. But A need not have a single ideal generator, so “cancel A in the quotient” is not a valid proof. We will construct a generator of this quotient from a finite inverse certificate.

A/PA is naturally a vector space over the field O/P: define (r+P)(x+PA)=rx+PA. If r changes by an element of P or x by an element of PA, the result changes by an element of PA, so this action is well-defined.

Experiment

Keep s²=−5 and I=(2,1+s). In O/(2), put ε=1+s. Express each residue a+bs using 1 and ε by substituting s=ε−1. Modulo 2, subtraction and addition agree. Predict the four residues and their multiplication before checking the table.

Here ε²=(1+s)²=−4+2s is zero modulo 2, but ε itself is nonzero because its two integer coordinates are odd. Thus the quotient is F₂[ε]/(ε²), with four elements. The chain O⊇I⊇I² has two successive quotients of size 2. Only the first quotient O/I is a field; the entire quotient O/I² is not.

Check the square-ideal certificate rather than assuming it

I² has generators 4, 2+2s and −4+2s. All lie in 2O, and (2+2s)−(−4+2s)−4=2 belongs to their span. Hence I²=2O. The map from F₂[X]/(X²) to O/2O sending X to 1+s is onto since s=X−1. Both sides have the unique two-coefficient forms over F₂, so the map is injective as well. This proves the ring description and its size directly, independently of the general norm theorem below.

Understand

Build a one-dimensional quotient from an inverse certificate

Work first in any domain R. Let A be an invertible integral ideal, let B=A⁻¹ be its fractional inverse, and let P be a maximal ideal with finite residue field R/P. The equality AB=R means there are finitely many aᵢ∈A and bᵢ∈B with Σaᵢbᵢ=1. Every product bᵢx with x∈A lies in R, even though bᵢ itself may be fractional.

Not every aᵢbᵢ can lie in P, since their sum is 1. Choose a,b from the certificate with ab∉P. Since R/P is a field, choose c∈R with cab≡1 modulo P. For any x∈A,

x−c(bx)a=(1−cab)x ∈ PA.

The coefficient c(bx) is in R. Therefore the class of a generates A/PA over R/P. The homomorphism R→A/PA sending r to ra+PA is surjective. Its kernel is exactly P: inclusion P⊆kernel is immediate; conversely, if ra∈PA, multiplying by b gives rab∈P because bA⊆R. Since ab∉P and P is prime, r∈P.

Thus A/PA≅R/P as R-modules and is one-dimensional over R/P. This is a module isomorphism, not generally a ring isomorphism: A/PA has no specified multiplicative identity. In particular [A:PA]=[R:P]. The generator a is selected from a certificate, and need not generate the ideal A itself.

Multiply indices along an actual finite chain

If C⊆B⊆A are additive groups and both [A:B] and [B:C] are finite, choose coset representatives aᵢ for A/B and bⱼ for B/C. The sums aᵢ+bⱼ form a complete, pairwise distinct representative list for A/C. Indeed reduce first modulo B and then modulo C; equality modulo C forces both choices equal. Hence [A:C]=[A:B][B:C]. This proves the index rule used here without treating quotient sets as numbers by analogy.

Now return to the full quadratic integer ring O. Factor a nonzero ideal B=P₁⋯Pᵣ by the preceding theorem, permitting repeated primes. Set A₀=A and Aⱼ=AP₁⋯Pⱼ. These are integral invertible ideals; Aⱼ⊆Aⱼ₋₁. The one-dimensional quotient result gives

[Aⱼ₋₁:Aⱼ]=N(Pⱼ).

The finite index rule implies [A:AB]=∏N(Pⱼ). Starting instead at O gives N(B)=∏N(Pⱼ). Finally the chain AB⊆A⊆O yields

N(AB)=N(A)[A:AB]=N(A)N(B).

The unit case B=O is the empty product and the chain has no steps; A=O is also included. No norm of a zero ideal has entered the argument. This proof uses factorization existence and invertibility; uniqueness ensures the prime data are intrinsic, but is not a substitute for the quotient-size argument.

In particular N(Pᵉ)=N(P)ᵉ for integers e≥0. For e≥2, O/Pᵉ is not a field: Pᵉ is strictly smaller than P because the index step is N(P)>1. More precisely choose x∈P∖Pᵉ; its nonzero residue modulo Pᵉ is nilpotent since xᵉ∈Pᵉ. This separates the field O/P from its thicker prime-power quotients.

The split quotient above 3: explicit CRT, not a splitting slogan

In ℤ[s], s²=−5, put P=(3,1+s) and Q=(3,1−s). The earlier lattice certificates give quotient maps a+bs↦a−b and a+b modulo 3. The two ideals are distinct and maximal. They are comaximal: their sum contains 3 and (1+s)+(1−s)=2, hence contains 1=3−2.

Their product has generators 9,3(1+s),3(1−s),6. All lie in 3O, and 9−6=3 proves PQ=3O. To justify the CRT kernel, comaximal ideals satisfy P∩Q=PQ: if u∈P,v∈Q and u+v=1, then x=xu+xv is in PQ for x∈P∩Q; the reverse containment always holds. Here u=3−(1+s)∈P and v=−(1−s)∈Q give such a pair.

The explicit map

a+bs ↦ (a−b,a+b) modulo 3

has kernel P∩Q=3O. It is onto: for a target (r,w), take a=2(r+w), b=2(w−r) modulo 3, since 2 is the inverse of 2 in F₃. Therefore

O/(3) ≅ F₃×F₃.

Its nine elements agree with N(3O)=|N(3)|=9 and N(P)N(Q)=3·3. It is not a field: (1,0)(0,1)=(0,0). In contrast with the quotient at 2, it has no nonzero nilpotents, since neither factor field does. These are two fully checked examples, not an unproved rule describing every rational prime.

Why the full-ring hypotheses cannot be discarded

In the nonmaximal order R=ℤ[√5], let J=(2,1+√5). Its integer basis has determinant 2, so [R:J]=2. Direct multiplication gives J²=2J. The basis of 2J is 4,2+2√5, with determinant 8. Thus [R:J²]=8, not 2². This does not contradict the theorem: J is not invertible in this ring, as the preceding inverse-ideal lesson proves. The finite-index condition alone never justified the one-dimensional quotient argument.

Apply

Change the full ring to O′=ℤ[ω], ω=(1+√5)/2 with ω²−ω−1=0. Determine O′/(2), its cardinality and whether it is a field. Compare its multiplicative behavior with the four-element quotient O/(2) from Try. Then compute N(I³) back in ℤ[√−5] and describe its additive group using I³=2I.

Check the changed ring and the higher power

Modulo 2 the defining polynomial is X²+X+1. It has neither 0 nor 1 as a root, so its degree-two irreducibility makes F₂[X]/(X²+X+1) a field. The integral basis gives O′/(2) exactly this quotient, with four elements. Every nonzero residue is invertible, unlike ε in the earlier four-element ring. Both cardinalities follow from a rank-two integer basis; the multiplication relations differ.

For I³=2I, the basis columns in 1,s coordinates are (4,0) and (2,2), with determinant 8. Smith theory gives diagonal entries 2,4, since the gcd of all entries is 2 and their product is 8. Its additive quotient is C₂⊕C₄, confirming N(I³)=2³=8. An additive-group description alone still does not specify its ring multiplication.

Continue

Return to ideal lattices and principal indices for exact membership certificates, and to fractional ideals and invertible classes for the inverse used in the proof. Norm multiplicativity is now justified for all nonzero integral ideals of the full quadratic integer ring. A complete class-number calculation still needs a proof that sufficiently small representatives cover every class; examples of particular classes do not supply that bound.