Prove unique factorization of nonzero ideals

Try

In O=ℤ[√−5], put s=√−5 and P=(2,1+s). Parentheses here mean the ideal generated by the listed elements. Multiplying generators gives P²=(4,2+2s,−4+2s). Each generator is divisible by 2 in O, and 2=2·4+(−4+2s)−(2+2s). Thus P²=(2). The ideal P is not principal, as proved in the preceding lesson. Ideal factorization can recover a unique multiplicative description even when element factorization fails.

A proposed shortcut says that every displayed generator is divisible by 2, therefore P²=(2). That proves only containment. Cover the explicit combination above and reconstruct coefficients proving the reverse containment.

Our goal is a general theorem: in a Dedekind domain, every nonzero proper ideal is a product of nonzero prime ideals, uniquely up to order. Here this term means a nonzero commutative domain R that is Noetherian, integrally closed in its fraction field K, and has every nonzero prime ideal maximal. Noetherian means ascending chains of ideals stabilize, equivalently every ideal is finitely generated. We will prove the factorization theorem from these hypotheses; invertibility is an outcome, not an extra assumption.

Predict

Suppose a nonzero prime ideal P has a fractional inverse B with PB=R. If I⊆P, predict why BI is an ordinary ideal inside R and why I=P(BI). Does that identity alone prove BI is strictly larger than I? Identify the extra argument that must rule out equality.

A fractional ideal is a nonzero finitely generated R-submodule of K with a common nonzero denominator. Review fractional ideals and inverses if needed. The colon (R:P) means all x∈K such that xP⊆R. Its name does not yet prove that it is an inverse.

Experiment

Work in O=ℤ[√−5] with P as above. The fractional ideal P/2 has product R with P by the repaired certificate. Starting from I=(2), multiply once by P/2 and then once more. Predict every row before comparing.

The inclusions (2)⊊P⊊O are strict: 1+s is not in (2), and the homomorphism O→𝔽₂ sending s to 1 has kernel P, so P is proper. Enlargement, not shrinking, is what lets the ascending-chain condition prove termination. Write the generator-product certificate and the two strictness witnesses separately.

Check the initial repair

The valid expression is 2=2·4+(−4+2s)−(2+2s). Therefore P² contains 2 and is contained in (2), proving equality. Containment in (2) alone would not prove equality: for example (4)⊊(2).

Understand

1. A prime product inside any nonzero ideal

Include R as the empty product of prime ideals. Suppose some nonzero ideal contains no finite product of nonzero prime ideals. By the ascending-chain condition choose an inclusion-maximal such counterexample I. It is proper because R contains the empty product. It cannot be prime, since a nonzero prime contains itself. Consequently there exist a,b∉I with ab∈I. Both I+(a) and I+(b) strictly contain I, and each contains a product of nonzero primes by maximality. Their product lies in I: expand (I+(a))(I+(b)) into I²+aI+bI+(ab). Multiplying the two contained prime products gives a prime product inside I, a contradiction. This also covers a larger ideal equal to R, using its empty product. Products of nonzero ideals are nonzero in a domain.

2. The colon of a prime really is larger than R

Fix a nonzero prime P and choose 0≠a∈P. The principal ideal (a) is nonzero and proper. Choose a product Q₁⋯Qᵣ⊆(a) of minimal length; r≥1. Since this product lies in P, one Qᵢ is contained in P. Indeed, if no factor were contained, choosing an element outside P from each would contradict primality. Both ideals are nonzero primes and hence maximal, so Qᵢ=P. Reorder it first.

The remaining product B=Q₂⋯Qᵣ is not contained in (a), or a shorter product would work. When r=1, B=R. Choose b∈B outside (a). Then x=b/a is not in R, but xP⊆R because bP⊆(a). Hence R⊊C=(R:P).

This colon is a fractional ideal, not merely a subset: it is an R-submodule, contains R, and aC⊆R because a∈P. The ideal aC is finitely generated by Noetherianity; dividing its generators by a generates C. Thus C⊆(1/a)R and has a common denominator.

3. Why the determinant trick applies to an ideal

We need a precise version of the finite-module criterion. If a nonzero finitely generated ideal J satisfies xJ⊆J for x∈K, choose R-generators m₁,…,mᵣ. Write xmᵢ=Σcᵢⱼmⱼ. Multiplying (xI−C)m=0 by its adjugate shows the monic polynomial f(X)=det(XI−C) satisfies f(x)mᵢ=0 for every generator. Some mᵢ is nonzero, and K is a field; therefore f(x)=0. Integral closure forces x∈R.

We did not assume 1∈J. Equivalently J is a faithful R[x]-module inside K: a scalar in R[x] annihilating a nonzero element must be zero. Faithfulness merely over R would not justify that conclusion for an arbitrary abstract module. Finite generation, a nonzero module, and its placement inside the fraction field are essential here.

4. Every nonzero prime is invertible

For C=(R:P), we have P⊆PC⊆R. Since P is maximal, PC is either P or R. If PC=P, the element x∈C outside R constructed above satisfies xP⊆P. Step 3 would force x∈R, a contradiction. Thus PC=R. This establishes invertibility without assuming the theorem we are proving.

5. Factorization by strict enlargement

Let I be a nonzero proper ideal. Choose a maximal ideal P containing I; such an ideal exists, for example by ACC applied to proper ideals containing I. It is nonzero and prime. Put I₁=P⁻¹I. Since I⊆P, we have I₁⊆P⁻¹P=R; because R⊆P⁻¹, also I⊆I₁. The product is a nonzero ideal and I=PI₁.

If I₁=I, every x∈P⁻¹ preserves I. But P⁻¹ contains an x outside R, contradicting Step 3. Thus I⊊I₁. Repeat whenever the current ideal is proper. An infinite repetition would be a strictly ascending chain of ideals, forbidden by Noetherianity. Therefore the process reaches R after finitely many steps. Reversing the identities gives I=P₁⋯Pᵣ. All nonzero integral ideals are now invertible, as products of invertibles. Clearing denominators gives the same conclusion for every nonzero fractional ideal.

6. Uniqueness and boundaries

If P₁⋯Pᵣ=Q₁⋯Qₛ with all factors nonzero prime ideals, the right side is contained in P₁. Primality forces some Qⱼ⊆P₁; maximality makes Qⱼ=P₁. Multiply by its proved fractional inverse and cancel. Repeat. A nonempty product of proper ideals lies in its first factor and cannot be R, so one side cannot run out of factors before the other. This proves equality of the multisets of primes.

The unit ideal has the empty factorization. In a field it is the only nonzero ideal, so the theorem is vacuous but correct. The zero ideal is excluded: a finite product of nonzero ideals cannot be zero in a domain. We make no claim that elements factor uniquely merely because ideals do.

7. Quadratic integer rings satisfy the hypotheses

For squarefree d≠0,1, the preceding integrality lesson proves that the full integer ring O in ℚ(√d) has a free ℤ-basis of length two and is integrally closed. It is a domain. Every ideal is an additive subgroup of ℤ², hence has finitely many integer generators by the subgroup theorem taught with integer presentations. These same generators generate the ideal over O: their O-span is contained in the ideal and contains its entire integer span. Thus O is Noetherian.

If P is a nonzero prime, choose 0≠α∈P. Its conjugate ᾱ is in O, and its nonzero integer norm αᾱ lies in P. Therefore O/P is a quotient of O/(N), a finite additive group for the nonzero integer N=αᾱ. Since P is prime, O/P is a finite domain. Every nonzero element of a finite domain has an inverse: multiplication by it is injective and hence surjective. Thus O/P is a field and P is maximal. All hypotheses are now checked for positive as well as negative squarefree d.

Apply

Change the scalar ring to R=ℤ[√5], put s=√5, and J=(2,1+s). Show J²=2J but J≠2R. Explain exactly why the proof above cannot conclude that J is invertible. Then replace R by the full integer ring O=ℤ[(1+s)/2] and describe the same subset J.

Compare a complete diagnosis

The square has generators 4,2+2s,6+2s. Their difference (6+2s)−(2+2s)=4 gives no new element, and 6+2s=4+(2+2s), so J²=(4,2+2s)=2J. Since 1+s∉2R, J≠2R. Invertibility would cancel J from J²=(2)J and force J=(2), impossible. Here ω=(1+s)/2 is outside R, yet 2ω=1+s and ω(1+s)=3+s=2+(1+s) show ωJ⊆J. The determinant trick correctly proves that ω is integral; R is not integrally closed, so its final inference fails. In O, the same subset is 2O, a principal invertible ideal. The ring is part of every ideal-factorization claim.

Continue

Reconstruct the theorem without looking: prime product inside I; an element of the prime colon outside R; determinant contradiction; strict enlargement; cancellation. At each arrow state which hypothesis is used. Return to Noetherian rings for ACC, or fractional ideal classes to interpret the resulting invertible ideals. Norm multiplicativity and exhaustive class-group computations need additional arguments; the theorem here does not enumerate the ideal classes of a quadratic ring.