Try
Let O=ℤ[s] with s²=−5, and let I=(2,1+s). Earlier exact calculations established I²=(2), although I is not principal. Predict what should multiply I to give the unit ideal O. Multiplying by I again gives 2O, not O; allowing coefficients in the fraction field suggests (1/2)I.
This is why fractional ideals are useful: the inverse need not be contained in the original ring. Review quadratic ideals as lattices for the quotient/index certificate, and integral elements and quadratic integers for the distinction between O and a nonmaximal quadratic order. We will prove a group law only for ideals that actually have inverses.
Predict
Let R be a commutative domain with identity 1≠0 and fraction field K. In this lesson a fractional ideal means a nonzero finitely generated R-submodule A of K for which some nonzero d∈R satisfies dA⊆R. Thus A consists of finite R-linear combinations of finitely many fractions, not necessarily of elements of R. A product AB means all finite sums of products ab with a∈A and b∈B, not merely the set of individual products.
For our finite-generation convention, a common denominator always exists: multiply denominators of a finite generator list. We state it explicitly because it explains why A becomes an integral ideal after scaling. Some texts allow fractional ideals that are not finitely generated; we use the finite-type convention throughout. Over a Noetherian domain the conventions agree, since dA is an ideal and multiplication by d is an R-module isomorphism A→dA.
Predict why A=(1/2)I is a legitimate fractional ideal of O even though (1+s)/2 is not in O. Then decide whether merely writing the candidate set (R:I)=❴x∈K:xI⊆R❵ already proves that it is an inverse. We will see a counterexample.
Experiment
Use the generators of I to compute the four generator products of I·(I/2). Their R-span is the product ideal; a single product equaling 1 is unnecessary.
| Generator of I | Generator of I/2 | Product |
|---|
| 2 | 1 | 2 |
| 2 | (1+s)/2 | 1+s |
| 1+s | 1 | 1+s |
| 1+s | (1+s)/2 | −2+s |
All four lie in O, and (1+s)−(−2+s)−2=1. Therefore the product ideal is exactly O. Record both containments: generator products give I(I/2)⊆O, while the displayed combination puts 1 in the product, forcing the reverse inclusion. This independently verifies the inverse rather than just naming it.
Understand
Product laws and their boundaries. If A is generated by a₁,…,aᵣ and B by b₁,…,bₘ, then AB is generated by the finitely many aᵢbⱼ. Distributing any product of two linear combinations proves this assertion in both directions. If dA⊆R and eB⊆R, then deAB⊆R, and de≠0 because R is a domain. Picking nonzero a∈A,b∈B shows AB≠0. Thus products remain fractional ideals.
Commutativity follows from multiplication in K. Associativity follows because both (AB)C and A(BC) consist of finite sums of triple products abc; expand the finite sums in either grouping. Also RA=A: absorption gives one containment and 1·a gives the other. These laws make the nonzero finite-type fractional ideals a commutative monoid, not necessarily a group.
A is invertible if a fractional ideal B satisfies AB=R. An inverse is unique: if AB=AC=R, then B=B(AC)=(BA)C=C. For invertible A its inverse is exactly the colon module
(R:A)=❴x∈K:xA⊆R❵.
Indeed, AB=R first implies B⊆(R:A). Write 1=Σaᵢbᵢ, possible by the definition of the product. For x∈(R:A), multiplication gives x=Σ(xaᵢ)bᵢ∈B, proving the reverse containment. This also proves finite generation and a common denominator for this colon module in the invertible case, because it equals B. We do not assume these properties or the inverse identity for a general colon module.
Cancellation is now justified when the factor being cancelled is invertible: AB=AC implies A⁻¹AB=A⁻¹AC and hence B=C. Without that hypothesis it may fail even for nonzero ideals in a domain.
The group and its quotient. Products of invertible ideals are invertible, with inverse A⁻¹B⁻¹. R is invertible and inverses remain invertible. The invertible fractional ideals therefore form an abelian group. For every nonzero x∈K, the principal fractional ideal xR is invertible with inverse x⁻¹R. Such ideals form a subgroup, so we may quotient by them. We call this the group of invertible ideal classes; no assertion that all nonzero ideals are invertible is built into this definition.
Two invertible fractional ideals A,B give the same class exactly when A=xB for some x∈K×. Indeed their classes agree iff AB⁻¹=xR, and multiplication by B gives the claimed equality; the reverse is immediate. The identity class consists precisely of principal fractional ideals.
An actual nonidentity class. The experiment shows I⁻¹=I/2. To recall nonprincipality, the quotient O/I has two elements: the map a+bs↦a+b modulo 2 is onto and its kernel is exactly I. If I=αO, then α∈I⊆O, even if it was initially described as a fractional generator. The principal-index formula from the preceding lesson gives |O/αO|=|N(α)|. For α=a+bs this norm is a²+5b², which cannot equal 2 for integers a,b. Thus I is not principal even as a fractional ideal. Since I²=2O is principal, its class has exact order two. This proves the class group contains an element of order two. It does not prove there are only two classes.
A noninvertible ideal in a different order. Now change both ring and relation: R=ℤ[s], s²=5, and J=(2,1+s). Its product generators are 4,2+2s and 6+2s. The last differs from the second by 4, so
J²=(4,2+2s)=2J.
But J≠2R, since 1+s has odd coordinates in the basis 1,s. If J were invertible, cancelling J from J·J=J·(2R) would force J=2R, a contradiction. This is also an explicit failure of cancellation without invertibility.
The integral element ω=(1+s)/2 does not lie in this R, but preserves J: ω·2=1+s and ω(1+s)=3+s=2+(1+s). For any invertible ideal A, xA⊆A implies xR=xAA⁻¹⊆AA⁻¹=R, and hence x∈R. Thus ω supplies a second obstruction to J being invertible. This makes the difference between the nonmaximal order ℤ[√5] and its full quadratic integer ring operational, without appealing to an unproved general theorem about all ideals.
Apply
Return to s²=−5. Let P=(3,1+s) and Q=(3,1−s). Compute PQ and find P⁻¹. Then decide whether the same formula “take the colon module” alone would establish invertibility of the preceding J in ℤ[√5].
Check the changed inverse and the failed shortcut
PQ is generated by 9,3(1+s),3(1−s),6. These all lie in 3O and 9−6=3 belongs to their span. Hence PQ=3O and P⁻¹=Q/3. P is nonprincipal too: its quotient has order3 by evaluating s at −1 modulo3, whereas a²+5b² cannot equal3. Moreover IP=(1+s)O: the generators 6,2(1+s),3(1+s),(1+s)² are all multiples of 1+s, and 3(1+s)−2(1+s)=1+s. Thus [P]=[I]⁻¹=[I], since [I] has order two. This identifies particular classes, not every class.
For J in ℤ[√5], the colon module is R+Rω=J/2. To check it, 2x∈R forces x=(a+bs)/2 with integers a,b; the other generator condition requires a+b even. Those are exactly the elements of ℤ+ℤω, which is also R+Rω because ω²=ω+1. But J(J/2)=J, not R, by J²=2J. The set of fractions sending J into R can exist with a finite generating set and still fail to be an inverse. Notice that J is proper: evaluation s↦1 modulo2 has kernel J.
Continue
Return to quadratic norm and ideal-product certificates to distinguish element norms, ideal indices and invertible classes. Unique factorization of all nonzero ideals, general multiplicativity of ideal norms and an exhaustive class-number computation require further proofs. The results here supply a group of verified invertible ideals, concrete inverse certificates and a genuinely nontrivial class; they do not supply those additional classification theorems.
Those further proofs begin with Dedekind ideal factorization, continue through multiplicativity of ideal norms, and culminate in a complete quadratic class-group calculation.
