Try
Let s=√(−5) inside C, and let R=Z[s]=❴a+bs:a,b∈Z❵. The coordinates a,b are unique. We have s²=−5, so multiplication stays in R. As a subring of C, R is a commutative integral domain with identity. In this ring,
6 = 2·3 = (1+s)(1−s).
Different-looking expressions need not violate unique factorization: perhaps a factor is reducible, or the factors differ by units. Predict which proofs would rule out those possibilities. A second question will concern the ideal I=(2,1+s), the set of all R-linear combinations of those generators. Can a nonprincipal ideal have a principal square?
Predict
Conjugation sends a+bs to a−bs. Define the norm N(a+bs)=(a+bs)(a−bs)=a²+5b². It is a nonnegative integer, and is zero exactly for the zero element. If a nonzero element factors as xy, its norm will factor as N(x)N(y). Predict whether a nonunit can have norm 2 or 3 by examining possible integer coordinates.
Recall that an irreducible is a nonzero nonunit whose every factorization has a unit factor. A prime element π is a nonzero nonunit such that π divides xy only if it divides x or y. In a domain every prime is irreducible, but the converse is not automatic. Reuse factorization definitions and ideals and quotients rather than assuming R is a PID.
Experiment
Complete the coordinate checks before using the norm as a factorization obstruction.
| Element or candidate norm | Calculation | Consequence to justify |
|---|
| 2 | N(2)=4 | A nontrivial factorization would need norms 2 and 2 |
| 3 | N(3)=9 | A nontrivial factorization would need norms 3 and 3 |
| 1+s and 1−s | Norm 6 each | Nontrivial factors would need norms 2 and 3 |
| Norm 2 or 3 | If b≠0, a²+5b²≥5; if b=0, a² is a square | Neither norm occurs |
| 1+s divisible by 2? | Both coordinates of 2(a+bs) are even | No |
For the ideal square, multiply its generators. The product of finitely generated ideals is generated by all pairwise products, so I²=(4,2+2s,−4+2s). Each displayed generator is divisible by 2 in R. Now seek an integer combination of the three generators equal to 2. Subtract the third from the second, then subtract the first. This gives
(2+2s) − (−4+2s) − 4 = 2.
The equation is a certificate, not just a norm comparison. It will prove the reverse containment needed for I²=(2).
Understand
Multiplicativity, units and the two irreducible factorizations
Conjugation preserves addition and multiplication, as can be checked from s²=−5. For x,y∈R, commutativity gives N(xy)=xy conjugate(x) conjugate(y)=N(x)N(y). If x is a unit with inverse y, its positive integer norm divides N(1)=1, so N(x)=1. The equation a²+5b²=1 forces b=0 and a=±1. Conversely these two elements are units. Thus the units of R are exactly 1 and −1.
A nonzero nonunit has norm at least 2. If 2 factored into two nonunits, their norms would be 2 and 2, impossible by the table. Similarly a nontrivial factorization of 3 would require norms 3 and 3. A nontrivial factorization of either 1+s or 1−s would require norms 2 and 3 in some order. Therefore all four factors are irreducible.
Associates differ by multiplication by a unit, here only a sign. The norms 4,9,6 already separate 2 and 3 from each other and from both 1±s. Finally 1+s is neither 1−s nor −(1−s), by comparing the unique integer coordinates. Hence no permutation and multiplication by units identifies the factors in the two factorizations of 6. This proves R is not a unique factorization domain; it does not merely exhibit two expressions for one element.
The element 2 is not prime: it divides (1+s)(1−s)=6, but neither factor has both coordinates even, so neither is divisible by 2. In quotient language, the nonzero classes of 1+s and 1−s multiply to zero modulo (2). This failure is compatible with 2 being irreducible because we have not assumed the PID hypotheses that make irreducibles prime.
The ideal I is proper and not principal
Define φ:R→F₂ by φ(a+bs)=a+b modulo 2. This is a unital ring homomorphism: it sends s to 1, and the relation s²=−5 becomes 1=1 in F₂. Explicitly, the constant and s coefficients of a product give the same result modulo two as multiplying a+b and c+d. Constants make φ surjective.
Its kernel is exactly I. Both 2 and 1+s map to zero, so I⊆ker φ. Conversely if a+b is even, then a−b is even, and
a+bs = (a−b) + b(1+s)
is a combination of 2 and 1+s with integer coefficients. Thus ker φ⊆I. The first ring isomorphism theorem gives R/I≅F₂, so I is proper (indeed maximal and prime). This is not the ideal (2); 1+s belongs to I but not to (2).
Suppose I=(α). Since I contains the nonzero element 2, α≠0. Divisibility of 2 and 1+s by α implies that the positive integer N(α) divides both 4 and 6. It is therefore 1 or 2. Norm 2 is impossible. Norm 1 would make α a unit and I=R, contradicting properness. This proves nonprincipality without assuming an index formula or any theorem about ideal factorization.
Prove both containments for the square
By definition, IJ consists of finite sums of products xy with x∈I and y∈J. If ideals are generated by finite lists, distributing those combinations proves that their product is generated by the pairwise generator products; conversely each such product belongs to IJ. Applying this to I gives the three generators in Experiment, since the two cross products coincide.
Each generator is 2 times an element of R: 4=2·2, 2+2s=2(1+s), and −4+2s=2(−2+s). Consequently I²⊆(2). The displayed integer combination proves 2∈I². As I² is an ideal it contains every multiple of 2, so (2)⊆I². Therefore I²=(2).
This computation does not say I is principal; we just proved otherwise. Nor does it establish that all ideals of R have unique factorizations, or compute a class number. Such general conclusions need additional definitions and proofs. Here we have verified a particular ideal identity directly.
Apply
Change the ideal to J=(3,1+s). Construct a quotient map to F₃, prove its kernel is J, and decide whether J is principal. Then predict IJ before multiplying its four generator products. Can the same irreducible element 1+s generate that product ideal?
Check the changed ideal and its product
Use ψ(a+bs)=a−b modulo 3, sending s to −1. Its square is 1, equal to −5 modulo 3, so this is a unital onto ring map. If a−b is divisible by 3, the identity a+bs=(a−b)+b(1+s) puts the element in J; the generators map to zero, proving the reverse inclusion. Thus R/J≅F₃ and J is proper.
If J=(α), then N(α) divides both N(3)=9 and N(1+s)=6, so is 1 or 3. Norm 3 does not occur, and norm 1 contradicts properness. Thus J is also nonprincipal.
Now IJ=(6,2(1+s),3(1+s),(1+s)²). All four are multiples of 1+s, since 6=(1+s)(1−s). Conversely 3(1+s)−2(1+s)=1+s belongs to IJ. Both containments give IJ=(1+s). Distinct nonprincipal ideals can therefore have a principal product. This is another exact identity, not a general classification of ideals.
For a norm check nearby, consider 2+s, whose norm is 9. Any nontrivial factorization would again require norm 3 factors, so it is irreducible. It is not associate to 3 despite having the same norm: ±3 have zero s coefficient. Equal norms alone do not imply association.
Continue
Return to prime and maximal ideals to connect the quotient fields with primality, and to Euclidean domains to identify why their stronger factorization theorems cannot be assumed here. The nonprincipal ideal already rules out R being a PID, and hence rules out R being Euclidean.
For a new norm argument, enumerate every possible product of positive integer norms and distinguish units, irreducibles, primes and associates. For an ideal equality, prove both containments with explicit generators or maps. These are different proof obligations: norms constrain element divisibility, while quotient maps and generator identities can expose ideal structure invisible in a factorization of one element.
Continue to the full ring of quadratic integers to justify the ambient ring, ideal lattices and indices to calculate quotient sizes, and invertible ideal classes to turn the product certificate into a nontrivial class.
