Prove existence and uniqueness separately

Try

The integer thirty can be factored as two times fifteen or three times ten. Continuing to prime factors gives the same three primes, up to their order and signs. Abstract unique factorization must explain two facts: the process eventually stops, and different completed factorizations agree up to units and order.

An irreducible is a nonzero nonunit whose factorizations always have a unit factor. A prime element is a nonzero nonunit p satisfying p divides ab only if p divides a or b. These definitions are not interchangeable in every domain.

Predict

Predict which property proves uniqueness: inability to split one factor further, or divisibility of a product forcing divisibility of one factor. The latter is exactly primality. A PID will let us prove irreducibles are prime using a gcd and a Bézout identity.

Experiment

Track the logical dependencies rather than only an integer factorization.

Goal	Needed mechanism
A nonunit eventually factors	No infinite strictly ascending chain of principal ideals
An irreducible divides a product factor	Bézout identity with a coprime element
Two completed factorizations agree	Prime divisibility and cancellation

For a concrete check, compare 84=2·42=3·28. Both routes finish at 2²·3·7. Write where cancellation appears when matching the first prime in one factorization to a prime in the other.

Understand

In a PID, an irreducible p is prime. If p does not divide a, a gcd of p and a must be a unit: a nonunit divisor of p is associated to p and would force p to divide a. Normalize a Bézout identity to rp+sa=1. If p divides ab, multiplying by b gives b=rpb+sab, both terms divisible by p. Hence p divides b.

Now prove ascending chains of ideals stabilize. The union of an ascending chain is an ideal: any two chosen elements lie together in some later member, where sums and scalar products remain. Since the ring is a PID, the union is (d). Its generator d belongs to some stage I_N, forcing the whole union into that stage. Every later ideal therefore equals I_N.

Suppose a nonzero nonunit a lacks a finite irreducible factorization. It cannot be irreducible, so write a=bc with both factors nonunits. At least one factor must also lack a factorization, or their factorizations would combine. Repeating with a bad factor creates strictly increasing ideals (a)⊊(b)⊊… . Strictness follows because if (a)=(b), then b=ar and a=bc imply b=bcr; domain cancellation would make c a unit. This contradicts stabilization, proving existence.

For uniqueness, let a have two irreducible factorizations. The first irreducible on one side is prime and divides the product on the other, so it divides one of those irreducibles. They are associates. Absorb the unit, cancel the common factor in the domain, and continue inductively. Thus the factors agree up to order and units.

We have proved Euclidean domain implies PID implies UFD. The reverse implications are not automatic. For instance, F[x,y] is a UFD but the ideal (x,y) is not principal: a generator would divide both x and y, forcing it to be a unit, while evaluation at zero shows the ideal is proper.

Apply

Mark the separate roles of domain cancellation, principal ideals, and units in the proof. Explain why a factorization claim that ignores units would falsely distinguish six from negative two times negative three.

Then repair “irreducible always means prime” by stating the PID or UFD hypothesis under which that implication has been justified. Abstract vocabulary should not import integer properties into an arbitrary domain without proof.

Continue

Polynomial division next establishes root bounds and quotient coordinates. Gauss’s lemma will then extend unique factorization from a UFD to its polynomial ring, even when that polynomial ring is not a PID.

Test the hypotheses in quadratic norms and factorization, where an irreducible is not prime. Then extend the ascending-chain argument in Noetherian rings and Hilbert basis.