Principal Ideal Domains and Factorization

A principal ideal domain (PID) is an integral domain R in which every ideal is principal, i.e. of the form

(a) = {ra|r ∈ R}

for some a ∈ R.

An element a ≠ 0 in a ring R is said to be irreducible if a is not a unit and whenever a = bc then one of b, c is a unit.

An element a ≠ 0 of a ring R is said to be prime if a is not a unit and whenever a|bc then either a|b or a|c.

Proposition 1. If R is an integral domain then a prime element is necessarily irreducible.

Proof. Take a prime in R, an integral domain. Then a is not a unit. Also, if a = bc then a|bc and so wlog we may assume a|b so b = ra for some r ∈ R and so a = arc; as R is an integral domain, 1 = rc and c is a unit.

Proposition 2. An irreducible element in a PID is prime.

Proof. Take a irreducible in a PID R. Suppose a|bc. Consider

(a, b) = {λa + μb|λ, μ ∈ R} ◁ R.

As R is a PID we have (a, b) = d for some d ∈ R. Now, d|a and so we can write a = de, say. As a is irreducible, EITHER d is a unit and so (a, b) = (1) = R and we can write 1 = λa + μb for some λ, μ ∈ R — but then c = λca + μbc, and as a|λca and a|μbc we have a|c; OR e is a unit, in which case d = ae^{−1} and a|d but then d|b (because b ∈ (d)) and so a|b. This shows that a is prime.

Proposition 3. Every PID is Noetherian.

Proof. Let R be a PID and suppose we have an ascending chain

(a_{1}) ⊂ (a_{2}) ⊂ (a_{3}) ⊂ …

of ideals in R. Then

⋃_{i=1}^{∞}(a_{i}) ◁ R.

So

⋃_{i=1}^{∞}(a_{i}) = (b)

for some b ∈ R. So b ∈ (a_{k}) for some k, and then

(b) ⊂ (a_{k}) ⊂ (a_{k+1}) ⊂ … ⊂ (b)

and so

(a_{k}) = (a_{k+1}) = (a_{k+2}) = ….

Proposition 4. In a PID, every non-zero element factorizes as a product of irreducible elements (“units are empty products”).

Proof. Suppose not. Then there is a non-factorizable element a, say. a is not irreducible and so we can write a = a_{1}b_{1} a proper factorization and where we have a_{1}, say, non-factorizable. Continuing this argument we get a_{1} = a_{2}b_{2}, a_{2} = a_{3}b_{3}, and so on, proper factorizations with each a_{i} a non-factorizable element, i.e. we have a sequence a, a_{1}, a_{2}, a_{3}, … with

(a) ⫋ (a_{1}) ⫋ (a_{2}) ⫋ (a_{3}) ⫋ …

contradicting Proposition 3.

Theorem 5. In a PID, any non-zero element can be factorized as a = up_{1} … p_{k} where u is a unit, p_{1}, p_{2}, …, p_{k} are prime, and this factorization is essentially unique in the sense that if a = up_{1} … p_{k} = vq_{1} … q_{l} are two prime factorizations then k = l and, after renumbering the q_{i}, we have p_{i} ∼ q_{i}, i.e. there exists a unit w_{i} such that p_{i} = w_{i}q_{i}.

Proof. In a PID we have factorization into irreducibles; but the irreducibles are prime; so any a ≠ 0 (whether a is a unit or not) can be written in the form a = up_{1} … p_{k}.

Suppose a = up_{1} … p_{k} = vq_{1} … q_{l}. Then p_{k}|vq_{1} … q_{l} so, as p_{k} prime, p_{k} must divide some q_{j} so by renumbering we can assume p_{k}|q_{l}. Then p_{k} ∼ q_{l}, say p_{k} = wq_{l} where w is a unit. Then up_{1} … p_{k−1}wq_{l} = vq_{1} … q_{l−1}q_{l} and cancelling q_{l}, we get uwp_{1} … p_{k−1} = vq_{1} … q_{l−1} and so we complete by induction.