every PID is a UFD

Theorem 1.

The first step of the proof shows that any PID is a Noetherian ringMathworldPlanetmath in which every irreduciblePlanetmathPlanetmathPlanetmathPlanetmathPlanetmath is prime. The second step is to show that any Noetherian ring in which every irreducible is prime is a UFD.

We will need the following

Lemma 2.

Every PID R is a gcd domain. Any two gcd’s of a pair of elements a,b are associatesMathworldPlanetmath of each other.


Suppose a,bR. Consider the ideal generated byPlanetmathPlanetmath a and b, (a,b). Since R is a PID, there is an element dR such that (a,b)=(d). But a,b(a,b), so da,db. So d is a common divisorMathworldPlanetmathPlanetmathPlanetmath of a and b. Now suppose ca,cb. Then (d)=(a,b)(c) and hence cd.

The second part of the lemma follows since if c,d are two such gcd’s, then (c)=(a,b)=(d), so cd and dc so that c,d are associates. ∎

Theorem 3.

If R is a PID, then R is Noetherian and every irreducible element of R is prime.


Let I1I2I3 be a chain of (principal) ideals in R. Then I=kIk is also an ideal. Since R is a PID, there is aR such that I=(a), and thus aIn for some n. Then for each m>n, Im=In. So R satisfies the ascending chain conditionMathworldPlanetmathPlanetmathPlanetmath and thus is Noetherian.

To show that each irreducible in R is prime, choose some irreducible aR, and suppose a=bc. Let d=gcd(a,b). Now, da, but a is irreducible. Thus either d is a unit, or d is an associate of a. If d is an associate of a, then adb so that ab and c is a unit. If d is itself a unit, then we can assume by the lemma that d=1. Then 1(a,b) so that there are x,yR such that xa+yb=1. Multiplying through by c, we see that xac+ybc=c. But axac and aybc=ya. Thus ac so that b is a unit. In either case, a is prime. ∎

Theorem 4.

If R is Noetherian, and if every irreducible element of R is prime, then R is a UFD.


We show that any nonzero nonunit is R is expressible as a product of irreducibles (and hence as a product of primes), and then show that the factorization is unique.

Let 𝒰R be the set of ideals generated by each element of R that cannot be written as a product of irreducible elements of R. If 𝒰, then 𝒰 has a maximal element (r) since R is Noetherian. r is not irreducible by construction and thus not prime, so (r) is not prime and thus not maximal. So there is a proper maximal idealMathworldPlanetmath (s) with (r)(s), and sr.

Since (r) is maximal in 𝒰, it follows that (s)𝒰 and thus that s is a product of irreducibles. Choose some irreducible as; then ar and


for some bR. If (b)𝒰 (note that this includes the case where b is a unit), then b and hence r is a product of irreducibles, a contradiction. If (b)𝒰 then (r)(b) (since br). (r)(b) since a is not a unit, and thus (r)(b). This contradicts the presumed maximality of (r) in 𝒰. Thus 𝒰= and each element of R can be written as a product of irreducibles (primes).

The proof of uniqueness is identical to the standard proof for the integers. Suppose


where the pi and qj are primes. Then p1q1qm; since p1 is prime, it must divide some qj. Reordering if necessary, assume j=1. Then p1=uq1 where u is a unit. Factoring out these terms since R is a domain, we get


We may continue the process, matching prime factorsMathworldPlanetmath from the two sides. ∎

Title every PID is a UFD
Canonical name EveryPIDIsAUFD
Date of creation 2013-03-22 16:55:51
Last modified on 2013-03-22 16:55:51
Owner rm50 (10146)
Last modified by rm50 (10146)
Numerical id 9
Author rm50 (10146)
Entry type Theorem
Classification msc 13F07
Classification msc 16D25
Classification msc 11N80
Classification msc 13G05
Classification msc 13A15
Related topic UFD
Related topic UniqueFactorizationAndIdealsInRingOfIntegers