unique factorization domain

1. Every element of R which is not zero or a unit can be written as a product of irreducible elements of R.
2. This expression is unique in the sense that if we write c=a₁...a_r=b₁...b_s in R, for some nonzero non-unit c, with each a_i and b_j irreducible, then we have r=s and after renumbering the a_i if necessary we have that each a_i is an associate of b_j.

Of course, the fundamental theorem of arithmetic asserts that the ring of integers Z is a UFD.

Proposition Any principal ideal domain is a UFD.

After the Proposition and the principal ideal domain writeup we know that besides the ring of integers we have several examples of UFDs. The ring of Gaussian integers Z[i] is a UFD and so is the polynomial ring k[x], for any field k.

Proof of the Proposition Let R be a PID. I first claim that every nonzero nonunit a in R can be factorized as product of irreducibles. If a is irreducible there is nothing to prove. If not then we can write a=a₁b₁ with a₁,b₁ non-units. If a₁,b₁ are both irreducible we are done. Suppose that a₁ is not. Then we factorize a₁=a₂b₂, as a product. We repeat this until we have expressed a as a product of irreducibles. Thus we will have a=a₁b₁=a₂b₁b₂...

We have to show that this process terminates. Write a₀=a. Then associated to our factorization attempt we have the increasing sequence of ideals a₀R < a₁R < a₁R < .... Our factorisation attempt terminates if this chain terminates.

Let I be the union of these ideals. Because this sequence is increasing it is not hard to see that I is also an ideal. Since R is a PID this means that I=bR. Now b has to be in one of the ideals a_iR (by the definition of I). Thus b=a_ir, say. But a_i must lie in I. It follows from this that b and a_i are associates. Clearly then a_jR=a_iR for any j>=i, as required to show the existence of factorizations.

We now have to show that factorizations are unique. But we know (since we have a PID) that the concepts of primeness and irreducibility coincide. Thus it is easy to adapt the familiar proof of the fundmental theorem of arithmetic to this case.