# Unique factorization domain that is not a field has either infinitely many units or infinitely many associate classes of irreducibles

## Statement

Suppose is a unique factorization domain that is not a field. Then, at least one of these is true for :

- has infinitely many units.
- has infinitely many associate classes of irreducible elements.

## Related facts

### Particular cases

- In the ring of rational integers, there are only finitely many units, and thus, there are infinitely many associate classes of primes. This is the well-known fact that there are infinitely many primes.
- In the formal power series ring over a field, there is only one associate class of irreducibles, but there are infinitely many units.

### Generalizations

The hypothesis can be weakened slightly: it works for any integral domain that is also a ring satisfying ACCP: this is precisely the condition needed to ensure that every element does have a factorization, not necessarily unique. `Further information: Integral domain satisfying ACCP that is not a field has either infinitely many units or infinitely many associate classes of irreducibles`

## Proof

**Given**: A unique factorization domain that is not a field.

**To prove**: either has infinitely many units or infinitely many irreducible elements.

**Proof**: For the proof, we assume that has only finitely many units, and prove that has infinitely many irreducible elements.

Since is not a field, it has at least one irreducible element, say . Suppose now that the set of associate classes of irreducible elements is finite, say with representatives for the associate classes. Let be the product of the s. Consider now the set:

cannot be equal to zero for any , because that would imply that is a unit, forcing to be units, a contradiction to irreducibility. For a similar reason, cannot ever be .

Since the are not zero, is not zero, and we have which is nonzero since both and are nonzero, and is an integral domain. Thus, all the are distinct. Thus, is an infinite set of distinct nonzero elements.

Since there are only finitely many units in , there exists such that is *not* a unit. Since is a unique factorization domain, has a factorization into irreducibles, and in particular, has at least one irreducible factor. However, if divides for any , we have that and , so , a contradiction. Thus, the assumption that there are only finitely many associate classes of irreducible elements is flawed.