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

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


 * $$R$$ has infinitely many units.
 * $$R$$ has infinitely many fact about::associate classes of fact about::irreducible elements.

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.

Proof
Given: A unique factorization domain $$R$$ that is not a field.

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

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

Since $$R$$ is not a field, it has at least one irreducible element, say $$p_1$$. Suppose now that the set of associate classes of irreducible elements is finite, say with representatives $$p_1,p_2, \dots, p_r$$ for the associate classes. Let $$a$$ be the product of the $$p_i$$s. Consider now the set:

$$S = \{ 1 + a, 1 + a^2, 1 + a^3, \dots \}$$

$$1 + a^n$$ cannot be equal to zero for any $$n$$, because that would imply that $$a$$ is a unit, forcing $$p_i$$ to be units, a contradiction to irreducibility. For a similar reason, $$a^n$$ cannot ever be $$1$$.

Since the $$p_i$$ are not zero, $$a$$ is not zero, and we have $$(1 + a^m) - (1 + a^n) = a^m(1 - a^{n-m})$$ which is nonzero since both $$a$$ and $$1 - a^{n-m}$$ are nonzero, and $$R$$ is an integral domain. Thus, all the $$1 + a^n$$ are distinct. Thus, $$S$$ is an infinite set of distinct nonzero elements.

Since there are only finitely many units in $$R$$, there exists $$n$$ such that $$1 + a^n$$ is not a unit. Since $$R$$ is a unique factorization domain, $$1 + a^n$$ has a factorization into irreducibles, and in particular, has at least one irreducible factor. However, if $$p_i$$ divides $$1 + a^n$$ for any $$i$$, we have that $$p_i | a^n$$ and $$p_i | 1 + a^n$$, so $$p_i | 1$$, a contradiction. Thus, the assumption that there are only finitely many associate classes of irreducible elements is flawed.