Integral domain satisfying ACCP that is not a field has either infinitely many units or infinitely many associate classes of irreducibles

Statement
Suppose $$R$$ is an fact about::integral domain that is also a fact about::ring satisfying ACCP: it satisfies the ascending chain condition on principal ideals. In particular, every nonzero non-unit of $$R$$ can be factorized (possibly non-uniquely) as a product of fact about::irreducible elements. Suppose, further, that $$R$$ is not a field. Then, $$R$$ satisfies at least one of these two conditions:


 * It has infinitely many associate classes of irreducible elements.
 * It has infinitely many units.

Note that since fact about::Noetherian domains and fact about::unique factorization domains in particular satisfy ACCP, this result applies to both.

Particular cases

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

Proof
Given: An integral domain $$R$$ satisfying the ascending chain condition on principal ideals. In particular, every non-zero non-unit of $$R$$ can be factorized as a product of irreducible elements.

To prove: $$R$$ either has infinitely many units or infinitely many associate classes of irreducibles.

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. Thus, by the hypotheses on $$R$$, $$1 + a^n$$ 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.