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

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Statement

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

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 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_is. 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.