Integral domains satisfying Bezout’s identity and having divisibility-minimal elements are PIDs

Let R be an integral domain. Prove that if

  1. Any two nonzero elements a,b \in R have a greatest common divisor d which can be written d = ax + by for some x,y \in R and
  2. For any sequence a : \mathbb{N} \rightarrow R such that a_{i+1} divides a_i, there exists M \in \mathbb{N} such that for all n \geq M there is a unit u_n such that a_n = u_na_M,

then R is a principal ideal domain.


Let I \subseteq R be a nonzero ideal. We can consider I to be partially ordered by divisibility as follows: say x \leq y if y|x. Let C \subseteq I be a chain; in fact C is a sequence with C_{i+1} dividing C_i for all i. By our hypothesis, there exists M such that for all n \geq M, C_n = u_nC_M for some unit u_n. That is, C_M divides C_n for all n \geq M. Certainly we also have that C_M divides C_n for n < M. Thus C_M is an upper bound of the chain C. By Zorn’s Lemma, I contains a maximal element with respect to this order; that is, I contains elements which are minimal with respect to divisibility.

Now let a,b \in I be two such minimal elements. By our hypothesis, a and b have a greatest common divisor d which we can write d = ax+by. In particular, d \in I. Since a and b are minimal with respect to divisibility, we have that a, b, and d are associates. Thus all divisibility-minimal elements of I are associates, and we have (for instance) I = (d). Thus I is principal, and R is a principal ideal domain.

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