## 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.