PlanetMath: properties of a gcd domain

Let

be a gcd domain. For any $a\in D$ , denote

the set of all elements in

that are associates of

, $\operatorname{GCD}(a,b)$ the set of all gcd's of elements

and

, and any $S\subseteq D$ , $mS:=\lbrace ms\mid s\in S\rbrace$ . Then

Proof. To aid in the proof of these properties, let us denote, for $a\in D$ and $S\subseteq D$ , $a\vert S$ to mean that every element of

is divisible by

, and $S\vert a$ to mean that every element in

divides

. We take the following four steps:

One direction is obvious from the definition. So now suppose $a\mid b$ . Then $a\mid\operatorname{GCD}(a,b)$ . But by definition, $\operatorname{GCD}(a,b)\mid a$ , so $[a]=\operatorname{GCD}(a,b)$ .
Suppose $d \in \operatorname{GCD}(a,b)$ . We want to show that $md\in \operatorname{GCD}(ma,mb)$ . In other words, if $x\in \operatorname{GCD}(ma,mb)$ , we want to show that and are associates. By assumption, $d\mid a$ and $d\mid b$ , so $md\mid ma$ and $md\mid mb$ , which implies that $md\mid x$ . Write for some $n\in D$ . Then $mn\mid ma$ and $mn\mid mb$ imply that $n\mid a$ and $n\mid b$ , and therefore $n\mid d$ since is a gcd of and . As a result, $mn\mid md$ , or $x\mid md$ , showing that and are associates.
On the other hand, if $x\in \operatorname{GCD}(ma,mb)$ , we want to show that for some $d\in \operatorname{GCD}(a,b)$ . Pick some $y\in \operatorname{GCD}(a,b)$ , then $my \in \operatorname{GCD}(ma,mb)$ by the previous paragraph. Therefore and are associates. Write for some unit $e\in D$ . Setting gives us the desired gcd of and .
If $d\mid a$ and $d\mid c$ , then $d\mid ab$ and $d\mid c$ . So $d\mid\operatorname{GCD}(ab,c)=[1]$ , hence is a unit and the result follows.
Suppose $d\mid a$ and $d\mid bc$ . Then $d\mid ab$ and $d\mid bc$ and hence $d\mid\operatorname{GCD}(ab,bc)=b\operatorname{GCD}(a,c)=[b]$ . But $d\mid a$ also, so $d\mid\operatorname{GCD}(a,b)=[1]$ and is a unit.
$\operatorname{GCD}(a,b)=[1]$ implies $[c]=\operatorname{GCD}(ac,bc)$ . Now, $a\mid ac$ and by assumption, $a\mid bc$ . Therefore, $a\mid\operatorname{GCD}(ac,bc)=[c]$ .

$\qedsymbol$