# properties of a gcd domain

Let $D$ be a gcd domain. For any $a\in D$, denote $[a]$ the set of all elements in $D$ that are associates of $a$, $\operatorname{GCD}(a,b)$ the set of all gcd’s of elements $a$ and $b$ in $D$, and any $S\subseteq D$, $mS:=\{ms\mid s\in S\}$. Then

1. 1.

$\operatorname{GCD}(a,b)=[a]$ iff $a\mid b$.

2. 2.

$m\operatorname{GCD}(a,b)=\operatorname{GCD}(ma,mb)$.

3. 3.

If $\operatorname{GCD}(ab,c)=[1]$, then $\operatorname{GCD}(a,c)=[1]$

4. 4.

If $\operatorname{GCD}(a,b)=[1]$ and $\operatorname{GCD}(a,c)=[1]$, then $\operatorname{GCD}(a,bc)=[1]$.

5. 5.

If $\operatorname{GCD}(a,b)=[1]$ and $a\mid bc$, then $a\mid c$.

###### Proof.

To aid in the proof of these properties, let us denote, for $a\in D$ and $S\subseteq D$, $a|S$ to mean that every element of $S$ is divisible by $a$, and $S|a$ to mean that every element in $S$ divides $a$. We take the following four steps:

1. 1.

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 $[a]=\operatorname{GCD}(a,b)$.

2. 2.

Pick $d\in\operatorname{GCD}(a,b)$ and $x\in\operatorname{GCD}(ma,mb)$. We want to show that $md$ and $x$ 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 $x=mn$ 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 $d$ is a gcd of $a$ and $b$. As a result, $mn\mid md$, or $x\mid md$, showing that $x$ and $md$ are associates. As a result, the map $f:m\operatorname{GCD}(a,b)\to\operatorname{GCD}(ma,mb)$ given by $f(d)=md$ is a bijection.

3. 3.

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 $d$ is a unit and the result follows.

4. 4.

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 $d$ is a unit.

5. 5.

$\operatorname{GCD}(a,b)=[1]$ implies $[c]=\operatorname{GCD}(ac,bc)$. Now, $a\mid ac$ and by assumption, $a\mid bc$. Therefore, $a\mid\operatorname{GCD}(ac,bc)=[c]$.

The second property above can be generalized to arbitrary integral domain: let $D$ be an integral domain, $a,b\in D$, with $\operatorname{GCD}(a,b)\neq\varnothing\neq\operatorname{GCD}(ma,mb)$, then $d\in\operatorname{GCD}(a,b)$ iff $md\in\operatorname{GCD}(ma,mb)$.

Title properties of a gcd domain PropertiesOfAGcdDomain 2013-03-22 18:18:44 2013-03-22 18:18:44 CWoo (3771) CWoo (3771) 9 CWoo (3771) Result msc 13G05