Sorry if this seem trivial, but im slightly stuck:
Let $a$ and $b$ be elements of an integral domain "$R$" and let $d$ be the $\gcd(a,b)$. Show that the set of GCD's of $a$ and $b$ is the set of associates of $d$.
If $a$ and $b$ are associates, show each is a GCD of $a$ and $b$.
Well: If $GCD(a,b)=d$ then $a=dx$ and $b=dy$ where $x,y \in \mathbb{Z^+}$
I know $a$ and $b$ are associate if $a|b$ and $b|a$.
Also if $a$ and $b$ are associate $a=bc$ and $b=ga$ where $g$ and $c$ are units in $R$.
Im just not sure how to fit this all together. Any help would be greatly appreciated!
Definition of GCD(a,b) is as follows. If $d$ is a $\gcd(a,b)$, then $d \mid a$ and $d\mid b$ and for any $d'\in D$ with $d'\mid a$ and $d' \mid b$, $d' \mid d$.
Let $d_1$ and $d_2$ be $\gcd(a,b)$. We prove they are associates. Certainly $d_1$ and $d_2$ both divide $a$ and $b$. Since $d_1$ is a $\gcd(a,b)$, $d_1 \mid d_2$. Similarly, $d_2 \mid d_1$. Thus they are associates. If $d_1$ is an associate of $d=\gcd(a,b)$, then $d_1=\lambda d$ for a unit $\lambda$. Thus $d$ is a divisor of $a$ and $b$, so $dk=a$ and $dl=b$. But then $d_1 \lambda^{-1}k=a$ and $d_1 \lambda^{-1}l=b$, so $d_1$ is a common divisor.
Moreover, if $d'$ is another element which divides both $a$ and $b$, we know that $d' \mid d$ since $d$ is a $\gcd(a,b)$. But then $d'm=d$ for some $m$. Then $d'm=\lambda^{-1}d_1$ which implies $d'(m\lambda)=d_1$, so $d' \mid d_1$ and this proves $d_1$ is a $\gcd(a,b)$.