GCDs in integral domains are unique up to unit factors (associates) The book I am reading mentioned that the greatest common divisor of $a_1, \cdots a_n$ in an integral domain $R$, whenever it exists, is unique up to arbitrary invertible factors. What does "up to arbitrary invertible factors" mean?
Definition: Let $a_1, \cdots, a_n$ be nonzero elements of the ring $R$. An element $d \in R$ is a greates comon divisor of $a_1, \cdots a_n$ if it possesses the properties:


*

*$d \vert a_i$, for $i=1, c\dots n$

*$c \vert a_i$, for $i=1, \cdots n$ implies $c \vert d$.

 A: $\newcommand{\Z}{\mathbb{Z}}$$\renewcommand{\epsilon}{\varepsilon}$It means 

if $d$ is a GCD, then for each unit $\epsilon$ we have that $\epsilon d$ is a GCD, and all GCDs have this form.

For instance in $\Z$ the pair $2, 3$ has the two GCDs $1, -1$.

I am taking the following definition. Let $R$ be an integral domain, and $a_{1}, \dots, a_{n} \in R$. An element $d \in R$ is said to be a GCD of $a_{1}, \dots, a_{n}$ if


*

*$d \mid a_{i}$ for all $i$;

*for each $e \in R$, if $e \mid a_{i}$ for all $i$, then $e \mid d$.


Of course a GCD need not exist for all choices of $a_{1}, \dots , a_{n}$.
A: Well, if $d$ is a common divisor of $a,b$ and $e$ is a unit, then $ed$ is a common divisor of $a,b$, since if $a=a'd$ and $b=b'd$ for some $a',b'$, then $a=(a'e^{-1})(ed)$ and $b = (b'e^{-1})(ed)$. This indeed means that the gcd is unique up to invertible elements.
A: By the universal $\rm\color{#c00}{def}$inition of the gcd, If $\,d,d'$ are both gcds of all of the $\,a_i\,$ then $\,c\mid d\!\!\!\overset{\rm\ \  \color{#c00}{def}}\iff\! c\mid a_1,\ldots ,a_n\!\!\!\overset{\rm\ \ \color{#c00}{def}}\iff\! c\mid d',\,$ so for $\,c=d\,$ we get $\,d\mid d'$ and for $\,c = d'\,$ we get $\,d'\mid d,\,$ so $\,d,d'$ divide each other so are associate (thus $\,d' = u d\,$ for some unit $u,\,$ being in a domain).
