UFD is GCD Domain
Theorem
Let $A$ be a unique factorisation domain.
Then $A$ is a GCD domain.
Proof 1
Let $x \divides y$ denote $x$ divides $y$.
Let $x, y \in A$, with complete factorizations:
- $x = u x_1 \cdots x_r$
- $y = v y_1 \cdots y_s$
where:
- $u, v$ are units
- the $x_i$, $y_i$ irreducible.
We arrange the complete factorizations as follows:
- $x = u \paren {x_1 \cdots x_t} x_{t + 1} \cdots x_r$
- $y = v \paren {y_1 \cdots y_t} y_{t + 1} \cdots y_s$
where:
- $t \le \min \set {r, s}$
- For $i = 1, \ldots, t$, $x_i$ and $y_i$ are associates
- For any $i \in \set {t + 1, \ldots, r}$, $j \in \set {t + 1, \ldots, s}$, $x_i$ and $y_j$ are not associates.
Let $d = x_1 \cdots x_t$ (recall that the empty product is $1$, i.e. $d = 1$ when $t = 0$).
We claim that $d$ is a greatest common divisor for $x$ and $y$.
Certainly $d \divides x$ and $d \divides y$.
So, let $f$ be another common divisor of $x$ and $y$.
We can find $w, z \in A$ such that $x = f w$, and $y = f z$.
If $f$ is a unit, then $f \divides d$ by definition.
Aiming for a contradiction, suppose $f \nmid d$.
Then the complete factorization of $f$ must contain an irreducible element that does not divide $d$.
Call this irreducible element $g$.
We have that:
- $g$ must divide some $x_j$ where $j > t$
and
- $g$ must divide some $y_k$ where $k > t$.
Either:
- $g$ is a unit, contradicting its irreducibility
or:
- $x_j$ and $y_k$ are not irreducible, which is a contradiction also.
Hence by Proof by Contradiction:
- $f \divides d$
and so $x$ and $y$ have a greatest common divisor.
$\blacksquare$