I am sorry if this is posted elsewhere, but I have not found a satisfying proof for UFD => GCD domain. Most proofs I found only give a potential candidate for gcd, but does not justify further. So I worked out the rest - and hope I could get some verification or advices on my proof. Thanks in advance.
Definition 0: Two elements $a,b$ in an integral domain $R$ with unit element are associates if $a = ub$ where $u$ is a unit in $R$. An element $a$ which is not a unit will be called irreducible if whenever $a=bc$ with $b,c \in R$, one of $b$ or $c$ must be a unit.
Definition 1: An integral domain, $R$, with unit element is a unique factorization domain if
a) Any nonzero element in $R$ is either a unit or can be written as the product of a finite number of irreducible elements of $R$.
b) The decomposition in part (a) is unique up to the order and associates of the irreducible elements.
Problem: If $R$ is a UFD and if $a,b$ are in $R$, then gcd $(a,b)$ exists in $R$. Moreover, if $a,b $ are relatively prime ($(a,b) =1$), whenever $a|bc$ then $a|c$.
Proof:
(i) We show the gcd exists. We may assume that both $a$ nor $b$ are non units. For if $a$ is a unit, then $b = 1 * b = axb$ so $a|b$ and $(a,b) = a $. Suppose $$ a = a_1a_2 \ldots a_n $$ $$b = b_1 b_2 \ldots b_m $$ where $a_i$ and $b_j$ are all irreducible elements using (a). Let $a_{i_1}$ be the first irreducible factor of $a$ such that there exists an irreducible factor of $b$ which is an associate. Wlog, we may assume it is $b_1$. Similarly, let $a_{i_2}$ be the next irreducible factor of $a$ that is an associate to one of $\{ b_2, \ldots, b_m \}$. Hence, we may assume, through relabelling,
$$ a = (a_1 \ldots a_s) a_{s+1} \ldots a_n $$ $$ b = (b_1 \ldots b_s ) b_{s+1} \ldots b_m $$ such that $a_i $ and $b_i$ are associates for $1 \le i \le s$ and $a_i$ and $b_j$ are not associates for $s+1 \le i \le n $ and $s+1 \le j \le m $. Now I claim that $d = a_1 \cdots a_s $ is $(a,b)$. Firstly, $d|a$ and $d|b$.
Let $h$ be another common divisor. If $h$ is a unit then $h|d$ by definition. So suppose $h$ is non unit. $$h = c_1 c_2 \ldots c_l $$ where each $c_k$ are irreducible elements. If $c_1$ is not an associate to any of the elements, $a_i$ for $1 \le i \le s$, then it must be an associate to one of the $a_i$ for some $s+1 \le i \le n$ and $b_j$ for some $s+1 \le j \le m $, using (b). However, as property of associates is an equivalence relation, $a_i$ and $b_j$ are associates, contradiction. Wlog assume $c_1$ is associate with $a_1$. We obtain, for some unit $u$, $$c_1 \ldots c_l x = uc_1(a_2 \ldots a_s) a_{s+1} \ldots a_n $$ $$ \Rightarrow c_1 (c_2 \ldots c_l xu^{-1} - (a_2 \ldots a_s) a_{s+1} \ldots a_n) = 0 $$ $$ \Rightarrow c_2 \ldots c_l xu^{-1} = (a_2 \ldots a_s) a_{s+1} \ldots a_n $$
by using property of integral domain. By induction, we deduce that each irreducible factor of $h$ is an associate to an irreducible factor of $d$. So $h | d $.
(ii) If $(a,b)=1$, and $a|bc$. So $s=0$ in the above scenario, giving that non of the irreducible factors of $a$ are associates with that of $b$. By uniqueness of decomposition (b), the irreducible factors of $a$, must be associates to that of $c$. Hence, $a|c$.