Gcd and lcm of $a_1, a_2, \dots,a_n$ exist in $R$ when $R$ is a UFD. Let $R$ be a commutative ring with identity and $a_1, a_2, \dots,a_n \in R - \{0\}$. Then to find the gcd and lcm of $a_1, a_2, \dots,a_n$ and they exist in $R$ when $R$ is a UFD.
We can express $a_1, a_2, \dots,a_n$ uniquely as finite product of irreducible elements as:
$a_1 \sim p_1^{r_1}p_2^{r_2} \dots p_n^{r_n} $, $a_2 \sim p_1^{s_1}p_2^{s_2} \dots p_n^{s_n}$ , $\dots,$ $a_n \sim p_1^{t_1}p_2^{t_2} \dots p_n^{t_n}$
Now, we define:
$\gcd (a_1, a_2, \dots,a_n ) = p_1^{c_1}p_2^{c_2} \dots p_n^{c_n}$ where $c_i = \min\{r_i,s_i,...,t_i\}$ and,
$lcm(a_1, a_2, \dots,a_n ) = p_1^{d_1}p_2^{d_2} \dots p_n^{d_n}$ where $d_i = \max\{r_i,s_i,...,t_i\}$ 
I can intuitively define them but how can I prove that they exist in $R$?
Please Help!
 A: By existence and uniqueness of prime factorizations, divisibility reduces to divisibility in each prime component, i.e. $\ p^{\large a} q^{\large a'}\!\cdots\mid p^{\large b} q^{\large b'}\!\cdots\!\iff p^{\large a}\mid p^{\large b},\,\ q^{\large a'}\!\mid q^{\large b'},\,\ldots\ $
So $\,\ a\mid b,c\iff p^{\large a}\mid p^{\large b},p^{\large c},\ \ q^{\large a'}\mid q^{\large b'},q^{\large c'},\ldots\ $  each which obey
$$\quad\ \ \ \ \ \ p^{\large a}\mid p^{\large b},p^{\large c}\! \iff  a\le b,c \iff a\le \min\{b,c\} \iff  p^{\large a}\mid p^{\large \min\{b,c\}}$$
$\!\begin{align}{\rm So}\ \ a\mid b,c&\iff p^{\large a}\mid p^{\large \min\{b,c\}},\ q^{\large a'}\mid q^{\large \min\{b',c'\}}\ldots\\[.5em]
&\iff a\mid  \color{#c00}{p^{\large \min\{b,c\}}  q^{\large \min\{b',c'\}}}\ldots \end{align}$
Remark $\ $ Above we employ the universal characterization / definition of  gcd and lcm, i.e.
$$\begin{align}   a\mid b,c\iff a\mid \color{#c00}{\gcd(b,c)}\\
b,c\mid a\iff {\rm lcm}(b,c)\mid a\end{align}\qquad\qquad$$
The $ $  lcm $ $  case is just the dual $\, \ b,c\mid a\iff \cdots,\ $ by reversing divisiblities in the gcd proof.
