Let $x,y \in R$, where $R$ is a unique factorization domain.
Let $P$ be the set of all representatives in each class of associate irreducible elements of $R$.
Then, suppose $x,y\in R$ are nonzero, nonunit elements (First Question: How would this proof change if either $x$ or $y$ is zero or is a unit of $R$?)
Then, since $R$ is a unique factorization domain, we can express $x$ and $y$ as the (unique) products (up to permutation of factors) $$ \displaystyle x = p_{1}^{t_{11}}p_{2}^{t_{12}}\cdots p_{k}^{t_{1k}}u_{1} $$
$$ \displaystyle y = p_{1}^{t_{21}}p_{2}^{t_{22}}\cdots p_{k}^{t_{2k}}u_{2} $$
adding in, if necessary, extra factors $p_{i}^{0}$ to the factorization of either $x$ or $y$.
(Also note that here $p_{i} \in P$, $u_{i} \in U(R)$, $t_{j} \geq 0$, $p_{i} \neq p_{j}$ for $i \neq j$.)
Now, for each $j \leq k$, let $s_{j} = \max(t_{1j}, t_{2j})$, and let $$ \displaystyle m = p_{1}^{s_{1}}p_{2}^{s_{2}}\cdots p_{k}^{s_{k}}. $$
Now, by construction, $m$ is a multiple of both $x$ and $y$.
We must prove that it is the least common multiple of $x$ and $y$ (up to associates):
Assume $m^{\prime}$ is another multiple of $x$ and $y$. Now, I was told to model this proof after a proof I have for the $\gcd$. For that proof, there was a $d$ constructed to be a divisor of both $x$ and $y$ by choosing $\min(t_{1j}, t_{2j})$ instead. Then, they showed that $d^{\prime}|d$, where $d^{\prime}$ was some other divisor of both $x$ and $y$. Second Question: Do I want to show here then that $m|m^{\prime}$? Is that sufficient for showing that the $m$ I came up with is the $lcm$?
If so, I'm going to need some help with that part. In the proof for the $\gcd$, they use the fact that since the factorizations of $x$ and $y$ are unique, $x$ and $y$ have no divisors $p \in P$ distinct from the $p_{j}$'s, and go on to say that since $d^{\prime}$ is a divisor of $x$, it has a factorization $$d^{\prime}=p_{1}^{r_{1}}p_{2}^{r_{2}}\cdots p_{k}^{r_{k}}v,\quad v \in U(R),\, r_{j} \geq 0$$.
From here they say that the fact that $d^{\prime}$ is a divisor of both $x$ and $y$ implies that $r_{j} \leq t_{ij}$ for $i = 1,2$, and all $j$, and hence $r_{j} \leq s_{j}$. Then, we have $$ d = d^{\prime}\cdot p_{1}^{s_{1}-r_{1}}p_{2}^{s_{2}}\cdots p_{k}^{s_{k}-r_{k}}v^{-1}$$ so that $d^{\prime}|d$.
Third Question: Can I just adapt this last part for $lcm$ instead of $\gcd$? Suppose I define
$$m^{\prime}=p_{1}^{r_{1}}p_{2}^{r_{2}}\cdots p_{k}^{r_{k}}v,\quad v \in U(R),\, r_{j} \geq 0 $$
Then, I can say that $m^{\prime}$ is a multiple of both $x$ and $y$. Would this mean then that $r_{j} \geq t_{ij}$ for $i = 1,2$ and all $j$, and hence that $r_{j} \geq s_{j}$?
If so, then how would I express $m^{\prime}$ in terms of $m$ like they expressed $d$ in terms of $d^{\prime}$? I'm getting a little confused by all the exponents... Do I just add the exponents instead of subtracting?? Can I really be that simple??
Thank you!
Note: The other questions that have thus far been posted (by other people) relating to this topic have not asked for differences in the proof for the case when either $x$ or $y$ is zero or a unit element (I have been told these cases are trivialities, but sometimes seeing trivialities completely worked out can be beneficial to gaining a true understanding of something). So, please do not mark this as a duplicate. I really need an answer to this question, and the other questions have not been helpful to me.