# Unique Factorization Domain and least common multiple

Let $R$ be a unique factorization domain and let $a,b \in R$ be nonzero, non-unit elements.

Let $a=up_1^{e_1}p_2^{e_2} \dots p_n^{e_n}, \,\,\,b = vp_1^{f_1}p_2^{f_2} \dots p_n^{f_n}$

And the primes $p_1,p_2,...,p_n$ are distinct and the exponents $e_i$ and $f_i$ are $\geq$ 0.

Prove that the element $l = p_1^{max(e_1,f_1)}p_2^{max(e_2,f_2)} \dots p_n^{max(e_n,f_n)}$.

is a least common multiple of a and b.

proof: Since the exponent of each of the primes occuring in $l$ are no smaller than the exponents occuring than the exponents occuring in the factorizations of both $a$ and $b$, $a|l$ and $b|l$.

Let $c = q_1^{g_1}q_2^{g_2} \dots q_m^{g_m}$

be the prime factorization of $c$, where $c$ is any common multiple of $a$ and $b$.

I am stuck! I am having trouble with the exponents.

I wanted to show $a|l$ and $b|l$ and $a|c$ and $b|c$, then $l|c$, so to conclude $l$ is the least common multiple.

First, note that $c=ax$ and $c=by$ for some $x,y$. Hence, $$uxp_1^{e_1}\cdots p_n^{e_n}=c=vyp_1^{f_1}\cdots p_n^{f_n}.$$ Now, for each $i$ you have that both $p_i^{e_i}$ and $p_i^{f_i}$ divide $c$, so $p_i^{\max(e_i,f_i)}$ divides $c$. You can now conclude that $l$ divides $c$ (to be completely formal, you should probably use induction on the number $n$ of prime factors).