R is a UFD, c|ab, gcd(a,c)=1, then c|b Let $R$ be a UFD and let $a,b,c \in R$. Prove that if $c|ab$ and $\gcd (a,c)=1$ then $c|b$.
This is easy to prove if $R$ is a Euclidean domain, but I'm having trouble proving this for UFDs. I have a whole bunch of problems to solve that are similar to this. I'm hoping that the proof isn't as long as what I have done so far. Feel free not to read what is below. I would just like to know if there is a shorter proof for this. Thank you.
Proof:
Since $R$ is a UFD, we can write 
$a=up_1^{e_{1}} \cdots p_n^{e_n}, \quad b=vp_1^{f_{1}} \cdots p_n^{f_n}, \quad \text{and} \quad c=wp_1^{g_{1}} \cdots p_n^{g_n},$
where $u,v,$ and $w$ are units, $p_1,\ldots , p_n$ are distinct irreducibles, and $e_i,f_i,g_i \geq 0$. Since $\gcd (a,c)=1$, $\min \left \{e_i,g_i\right \}=0$ for all $i$. Also, since $c|ab$, $g_i \leq e_i+f_i$ for all $i$. Let $M_i=\max \left \{e_i,g_i \right \}$. Then by reindexing, we have
$a=up_1^{M_{1}} \cdots p_m^{M_m} p_{m+1}^0\cdots p_n^0 \quad \text{and} \quad c=wp_1^0 \cdots p_m^0 p_{m+1}^{M_{m+1}}\cdots p_n^{M_n}.$
Since $c|ab$, then $ab=cr$ for some $r\in R$. Thus, 
$ab= up_1^{M_{1}} \cdots p_m^{M_m} \cdot vp_1^{f_{1}} \cdots p_n^{f_n} = wp_{m+1}^{M_{m+1}}\cdots p_n^{M_n} r = cr.$
Since $g_i \leq e_i +f_i$ and $f_i=0$ for each $i \in \left \{m, m+1, \ldots , n\right \}$, then for such $i$, we have that $M_i \leq e_i$. Then we can write...I stopped here
 A: I am going to attempt to write a proof that is much shorter than my original proof. I would greatly appreciate some feedback! Thank you!
Let $c=p_1\cdots p_n$ where each $p_i$ is an irreducible in $R$. Since $R$ is a unique factorization domain, each $p_i$ is prime. Thus, since $c|ab \Rightarrow (p_1\cdots p_n) | ab \Rightarrow p_i|ab$, meaning $p_i|a$ or $p_i|b$. However, since $\gcd (a,c)=1 \Rightarrow \gcd (a,p_i)=1$, thus each $p_i$ must divide $b$, implying that $c|b$. 
A: Since R is a UFD, $a = a_1^{k_1} \cdots a_n^{k_n} $ and $c = c_1^{p_1} \cdots  c_m^{p_m}$. Given that $gcd(a,c) = 1$, it follows that $a_i, c_j$ are relatively prime for all $i, j$. 
Suppose there are $i, j$ such that $a_i$ and $c_j$ are not relatively prime. Then, there is $d \in R$ such that $d|a_i$ and $d|c_j$ and $d = gcd(a,c)$. Hence, $dk_i = a_i$ and $dk_j = c_j$. 
It follows that $a = \cdots dk_i \cdots$ and $c = \cdots ck_i \cdots $, implying that $d|a$ and $d|c$, contradicting the fact that $gcd(a,c) = 1$. 
Therefore, $c|b$. 
