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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

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  • $\begingroup$ Yes, I did. Sorry, I'll edit that now. Thank you. $\endgroup$ – Sarah Oct 6 '13 at 20:32
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    $\begingroup$ Use the fact that irreducibles are prime $\endgroup$ – Cocopuffs Oct 6 '13 at 20:40
  • $\begingroup$ Does it hold that $a$ and $c$ share no common factor? $\endgroup$ – KirkLand Oct 25 '17 at 18:59
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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 $c|b$.

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  • $\begingroup$ @Cocopuffs What do you think of this proof? $\endgroup$ – Sarah Oct 7 '13 at 2:17
  • $\begingroup$ You need to use the unique factorization of $b$ -- just because each $p_i$ divides $b$ does not mean their product does. $\endgroup$ – vadim123 Oct 7 '13 at 3:09
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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, $c|b$.

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  • $\begingroup$ How does the $a_i$ and $c_j$ being relatively prime imply $c \mid b$? I don't see how that follows. $\endgroup$ – user193319 Jan 1 '18 at 16:24
  • $\begingroup$ That is not true, take $a=9, b=15$ relatively prime but their factorization similar. $\endgroup$ – Ahmed Nov 6 '19 at 20:23

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