Let $D$ be a UFD (unique factorization domain) , if $a,b,c \in D$ and $a \mid c , b \mid c$ and $a$ and $b$ are relatively prime then $ab \mid c$. I started by factorizing $c$ into irreducibles, then I said $a = c_k \ldots c_l$ and $b = c_h \dots c_q$ . All that remains is to show that all $c_i$'s are distinct but I can't seem to prove it, how is it done? Or is this method useless?
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Do you mean $a=c_k\cdots c_l$ and $b=c_h\cdots c_q$? – Zev Chonoles Apr 02 '13 at 22:51
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@ZevChonoles I did not mean that but you are right I'll fix it. – user10444 Apr 02 '13 at 22:53
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1Which condition have you not used yet? It is not necessary that the $c_i$ from $a$ be distinct, only that no $c_i$ is in both $a$ and $b$ – Thomas Andrews Apr 02 '13 at 22:56
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@ZevChonoles Please don't tell me it becomes as easy as saying that if such a $c_j$ exists then $\gcd(a,b)=c_j$ which is a contradiction – user10444 Apr 02 '13 at 22:57
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Yes, that's it. Just like primes in $\Bbb Z$. – Berci Apr 02 '13 at 22:58
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What is the future of this question? Do I delete it or should someone answer and I'll acccept? – user10444 Apr 02 '13 at 23:00
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1@user10444: In fact, you can write up your solution (and accept it, if you'd like). This is explicitly encouraged by the SE network of sites; see here and here. This is a good way of checking for yourself whether you understand the details of the argument. – Zev Chonoles Apr 02 '13 at 23:07
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Showing the uniqueness of each $c_i$: Assume one of them is not unique call it $c_j$ , then we have $c_j \mid a$ and $c_j \mid b$ but $\gcd(a,b)=1$ , which is a contradiction since $c_j \nmid 1$ (since $c_j$ is not a unit) and $1$ and $c_j$ are not associates hence all the $c_i$'s are distinct and $ab=c_k \dots c_lc_h \dots c_q$ which divides $c$ hence $ab \mid c$ .
user10444
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Yes, you can prove $\rm\:(a,b)=1\:\Rightarrow\:lcm(a,b) = ab\:$ in any UFD by using prime factorizations. More generally one can prove it in any gcd domain using Euclid's Lemma $\rm\color{#C00}{(EL)}\,$ as follows
$$\rm (a,b)=1,\,\ a,b\mid c\:\Rightarrow\: a\mid b(c/b)\color{#C00}{\stackrel{(EL)}{\Rightarrow}}a\mid c/b\:\Rightarrow\: ab\mid c$$
Alternatively, one may proceed using the gcd distributive law $\rm\,\color{#0A0}{(DL)}\,$ as follows
$$\rm (a,b)=1,\,\ a,b\mid c\:\Rightarrow\: ab\mid ac,bc\:\Rightarrow\:ab\mid (ac,bc)\stackrel{\color{#0A0}{(DL)}}{=}(a,b)c = c$$
