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If $R$ is a UFD and $a,b \in R$ are relatively prime and $a|bc$, then $a|c$.

we know if a and b relatively prime (a,b)=1 so $a\bot b$. since $a|bc$ and $a\bot b$ doesn't $a$ has to divide $c$? I dont understand why do we need UFD?

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or since it is ufd, let's assume $\displaystyle a \;=\; p_1\cdots p_l, \quad b \;=\; q_1\cdots q_m, \quad c \;=\; r_1\cdots r_n$ $\displaystyle p_1\cdots p_l\,=\; q_1\cdots q_m \; r_1\cdots r_n\,k$ , $k\in R$ . and since $a\bot b$ , $q_j$ cant be associate of one of $ p_i$'s . so each one of $ p_i$ must be associate to $q_j$'s and k, then what? :) – emmett Dec 13 '12 at 20:08
What do yu mean by $a\bot b$? How is it different from $(a,b)=1$? – Thomas Andrews Dec 13 '12 at 20:08
I meant a cant divide b – emmett Dec 13 '12 at 20:09

Let $A$ be the ring of polynomials in $4$ variables $x,y,z,w$ with integer coefficients, and let $R$ be the quotient of $A$ by the ideal generated by $xy-zw$. $R$ is an integral domain, but not a UFD. $x$ and $z$ are relatively prime in $R$, and $x$ divides $zw$ in $R$ (because $zw$ is $xy$ in $R$), but $x$ doesn't divide $w$.

A similar example can be produced in any non-UFD. If you have two factorizations into distinct irreducibles, $r=p_1p_2\cdots p_m=q_1q_2\cdots q_n$, you get $p_1$ relatively prime to $q_1$, $p_1$ divides $q_1(q_2\cdots q_n)$, $p_1$ doesn't divide $q_2\cdots q_n$.

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Because if you had different factorizations or it weren't a domain, then it wouldn't be true. Try $\mathbb{Z}_6$. Then $5 |0 = 3*2$, but it doesn't divide 3 or 2.

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When proving "If A then B", starting off with "If A is false..." is not getting you anywhere. – rschwieb Dec 13 '12 at 20:55

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