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Let $R$ be a UFD. Let $r,a,b \in R$. Then how can I show that $\operatorname{gcd} (ra,rb) \sim r \operatorname{gcd} (a,b)$?

If $\operatorname{gcd} (a,b) = d$ and $\operatorname{gcd} (ra,rb) = d'$.Then $d|a$ and $d|b$ $\implies$ $rd|ra$ and $rd|rb$ i.e. $rd$ becomes a common divisor of $ra$ and $rb$. Hence we must have $rd|d'$. But I fail to show the converse part i.e. $d'|rd$.

How can I show this? Please help me.

Thank you in advance.

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    See the 2nd and 3rd [proofs here](http://math.stackexchange.com/a/785544/242) of this **GCD Distributive Law**. – 2017-01-14
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    $d=ax+by \Rightarrow rd=rax+rby=(ra)x+(rb)y$ then $d'|(ra)x+(rb)y\Rightarrow d'|rd$ – 2017-01-14
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    @A.Chattopadhyay. You are right. That only works if $R$ is a bezout domain. – 2017-01-14
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    No.As far as I know this thing is true in PID and hence in ED.But how can you prove it in a UFD?Please tell me. – 2017-01-14
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    I said it to Mustafa not you @mathchat. – 2017-01-14
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    @MathChat Correct, UFDs need not be Bezour domains, e.g. in $\,R = \Bbb Z[x,y]\,$ we have $\,\gcd(x,y)=1\,$ but there are no $\,f,g\in R\,$ with $\,x f + y g = 1\,$ (else eval at $\,x=0=y\,$ yields $0 = 1).\,$ That's why I mentioned only the proofs that do not use Bezout / Euclid in my linked answer. – 2017-01-14
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    @ Bill Dubuque. Yes. You are absolutely right. – 2017-01-14
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    I did notice you used $\sim$ rather than $=$, but I still feel a bit uneasy about the possibility, for example, that $r$ is negative while $a$ and $b$ are positive. – 2017-01-14

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