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Let $D$ be a GCD domain, and let $a,b,x \in D \setminus \{0\}$. Then is it true that $\gcd (ax,bx)=x \cdot \gcd (a,b)$ ?

Let $c=\gcd (a,b)$ and $d=\gcd(ax,bx)$, then as $cx|ax$ and $cx|bx$ so $cx|d$. We would be done if we could show $d|cx$, but I am unable to show that. Please help me to solve this problem.

darij grinberg
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2 Answers2

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Hint $\ $ By below $\,(a,b)\,$ and $\,(ax,bx)/x\,$ divide each other. See here for a few more proofs.

$\ c\mid(a,b)\!\iff\! c\mid a,b \!\iff\! cx\mid ax,bx \!\iff\! cx\mid (ax,bx) \!\iff\! c\mid (ax,bx)/x,\ $ for any $\,c.\,$

Bill Dubuque
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Since $x \mid ax $ and $x \mid bx$ it follows that $x \mid d $. The codivisor $dx^{-1}$ of $x$ in $d$, divides $(ax)x^{-1}=a$ and $(bx)x^{-1}=b$, thus $dx^{-1}$ divides $\gcd(a,b)=c$. Thus $d=(dx^{-1})x$ divides $cx$.

darij grinberg
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quid
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    $x$ is not necessarily a unit ... –  Sep 19 '16 at 09:05
  • Yes I know that $x$ is not necessarily a unit in $D$. This is why I made a point of saying "[t]he codivisor $dx^{−1}$ of $x$ in $d$." Since $x \mid d$, there is some element $m \in D $ such that $mx = d$. This $m$ is $dx^{-1}$. You can take this as notation, or you can think of $x^{-1}$ being an element of the quotient field of $K$. And you know that $dx^{-1}$ is an element of $D$. It is like writing $\frac{12}{2}$, which is an integer, while $2$ is not a unit in the integers. – quid Sep 19 '16 at 12:41