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Let $D$ be an integral domain and $a,b,x \in D$.

If $d$ is a greatest common divisor ( gcd ) of $a$ and $b$, is it true that $dx$ is a gcd of $ax$ and $bx$?

Note that $D$ is an integer domain, not a GCD domain. So this question is not the same as any of the following questions：1 2 3 etc. More specifically, the existence of gcd of $ax$ and $bx$ is not guaranteed.

Any insights are much appreciated.

BTW: To avoid ambiguity, I haven't used notations such as $(a,b)$ or $(ax,bx)$, which makes me verbose. You may use it at will.

Sky subO
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1 Answers1

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This is not true. Take $D=\Bbb Z[\sqrt{-5}]$, $a=2,b=1+1\sqrt{-5},x=3$. Using the norm we see that $d=1$ is a gcd of $a,b$. However, $3d=3$ is not a gcd of $ax=6$ and $bx=3+3\sqrt{-5}$. This is because $1+\sqrt{-5}\mid ax,bx$ but $1+\sqrt{-5}\nmid 3$. (consequently $ax$ and $bx$ don't have a gcd.)

leoli1
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  • Thanks for the answer. I feel that $\Bbb Z[\sqrt{-5}]$ is important. – Sky subO Sep 04 '21 at 16:29
  • Excuse me, may I ask one more question? Why do you say that $ax$ and $bx$ don't have a gcd? I know that $3$ and $1+\sqrt{-5}$ are not gcds of $6$ and $3+3\sqrt{-5}$. But maybe other elements? Thank you. – Sky subO Sep 15 '21 at 15:38
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    @SkysubO the questions/answers you linked in your question shows that if $ax$ and $bx$ have a gcd, then it has to be (up to units) $dx$. But here $dx=3$ is not a gcd of $ax$ and $bx$, hence they cannot have a gcd. One can also see this directly: suppose they had a gcd $e$. By taking norms we would have $N(e)\mid 18$ and $9,6\mid N(e)$ (because $e\mid ax,bx$ and $3, 1+\sqrt{-5}\mid e$). But then $N(e)=18$ and you can verify that there is no solution to $N(e)=18$ in $\Bbb Z[\sqrt{-5}]$. – leoli1 Sep 15 '21 at 15:51
  • Ah, I see. The first way is ingenious. Thank you very much! – Sky subO Sep 15 '21 at 16:29