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Is it true that in a domain $R$, $\gcd(ta,tb)=t\gcd(a,b)$?

I know how to prove it in $\mathbb{Z}$, but what about if it's an arbitrary domain $R$?

user26857
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Qian Huang
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    No, it's not! It's possible to exist $\gcd(a,b)$, but not $\gcd(ta,tb)$. If assume that both exist, then the claim holds. – user26857 Dec 05 '16 at 09:37
  • Thanks. Is there any example where $\gcd(ta,tb)$ does not exist? Also, it's obvious that $t\gcd(a,b)|\gcd(ta,tb)$, but I still have no idea on the opposite direction, i.e. $\gcd(ta,tb)|t\gcd(a,b)$. – Qian Huang Dec 07 '16 at 18:14
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    If $\gcd(ta,tb)$ exists you can find a proof in the below terse answer. And yes, there are such examples in the ring $K[X^2,X^3]$. – user26857 Dec 07 '16 at 18:52

2 Answers2

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Theorem $\rm\ \ (a,b)\ =\ (ac,bc)/c\ \ $ if $\rm\ \ (ac,bc)\ $ exists $\rm\quad$ [GCD distributive law]

Proof $\rm\quad d\mid a,b \iff dc\mid ac,bc \color{#c00}\iff dc\mid (ac,bc) \iff d\mid (ac,bc)/c$

where above we have $\rm\color{#c00}{used}$ the universal definition of GCD.

Community
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Bill Dubuque
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Thanks to all replies and comments. I have come to realize that it is actually rather simple, although it seems to puzzle some new learners like me...

The most naive solution is as follows (We need $t,a,b\not =0.$)

Let $B=\gcd(a,b), A=\gcd(ta,tb)$. We have $B|a,b \Rightarrow tB|ta,tb \Rightarrow tB|A$. On the other direction, we note that $t|ta,tb \Rightarrow t|A$. Suppose $A=tu$, $ta=\beta A=\beta tu$ $\Rightarrow u|a$. Similarly, $u|b$. Hence, $u|a,b \Rightarrow u|B \Rightarrow tu|tB$. Therefore, $A$ and $tB$ are associates.

PS. I feel like that sometimes a lack of example makes me uncomfortable. There are domains that gcd may not exist. There must also be domains that some element cannot be expressed as the product of irreducible elements...

Qian Huang
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