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I seem to be stumped on this question. For the setting, let $R$ be an integral domain and let $r, s, t \in R$. The question asks

Show that $r \gcd(s, t)$ is associate to $\gcd (rs, rt)$

To start, let $d$ be some $\gcd$ of $s$ and $t$, and let $\overline{d}$ be some $\gcd$ of $rs$ and $rt$. Since $d \mid s$ and $d \mid t$, we see that $rd \mid rs$ and $rd \mid rt$. By definition, then $rd \mid \overline{d}$. It is here that I am stuck. I would like to show that $\overline{d} \mid rd$ to conclude that $rd$ and $\overline{d}$ are associate, but I cannot see how to get there. I know that I can write $s = di$ and $t = dj$ for some $i, j \in R$. I can also write $rt = \overline{d}n$ and $rs = \overline{d}m$ for some $m, n \in R$. But all this gets me is $rd(i-j) = \overline{d}(m-n)$, which is not quite what I am after. Any helpful hints would be greatly appreciated.

tylerc0816
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1 Answers1

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It suffices to show $r\mid\bar{d}$, since then using cancellation ($R$ is an integral domain) we get

$$\bar{d}\mid rs,rt\implies (\bar{d}/r)\mid s,t\implies (\bar{d}/r)\mid d\implies \bar{d}\mid rd.$$

How to show $r\mid\bar{d}$? Easy: $r\mid rs,rt\implies r\mid\bar{d}$.

anon
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  • What if $r$ is not a unit? – tylerc0816 Nov 04 '13 at 01:02
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    @tylerc0816 $\bar{d}/r$ is simply shorthand for the element which multiplies by $r$ to get $\bar{d}$. (This is part of why I put parentheses around it.) For example, $4/2$ is an integer even though $2$ is not a unit in $\Bbb Z$. If you want, you can think of the division as taking place in the fraction field. Alternatively, you can write "let $\bar{d}=rn$, ..." and then replace the occurrence of $\bar{d}/r$ in this answer with the letter $n$. – anon Nov 04 '13 at 01:04