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Let $R$ be a subring of an integral domain $S$, and suppose $R$ is a PID. Then it follows that if $r\in R$ is a gcd of $r_1$ and $r_2$ in $R$, where $r_1$ and $r_2$ are not both zero, then $r$ is a gcd of $r_1$ and $r_2$ in $S$.

What I would like to know why the same conclusion would fail if $R$ is only an UFD instead of a PID. I have been trying to think of counter-examples of R, like for instance, $\mathbb{Z}[X]$ and etc., but to no avail. Is there any counter-examples to disprove the statement when $R$ is an UFD but not a PID?

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but doesn't 2 divide 1 in $\mathbb{Q}[X]$? Or did I misinterpret something? –  yoshi Mar 9 '12 at 3:22
    
yeah, sorry for the noise. –  lhf Mar 9 '12 at 3:23
    
See also this answer. –  Bill Dubuque Mar 9 '12 at 3:29
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1 Answer

Hint $\rm\ gcd(2,x) = 1\:$ in $\rm\mathbb Z[x]\:$ vs. $2$ in $\rm\:\mathbb Z[x/2]$

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