greatest common divisor in a PID the same as in some domain that contains it

Question

If $R$ is PID and $S$ is an integral domain that contains $R$, why is the gcd of $a$ and $b$ in $R$ the same as the gcd of a and b in $S$ ? Is this a standard result? where can I find the proof? Thanks.

score 2 · Accepted Answer · edited Apr 01 '11 at 23:48

2

HINT $\ $ In a PID, $\rm\ \ gcd(a,b)\: =\: c\ \iff\ (a,b)\: =\: (c)\:.\:$ The latter lifts up to superrings.

edited Apr 01 '11 at 23:48

answered Apr 01 '11 at 22:49

Bill Dubuque

272,048

For much further detail see this answer. – Math Gems Feb 18 '12 at 21:51

greatest common divisor in a PID the same as in some domain that contains it

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