1
$\begingroup$

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.

1 Answers 1

2

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

  • 0
    For much further detail see [this answer.](http://math.stackexchange.com/a/20136/23500)2012-02-18