Extending ideals from principal ideal domains Let $D$ be a PID, $E$ a domain containing $D$ as a subring. Is it true that if $d$ is a gcd of $a$ and $b$ in $D$, then $d$ is also a gcd of $a$ and $b$ in $E$?
 A: In any domain $D$, for $a,b \in D \setminus \{0\}$, if the ideal $\langle a,b \rangle_D = \{xa + yb \ | \ x,y \in D\}$ of $D$ is principal, then any generator $d$ of the ideal is a gcd of $a$ and $b$.  (Note that in general gcd's are unique precisely up to units, i.e., the corresponding principal ideal is unique.)
So if $D$ is a PID, there is $d \in D$ such that $\langle a,b \rangle_D = dD$.  Now push forward these ideals to $E$:
$\langle a,b \rangle_E = \langle a,b \rangle_D E = (d D) E = dE$.
Thus the ideal of $E$ generated by $a$ and $b$ is still principal and still generated by the same element $d$ (now well-determined up to a unit of $E$; note that the unit group of $E$ could be larger than the unit group of $D$).
So in summary: yes.
A: Gcds in a PID D persist in extension rings because the gcd may be specified by the solvability of (linear) equations over D,  and such solutions always persist in extension rings, i.e. roots in D remain roots in rings $\rm\,R \supset D.\:$ More precisely, the Bezout identity for the gcd yields the following ring-theoretic equational specification for the gcd
$$\begin{eqnarray} \rm\gcd(a,b) = c &\iff&\rm (a,b) = (c)\ \ \ {\rm [equality\ of\  ideals]}\\
&\iff&\rm  a\: \color{#C00}x = c,\  b\:\color{#C00} y = c,\,\ a\:\color{#C00} u + b\: \color{#C00}v = c\ \  has\ roots\ \ \color{#C00}{x,y,u,v}\in D\end{eqnarray}$$
Proof $\ (\Leftarrow)\:$ In any ring $\rm R,\:$  $\rm\:a\: x = c,\ b\: y = c\:$ have roots $\rm\:x,y\in R$ $\iff$ $\rm c\ |\ a,b\:$ in $\rm R.$ Further if $\rm\:c = a\: u + b\: v\:$ has roots $\rm\:u,v\in R\:$ then $\rm\:d\ |\ a,b$ $\:\Rightarrow\:$ $\rm\:d\ |\ a\:u+b\:v = c\:$ in $\rm\: R.\:$ Hence we infer $\rm\:c = gcd(a,b)\:$ in $\rm\: R,\:$ being a common divisor divisible by every common divisor. $\ (\Rightarrow)\ $ If $\rm\:c = gcd(a,b)\:$ in D then the Bezout identity  implies the existence of such roots $\rm\:u,v\in D.$
