GCD in a PID persists in extension domains Let $R$ be subring of integral domain $S.$ Suppose $R$ is $\text{PID}.$ Let $a\in R$ be a greatest common divisor of $r_1,r_2$ in $R$. ($r_1,r_2 \in R$, not both zero). Could anyone advise me on how to prove $a$ is greatest common divisor of $r_1,r_2$ in $S?$ 
Hints will suffice, thank you. 
 A: Hint $ $ gcds in PIDs 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, PIDs enjoy  Bezout identity for their gcds, which yields the following ring-theoretic equational specification for the gcd
$$\begin{eqnarray} \rm\gcd(a,b) = c &\iff&\rm (a,b) = (c)\\
&\iff&\rm  a = c\:\color{#C00}x ,\  b= c\:\color{#C00} y ,\,\ a\:\color{#C00} u + b\: \color{#C00}v = c\ \  has\ roots\ \ \color{#C00}{x,y,u,v}\in D\end{eqnarray}$$
Proof $\, $ The first equivalence  is proved here. For the second: 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 $\rm R\!$ $\rm \iff\! (c)\supseteq (a,b).\,$ Reversely $\,\rm(c)\subseteq (a,b)\iff$ $\rm c\in (a,b) \iff c = a\,u + b\,v\,$ for some $\rm \,u,v\in R.\ $ QED
Rings with such linearly representable gcds are known as Bezout rings. As above, gcds in such rings always persist in extension rings. In particular, coprime elements remain coprime in extension rings (with same $1$). This need not be true without such Bezout linear representations of the gcd. For example, $\rm\:\gcd(2,x) = 1\:$ in $\rm\:\mathbb Z[x]\:$ but the gcd is the nonunit  $\:2\:$ in $\rm\:\mathbb Z[x/2]\subset \mathbb Q[x]$. The above proof fails because there is no Bezout equation $\rm\, 2\,f + x\,g = 1,\,$ else $\rm\,2\, f(0) = 1\,$ by eval at $\rm\,x = 0$.
A: Here is another proof.
Let $()_S$ and $()_R$ denote the ideal generated by the thing inside the parentheses in $S$ and $R$ respectively. As $a$ is a GCD of $r_1$ and $r_2$ in $R$, $(a)_R = (r_1,r_2)_R$. Now if $e\mid r_1$ and $e\mid r_2$ in $S$, then $(r_1,r_2)_S \subset (e)_S$. Obviously, $(r_1,r_2)_R \subset (r_1,r_2)_S$, so we get $(a)_R \subset (e)_S$, thus $a \in (e)_S$ so $e \mid a$ in $S$. And of course $a\mid r_1$ and $a\mid r_2$ in $S$. Therefore $a$ is a GCD in $S$ as well. 
