GCD in a subring is GCD in a bigger ring?

Question

Let $R$ be a UFD which is a subring of an integral domain $S$. If $r_1$ and $r_2$ are two nonzero elements of $R$ with GCD $d$, is it true that $d$ is also a GCD of $r_1$ and $r_2$ in $S$?

I know this is true if $R$ is a PID.

Answer 1

Let $K$ be a field. First note, that $K[X,Y] \cong K[X,XY]$, via $X \mapsto X, \, Y \mapsto XY$, so $\gcd_{K[X,XY]}(X,XY) = 1$. Now consider the canonical inclusion $\iota \colon K[X,XY] \hookrightarrow K(Y)[X]$. Obviously $\gcd_{K(Y)[X]}(X,XY) = X \not \sim_{K(Y)[X]} 1$. Note that $K[X,XY]$ is a UFD and $K(Y)[X]$ is even a PID.

Answer 2

No, e.g. $\rm\:\gcd(2,x) = 1\:$ in $\rm\:\mathbb Z[x]\:$ but the gcd is the nonunit $\:2\:$ in $\rm\:\mathbb Z[x/2]\subset \mathbb Q[x]$.

But 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, see here.