Exercise A0.V.2.22

Proposition

Let $R \subseteq S$ be an inclusion of integral domains, with $R$ a PID. Let $a, b \in R$ and let $d = \gcd(a,b) \in R.$ Then $d$ is also a $\gcd(a,b)$ in $S.$

Proof

Observe that, by definition of GCD, $(a) + (b) \subseteq (d).$ Hence, we have $p, q \in R$ such that $dp = a$ and $dq = b.$ Since $R$ is a PID, we also have $u,v \in R$ such that $d = au + bv.$ Hence, $(d) \subseteq (a) + (b).$ Thus $(d) = (a) + (b).$

These equalities also hold in $S,$ which is sufficient to demonstrate that no ideal smaller than $(d)$ can contain $(a)+(b).$

$\blacksquare$