Prove that GCD domain is integrally closed In the case $R$ is a UFD, the proof of that $R$ is integrally closed essentially uses a key fact: If gcd($a,b)=1$, then gcd$(a^n,b)=1$. The proof of this relies on UFD prop...--prophetes.ai

Prove that GCD domain is integrally closed In the case $R$ is a UFD, the proof of that $R$ is integrally closed essentially uses a key fact: If gcd($a,b)=1$, then gcd$(a^n,b)=1$. The proof of this relies on UFD property: if $a,b$ are coprime, then they do not have common prime factors, so the same holds for their powers. However, it is stated in wikipedia that GCD domain is also integrally closed. Can we prove it using the same fact above? If yes, how do we prove it in GCD?

$\color{#0a0}{(a,b)=1}\,=\,(c,\,b)\,\Rightarrow\,\color{#c00}1 = (a,b)(c,b) = (ac,ab,bb,bc)= (ac,\color{#0a0}{(a,b},c)b) = \color{#c00}{(ac,b)},\ $ therefore

$\color{}{(a,b)=1}=(a^n,b)\,\Rightarrow\,\color{#c00}{1=(a^{n+1},b)}\ $ yields the induction step.