Let $R$ be an integrally closed domain and $S$ be an integral domain that contains $R$. Assume that $a\in S$ is integral over $R$. Prove that $I=\left\{ f\left(x\right)\in R\left[x\right]\mid f\left(a\right)=0\right\} $ is a principal ideal of $R[x].$
I only know that $R[x]$ is also an integrally closed integral domain and $a$ is a root of a monic polynomial of $R[x]$. So if $g(x)$ is a monic polynomial of $R[x]$ then $g(a)=0$.
Help me a hint.
Thank for any insight.
Let $p$ be the minimal polynomial of $a$ over $K$, the field of fractions of $R$, and $f\in I$ monic. We should have $p\mid f$ in $K[x]$, and by Gauss' lemma $p\mid f$ in $R[x]$. Now it's easy to show that $I=(p)$.