Principal ideal of an integrally closed domain

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.

• You know the proof when $R$ is a field. Try to imitate it. – Martin Brandenburg Nov 20 '13 at 10:19
• Ehm, if $R$ is a field, $R[x]$ is PID. It's easy to prove that $I$ is an ideal of $R[x]$. Since $R[x]$ is PID, $I$ is a principal ideal of $R[x]$. – user109584 Nov 20 '13 at 10:25
• I think this case is so trivial. Can you give me another hint? – user109584 Nov 20 '13 at 10:39
• Help me. Please. – user109584 Nov 20 '13 at 10:49
• Yes and how do you prove that the polynomial ring is a PID? – Martin Brandenburg Nov 20 '13 at 11:51

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 $I=(p)$.
• If a product of two monic polynomials from $K[x]$ is in $R[x]$, then both polynomials are in $R[x]$ (of course, $R$ is integrally closed and $K$ field of fractions of $R$). See here, for example. (Btw, normal = integrally closed.) – user89712 Nov 20 '13 at 11:14
• @user109584: No, only nearly. If $R$ is integrally closed domain we know that $R[x]$ is also. Let $P$ be a prime ideal of $R[x]$ such that $P\cap R=\left\{ 0\right\}$ and $P$ contains a monic polynomial. You can check that $P$ is principal. – chuyenvien94 Nov 20 '13 at 11:30