Polynomial ring over integrally closed subring is integrally closed in polynomial ring

Statement

Suppose $R$ is an integrally closed subring of a commutative unital ring $S$. Then, the polynomial ring $R[x]$ is an integrally closed subring of $S[x]$.

Proof

Given: A ring $S$, an integrally closed subring $R$.

To prove: $R[x]$ is an integrally closed subring of $S[x]$.

Proof: Suppose $f\in S[x]$ satisfies a monic polynomial over $R[x]$:

$f^{n}+g_{n-1}f^{n-1}+g_{n-2}f^{n-2}+\dots +g_{0}=0$.

Here, $g_{i}\in R[x]$ for all math>i</math>. Write $f=h+x^{r}$, for $r$ greater than the maximum of the degrees of the $g_{i}$. Note that $f\in R[x]$ if and only if $h\in R[x]$, so it suffices to show that $h\in R[x]$.

$h$ satisfies the monic polynomial:

$(h+x^{r})^{n}+g_{n-1}(h+x^{r})^{n-1}+g_{n-2}(h+x^{r})^{n-2}+\dots +g_{0}=0$.

The constant term of this, viewed as a polynomial in $h$, is:

$(x^{r})^{n}+g_{n-1}(x^{r})^{n-1}+\dots +g_{0}$.

Since all the $g_{i}$ are in $R[x]$, this whole constant term is an element of $R[x]$.

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