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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