Ring of integer-valued polynomials over normal domain is normal

Statement

Suppose $R$ is a normal domain. Let $\operatorname {Int} (R)$ denote the ring of integer-valued polynomials over $R$. Then, $\operatorname {Int} (R)$ is also a normal domain.

Definitions used

Ring of integer-valued polynomials

Further information: Ring of integer-valued polynomials

Normal domain

Further information: Normal domain

Facts used

1. polynomial ring over a field is normal

Proof

Given: A normal domain $R$. $\operatorname {Int} (R)$ is the ring of integer-valued polynomials over $R$.

To prove: $\operatorname {Int} (R)$ is also a normal domain.

Proof: Let $K$ be the field of fractions of $R$. Now, $\operatorname {Int} (R)$ is a subring of $K[x]$. We thus have:

$\operatorname {Int} (R)\subseteq K[x]\subseteq K(x)$

where $K(x)$, the field of rational functions, is the field of fractions of both $\operatorname {Int} (R)$ and $K[x]$. By fact (1), $K[x]$ is normal, so the integral closure of $\operatorname {Int} (R)$ in $K(x)$ is contained in $K[x]$. Thus it suffices to show that elements of $K[x]$ that are integral over $\operatorname {Int} (R)$ are in $\operatorname {Int} (R)$.

Suppose now that $f\in K[x]$ is integral over $\operatorname {Int} (R)$. In other words, we have a monic polynomial that $f$ satisfies with coefficients in $\operatorname {Int} (R)$:

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

where $g_{i}\in \operatorname {Int} (R)$.

Our goal is to show that $f(a)\in R$ for any $a\in R$.

For any $a\in R$, evaluation at $a$ gives:

$(f(a))^{n}+g_{n-1}(a)f(a)^{n-1}+\dots +g_{0}(a)=0$.

Since all the $g_{i}$ are in $\operatorname {Int} (R)$, this is a monic polynomial with coefficients in $R$. Further, since $f\in K[x]$, $f(a)\in K$ satisfies a monic polynomial with coefficients in $R$. This forces $f(a)\in R$ since $R$ is normal.