Integral closure

The integral closure of $R$ in $A$ is the subset

It is a subring of $A$ containing the image of $R$.

When $R$ is a domain with fraction field $K$, the integral closure of $R$ in $K$ is often called the normalization of $R$. The domain $R$ is integrally closed precisely when its integral closure in its fraction field equals $R$.

• If $R\subseteq A$ is an integral extension, then every element of $A$ is integral over $R$, hence $\overline{R}^{\,A}=A$.
• If $B\subseteq A$ is any subring containing $R$ and consisting of elements integral over $R$, then $B\subseteq \overline{R}^{\,A}$ (maximality of the integral closure).

1. Integers inside rationals. Take $R=\mathbb{Z}$ and $A=\mathbb{Q}$. If $a/b\in \mathbb{Q}$ (in lowest terms) is integral over $\mathbb{Z}$, then it satisfies a monic polynomial with integer coefficients, forcing $b=\pm 1$. Hence $\overline{\mathbb{Z}}^{\,\mathbb{Q}}=\mathbb{Z}$.

1. A non-normal affine subring. Let $k$ be a field and consider $R=k[x^2,x^3]\subseteq A=k(x)$ (the rational function field in $x$). The element $x\in k(x)$ satisfies the monic equation $T^2-x^2=0$ with $x^2\in R$, so $x$ is integral over $R$. Thus the integral closure contains $k[x]$. In fact one checks $\overline{R}^{\,k(x)}=k[x]$.

1. Localizing does not change “integrality in the fraction field.” If $R$ is a domain and $S$ is a multiplicative set of nonzero elements, then $S^{-1}R$ sits in the same fraction field as $R$. Elements of the fraction field integral over $S^{-1}R$ are exactly those integral over $R$ after clearing denominators, so integral closure interacts naturally with localization.