Integrally closed domain

Equivalently, if one forms the integral closure of $R$ inside $K$, then $R$ is integrally closed exactly when

This condition is often phrased as: “$R$ has no new integral elements in its fraction field.”

Because $K$ is the localization $S^{-1}R$ with $S=R\setminus\{0\}$, integrally closedness can be viewed as a statement about integrality after inverting all nonzero elements.

In many situations, integrally closedness behaves well under localization at primes: roughly, $R$ is integrally closed if and only if all localizations $R_\mathfrak p$ are integrally closed.

1. Principal ideal domains (e.g. $\mathbb{Z}$). The ring $\mathbb{Z}$ is integrally closed in $\mathbb{Q}$: any rational number integral over $\mathbb{Z}$ must be an integer (as in the integral closure example).

1. Polynomial rings over a field. If $k$ is a field, then $k[x_1,\dots,x_n]$ is integrally closed in its fraction field $k(x_1,\dots,x_n)$. (More generally, unique factorization domains are integrally closed.)

1. Discrete valuation rings. Any discrete valuation ring is integrally closed. Concretely, $k[[t]]$ is integrally closed in $k((t))$.

• A cusp subring. $R=k[x^2,x^3]\subset k(x)$ is not integrally closed because $x\in k(x)$ is integral over $R$ (it satisfies $T^2-x^2=0$ with $x^2\in R$) but $x\notin R$. Its integral closure in $k(x)$ is $k[x]$.

• A classical quadratic example. $R=\mathbb{Z}[\sqrt5]$ is not integrally closed in $\mathbb{Q}(\sqrt5)$: the element $\frac{1+\sqrt5}{2}$ is integral over $R$ (it satisfies $T^2-T-1=0$) but is not in $R$.