Integrally closed domain

Equivalently, the integral closure of $R$ in $K$ is $R$:

This condition is often phrased by saying that $R$ has no new integral elements in its fraction field.

Because $K=S^{-1}R$ for $S=R\setminus\{0\}$, integral closedness concerns elements obtained after inverting all nonzero elements.

Integral closedness is local: $R$ is integrally closed if and only if $R_{\mathfrak p}$ is integrally closed for every prime ideal $\mathfrak p$.

1. Principal ideal domains, such as $\mathbb Z$. Every PID is integrally closed. In particular, any rational number integral over $\mathbb Z$ is an integer.

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

1. Discrete valuation rings. Any discrete valuation ring is integrally closed. For example, $k\llbracket t\rrbracket$ is integrally closed in $k((t))$.

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

• A classical quadratic example. The ring $\mathbb Z[\sqrt5]$ is not integrally closed in $\mathbb Q(\sqrt5)$: the element $(1+\sqrt5)/2$ satisfies $T^2-T-1=0$ but does not belong to $\mathbb Z[\sqrt5]$.