Normal scheme

Remarks

An affine scheme $\operatorname{Spec} A$ is normal if and only if $A_{\mathfrak p}$ is an integrally closed domain for every prime $\mathfrak{p}$, that is, $A$ is a normal ring. For an integral domain $A$, this holds if and only if $A$ is integrally closed. Example: $\operatorname{Spec} \Z$ is normal, since $\Z$ is a unique factorization domain and hence integrally closed. Non-example: let $A = k[x, y]/(y^2 - x^3)$ for a field $k$; it is a domain. The element $t = y/x$ of its field of fractions satisfies $t^2 - x = 0$, which is monic over $A$, but $t \notin A$. So $A$ is not integrally closed, and the cusp $\operatorname{Spec} A$ is not normal; the failure is at the origin. A normal scheme is reduced, but it need not be irreducible: a disjoint union of two normal schemes is normal.