Serre's criterion for normality

Remarks

The proof works locally. A normal local ring of dimension $1$ is a discrete valuation ring, hence regular, which gives $(R_1)$. For $(S_2)$, let $\dim A_{\mathfrak{p}} \ge 2$ and let $x \in \mathfrak{p}A_{\mathfrak{p}}$ be a nonzerodivisor. Suppose the maximal ideal were associated to $A_{\mathfrak{p}}/(x)$. Then some quotient $y/x$ would be integral but not in the ring, or it would make the maximal ideal principal and the dimension $1$. Conversely, $(S_2)$ says that $A$ is the intersection of its localizations at height one primes, and $(R_1)$ says those localizations are discrete valuation rings, hence integrally closed. The companion statement says that $A$ is reduced if and only if it satisfies $(R_0)$ and $(S_1)$.

Example: $A = k[x, y, z]/(z^2 - xy)$ over a field $k$ of characteristic not $2$ is normal. It is a hypersurface, hence Cohen–Macaulay, so $(S_2)$ holds; its only singular point is the origin, whose prime has height $2$, so $(R_1)$ holds.

Both conditions are needed. The cusp $k[x, y]/(y^2 - x^3)$ is Cohen–Macaulay but fails $(R_1)$ at the origin, a prime of height $1$; indeed $t = y/x$ is integral, since $t^2 = x$, but not in the ring. Two planes meeting in one point, $k[x, y, u, v]/((x, y) \cap (u, v))$, satisfy $(R_1)$. But they have depth $1$ at the origin, where the dimension is $2$, so $(S_2)$ fails. The ring is not normal, because its local ring at the origin is not a domain.

Going deeper

For a student

In practice, $(S_2)$ is often automatic: it holds for every Cohen–Macaulay ring, in particular for complete intersections. Then normality reduces to $(R_1)$, which one checks with the Jacobian criterion by computing the codimension of the singular locus. This is the standard way to prove that a hypersurface with isolated singularities in dimension at least $2$ is normal. The criterion also explains the algebraic Hartogs property of normal rings: a normal Noetherian domain is the intersection of its localizations at height one primes.