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  "body": "\n## Statement\n\nLet $A$ be a [[Noetherian ring|0670T59HGE]].\nFor a [[prime ideal|00MHHZ9SNE]] $\\mathfrak{p}$, write $A_{\\mathfrak{p}}$ for the [[localization|X6JD4AG973]] at $\\mathfrak{p}$, $\\operatorname{ht} \\mathfrak{p}$ for its [[height|87WH65FJB8]], and $\\operatorname{depth} A_{\\mathfrak{p}}$ for the [[depth|M2RADBFW99]] of $A_{\\mathfrak{p}}$.\nConsider the conditions:\n\n- $(R_1)$: $A_{\\mathfrak{p}}$ is a [[regular local ring|63XXTF6B5M]] for every prime $\\mathfrak{p}$ with $\\operatorname{ht} \\mathfrak{p} \\le 1$.\n- $(S_2)$: $\\operatorname{depth} A_{\\mathfrak{p}} \\ge \\min(2, \\dim A_{\\mathfrak{p}})$ for every prime $\\mathfrak{p}$.\n\nCall $A$ a [[normal ring|KR7SWBYHRQ]] if $A_{\\mathfrak{p}}$ is an [[integrally closed domain|H20P600HQS]] for every prime $\\mathfrak{p}$.\nThen $A$ is normal if and only if $A$ satisfies $(R_1)$ and $(S_2)$.\n\n## Remarks\n\nThe proof works locally.\nA normal local ring of dimension $1$ is a discrete valuation ring, hence regular, which gives $(R_1)$.\nFor $(S_2)$, let $\\dim A_{\\mathfrak{p}} \\ge 2$ and let $x \\in \\mathfrak{p}A_{\\mathfrak{p}}$ be a nonzerodivisor.\nSuppose the maximal ideal were associated to $A_{\\mathfrak{p}}/(x)$.\nThen some quotient $y/x$ would be integral but not in the ring, or it would make the maximal ideal principal and the dimension $1$.\nConversely, $(S_2)$ 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.\nThe companion statement says that $A$ is reduced if and only if it satisfies $(R_0)$ and $(S_1)$.\n\nExample: $A = k[x, y, z]/(z^2 - xy)$ over a field $k$ of characteristic not $2$ is normal.\nIt is a hypersurface, hence [[Cohen–Macaulay|N1QJQEGBFC]], so $(S_2)$ holds; its only singular point is the origin, whose prime has height $2$, so $(R_1)$ holds.\n\nBoth conditions are needed.\nThe 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.\nTwo planes meeting in one point, $k[x, y, u, v]/((x, y) \\cap (u, v))$, satisfy $(R_1)$.\nBut they have depth $1$ at the origin, where the dimension is $2$, so $(S_2)$ fails.\nThe ring is not normal, because its local ring at the origin is not a domain.\n\n## Explanations\n\n### L1 human\n\nA normal ring is one with no \"missing fractions\", which in geometry means mild singularities.\nSerre's criterion splits normality into two simpler checks.\nThe first check says the space has no singularities in codimension one, so it is smooth away from a small set.\nThe second says that the ring has enough depth, so functions defined away from that small set extend over it.\n\n### L2 human\n\nIn practice, $(S_2)$ is often automatic: it holds for every [[Cohen–Macaulay ring|N1QJQEGBFC]], in particular for complete intersections.\nThen normality reduces to $(R_1)$, which one checks with the Jacobian criterion by computing the codimension of the singular locus.\nThis is the standard way to prove that a hypersurface with isolated singularities in dimension at least $2$ is normal.\nThe criterion also explains the algebraic Hartogs property of normal rings: a normal Noetherian domain is the intersection of its localizations at height one primes.\n",
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