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  "body": "\n## Definition\n\nLet $X$ be an [[integral scheme|17NVR4XBHK]] with function field $K$.\nFor each open affine subscheme $U = \\operatorname{Spec} A$ of $X$, let $\\tilde{A}$ be the [[integral closure|G0V86SJBQW]] of $A$ in its [[field of fractions|0F2PTGX1A2]] $K$.\nThe *normalization* of $X$ is the integral scheme $\\tilde{X}$ with a [[morphism|JXRKH6P1Z4]] $\\nu \\colon \\tilde{X} \\to X$ such that, for every such $U$:\n\n1. $\\nu^{-1}(U)$ is isomorphic to $\\operatorname{Spec} \\tilde{A}$;\n2. under this isomorphism, $\\nu$ restricted to $\\nu^{-1}(U)$ is the morphism $\\operatorname{Spec} \\tilde{A} \\to \\operatorname{Spec} A$ induced by the inclusion $A \\subseteq \\tilde{A}$.\n\nIt is obtained by [[gluing|ABVBHCA9VW]] the schemes $\\operatorname{Spec} \\tilde{A}$, and it is unique up to unique isomorphism over $X$.\n\n## Remarks\n\nThe scheme $\\tilde{X}$ is [[normal|EK261G1XP9]].\nFor every normal integral scheme $Z$ and every dominant morphism $f \\colon Z \\to X$, there is a unique morphism $g \\colon Z \\to \\tilde{X}$ with $f = \\nu \\circ g$.\nExample: let $A = k[x, y]/(y^2 - x^3) \\cong k[t^2, t^3]$ over a field $k$, with $x = t^2$ and $y = t^3$.\nIts fraction field is $k(t)$, and $t$ is integral over $A$ since $t^2 = x$; also $k[t]$ is integrally closed, so $\\tilde{A} = k[t]$.\nThe normalization of the cusp is $\\mathbb{A}^1_k \\to \\operatorname{Spec} A$, $t \\mapsto (t^2, t^3)$.\nNon-example: the identity morphism of the cusp is not its normalization, since the cusp is not normal and $A \\ne \\tilde{A}$.\nIf $X$ is normal, then $\\nu$ is an isomorphism.\n",
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