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  "body": "\n## Definition\n\nLet $B$ be a commutative ring and let $A$ be a [[subring|0GNCRPYXYV]] of $B$.\nThe *integral closure of $A$ in $B$* is the set\n$$\\overline{A} = \\set{b \\in B : b \\text{ is integral over } A}$$\nof the elements of $B$ that are [[integral|1EHN4KPSFS]] over $A$.\nThe ring $A$ is *integrally closed in $B$* if $\\overline{A} = A$.\nFor an [[integral domain|0QVD7WBFGN]] $A$, the integral closure of $A$ in its [[field of fractions|0F2PTGX1A2]] is the *normalization* of $A$.\n\n## Remarks\n\nThe set $\\overline{A}$ is a subring of $B$ that contains $A$, and it is integrally closed in $B$ by [[transitivity of integrality|ZXZMQM818D]].\n\nExample: the integral closure of $\\Z$ in $\\Q(i)$ is $\\Z[i]$.\nIf $a + bi$ with $a, b \\in \\Q$ is integral, then so is its conjugate, so $2a$ and $a^2 + b^2$ are rational and integral over $\\Z$, hence integers, and this forces $a, b \\in \\Z$.\n\nNon-example: $\\Z[\\sqrt{5}]$ is not integrally closed in $\\Q(\\sqrt{5})$.\nThe element $(1 + \\sqrt{5})/2$ is a root of $x^2 - x - 1$, so it is integral over $\\Z$, but it is not in $\\Z[\\sqrt{5}]$.\nThe integral closure of $\\Z$ in $\\Q(\\sqrt{5})$ is $\\Z[(1 + \\sqrt{5})/2]$.\n",
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