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  "body": "\n## Definition\n\nLet $k$ be a field, let $n \\ge 0$, and let $X \\subseteq k^n$ be any subset.\nThe *ideal of $X$* (or vanishing ideal of $X$) is\n$$I(X) = \\set{f \\in k[x_1, \\dots, x_n] : f(a) = 0 \\text{ for all } a \\in X},$$\nwhere $k[x_1, \\dots, x_n]$ is the [[polynomial ring|0SDC934R52]].\nIt is used mostly when $X$ is an [[algebraic set|VHFMQQYSTN]].\n\n## Remarks\n\n$I(X)$ is an [[ideal|096Z9MP2ME]], and it equals its own [[radical|437YRCNYRJ]]: if $f^m$ vanishes at a point $a$, then $f(a)^m = 0$, so $f(a) = 0$.\n\nExample: for the parabola $X = V(y - x^2) \\subseteq k^2$ with $k$ infinite, $I(X) = (y - x^2)$.\nCheck: dividing $f(x, y)$ by $y - x^2$ as a polynomial in $y$ gives $f = (y - x^2) q + r(x)$; if $f$ vanishes on $X$, then $r(t) = 0$ for all $t \\in k$, so $r = 0$.\n\nNon-example: the ideal $(x^2)$ in $k[x]$ is not the ideal of any set.\nIt is not radical, since $x \\notin (x^2)$ while $x^2 \\in (x^2)$; in fact $I(V(x^2)) = I(\\set{0}) = (x)$.\n\nFor an algebraic set $X$ one has $V(I(X)) = X$, and $X \\subseteq Y$ if and only if $I(Y) \\subseteq I(X)$.\nIf $k$ is [[algebraically closed|0PF6TM8TWT]], [[Hilbert's Nullstellensatz|D85VZWS9QK]] gives $I(V(J)) = \\sqrt{J}$ for every ideal $J$.\nSo $X \\mapsto I(X)$ is then an inclusion-reversing bijection from algebraic sets in $k^n$ to radical ideals of $k[x_1, \\dots, x_n]$.\n",
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