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  "body": "\n## Definition\n\nLet $k$ be a [[field|04C5ZBG1WT]], let $n \\ge 0$ be an integer, and let $\\mathbb{A}^n = k^n$ be the affine $n$-space over $k$.\nLet $k[x_1, \\dots, x_n]$ be the [[polynomial ring|0SDC934R52]] in $n$ variables over $k$.\nFor a subset $S \\subseteq k[x_1, \\dots, x_n]$, its zero set is\n$$V(S) = \\set{a \\in k^n : f(a) = 0 \\text{ for all } f \\in S}.$$\nA subset $X \\subseteq k^n$ is an *algebraic set* (or affine algebraic set) if $X = V(S)$ for some subset $S \\subseteq k[x_1, \\dots, x_n]$.\n\n## Remarks\n\nExample: in $\\R^2$, the unit circle is $V(x^2 + y^2 - 1)$, so it is an algebraic set.\nEach point $a = (a_1, \\dots, a_n)$ is algebraic, since $\\set{a} = V(x_1 - a_1, \\dots, x_n - a_n)$.\n\nNon-example: the half-line $H = \\set{(t, 0) : t \\ge 0}$ in $\\R^2$ is not an algebraic set.\nIf a polynomial $f$ vanishes on $H$, then $t \\mapsto f(t, 0)$ is a polynomial in one variable with infinitely many roots, so it is zero.\nThen $f$ also vanishes at $(-1, 0)$, so every $V(S)$ that contains $H$ contains the whole $x$-axis.\n\nThe zero set only depends on the [[ideal|096Z9MP2ME]] $I$ generated by $S$: $V(S) = V(I)$.\nBy [[Hilbert's basis theorem|0YKA9J3XEE]], $I$ is finitely generated, so every algebraic set is the zero set of finitely many polynomials.\nThe rules $V(I) \\cup V(J) = V(IJ)$ and $\\bigcap_i V(I_i) = V\\left(\\sum_i I_i\\right)$ show that the algebraic sets are the closed sets of a topology on $k^n$.\nThis topology is the classical form of the [[Zariski topology|8SQK7RTB63]].\n",
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