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  "body": "\n## Definition\n\nLet $K$ be a [[field|04C5ZBG1WT]] and let $X$ be a [[scheme|5A35TYZD02]] over $K$, for example an [[algebraic variety|HV6YBS0YDH]] over $K$.\nA *$K$-rational point* of $X$ is a morphism of $K$-schemes $\\operatorname{Spec} K \\to X$.\nThe set of $K$-rational points is written $X(K)$.\n\n## Remarks\n\nIf $X \\subseteq \\mathbb{A}^n_K$ is the affine variety defined by polynomials $f_1, \\dots, f_m \\in K[x_1, \\dots, x_n]$, then $X(K)$ is the set of $a \\in K^n$ with $f_1(a) = \\cdots = f_m(a) = 0$.\nIf $X \\subseteq \\mathbb{P}^n_K$ is defined by homogeneous polynomials, then $X(K)$ is the set of points $[a_0 : \\cdots : a_n]$ with all $a_i \\in K$, not all zero, at which the polynomials vanish.\nEquivalently, when $X$ is of finite type over $K$, a $K$-rational point is a closed point of $X$ whose residue field is $K$.\n\nExample: $(3/5, 4/5) \\in X(\\Q)$ for $X = V(x^2 + y^2 - 1) \\subseteq \\mathbb{A}^2_\\Q$, since $9/25 + 16/25 = 1$.\n\nNon-example: $(1, \\sqrt{2})$ satisfies $x^2 + y^2 = 3$, but it is not a $\\Q$-rational point, because $\\sqrt{2} \\notin \\Q$.\nIn fact $x^2 + y^2 = 3$ has no $\\Q$-rational point: clearing denominators gives $a^2 + b^2 = 3c^2$ in integers with $\\gcd(a, b, c) = 1$, and reducing modulo $3$ forces $3$ to divide $a$, $b$ and then $c$.\n",
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