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  "body": "\n## Definition\n\nLet $K$ be a [[field|04C5ZBG1WT]] with [[algebraic closure|0X6AB5P3G1]] $\\bar K$.\nAn affine variety $V \\subseteq \\bar K^n$ defined over $K$ is the set of common zeros in $\\bar K^n$ of a set of polynomials in $K[X_1, \\dots, X_n]$.\nIt is *absolutely irreducible* if it is irreducible as a variety over $\\bar K$.\nThis means that it is not the union of two proper closed subsets defined over $\\bar K$.\nA point of $V$ is $K$-rational if all its coordinates lie in $K$.\nThe field $K$ is *pseudo algebraically closed* (PAC) if every nonempty absolutely irreducible affine variety defined over $K$ has a $K$-rational point.\n\n## Remarks\n\nIt is equivalent to ask the condition only for curves, that is, for varieties of dimension $1$.\n\nExample: every [[algebraically closed field|0PF6TM8TWT]] $K$ is PAC, since then $\\bar K = K$ and every point of a nonempty variety is $K$-rational.\nEvery infinite algebraic extension of a [[finite field|0ZXZ82FTX2]] is PAC.\n\nNon-example: $\\Q$ and $\\R$ are not PAC.\nThe curve $V$ defined by $x^2 + y^2 + 1 = 0$ is absolutely irreducible, since $x^2 + y^2 + 1$ is irreducible over $\\C$, and $V$ is nonempty, since it contains $(i, 0)$.\nBut it has no point with real coordinates, because $x^2 + y^2 + 1 \u003e 0$ for real $x, y$.\n\nNon-example: a finite field $\\F_q$ is not PAC.\nFor instance, the curve $y^q - y = x^{q+1} - x^2 + 1$ is absolutely irreducible, because the degree $q + 1$ of the right side is prime to the characteristic of $\\F_q$.\nIt has no $\\F_q$-rational point, because $y^q - y = 0$ for $y \\in \\F_q$, while $x^{q+1} - x^2 + 1 = 1$ for $x \\in \\F_q$.\n",
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