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  "body": "\n## Statement\n\nLet $K$ be an [[algebraically closed field|0PF6TM8TWT]], let $n \\ge 1$, and let $f_1, \\dots, f_n \\in K[x_1, \\dots, x_n]$ be [[polynomials|1G0B73WPX8]].\nLet $F \\colon K^n \\to K^n$ be the map $F(a) = (f_1(a), \\dots, f_n(a))$.\nIf $F$ is [[injective|09NWETVTNY]], then $F$ is [[surjective|01RVCMSN5N]].\n\n## Remarks\n\nFor fixed $n$ and a bound $d$ on the degrees of the $f_i$, the statement is a single first-order sentence in the language of rings.\nIt holds for every [[finite field|0ZXZ82FTX2]], because an injective map from a finite set to itself is surjective.\nIt also holds in the algebraic closure $\\overline{\\F_p}$, the union of the finite fields $\\F_{p^k}$.\nGiven an injective $F$ over $\\overline{\\F_p}$ and a point $b$, the coefficients of the $f_i$ and the coordinates of $b$ lie in one finite field $\\F_{p^k}$.\nThe map $F$ sends $\\F_{p^k}^n$ injectively into itself, so it is onto $\\F_{p^k}^n$, and $b$ has a preimage.\nBy the [[Lefschetz principle|SKTMN14X5P]], it then holds in every algebraically closed field of characteristic $0$, such as $\\C$.\n\nExample: over $\\C$, the map $F(x, y) = (x, y + x^2)$ is injective, and it is surjective, since $(a, b) = F(a, b - a^2)$.\n\nThe hypothesis that $K$ is algebraically closed is necessary.\nOver $\\Q$, the polynomial map $x \\mapsto x^3$ is injective but not surjective, since $2$ is not the cube of a rational number.\n\n## Explanations\n\n### L1 human\n\nTake a map from $n$-dimensional space to itself given by polynomials, over the complex numbers.\nIf the map never sends two different points to the same point, then it reaches every point.\nFor finite sets this is obvious, and the theorem carries this simple fact over to the complex numbers.\n\n### L2 human\n\nThe theorem was proved by James Ax with model theory and by Alexander Grothendieck with algebraic geometry.\nThe model-theoretic proof is the standard showcase of the Lefschetz principle: a statement checked over finite fields transfers to $\\C$.\nThe converse fails: a surjective polynomial map need not be injective, as $z \\mapsto z^2$ on $\\C$ shows.\nIt is related to the open Jacobian conjecture, which asks whether a polynomial map $\\C^n \\to \\C^n$ with constant nonzero Jacobian determinant has a polynomial inverse.\n",
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