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  "body": "\n## Definition\n\nLet $K$ be a [[field|04C5ZBG1WT]], and let $K[T, X]$ be the [[polynomial ring|07GNX8GCYP]] in two variables over $K$.\nConsider [[irreducible polynomials|03FKX7Z5DC]] $f_1, \\dots, f_m \\in K[T, X]$, each of positive degree in $X$ and separable in $X$, and a nonzero $g \\in K[T]$.\nThe field $K$ is *Hilbertian* if, for all such data, there is $a \\in K$ such that:\n\n1. $g(a) \\ne 0$;\n2. for each $i$, the polynomial $f_i(a, X) \\in K[X]$ is irreducible over $K$ and has the same degree in $X$ as $f_i$.\n\n## Remarks\n\nIf $K$ is Hilbertian, then for such data there are infinitely many $a \\in K$ with these properties, since one can add to $g$ factors $T - a$ for the values already found.\nA Hilbertian field is infinite, and the same condition then holds with $T$ replaced by several variables $T_1, \\dots, T_r$.\nIn characteristic $0$, the separability condition is automatic, and $K$ is Hilbertian exactly when $K$ itself is not a thin subset of the affine line over $K$.\n\nExample: $\\Q$ is Hilbertian by [[Hilbert's irreducibility theorem|N1V3YEYHNP]], and so is every number field.\nFor any field $k$, the rational function field $k(t)$ is Hilbertian.\n\nNon-example: $\\R$ is not Hilbertian.\nThe polynomial $f = X^2 - T^2 - 1$ is irreducible in $\\R[T, X]$, because $T^2 + 1$ is not a square in $\\R(T)$.\nBut for every $a \\in \\R$ one has $a^2 + 1 \u003e 0$, so $f(a, X) = (X - \\sqrt{a^2 + 1})(X + \\sqrt{a^2 + 1})$ factors over $\\R$.\nAn algebraically closed field is not Hilbertian either, because every polynomial in $X$ of degree at least $2$ factors over it.\n",
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