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  "body": "\n## Statement\n\nLet $k$ be an [[algebraically closed field|0PF6TM8TWT]] and let $n \\ge 1$.\n\n1. The [[maximal ideals|0PV74TR5FP]] of the [[polynomial ring|0SDC934R52]] $k[x_1, \\ldots, x_n]$ are exactly the ideals $\\mathfrak{m}_a = (x_1 - a_1, \\ldots, x_n - a_n)$ for $a = (a_1, \\ldots, a_n) \\in k^n$.\n2. If $I$ is an [[ideal|096Z9MP2ME]] of $k[x_1, \\ldots, x_n]$ with $I \\ne k[x_1, \\ldots, x_n]$, then there is $a \\in k^n$ with $f(a) = 0$ for all $f \\in I$.\n\n## Remarks\n\nEach $\\mathfrak{m}_a$ is maximal, since it is the kernel of the surjective evaluation map $f \\mapsto f(a)$ onto the field $k$.\nConversely, let $\\mathfrak{m}$ be maximal; then $k[x_1, \\ldots, x_n]/\\mathfrak{m}$ is a field finitely generated as a $k$-algebra.\nSo it is finite over $k$ by [[Zariski's lemma|YBJ2VQHFDQ]], hence equal to $k$ as $k$ is algebraically closed.\nIf $a_i \\in k$ is the image of $x_i$, then $\\mathfrak{m} \\supseteq \\mathfrak{m}_a$, so $\\mathfrak{m} = \\mathfrak{m}_a$.\nPart 2 follows from part 1: a proper ideal lies in a maximal ideal $\\mathfrak{m}_a$, and then every $f \\in I$ vanishes at $a$.\n\nExample: over $\\C$, the ideal $(x^2 + y^2 - 1, x - 1)$ of $\\C[x, y]$ is proper and the point $(1, 0)$ is a common zero.\n\nThe field must be algebraically closed: over $\\R$, the ideal $(x^2 + 1)$ of $\\R[x]$ is maximal, but it has no zero in $\\R$ and is not of the form $(x - a)$.\n\n## Explanations\n\n### L1 human\n\nOver the complex numbers, a system of polynomial equations has no solution only when the equations imply $1 = 0$ by algebra.\nIn other words, the points of $\\C^n$ are the same thing as the maximal ideals of the polynomial ring.\n\n### L2 human\n\nThe weak form is the step from which the full [[Nullstellensatz|D85VZWS9QK]] follows by the Rabinowitsch trick.\nIt identifies the points of affine space with the maximal ideals of the coordinate ring.\nThis is the starting point for the maximal spectrum and, later, for schemes.\n",
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