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  "body": "\n## Problem\n\nLet $k$ be a [[field|04C5ZBG1WT]] and let $K$ be a field with $k \\subseteq K \\subseteq k(x_1, \\ldots, x_n)$.\nIs the $k$-algebra $K \\cap k[x_1, \\ldots, x_n]$ always finitely generated?\n\n## Remarks\n\nThe motivating case is a [[ring of invariants|6VTR5EBNYC]]: if a group $G$ acts linearly on $k^n$, then $k[x_1, \\ldots, x_n]^G = k(x_1, \\ldots, x_n)^G \\cap k[x_1, \\ldots, x_n]$.\nThe answer is no in general.\nNagata gave a counterexample in 1959, as the ring of invariants of a linear action of a group on a polynomial ring.\nThe answer is yes in several important cases.\nIf $G$ is a reductive group, the ring of invariants of a rational action on $k[x_1, \\ldots, x_n]$ is finitely generated.\nIn characteristic $0$ this goes back to Hilbert's finiteness theorem of 1890.\nZariski proved that the answer is yes when $K$ has [[transcendence degree|D7TR0YWR9H]] at most $2$ over $k$.\nWeitzenböck proved that, in characteristic $0$, the invariants of a linear action of the additive group of $k$ are finitely generated.\n\n## Explanations\n\n### L1 human\n\nHilbert asked whether the polynomials that stay unchanged under a group of linear substitutions can always be built from finitely many of them.\nFor many groups, such as finite groups and the general linear group, they can.\nNagata showed that for some groups they cannot, so the answer to Hilbert's question in its general form is negative.\n",
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