---
tag: M5N08HD75V
kind: problem
name: Hilbert's fourteenth problem
msc: [13A50, 14L24, "13E15"]
uses: [04C5ZBG1WT, 0SDC934R52, 6VTR5EBNYC, D7TR0YWR9H]
about: [6VTR5EBNYC]
sources:
  - url: https://en.wikipedia.org/wiki/Hilbert's_fourteenth_problem
same_as:
  wikidata: Q4515163
  wikipedia: Hilbert's fourteenth problem
history:
  posed_by: "David Hilbert"
  first_stated: "1900"
---

## Problem

Let $k$ be a [[field|04C5ZBG1WT]] and let $K$ be a field with $k \subseteq K \subseteq k(x_1, \ldots, x_n)$.
Is the $k$-algebra $K \cap k[x_1, \ldots, x_n]$ always finitely generated?

## Remarks

The motivating case is a [[ring of invariants|6VTR5EBNYC]]: if a group $G$ acts linearly on $k^n$, then $k[x_1, \ldots, x_n]^G = k(x_1, \ldots, x_n)^G \cap k[x_1, \ldots, x_n]$.
The answer is no in general.
Nagata gave a counterexample in 1959, as the ring of invariants of a linear action of a group on a polynomial ring.
The answer is yes in several important cases.
If $G$ is a reductive group, the ring of invariants of a rational action on $k[x_1, \ldots, x_n]$ is finitely generated.
In characteristic $0$ this goes back to Hilbert's finiteness theorem of 1890.
Zariski proved that the answer is yes when $K$ has [[transcendence degree|D7TR0YWR9H]] at most $2$ over $k$.
Weitzenböck proved that, in characteristic $0$, the invariants of a linear action of the additive group of $k$ are finitely generated.

## Explanations

### L1 human

Hilbert asked whether the polynomials that stay unchanged under a group of linear substitutions can always be built from finitely many of them.
For many groups, such as finite groups and the general linear group, they can.
Nagata showed that for some groups they cannot, so the answer to Hilbert's question in its general form is negative.