Let $k$ be an algebraically closed field of characteristic zero and let $X$ be a toric variety over $k$, i.e. $X$ is a normal, irreducible $k$-variety and it admits an algebraic action of a torus with an open orbit.

Now comes the question: If $X$ is quasi-affine and smooth, is the $k$-algebra $k[X]$ of regular functions on $X$ always finitely generated (over $k$)? Is this true for arbitrary toric $X$?

I know, that in general (i.e. when $X$ is not necessarily toric), then $k[X]$ can be non-finitely generated, see e.g. the article of Winkelmann [Win03].

Any proof, counter-example or reference to my question would be very welcome!

[Win03] Winkelmann, J\"org, Invariant rings and quasiaffine quotients, Math. Z. 244, No.1, 163-174 (2003). ZBL1019.13003.

  • 2
    $\begingroup$ If $X$ is affine, corresponding to a rational cone $\sigma$ in $N_{\mathbf{R}}$, then $\Gamma(X, \mathcal{O}_X)$ is the semigroup algebra of the monoid $P_\sigma = \sigma^\vee\cap M$ ($M=N^\vee$) which is of course finitely generated. It is a subalgebra of the group algebra of $M$, which is the character group of the torus. A general toric $X$ corresponds to a fan $\Sigma$ which is a collection of such cones $\sigma$, and $\Gamma(X, \mathcal{O}_X)$ is the semigroup algebra of the intersection of all $P_\sigma$, $\sigma\in\Sigma$ inside $M$, which is finitely generated for the same reasons. $\endgroup$ – Piotr Achinger Apr 28 '17 at 15:18
  • $\begingroup$ @PiotrAchinger. You beat me to it. I will post my answer anyway in a moment. $\endgroup$ – Jason Starr Apr 28 '17 at 15:19

One can show that for any normal spherical variety $X$ the algebra $k[X]$ is finitely generated. No quasi-affinity or smoothness needed. This holds more generally for $G$-varieties ($G$ connected reductive) where $G$ has an open orbit and a Borel subgroup has an orbit of codimension $\le1$. The latter condition is optimal. See the paper Knop, F: Über Hilberts vierzehntes Problem für Varietäten mit Kompliziertheit eins, Mathematische Zeitschrift 213 (1993) 33-35.


You are essentially asking whether or not the intersection in $\mathbb{Z}^n$ of finitely many finitely generated subsemigroups is again finitely generated. That is true. By induction on the number of semigroups, it suffices to consider the case of two semigroups, $\Gamma,\Delta.$ For surjections of finitely generated semigroups, $$\gamma:\mathbb{Z}_{\geq 0}^m \to \Gamma, \ \ \delta:\mathbb{Z}_{\geq 0}^\ell \to \Delta,$$ there is an associated homomorphism, $$(\gamma,-\delta):\mathbb{Z}^m\times \mathbb{Z}^\ell \to \mathbb{Z}^n.$$ The kernel of $(\gamma,-\delta)$ is finitely generated, i.e., there exists a surjection, $$\epsilon:\mathbb{Z}^k\to \text{Ker}(\gamma,-\delta).$$ For each coordinate projection $$\pi_i:\mathbb{Z}^{m+\ell}\to \mathbb{Z},$$ consider the half-space $$H_i=\{v\in \mathbb{Z}^k : \pi_i(\epsilon(v))\geq 0\}.$$ The finite intersection of these half-spaces is a finitely generated subsemigroup of $\mathbb{Z}^k$ that surjects under $\epsilon$ to the intersection of $\Gamma$ and $\Delta$. Apparently it is a difficult computational problem to pass from the finitely many half-spaces $H_i$ to generators of the intersection, cf. Algorithm for the intersection of a vector subspace with a cone of non-negative vectors

  • $\begingroup$ Hi, why is the finite intersection of half-spaces a finitely generated semigroup? $\endgroup$ – user2831784 Jan 23 '19 at 5:24
  • $\begingroup$ @user2831784 There is an algorithm for producing the finite list of generators in the link in my answer. $\endgroup$ – Jason Starr Jan 23 '19 at 14:31

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