Let $k$ be an algebraically closed field. Let $\sigma$ be a 3-dimensional simplicial cone in $\mathbb{R}^3$ and let $\rho$ be a ray in $\sigma$. Let $U_{\rho}$ be defined as $\operatorname{Spec}(k[\rho^{\vee}\cap\mathbb{Z}^3])$. Is $U_{\rho}$ always a smooth toric variety? In the special case where $\sigma$ is the first octant and $\rho$ is the non-negative z-axis, we have $U_{\rho}$ isomorphic to $\operatorname{Spec}(k[x,x^{-1},y,y^{-1},z])$, so that $U_{\rho}$ is smooth in this case, since $k[x,x^{-1},y,y^{-1},z])$ is a regular ring and the field $k$ is algebraically closed.
2
-
1$\begingroup$ I don’t see the relevance of $\sigma$. Forgetting $\sigma$, the “special case” you discuss is not so special: there exists an automorphism of $\mathbb{Z}^3$ transforming $\rho$ into one of the axes. $\endgroup$– Piotr AchingerCommented Dec 21, 2023 at 7:30
-
$\begingroup$ @PiotrAchinger Thank you very much for your kind help. Could you explain how to construct this automorphism of $\mathbb{Z}^3$? Thank you very much. $\endgroup$– BorisCommented Dec 21, 2023 at 14:24
Add a comment
|
2 Answers
$\begingroup$
$\endgroup$
Anna is basically correct, except that $ U_{\sigma} $ is smooth if and only if the set of its minimal ray generators form a subset of a $ \mathbb{Z} $-basis.
Therefore, since $ \mathbb{Z}u_{\rho} $ is linearly independent the variety $ U_{\rho} $ is smooth.
$\begingroup$
$\endgroup$
$U_\sigma$ is smooth if and only if the generating set of $\sigma$, that is the minimal $\{u_\rho| \rho \in \sigma(1)\}$ is a $\mathbb{Z}$-basis for $N$, the lattice you are considering (which seems like here is $\mathbb{Z}^3$). You can find all the details you need on this in Cox, Schenck and Little's book "toric varieties".
