Is the quotient of a toric variety by a finite group still toric

I have asked this question on Math.StackExchange, but haven't got any reply.

Suppose $X$ is a toric variety with fan $\Sigma$, and the lattice of one-parameter subgroups of its torus is $N$. Suppose $\phi$ is a representation of finite group $G$ in $N \otimes_{\mathbb{Z}} \mathbb{R}$, $$\phi:G \rightarrow \text{GL}( N \otimes_{\mathbb{Z}} \mathbb{R})$$ which sends the lattice $N$ to $N$, and the cones of $\Sigma$ to cones of $\Sigma$, therefore each $\phi(g)$ defines a toric automorphism of $X$.

Question 1: Is the quotient variety $X/G$ (which exists from GIT) still a toric variety? Is the quotient morphism $X \rightarrow X/G$ toric?

Question 2: If so, how to construct the fan of $X/G$ and the toric morphism $X \rightarrow X/G$?

Any references on the two questions?

• The quotient morphism is certainly never toric unless $\phi$ is trivial. Toric morphisms between toric varieties are those that commute with the torus action, while $\phi$ obviously doesn't. If $X/G$ is toric, it may be toric in more than one way, so there might be no construction of the toric fan. – Will Sawin Apr 10 '18 at 12:28
• @WillSawin Sorry for bothering you with a question again after 10 days. Since $X$ is toric, it's defined over $\mathbb{Q}$. Even the quotient $X/G$ may not be toric, I guess it is still a variety defined over $\mathbb{Q}$? But I don't know how to prove this! – Wenzhe Apr 20 '18 at 11:56
• Take a $G$-invariant ample line bundle. This is also defined over $\mathbb Q$. Then the quotient is the Proj of the ring of $G$-invariant sections. Because everything in sight is defined over $\mathbb Q$, this ring is defined over $\mathbb Q$. – Will Sawin Apr 20 '18 at 13:49

No. If $X$ is the 47-dimensional torus and $G=\mathbb{Z}/47\mathbb{Z}$ acting by the permutation representation, then $X/G$ is not a rational variety (R. Swan, Inv. math. 7, 148-158 (1969)), therefore not toric.