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Let $G \subset SL(n,\mathbb{C})$ be a cyclic subgroup of finite order, Is it true that $\mathbb{C}^n /G$ is toric ? If not then when it is ?

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The "cyclic" condition is redundant, the quotient is toric for any finite Abelian group. – Dmitri Feb 26 '12 at 0:21
I kind of knew that but I just simplified the problem. I could not remember any proof or procedure. How is the the construction? – Mohammad F. Tehrani Feb 26 '12 at 0:34
The action of any finite Abelian group on $\mathbb C^n$ is diagonalisable - this is just linear algebra. So the action commutes with diagonal $(\mathbb C^*)^n$, this makes your quotient toric. – Dmitri Feb 26 '12 at 1:26
up vote 6 down vote accepted

Yes, it is true. Let $G \cong \mathbb{Z}/m$ act by $\mathrm{diag}(\zeta^{a_1}, \zeta^{a_2}, \ldots, \zeta^{a_n})$, where $\zeta$ is a primitive $n$-th root of unity. Let $S$ be the semigroup $\{ (b_1, \ldots, b_n) \in \mathbb{Z}_{\geq 0}^n : \sum a_i b_i \equiv 0 \mod m \}$. Then the quotient is Spec of the semigroup ring $\mathbb{C}[S]$. Since the semigroup is torsion free, finitely generated and saturated, the corresponding Spec is an affine toric variety.

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