Clarification on the definition of a quotient singularity I am working on the quotient construction of a simplicial toric variety as described in chapter 5 of this book. I have tried the following two examples -


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*The fan $\Delta$ in $\mathbb R^2$ consists of the cone $\sigma=\langle e_1,e_1+2e_2\rangle$ and its faces.

*The fan $\Delta'$ in $\mathbb R^2$ consists of the edge vectors $v_1=e_1,\ v_2=e_1+2e_2,\ v_3=-e_1+3e_2,\ v_4=-e_1-e_2$ and the cones generated by successive pairs.
For 1 the corresponding toric variety is $\mathbb C^2/G$ where $G=\{(t,t)\in(\mathbb C^*)^2\ | \ t^2=1\}\cong\mathbb Z_2$ and for 2 the corresponding toric variety is $(\mathbb C^4\setminus\mathcal Z)/G$ where $G=\{(t_2t_3^4,t_2,t_3,t_2^2t_3^3)\ |\ t_2,t_3\in\mathbb C^*\}$ and $\mathcal Z=(0\times0\times0\times\mathbb C) \cup (0\times0\times\mathbb C\times0)\cup(0\times\mathbb C\times0\times 0)\cup(\mathbb C\times0\times0\times0)$
Fulton's Introduction to Toric Varieties calls the variety in example 1 a cyclic quotient singularity.
Googling "quotient singularity" gave that it is the quotient of an affine variety $V$ by a finite group $G\subseteq \text{Aut }(V)$ and if $G=\mathbb Z/r$ then it is a cyclic quotient singularity. 
The second example however has $G\cong\mathbb{C^*\times C^*}$ which is not finite and $V=\mathbb C^2\setminus\mathcal Z$ is a quasi affine variety. Is there a similar name for such a toric variety $V/G$?
Also, in both examples the fan is simplicial and hence the associated toric variety has singular points. Does "singularity" refer to the fact that the toric variety has singular points? 
Thank you.
 A: The connection of toric varieties with quotient singularities is actually quite easy to describe. Let $\Delta\subseteq \mathbb R^n$ be a convex cone whose extremal rays are generated by $v_1,\ldots,v_d\in\mathbb Z^n$. We may assume that the $v_i$ are primitive. Let $X$ be the toric variety attached to $\Delta$. If $d$ is bigger than the dimension of $\Delta$ then $X$ is very singular in the closed orbit (e.g., the divisor class group is infinite). On the other side, if the $v_i$ are linearly independent then the singularieties are very mild, namely abelian quotient singularities.
This can be sees as follows. Assume for simplicity that $d=n$, i.e., that the $v_i$ constitute a basis. Let $\Lambda=\langle v_1,\ldots,v_n\rangle_{\mathbb Z}$. That is a subgroup of $\mathbb Z^n$ of finite index. The torus to be embedded is $T=(\mathbb C^*)^n$ which can also be written as $T=\mathbb C^n/\mathbb Z^n$. Now take the larger torus $\tilde T=\mathbb C^n/\Lambda$. Then there is a surjective homomorphism $\phi:\tilde T\to T$ with kernel $K:=\mathbb Z^n/\Lambda$.
The cone $\Delta$ induces also an embedding $\tilde X$ of $\tilde T$ which this time is isomorphic to $\mathbb C^n$ because the $v_i$ do from a $\mathbb Z$-basis of $\Lambda$. Then $\phi$ extends to a map $\tilde X\to X$ which is just the quotient map by $K$. Thus we realized $X$ as a quotient of the vector space $\tilde X$ by a finite abelian group $K$, i.e. $X\cong\mathbb C^n/K$. Since the $v_i$ were chosen to be primitive one can show that $\mathbb C^n/K$ is singular unless $K=1$.
The group $K$ is easy to compute: form the $n\times n$-matrix $M=[v_1,\ldots,v_n]$ and compute its elementary divisors $d_1,\ldots,d_n$. Then $$K\cong\mathbb Z/d_1\mathbb Z\times\ldots\times\mathbb Z/d_n\mathbb Z.$$
For $n=2$ this is  particularly easy: Again because the $v_i$ are primitive we have $d_1=1$. On the other hand $d_1\ldots d_n=\det M$. So $d_2=\det M$ if $n=2$. Using this one quckly sees that $K$ in the four fixed points of your second example is cyclic of order $2$, $5$, $3$, and $1$. So the last fixed point is actually not a singularity.
One final remark: The isomorphism type of $K$ does not suffice to completely determine the isomrophism type of the singularity. For that one also needs to know how $K$ acts on $\mathbb C^n$.
