This is exercise 1.10 from Reid's Young person's guide to canonical singularites.


Let $X=\mathbb{C}^3/ \mu\_3$ where $\epsilon \in \mu\_3$ acts by
$$ (x,y,z) \to (\epsilon x, \epsilon y, \epsilon^2 z).$$
Then blowing up the origin gives us $E\_1 \cup E\_2$ where $E\_1$ is a plane and $E\_2$ is a quartic scroll.


We know $X=\text{Spec}(A)$ where
$$A = \mathbb{C}[x^3, x^2 y, x y^2, y^3, xz , yz, z^3]= \mathbb{C}[u\_0, u\_1, u\_2, u\_3, v\_0, v\_1, w]/I$$
$$I=\langle u\_0u\_3-u\_1 u\_2, u\_0u\_2-u\_1^2, u\_1u\_3-u\_2^2, u\_2 v\_1- u\_3 v\_0, u\_1 v\_1 - u\_2 v\_0, u\_0 v\_1- u\_1 v\_0, \\ v\_0^3 - u\_0 w, v\_0^2 v\_1 - u\_1 w, v\_0 v\_1^2 - u\_2 w, v\_1^3 - u\_3 w\rangle$$


If I consider the map
$$(s, t, p) \to (s, st , st^2, st^3, sp, stp, s^2p^3),$$
it seems that $s=0$ corresponds to $E\_1$ and the pullback of
$$s = \frac{1}{u\_0^5}(du\_0 \wedge du\_1 \wedge du\_2)^{\otimes 3 }$$
has zeros of order 1 along $E\_1$.


I'm not sure how to get $E\_2$. If I blow up the origin of $\mathbb{C}[u\_0, u\_1, u\_2, u\_3, v\_0, v\_1, w],$ how can I identify $\mathbb{P}^6$ with $E\_1 \cup E\_2$? Why the pullback of $s$ has has zeros of order 2 along $E\_2$.



Resolution of 3-fold quotient singularities
The quotients $\mathbf{C}^n / G$ with $G$ finite abelian group (acting linearly) are toric varieties. I present the toric description of the resolution and the discrepancies. If one needs to, one could get differential forms and coordinates from this description. The reference [CLS] is the book "Toric Varieties" by Cox, Little, Schenck.


**The $\frac1{n}(1,a,b)$ singularity.** Let us consider $n = 3$ and $G = \mathbf{Z}/n$ acting on $\mathbf{C}^3$ by $$(x,y,z) \mapsto (\zeta x, \zeta^a y, \zeta^b z)$$ where $\zeta$ is the primitive $n$-th root of unity and $a$, $b$ are coprime to $n$. Let $X = \mathbf{C}^3 / G$. This is the $\frac1{n}(1,a,b)$ singularity. By [CLS, Exercise 11.4.8] it is an affine toric variety with the fan generated by the cone spanned by vectors
$$
v\_1 = (1,0,0), \; v\_2 = (0,1,0), \; v\_3 = (-a,-b,n) \in \mathbf{Z}^3.
$$
The determinant of the resulting $3 \times 3$ matrix equals $n$. To resolve singularities of this toric variety, one needs to perform some star-subdivisions adding new rays so that the resulting determinants are equal to one.


**Resolution of $\frac13(1,1,2)$.** We take $n = 3$, $a = 1$, $b = 2$, in which case we can first add the vector
$
v\_4 = (0,0,1)
$
which subdivides the cone into three cones, with $v\_1, v\_2, v\_4$ and $v\_2, v\_3, v\_4$ smooth, but
$
v\_1, v\_3, v\_4
$
still singular as the corresponding determinant equals $2$. One subdivides this cone by adding $v\_5 = (0,-1,2)$, and we obtain five smooth cones. This describes the toric resolution $\pi: \widetilde{X} \to X$.


The new rays $v\_4$, $v\_5$ give two exceptional divisors and using [CLS, Proposition 3.2.7] one checks that one of them is isomorphic to $\mathbf{P}^2$, and the other is isomorphic to $\mathbf{F}\_2$ (the Hirzebruch surface); they intersect along a $\mathbf{P}^1$ (which is a line on $\mathbf{P}^2$ and the negative section on $\mathbf{F}\_2$). Geometrically $\widetilde{X}$ is obtained as a composition of two blow ups of $X$: the first is a weighted blow up of the origin, and the second is the usual blow up of the resulting singular point. At this point I am confused why Miles Reid says that blowing up the singular point of $X$ gives resolution of singularities as blowing up a torus-invariant point adds just one divisor.


**Discrepancies.** To compute the discrepancies, one needs to express each $v\_4$, $v\_5$ as
$$
\alpha\_1 v\_1 + \alpha\_2 v\_2 + \alpha\_3 v\_3,
$$
and the corresponding discrepancy equals $\alpha\_1 + \alpha\_2 + \alpha\_3 - 1$ [CLS, Lemma 11.4.10].
Using this one computes that
$$
K\_{\widetilde{X}} = \pi^\*(K\_X) + \frac13 E\_4 + \frac23 E\_5.
$$


In particular, the singularity $\frac13(1,1,2)$ is terminal.


**Terminal toric singularities.** One can use this method to verify that $\frac12(1,1,1)$ is terminal, $\frac13(1,1,1)$ is canonical non-terminal, and all $\frac1{n}(1,1,1)$ are not canonical for $n > 3$; this is very easy as only one ray will be added, and its discrepancy is $\frac{3}{n} - 1$. See [CLS, Proposition 11.4.12 and Theorem 11.4.21] for a description and classification of terminal threefolds. The upshot is that toric terminal threefolds are $\frac1{n}(1,a,-a)$ and $xy = zw$. Much of this actually goes back to Miles Reid.


**Credits**: this question was discussed as an exercise in toric geometry in the Sheffield Algebraic Geometry learning seminar.


