I've been working on some varieties defined by taking some quotients of group actions, and the resolutions have been straightforward... until now.


E.g., consider $\mathbb{C}^2$ with the action $(x,y)\mapsto(-x,-y)$. Above the origin you get a fixed $\mathbb{P}^1$, and everything is resolved. Consider instead, the action $(x,y)\mapsto (-x,iy)$ and above the origin you get a fixed point with induced action $(x,y)\mapsto (-x,-iy)$. Blowing up once more gives a fixed point with the original action $(x,y)\mapsto(-x,iy)$ again, and this would continue ad infinitum.


The problem is presumably that instead of blowing up the quotient itself, I am blowing up $\mathbb{C}^2$ and lifting the action, hoping it lifts well. So, to blow up the quotient right away, we look at
$$ \operatorname{Spec} \mathbb{C}[x^2,y^4,xy^2] \simeq \operatorname{Spec} \mathbb{C}[u,v,w]/(w^2 - uv) $$
and this may not be so hard to resolve, but I'd like to do higher dimensional examples as well, for instance,
$$ \operatorname{Spec} \mathbb{C}[x^2,y^4,z^4,xy^2,xz^2,y^2z^2] \simeq \operatorname{Spec} \mathbb{C}[u,v,w,s,t]/(s^2 - uv,t^2 - uw) $$
and resolving this (and other messier examples) looks unpleasant, so I'd love to avoid this if possible.


I've got the Hodge numbers from orbifold cohomology, and I've seen Hizebruch-Jung resolutions on surfaces, though the higher dimensional case seem to, again, be more messy, so I was hoping to avoid a lot of technical overhead if a simpler, more elementary method exists.



Resolution of unpleasant singularity
Note that $-1 = i^2$. Thus your action is $(x,y)\mapsto (i^2x,iy)$. You can interpret this in the following way. Let $\epsilon\in \mu\_4$ and consider tha action given by $(x,y)\mapsto (\epsilon^2 x,\epsilon y)$. The quotient $X = \mathbb{C}^2/\mu\_4$ is given by $X = (w^2-uv = 0)\subset\mathbb{A}^3$. This is because as you observed the invariant polynomials are $x^2,y^4,xy^2$.


Now, the singularity of $X$ at the origin is a cyclic quotient singularity of type $\frac{1}{4}(1,2) = \frac{1}{2}(1,1)$. That is a $A\_1$ singularity. In other words $X$ is a quadric cone and you can solve the singularity at the vertex $v\in X$ just by blowing-up $v$. You get a smooth surface $Y=Bl\_vX\rightarrow X$ and the exceptional divisor $E$ over $v$ is a $(-2)$-curve. That is $E\cong\mathbb{P}^1$ and $E^2 = -2$.


You can see it in another way. Consider the action of $\mathbb{C}^{\*}$ on $\mathbb{C}^{3}\setminus\{0\}$ given by 
$$\lambda (x,y,z)\mapsto (\lambda x,\lambda^2 y, \lambda^4 z).$$
The quotient is the weighted projective plane $\mathbb{P}(1,2,4)$. The singularity at $[0:0:1]\in\mathbb{P}(1,2,4)$ is exactly the singularity you are looking at. Now, the second truncation of the polynomial ring $\mathbb{C}[x,y,z]$ is $\mathbb{C}[x^2,y,z]$ and $\mathbb{P}(1,2,4)$ is isomorphic to $\mathbb{P}(1,2/2,4/2) = \mathbb{P}(1,1,2)$. Finally, you see that your singularity is the one given by the action $(x,y)\mapsto (-x,-y)$, and that $\mathbb{P}(1,1,2)$ can be embedded in $\mathbb{P}^3$ as a quadric cone.


The second example you are looking at is a complete intersection of two quadrics in $\mathbb{P}^4$. Both the quadrics are singular along double lines $L,R$. Therefore the surface $X$ has an ordinary singularity of multiplicity $4$ at the point $p = L\cap R$. Since $p$ is an ordinary singularity you can resolve it just by blowing-up once.


