Resolution of unpleasant singularity 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.
 A: 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.
A: The problem in its generality does not look easy.
In the two-dimensional case, however, everything is known. 
Let us consider your action of $\mathbb{Z}/4 \mathbb{Z}$ over $\mathbb{C}^2$ given by $(x, \, y ) \mapsto (-x, \; iy)$.
Using the notations of [Barth-Peters-Van de Ven, Compact Complex Surfaces (1984)], Chapter III, Section 5, page 84,  the numerical data attached to this cyclic action  are $$n=4, \quad q_1=2, \quad q_2 = 1.$$
In other words, you are acting by multiplication with a primitive fourth root of unity on the second coordinate and with its square on the first coordinate. Then, looking further in the same page, with easy numerical manipulations one computes the integers $m, \, q$ finding $m=2, \, q=1$. 
Now Proposition 5.3 implies that your singularity is of type $A_{m, \, q}=A_{2, \, 1}$, in other words it is equivalent to the (normalization of the) hypersurface singularity of $\mathbb{C}^3$ defined by the equation $z^m =w_1w_2^{m-q}$, that is $z^2 =w_1w_2$ (see page 81).  
But this is simply a quadric cone, hence the singularity is an ordinary double point, i.e. it is isomorphic to the quotient of $\mathbb{C}^2$ by the involution $(x, \, y) \mapsto (-x, \, -y)$. It can be resolved with a single blow-up and the exceptional divisor is a unique $(-2)$-curve.
