My interest is in the "finished product", the desingularization $\tilde{X}$ of a variety (or scheme) $X$. It is widely described how to get to the different resolved charts but not how they glue together.
I think I understood that if I find an isomorphism between two exceptional divisors, I can glue the charts via this isomorphism. I just do not understand how I could find this isomorphism, or how I know they are the same. I'm working in the embedded case. My understanding so far:
Case 1: Take the variety $X=V(x^2-y^3)$. Wikipedia shows an image of the non-resolved charts and one final chart. That's enough because we don't have any new information in the other charts about the exceptional divisors.
Case 2: Say, we have the variety $V(x^2+y^2-z^3)$. Blow it up to get chart 1 and 2 (both resolved) and chart 3 with exceptional divisor $D$ from the first blowup. Get charts 3.1, 3.2 and 3.3 (3.3 resolved) with exceptional divisors $D$ and $E$ in both non-resolved charts from blowing up chart 3. Now we blow both of these up in their center $D \cap E$. Since the blowup is determined by its center, we know that the resulting divisors must be isomorphic, although I don't know how to compute or glue that explicitly.
Case 3 We have a variety where the same exceptional divisor appears in different blowups, i. e. there are two charts with isomorphic centers, without it being so "obvious". We could use the previous example, assuming that we don't see that it's both $D \cap E$. I have done the following calculations for attaining the charts:
Variety $V(x^2+y^2-z^3)$. Use variables $(x,y,z,u:v:w)$, blowing up in $(0,0,0)$.
Get chart 3 $(w=1)$ via $y=zv, x=zu$:$V(u^2+v^2-z)$ and $D=V(z)$. Next, use variables $(u,v,z,r:s:t)$, blowing up in $(0,0,0)$.
Get chart 3.1 $(r=1)$ via $v=us,z=ut$: $V(u+us^2-t)$ and $\tilde{D}=V(t)$, $E=V(u)$ and center $V(u,t)$.
Get chart 3.2 $(s=1)$ via $z=vt, u=vr$: $V(vr^2+v-t)$ and $\tilde{D}=V(t)$ and $E=V(v)$ with center $V(t,v)$ for the following blowup.
I appreciate any help!
Direct calculation - is that correct and sufficient?
Let $U_{31}$ be the affine chart 3.1 with exceptional divisors $D_{31}$ and $E_{31}$ and coordinates $(u,v,z,s,t)$, same with 3 and 3.2.
$$\pi_{31}: U_{31} \rightarrow U_{3}: (u,v,z,s,t) = (u,us,ut,s,t) \mapsto (u,s,t)$$
which gives us the following map for the center $C_{31}=V(u,t)$:
$$\pi_{31}(C_{31})=\pi_{31}(V(u,t))=\pi_{31}(\{(0,v,z,s,0)\})=\{(0,r,0)\}$$
as well as, for the other chart,
$$\pi_{32}: U_{32} \rightarrow U_{3}: (u,v,z,r,t) = (vr,v,vt,r,t) \mapsto (v,r,t), C_{32}=\{(0,v,z,s,0)\} \mapsto \{(0,s,0)\}$$
Now, $\{(0,r,0)\}$ and $\{(0,s,0)\}$ are isomorphic in $U_{3}$. But that's the same for points that are not the same under isomorphism?