I'm interested in having especially simple embedded resolutions of a closed subscheme $Z \subset \mathbb{A}^{n}$ in characteristic zero. 


Hironaka gives the existence of a smooth variety $M$ and a proper map $p: M \rightarrow \mathbb{A}^{n}$ that is an isomorphism away from $D=p^{-1}(Z)$ and $D$ has simple normal crossings. 


I would like to be able to assume that the intersection of any two irreducible components of $D$ is either empty or connected. If two irreducible components did have disconnected intersection, then perhaps this could be achieved by blowing up $M$ along the connected components of the intersection.


Is this too much to ask for? 



Simple normal crossings divisor with connected intersections
Yes this is always possible. What you do is you start with a simple normal crossing resolution $M$. Let $D\_1, \ldots, D\_k$ be the irreducible components of the exceptional divisor. Then, by the simple normal crossing assumption, for any set of indices $I$ of size $n$, the intersection $\cap\_{i \in I} D\_i$ is a finite (possibly empty) set of points. Blow up all of these points as $I$ ranges over all sets of size $n$. Now take the strict transforms of the $D\_i$ and the $(n-1)$-way intersections of these will be a disjoint union of smooth curves, which we now blow-up. Repeat by blowing up the intersections of the strict transforms of the original divisors in higher and higher dimensions. The final result will have the property that the intersections of any set of divisors (not just pairs) will be either empty or connected. I learned this procedure from Remark 2.1 in [a paper by Sam Payne](http://arxiv.org/abs/1109.4286).


