TITLE: When is a blow-up non-singular?
QUESTION [21 upvotes]: Suppose that $X$ is a non-singular variety and $Z \subset X$ is a closed subscheme.  When is the
blow-up $\operatorname{Bl}_{Z}(X)$ non-singular?  
The blow-up of a non-singular variety along a non-singular subvariety is well-known to be non-singular, so the real question is ``what happens when $Z$ is singular?"  The blow-up can be singular as the case when $X = \mathbb{A}^{2}$and $Z$ is defined by the ideal $(x^2, y)$ shows. On the other hand, the example where $Z$ is defined by the ideal $(x,y)^2$ shows that the blow-up can be non-singular.
Edit:  Based on the comments of Karl Schwede and VA, I think that it would also be interesting to find non-trivial examples of appropriate $Z$'s.  I am splitting this off as a  separate question.  In the comments there, the users quim and Karl Schwede say a bit about what can be said about this question using Zariski factorization.

REPLY [2 votes]: A topological remark:
If $E \subset Y$ is a closed analytic subspace of a smooth space $X$, then the boundary
of a tubular neighbourhood is an (odd dimensional, real) sphere bundle over $E$.
Thus if the blowup of $Z \subset X$ is smooth then the boundary of a tubular 
neighbourhood of $Z \subset X$ can be written as a sphere bundle over the exceptional 
divisor $E = P(N_Z X)$.
One might use this to arrive at a contradiction e.g. by calculating cohomology.
Similar ideas were used by Mumford in his study of normal surface singularities.
Mumford: The topology of normal singularities of an algebraic surface and a criterion for simplicity.