# When is a blow-up non-singular?

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.

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Craig Huneke told me about this paper: "On the smoothness of blow-ups" (MR1446135, by O'Carroll and Valla). The title alone seems to suggest it might be useful for you.

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@Hailong Dao, Thanks! My previous comment summarizing the paper was inaccurate. The authors prove that if $Z$ is a complete intersection, then the blow-up along $Z$ is non-singular iff $Z$ is non-singular (Thm 2.1). They give similar results for other special cases. For example, they prove that the blow-up of $A^3$ along a non-degenerate monomial curve is singular (Cor. 3.3). They also have results about blowing up determinantal loci. – jlk Apr 27 '10 at 16:10

There is no general criterion, as far as I know, it is all try and see.

Any projective birational morphism $f:X\to Y$ between varieties is the blowup of some sheaf of ideals $I$ on $Y$, so you can see that anything can happen.

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Although there is a good reason that $(x,y)^2$ has a smooth blow-up. It is a power of an ideal which itself has a smooth blowup. See for example Hartshorne, Algebraic Geometry, Chapter II, Section 7, Exercise 7.11.

I suspect that, on smooth surfaces, one can probably say more, via "Zariski-factorization" type ideas, but I'm not sure what the right answer would be.

Edit: I've looked around for a good reference on "Zariski-Factorization", but I'm not sure what a good one is. Does someone know?

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I remember reading about them in Dale's "Complete ideals in algebra and geometry" paper. – Hailong Dao Apr 16 '10 at 16:54

For monomial ideals there is a combinatorial smoothness criterion, see "Blowups in tame monomial ideals" http://arxiv.org/abs/0905.4511

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For the case of singular plane curves though $C\subset \mathbb{P}^2$, the result is well known. The blowup of the projective plane itself is non singular yeah, but what about the image of the curve $C$ in the blowup? That is, the proper transform of $C$? Is this proper transform $\tilde{C}$ smooth?. Such an information lies in the intersection of $\tilde{C}\cap E$, where $E$ is the exceptional divisor of the blowup. Basically, if this intersection is still singular, you will need to blowup again up to the point of getting smooth point within such intersections. This is a finite process due to the fact that every single time you blowup the curve $C$, the arithmetic genus drops by one. Since there is no negative arithmetic genus, that means the process needs to finish eventually. Notice though, that the (potential) smooth curve $\tilde{C}$ no longer live in the projective plane. You can send it back to $\mathbb{P}^2$ though, but the price to pay is that you can get singular points again. These singular points though, sometimes are milder than the original ones. But then we have got something after all!. This can be understood in terms of Cremona transformations of the plane. Here is an example of this phenomena:

$$X^2Y^2+X^2+Y^2=0$$ is singular at $[0:0:1]$. Apply the cremona transformation (which is a rational map from $\mathbb{P}^2$ to itself) $$\phi:[x:y:z]\mapsto [\tfrac{1}{x}:\tfrac{1}{y}:\tfrac{1}{z}]$$. Under such a map the curve $C$ becomes the conic $Z^2+X^2+Y^2=0$ which is indeed smooth. According to another answer above, roughly speaking, this rational map corresponds to blowup $\mathbb{P}^2$ at $[1:0:0],[0:1:0],[0:0:1]$. That's why under $\phi$ the singular curve $C$ maps to a smooth one.

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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.

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