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This question arose from the responses to this question. The references to the comments of Karl Schwede and VA are to comments made there.

The blow-up of the variety $X=\mathbb{A}^2$ along the closed subscheme $Z$ defined by $(x,y)^2$ is non-singular. As Karl Schwede points out, this example is trivial in the sense that the blow-up along the power of a maximal ideal is naturally isomorphic to the blow-up of the maximal ideal. VA's comment, on the other hand, suggests that perhaps singular closed schemes $Z$ with $\operatorname{Bl}_{Z}(X)$ non-singular are ubiquitous.

This suggests a question: what are non-trivial examples of a singular closed subscheme $Z$ of a non-singular variety $X$ with $\operatorname{Bl}_{Z}(X)$ non-singular. Here "non-trivial" means the ideal of $Z$ is not a power of the ideal of a non-singular subvariety.

Particularly interesting would be such a $Z$ such that

$\operatorname{Bl}_{Z}(X)$

is not isomorphic (as a scheme over $X$) to $\operatorname{Bl}_{Z'}(X)$ for any non-singular subvariety $Z'$ of $X$.

Edit: I have not been able to access the paper "On the smoothness of blow-ups" (MR1446135, by O'Carroll and Valla) yet, but the mathsci review states that they prove that the blow-up of a regular local ring $A$ along an ideal generated by a subset of a regular system of parameters is smooth. Let's also consider those examples to be trivial.

Edit: I added "of a non-singular variety" to the title to emphasize that I am interested in examples where the ambient space is non-singular.

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2 Answers 2

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Suppose that $X = \mathbb{A}^2$. Let $Y$ be the blow up of $X$ at the maximal ideal $(x,y)$ and let $W$ be the blow up of $Y$ at a point on the exceptional divisor of $Y$ over $X$. Of course, the composition $f: W \rightarrow X$ is birational and an isomorphism away from the origin. The fiber of $f$ over the origin is the union of two $\mathbb{P}^1$'s meeting at a single point, but the total space $W$ is non-singular. The map $f$ is identified with the blowup of $X$ along some closed subscheme $Z$ of $X$ supported only at the origin. I believe an example of an ideal defining such a $Z$ is $(x^3, xy, y^2)$.

By taking the composition of blowups along smooth centers, there is some ideal sheaf on the base giving the composition in "one step". In theory, you can compute this ideal by tracing through the proof that every birational morphisms is a blow up - but in practice I think this is usually difficult.

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    $\begingroup$ If you'll notice, $(x,y)(x^2,y) = (x^3, x^2y, xy, y^2)$ which is the integral closure of the ideal you mentioned $(x^3,xy, y^2)$. Generally speaking, we should probably limit ourselves to blowing up integrally closed ideals (as ideals have the same normalized blow-up as their integral closures). Finally, blowing up products of ideals is essentially like blowing up a series of ideals in succession. In this case, blowing up something like $(x,y)(x^2,y)$ is like blowing up the origin $(x,y)$ and then a point on the exceptional $P^1$ (this is related to "Zariski factorization") $\endgroup$
    – Karl Schwede
    Commented Apr 17, 2010 at 19:51
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    $\begingroup$ Bah, ignore the first sentence of my previous comment! The two ideals are already the SAME (no need to take the integral closure). In particular, if an ideal contains $xy$, then it also contains $x^2 y$. Sorry about that. $\endgroup$
    – Karl Schwede
    Commented Apr 17, 2010 at 22:46
  • $\begingroup$ @Karl, if you figure out what Zariski factorization says about this question generally, then I'd be interested to hear. $\endgroup$
    – jlk
    Commented Apr 19, 2010 at 1:29
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    $\begingroup$ Let I be an integrally closed ideal, and let $J_1$, ... $J_r$ be the irreducible complete ideals associated to the base points of I (ie, blow up each point on the cosupport of I and take the strict transform; iterate until I becomes trivial). I believe that the surface obtained blowing up I an integrally closed ideal is nonsingular if and only if each of the $J_i$ appears in the Zariski factorization of I. The proof should be in some paper by Spivakovsky, maybe MR1053487. $\endgroup$
    – quim
    Commented Apr 19, 2010 at 8:23
  • $\begingroup$ By the way, the surface singularities obtained blowing up integrally closed ideals are called "sandwiched," because they lie between two smooth surfaces, the resolution on top and the surface before blowup below. I don't know if "sandwiched" singularities of higher dimensional varieties have been studied, maybe a MathSciNet search is in order... $\endgroup$
    – quim
    Commented Apr 19, 2010 at 8:26
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For $X=\mathbb{A}^3$ you may take as $Z$ the three coordinate axes defined by $(x,y)(x,z)(y,z)$: The blowup $Bl_Z(X)$ of $X$ in $Z$ is nonsingular. However $Bl_Z(X)$ is isomorphic to a composition of blowups in smooth centers, namely, first blowup $X$ in the origin $(x,y,z)$, which separates the three coordinate axes and then you blow them up separately.

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  • $\begingroup$ Why is this true? It is seems to me that if we blow-up the three lines the fiber over the origin is $\mathbb{P}^3$, while if we first blow-up the origin and the three lines the fiber over the origin is $\mathbb{P}^2$ blown-up in three points. $\endgroup$
    – Puzzled
    Commented Sep 9, 2015 at 22:37

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