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In his "Murphy's Law" paper, Vakil gives a definition equivalent to the following:

The singularity type of a pointed scheme $(X,p)$ its equivalence class, under the following equivalence relation: $(X,p) \sim (Y,q)$ if there exists a pointed scheme $(Z,r)$ admitting smooth morphisms $(Z,r) \to (X,p)$ and $(Z,r) \to (Y,q)$.

This relation is obviously reflexive and symmetric; to check transitivity, use the fact that pullbacks of smooth morphisms are smooth.

On its surface, this appears, at least to me, to have a different flavor from other ways in which singularities are studied. (I'm thinking in particular of the study of deformations of singularities, but I know very little about this subject, so my comment should not be taken too seriously.) Thus, I'd like to know what is known about the "singularity types" of this definition, other than the fact that all of them show up in each of a large number of moduli spaces.

I'm really interested in getting a general "flavor" of the theory, but since this is a question-and-answer site, here's a specific question:

Can infinitely many singularity types occur on a single scheme $X$ that is of finite type over a field?

Obviously, infinitely many singularity types show up in the moduli spaces for which Vakil proves Murphy's Law. But if I understand correctly, these spaces are only locally noetherian, so it is still possible that only finitely many singularity types show up on each connected component.

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Charles my first thought is that maybe the following family has a different singularity type at each singular point? $$ k[x,y,z,t]/\langle y^2z - x(x+z)(x+tz) \rangle. $$ The singular locus is a $\mathbb{A}^1$, it is just $V(x,y,z)$. The reason is that the singularities are changing at each point in this locus. In particular, if one blows up the singular locus $\pi : Y \to X$, the pre-image of each closed point on the singular locus is a different elliptic curve. I don't see how to prove that these really are different in the above sense right now though... – Karl Schwede Apr 29 '12 at 23:34
@Karl: Since blowups commute with smooth base change isn't what you said enough to conclude that you get infinitely many singularity types? To be more precise, for any singularity type $(X,p)$, the inverse image of $p$ in the blowup of $X$ at its singular locus is a well defined scheme, independent of the chosen representative. – ulrich Apr 30 '12 at 3:54
I'll have to think about this to be sure, but if ulrich's comment is correct, then ulrich and Karl seem to have produced a more-or-less computable invariant of "singularity types." – Charles Staats Apr 30 '12 at 22:21
You don't even need elliptic curves / the Whitney umbrella. As any book on singularity theorist will tell you, the cross-ratio of four concurrent lines in the plane gives a singularity invariant. Just consider the tangent cone of your singularity. – Jason Starr May 1 '12 at 12:10
Jason: Perhaps I'm being dense, but I don't see how your comment applies. In the example at hand, we are looking at the "singularity types" of points on a three-fold. Also, the tangent cone is itself not invariant under "singularity type;" if nothing else, taking the product with $\mathbb{A}^n$ always produces a singularity of the same type. – Charles Staats May 2 '12 at 1:26

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