Let $X$ be a surface with a non isolated singularity $C = Sing(X)$ such that the curve $C$ has singularities itself. We can solve $Sing(X)$ by blowing up close points and by normalizing. Indeed, we can first solve the 1-dimensional singularities with only normalizations $\pi_1: X_1 \to X$ . In the surface $X_1$ the preimage of $C$ has at most isolated singularities which we can solve by $\pi_2:X_2 \to X_1$.

Another approach is to solve the 0-dimensional singularities by $\varphi_1:Y_1 \to X$, and then to finish with a one dimensional singular locus on $Y_1$ that we can be solve by normalization $\varphi_2:Y_2 \to Y_1$ such that $Y_2$ is smooth (Maybe some ADE singularities, but it would not matter).

I am wondering if the second approach is always possible, and if "commutes" with the first one in some way. I have this idea that normalization remove the 1-dimensional singularity $C$ without affecting the 0-dimensional ones, even if they are supported on $C$. Is this true? In that fantasy, we can "commute" those processes of solving 0-dimensional singularities, and solving 1-dimensional singularities.

I will appreciate any enlighting


2 Answers 2


Normalization can definitely change isolated singularities as we well.

For example, $k[x,y,u,v]/(x,y)\cap(u,v)$ has an isolated singularity (this is not irreducible, but there are irreducible versions with analytically isomorphic singularities). However, it is not normal and the normalization is smooth.

On the other hand, if you first resolve the 0-dimensional singular locus (ie, the singular closed points) by blowing up and then do a normalization, you can have much worse than ADE singularities. In fact, you can have a surface with no isolated singularities whose normalization has any possible isolated singularity.

However, something like your thought should be possible I think, you just might have to do blow-ups of points over the 1-dimensional singular locus as well.

For example, see the notion of semiresolution especially in the works of János Kollár.

  • $\begingroup$ Yes, sorry about that. $\endgroup$ Mar 22, 2012 at 11:28

What exactly do you mean by "solve the $0$-dimensional singularities"? If you mean

  • to resolve the isolated locus only, then you get nowhere as you can have arbitrarily complicated singularities in codimension two that hide inside a mild singular locus.

  • to partially resolve the locus where the singularity is such that normalization will not resolve it. This might work (modulo what Karl remarked), but how do you identify these points without knowing the resolution or the normalization?
    Here is an example to contemplate: Let $X=Z(xy^2-z^2)\subset \mathbb A^3$. This is normal crossing away from the origin, but a little worse than that at the origin. It turns out that the normalization of this is smooth, but for now suppose we don't know that. (My point is that you don't necessarily know that the normalization is smooth just by looking at the definition.)
    If you try to partially resolve the point where it is not a normal crossing singularity, then you are in for a pretty long computation. Do try to compute the blow up of this at the origin, it is instructional. You should get a few new singular points and on one chart get back a singularity that's locally isomorphic to this one. In other words, you can't make this point "better" without normalization.
    So, the solution seems to be, as Karl already suggested, to look at semiresolutions and János Kollár's work is indeed a good place to start.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.