# commuting the resolution of 1-dim singular locus and 0-dim singularities in a non isolated singularity of a surface

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

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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.

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Yes, sorry about that. –  Karl Schwede Mar 22 '12 at 11:28
What exactly do you mean by "solve the $0$-dimensional singularities"? If you mean
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.)