What does being Analytically Isomorphic imply for classification of singularities on curves? - MathOverflow

What does being Analytically Isomorphic imply for classification of singularities on curves?

2010-03-18T21:55:06Z

Hartshorne I.5 mentions the definition of being analytically isomorphic: P on X and Q on Y are analytically isomorphic iff the completion of O_P is isomorphic to the completion of O_Q where the completion is according to their maximal ideal in that local field.

For any pair of regular points, they are always analytically isomorphic as long as they are of the same dimension according to Cohen Structure Theorem.

But for singularities, they might be different.

My question is: Is classification of the completion of O_P a first step to classify all singularities(on curves in particular)? What are known results about that classification? What does that imply if the classification of rings after completion was complete? WHAT IS THE CORRECT WAY TO CLASSIFY SINGULARITIES?

Answer by Charles Siegel for What does being Analytically Isomorphic imply for classification of singularities on curves?

2010-03-18T22:07:06Z

It can be seen as a first step in a classification of singularities. In particular, looking at the completions there can give you quite a lot of information about the singularity itself, and is comparable to looking in a small open set on a complex variety. For instance, I believe that problem involves a classification of all double points of plane curves, and you can use the isomorphism type of the completion to tell things like how many branches the curve has at that point, are they tangent, to what order, etc.

Answer by quim for What does being Analytically Isomorphic imply for classification of singularities on curves?

2010-03-22T10:48:35Z

Analytically equivalent (over C) implies, for instance, topologically equivalent. All information items Charles Siegel mentions (number of branches, tangencies, etc, and also the classification of double points) are topological invariants so in particular they are analytic invariants and can be read off the completion indeed.

I would say that the first step to classify singularities is equisingularity, which over C is the same as topological classification. The analytic classification of is extremely complicated, even for plane curves. For each topological class there is an analytic moduli space, but it need not be irreducible, equidimensional, or even separated.

For plane curve singularities, there are a few cases where analytically equivalent and topologically equivalent are the same thing. These are the so-called simple, or Du Val, singularities, namely "A-singularities" or double points, whose equations have the form $y²-x^n=0$, "D-singularities" or $x(y^2-x^n)=0$ and $E_6:y^3-x^4=0$, $E_7:y(y^2-x^3)=0$, $E_8:y^3-x^5=0$.

A good old reference for the analytic classification of plane curves is Zariski's booklet ''Le problème des modules pour les branches planes.'' As introduction to the theory of plane curve singularities you have Casas-Alvero ''Singularities of plane curves'', and Laudal-Pfister's LNM ''Local moduli and singularities'' is also recommendable.

Answer by Karl Schwede for What does being Analytically Isomorphic imply for classification of singularities on curves?

2011-02-05T19:02:29Z

Let me first give a quick example of why it is reasonable to consider only the completion and then suggest a reasonable way to classify curve singularities.

For example, fix a smooth curve $C$ and two closed points $p, q \in C$ and fix $Z = \text{Spec} \mathbb{C}$ and consider the obvious map ${p, q} \to Z$.

Then, consider the pushout {$ C \leftarrow {p, q} \rightarrow Z $}, which does exist in the category of schemes in this case. This pushout will always create a node singularity -- the points $p$ and $q$ will be identified. However, Zariski local neighborhoods, or even stalks, at those singularities of that nodal point won't be isomorphic for different curves because they still know the isomorphism class of the curve -- the fraction field. However, the completions will all be isomorphic.

I've never checked the following carefully but I believe it is correct:

Every curve singularity is the pushout of an appropriate diagram. However ${p,q}$ and $Z$ need to be replaced by artinian schemes say $V$ and $W$ respectively. We also assume $C \leftarrow V$ is a closed embedding.
The completion of a curve singularity has exactly the same data as that pushout diagram (ignoring the actual curve).

For one more example, { $\text{Spec } k[x] \leftarrow \text{Spec } k[x]/x^2 \rightarrow \text{Spec } k$ }