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Let $f(x,y)\in \mathbb{C}[x,y]$ define a curve $C$ which is singular at the origin. By successively blowing-up the origin, we can resolve the singularities of $C$. Of course to make sure this process terminates, we need a measurable way of seeing an "improvement" in the singularity.

Define the Milnor number of $f$ as $\mu(f) = dim_{\mathbb{C}} \frac{\mathbb{C}[[x,y]]}{<\frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}>}$,

i.e. the dimension of this ring as a complex vector space.

I believe I once heard that we can use the Milnor number of $f$ to resolve singularities, which I suppose should mean that the Milnor number decreases after blowing-up until it's $0$ (where we then have a smooth curve).

Is this true? If so could you provide a reference, and if not a quick counter-example?

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"to make sure this process terminates": do not we know it terminates, by Hironaka's theorem? Also, I do not understand: when you say "the Milnor number decreases", what is the new $\mu(f')$ which you compare to the first $\mu(f)$? Sorry for being pedant but I would like to understand your question better, I find it interesting :) – Brenin Mar 23 '14 at 22:48
@Brenin, I think Sergio must be wondering how to replace other invariants to prove Hironaka's theorem, i.e., apply induction on the Milnor number. The new Milnor number must be the Milnor number on any singularity of the fibre above the origin. – Andrew Mar 23 '14 at 23:06
Andrew, this is correct. I already know how to compute Hironaka's invariant (using orders and exceptional divisors), but for curves there are other methods. The new Milnor number is as you say. I basically wanted a simple way of explaining to my student's why blowing-up finitely many times works - the hope being that the Milnor number decreases with each successive blow-up. – Sergio Da Silva Mar 24 '14 at 1:13

1 Answer 1

up vote 1 down vote accepted

I actually stumbled upon the answer in an article entitled singularities of plane algebraic curves by Jonathan Hillman:

He defines $\delta_A = (\mu(f) +1-r)/2$ where $r$ is the number of irreducible factors of $f$. Later on page 249, we see the formula $\delta_B = \delta_A - \frac{n(n-1)}{2}$ where $\delta_B$ is computed for the strict transform of $f$. When we blow-up a plane curve (in $x$ and $y$ say) at the origin (after a possible change of coordinates), we can look at the $x$-chart for example, and $x^{n}$ can be factored out for some $n$. This is the $n$ appearing in the formula. The $A$ and $B$ are the completions of the coordinate rings of $f$ and it's strict transform respectively.

For example, take the cusp $x^2-y^3=0$. Here $\mathbb{C}[[x,y]]/<2x,-3y^2>$ has dimension 1 as a complex vector space. Since $r=1$, we have $\delta_A = \frac{1}{2}$. If we blow-up the origin and look in the $y$-chart, we get a strict transform of $y^2(x^2-y)=0$ so $n=2$. Then $\delta_B= \frac{-1}{2}$ which matches the computation for $\delta_B$ directly.

In the end, we needed to write down the strict transform in coordinates and differentiate to compare the Milnor number for $f$ and it's strict transform. The key that was missing was that we should include the number of components in the computation also. This happens in other areas too. For example, in the paper below, the desingularization invariant alone is not enough to determine whether a point is simple normal crossings. For that we also need the number of irreducible components.

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