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?
I actually stumbled upon the answer in an article entitled singularities of plane algebraic curves by Jonathan Hillman:
http://www.sciencedirect.com/science/article/pii/S0723086905000083
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.
http://arxiv.org/abs/1206.5316