# Detecting the type of singularity with the Jacobian

Say we have a plane curve $\mathcal{C} = V(f(x,y)) \subset \mathbb{A}^2_{\mathbb{C}}$. The partial derivatives tell us about the singularities: if they all vanish at a point $p \in\mathcal{C}$ then the curve is singular at this point. My question is: do the partials tell us what kind of singularity there is? That is, would we be able to detect a cusp, node, etc. just by looking at the partials?

More generally, if we have some space curve, $\mathcal{C} \subset \mathbb{A}^n_{\mathbb{C}}$, the minors of the Jacobian cut out the singular locus. Can they tell us what kind of singularity we have?

No you can't detect the singularity type with the Jacobian. Did you try with the simplest example of $y^2-x^3$ and $y^2-x^2(x+1)$ ?
• So in the first case we have a cusp and in the second we have a node. The Jacobians are $\begin{bmatrix} -3x^2 & 2y \end{bmatrix}$ and $\begin{bmatrix} -3x^2-2x & 2y \end{bmatrix}$, respectively, while the Hessians are $\begin{bmatrix} -6x & 0 \\ 0 & 2 \end{bmatrix}$ and $\begin{bmatrix} -6x-2 & 0 \\ 0 & 2\end{bmatrix}$, respectively. The only thing I notice is that the determinant of the Hessian in the first case at $(0,0)$ is $0$ and in the second case is $-4$. Is this relevant? Oct 11, 2012 at 15:53
• @unit3000-21, the singularity (cusp) in the first case is in some sense worse than the second one (ordinary double point). This is as for one-variable function: the first has vanish order $\ge 3$ and the second has order $=2$. This difference is detected by the Hessian.