# 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?

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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? –  Derek Allums Oct 11 '12 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. –  user18119 Oct 11 '12 at 19:27