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Let $I = (f_1,...,f_m) \subset k[x_1,...,x_n]$ be a prime ideal where $k$ is an algebrically closed field and let $V = \text{Spec}(k[x_1,...,x_n]/I)$ be the corresponding algebraic variety. If we take any $p \in V$, we can define the Jacobian matrix $$J_p = \begin{pmatrix}\frac{\partial f_i}{\partial x_j}(p)\end{pmatrix}.$$ The Jacobian criterion tells us that $V$ is smooth at $p$ if and only if $\text{rk}(J_p) = n - \dim V$. Certainly, if $rk(J_p) < n - \dim V$ then $p$ is singular.

However, is there a way of using the Jacobian matrix to tell what type of singularity $p$ is if $p$ was to be a singular point? In particular, can we tell if $p$ is a nodal singularity or not using the Jacobian matrix? By definition $p$ is nodal if and only if $\hat{\mathcal{O}}_{V,p} \cong k[[x,y]]/(xy)$ but this completion might be hard to compute if $I$ was complicated.

But even for a simple example $V:y^2 - x^2 - x^3$, is there a way of using the partial derivatives to tell that $V$ the singularity at the origin is nodal? If there is such a criterion, a explanation or a link to a reference would be appreciated.

KSAKY
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Let's look at the nodal cubic $y^2=x^3+x^2$ and the cuspidal cubic $y^2=x^3$.

The point is that the nodal cubic has two tangent lines $y=\pm x$ at $(0,0)$ whilst the cuspidal cubic has $y=x$.

The way we obtain these tangent lines is to take Taylor expansions of $f(x,y)=y^2-x^3-x^2$ and $g(x,y)=y^2-x^3$.

$f(0,0)=0$ is telling us $(0,0)$ is a point on the curve and the first derivatives both vanishing then tells us that this point $(0,0)$ is singular. The higher derivatives, tell us the quadratic approximation of these curves.

$f(x,y)=y^2-x^2+$(higher order terms) and observe that $y^2-x^2=(y-x)(y+x)$ which tells us that $y^2-x^2=0$ is tangential to the nodal curve $f(x,y)=0$.

Likewise $g(x,y)=y^2+$(higher order terms) so we get that $y=0$ is tangential to $g(x,y)=0$.

So to determine whether a cubic affine curve $f(x,y)=0$ is a nodal or cuspidal curve, we are essentially interested in the quadratic polynomial,

$ax^2+bxy+cy^2=0$ where $a=f_{xx}(P), b=f_{xy}(P), c=f_{yy}(P)$ and $P$ is the point of singularity. In particular, we are interested whether the discriminant of this quadratic polynomial is exactly $0$ or not.

i.e. if $b^2-4ac=0$, we have a cuspidal curve and if $b^2-4ac\neq 0$, we have a nodal curve.

This is covered somewhat in Shafarevich's book. It is also covered in Silverman's book.

daruma
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  • Thanks for the answer. Does such a criterion exist for affine scheme with more than one generator? Suppose $V = V_1 \cup V_2$ and $V_1,V_2$ intersect at a singular point $p \in V$ where $V_1 = \mathbb{V}(f_1,...,f_s)$ and $V_2 = \mathbb{V}(g_1,...,g_t)$. Is there a way to tell $p$ whether is nodal or not using the Jocobian matrices $\begin{pmatrix}\frac{\partial f_i}{\partial x_j}(p)\end{pmatrix}$ and $\begin{pmatrix}\frac{\partial g_i}{\partial x_j}(p)\end{pmatrix}$? E.g. from the kernels or determinants of these matrices? – KSAKY Jun 02 '23 at 14:58
  • @KSAKY I don't know if there is a nice description of the types of singularity for more general affine $k$-schemes. But it seems like the Jacobian $\left(\frac{\partial f_i}{\partial x_j} (p)\right)$ should only tell us whether it is singular or not. The higher order partials should really be giving the information about how it behaves near the singular point. – daruma Jun 02 '23 at 22:47
  • My apologies for not specifying my question in the first place. $V_1$ and $V_2$ are smooth, irreducible curves defined by $f_1,...,f_s$ and $g_1,...,g_t$ respectively. Now suppose $V$ is the union of $V_1$ and $V_2$ where $V_1$ and $V_2$ intersects at a single point $p$. We know that $\dim_k \ker\begin{pmatrix} \frac{\partial f_i}{x_j}(p) \end{pmatrix} = \dim_k \ker\begin{pmatrix} \frac{\partial g_i}{x_j}(p) \end{pmatrix} = 1$ and let $v_1$ and $v_2$ be the corresponding spanning vectors for the two kernels. Now assume $v_1$ and $v_2$ are orthogonal, can we show that $p$ is a node of $V$? – KSAKY Jun 02 '23 at 23:57