I am trying to show that $V_p(x^2 y - z^3)$ is singular, however I am somehow stuck.
I haved tried
- Restriction to the affine patches: I tried to intersect with $\mathbb A^2 \subseteq \mathbb P^3$ by setting $z = 1$ and considering $V_a(x^2 y - 1)$ (singularity of this should imply singularity of $V_p(x^2 y - z^3)$, since it embeds into the latter as a closed subvariety, correct?), but it doesn't appear that this is singular, as far as I can tell.
- Applying the projective Jacobi criterion. Here I get the matrix $(2xy, x^2, -3z^2)$, which is of rank greater than or equal to $1$ whenever $(x, y, z) \neq 0$, which is insufficient for the theorem, because $V_p(x^2 y - z^3)$ is irreducible of dimension $2$, unless I am mistaken.
Since the projective Jacobi criterion is "if and only if", I must be missing something obvious.
$\frac{\partial f}{\partial x}= 2xy$, $\frac{\partial f}{\partial y}= x^2$, and $\frac{\partial f}{\partial z}= -3z^2$. Thus for the point $P=(0:1:0)$, $\frac{\partial f}{\partial x} (P)=0$, $\frac{\partial f}{\partial y} (P)=0$, and $\frac{\partial f}{\partial z}(P)=0$. Hence, the curve is singular at $P$.
This is the only singularity, since for each point of the form $P=(a:b:1)$, $\frac{\partial f}{\partial z}(P)\ne 0$, for each point of the form $P=(a:1:0)$ with $a\ne 0$, $\frac{\partial f}{\partial y}(P)\ne 0$ and for the point $P=(1:0:0)$, $\frac{\partial f}{\partial y}(P)\ne 0$.