Questions about singularity of a projective cubic in the projective plane.

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I'm studying algebraic geometry, in particular the singularity of certain affine and projective varieties and differential forms in said sets.

In particular I have, in $P^2$ (over complex numbers) with coordinates $[z_{0}, z_{1}, z_{2}]$, $X = V(z_{1}^{2}z_{2} - z_{0}^{3} + z_{2}^{3})$, that makes sense because $f = z_{1}^{2}z_{2} - z_{0}^{3} + z_{2}^{3}$ is an homogeneous polynomial in $z_{0}, z_{1}, z_{2}$.

Now, the course's notes say that this is a smooth cubic with a well defined regular volume form over $X$. The proof given states the following:

  • Let's define the affine sheets $U_{1}$ and $U_2$ in the usual way, so $U_{1} = \{[z_{0}, z_{1}, z_{2}] \in P^{2} | z_1 \neq 0\}$ and $U_2$ similarly. I can imagine $U_1$ and $U_2$ as two-dimensional affine spaces by defining the two coordinates (let's do it in $U_1$ but it works the same for $U_2$) $x_1 = \frac{z_0}{z_1}$ and $x_2 = \frac{z_2}{z_1}$.
  • We observe that when we restrict $U_1$ to $X$, we can express the new algebraic set as the set of zeroes of the polynomial $f_1$ obtained by de-homogenizing f by the variable $z_1$, so we obtain $U_{1}\big|_{X} = V(f_1)$ with $f_1 = x_2 - x_1^{3} + x_2^{3}$.

This is useful because we proved earlier in the course that any non singular affine variety has a regular volume form defined in all $X$. So, we want to prove $f_1$ is NON singular in any point of $X$.

Here lies my problem with it: if we try to erase the partial derivatives of $f_1$, we get $\frac{\partial{f_1}}{\partial{x_1}} = -3x_1^{2}, \frac{\partial{f_1}}{\partial{x_2}} = 1 + 3x_2^{2}$ and they are null when $x_1 = 0$, so $z_0 = 0$, and $x_2 = \pm \sqrt{-\frac{1}{3}}$, so $z_2 = \pm \sqrt{-\frac{1}{3}}$ $z_1$. This gives us two singular points for $f_1$, that surely are in $X$. So $X$ is not non-singular. The teacher stated that it was indeed non singular and used this very method to determine why.

Am I wrong about any of the things I stated here? How can I determine whether the variety is singular or not? Is not being singular a necessary condition for the volume form to be defined over $X$? Thank you in advance.

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$V(x_2-x_1^3+x_2^3)$ is nonsingular. The partial derivatives are $-3x_1^2$ and $1+3x_2^2$, so if the characteristic is not 3, then these derivatives vanish when $x_1=0$ and $x_2^2=\frac{-1}{3}$. But these points do not satisfy $x_2-x_1^3+x_2^3$, as you can check from plugging in. (You wrote "surely these are in $X$" which makes me think you either didn't check this or you made an error here.)

(When the characteristic is 3, then $1+3x_2^2=1$ and so the partial derivatives don't vanish anywhere, so the affine variety $V(f_1)$ is nonsingular in all characteristics.)

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You found the points that make the partial derivatives vanish, the points $(0, \pm\sqrt{\frac{-1}{3}})$

Plugging those into the original de-homogenized $f_{1}$ we have that

$f_{1}(0,\sqrt{\frac{-1}{3}})=\sqrt{\frac{-1}{3}}+\sqrt{\frac{-1}{3}}^{3}=\frac{2i}{3\sqrt{3}}\neq 0$

$f_{1}(0,-\sqrt{\frac{-1}{3}})=-\sqrt{\frac{-1}{3}}-\sqrt{\frac{-1}{3}}^{3}= \frac{-2i}{3\sqrt{3}}\neq 0$

So the possible points that would make the partial derivatives vanish, don't actually lie on the curve.