# Prove complex algebraic curve is a smooth surface

Show that the complex algebraic curve $\{(x,y)\in \mathbb{C^2}:y^2+x^2y+x=0\}$ is a smooth surface in $\mathbb{C^2}$.

Let $p(x,y)=y^2+x^2y+x$

$\nabla p=(2xy+1,2y+x^2)=(0,0)$

$\Rightarrow 2xy+1=0, 2y+x^2=0$

$\Rightarrow y=\frac{-x^2}{2}$

$\Rightarrow 2y+x^2=2(\frac{-x^2}{2})+x^2=-x^3+1=0 \Rightarrow x=1,-\frac{1}{2}+\frac{\sqrt3}{2}i,-\frac{1}{2}-\frac{\sqrt3}{2}i$

$x=1\Rightarrow y=-\frac{1}{2}:p(1,-\frac{1}{2})=\frac{3}{4}\neq0$ and similarly for the other two $x$ values (which I'm pretty sure don't amount to $0$ either).

I'm not sure how this helps show whether we have a smooth surface or not. Can someone please tell me what I'm supposed to do with this information/what to do next to show the algebraic curve is a smooth surface or not?

EDIT: Also how do I go about identifying it as a closed surface with a finite number of points removed?

• This shows that the curve is smooth because the points that make the gradient equal to zero don't belong to the curve. Sep 24, 2017 at 9:15
• Ooohh okay I'm so sorry! @Federico thank you so much!
– user475596
Sep 24, 2017 at 9:22
• Can I quickly ask you another question about the above algebraic curve? (which is how you would identify it as a closed surface with a finite number of points removed)
– user475596
Sep 24, 2017 at 9:24
• @Luke : take the closure of $C$ in $\Bbb P^2$, it will be closed surface, and the point you did removed are the point in $\overline{C} \cap L$ which is a finite number by Bézout's theorem. Sep 24, 2017 at 9:25

This is a curve, not a surface in $\mathbb C^2$. The work you have done is called the the Jacobi criterion. You have $C=V(f)=V(y^2+x^2y+y)$ is of dimension one. The rank of $J((f))$ is one too, this is a the smoothness you want. The second part, I believe Nicolas’s comment above is correct.