According to this question: Nonsingular projective variety of degree $d$, the curve $x^d+y^d+z^d=0$ in $\mathbb{P}^2$ is nonsingular. I'm trying to prove this.
Hartshorne defines nonsingular points using the Jacobian matrix in $\mathbb{A}^n$ and proves proves that this is the same as the local ring being regular. Then the book proceeds to define nonsingular points for arbitrary varieties as having regular local rings.
Since checking the Jacobian matrix is easier, I'm trying to show that we can do this in $\mathbb{P}^2$ too. Let $P=(1:0:0)$. Let $U_0$ be the open subset of $\mathbb{P}^2$ with $x\neq 0$. Since the local ring of $P$ in $U_0$ is the same as that in $\mathbb{P}^2$, it is sufficient to consider the image of the curve under the isomorphism in $U_{0}\to\mathbb{A}^2$. According to theorem 2.2 in the book, the image of $(1:0:0)$ is $(0, 0)$ and the image of $C\cap U_{0}$ (C is the curve) is $Z(1+y^d+z^d)$. The Jacobian matrix is $0$ at this point so my reasoning must be flawed.
- Where is the flaw?
- How can I apply the Jacobian matrix criteria in projective space and why?
It's because $(0,0)$ isn't on that curve. :)
The singular locus for a plane curve $V(f)\subseteq \mathbb{A}^2$ is the solution to the following system of equations:
$$\begin{cases}f= 0\\ f_x =0\\ f_y=0\end{cases}$$
In your case, with $f(x,y)=1+x^d+y^d$, you have that $f_x=f_y=0$ if and only if $(x,y)=(0,0)$, but $(0,0)$ doesn't kill $f$.