Differential of Polynomial in $\mathbb{P}^2$ Lets $f(x, y, z)$ is homogeneous polynomial that equation
$$f(x, y, z)=0$$
determines some curve $C_f$ in $\mathbb{P}^2$. By definition: point $P\in C_f$ is singularity point if $df_p=0$.
Question:
Am I right that $df_p$ written in coordinates in $\mathbb{P}^2$ as:
$$\left(\frac{\partial f}{\partial x}(P), \frac{\partial f}{\partial y}(P), \frac{\partial f}{\partial z}(P)\right)$$
(i.e. such in $\mathbb{R}^3$)
Thanks.
 A: I think you get carried away by the powerful formalism of projective geometry. 
Actually the polynomial $f(x,y,z)$ does not correspond to a function $\mathbb    P^2_k\to k$ , so that you can't define its differential. This is because, if the degree of $f$ is $d$, we have for all $\lambda\in k$ the equality $f(\lambda x,\lambda y,\lambda z)=\lambda^d f(x,y,z)$, and since 
$[x:y:z]=[\lambda x:\lambda y:\lambda z]\in \mathbb P^2_k$ there is no reasonable way to define $f(x,y,z)\in k$.
However the polynomials $\frac{\partial f}{\partial x}, \frac{\partial f}{\partial y},\frac{\partial f}{\partial z}$ have  three perfectly well defined zero sets, curves whose common  points  indeed are the singularities of $C(f)$.
But all is not lost. To $f$ you can associate an affine cone $V=V(f)\subset \mathbb A^3_k$ whose equation is $f(x,y,z)=0$. The polynomial $f$ defines a perfectly regular function $f:V\to k$ i.e. $f\in \mathcal O (\mathbb A^3_k)$.    Then for all $P\in \mathbb A^3_k$ we have indeed $df_P=\left(\frac{\partial f}{\partial x}(P), \frac{\partial f}{\partial y}(P), \frac{\partial f}{\partial z}(P)\right):k^3=T_P(V)\to k$, and for $P\in V$, the vanishing of $df_P$ tells you whether $P$ is singular on $V$.
 For example if $degree(f) \gt 1$, the origin is a singular point of the cone $V$, as it should be.
