# Relationship between Affine definition of singular point and projective definition

Let $$C : F(X,Y,Z) = 0$$ be a projective curve given by a homogeneous polynomial $$F \in \mathbb{C}[X,Y,Z]$$, and let $$P \in \mathbb{P}^2$$ be a point.

Prove that $$P$$ is a singular point of $$C$$ if and only if $$\frac{\partial F}{\partial X}(P) = \frac{\partial F}{\partial Y}(P) = \frac{\partial F}{\partial Z}(P) = 0.$$

I'm very lost on this problem. I was trying to prove the forward direction by using the fact that if $$P = [a,b,c]$$ is a singular point on $$C$$ then $$(a/c,b/c)$$ is a singular point on the curve $$C_0 : F[x,y,1] = 0$$ but I have no idea how to relate the two.

Write $$f(x,y)=F(x,y,1).$$ In the affine plane, $$(a,b)$$ is a singularity of $$f=0$$ iff $$f(a,b)=\frac{\partial f}{\partial x}(a,b)=\frac{\partial f}{\partial y}(a,b) =0.$$ That is equivalent to $$F(a,b,1)=\frac{\partial F}{\partial X}(a,b,1) =\frac{\partial F}{\partial Y}(a,b,1)=0.\tag{*}$$ But $$F$$ must be homogeneous of degree $$n$$ say, so it satisfies Euler's identity $$X\frac{\partial F}{\partial X}+Y\frac{\partial F}{\partial Y} +Z\frac{\partial F}{\partial Z}=nF.$$ Using Euler's identity, (*) is equivalent to $$F(a,b,1)=\frac{\partial F}{\partial X}(a,b,1) =\frac{\partial F}{\partial Y}(a,b,1) =\frac{\partial F}{\partial Z}(a,b,1)=0,$$ that is to $$F(P)=\frac{\partial F}{\partial X}(P) =\frac{\partial F}{\partial Y}(P) =\frac{\partial F}{\partial Z}(P)=0.$$
In characteristic zero, Euler's identity means that $$\partial F/\partial X=\partial F/\partial Y=\partial F/\partial Z=0$$ implies $$F=0$$, so that singularity is equivalent to $$\frac{\partial F}{\partial X}(P) =\frac{\partial F}{\partial Y}(P) =\frac{\partial F}{\partial Z}(P)=0.\tag{\dagger}$$ But in characteristic $$p$$ one can have ($$\dagger$$) holding without $$F$$ being zero.