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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.

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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.

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