Suppose that $F\in \mathbb{C}[x, y, z]$ is a homogeneous polynomial defining a curve $F = 0$ in the complex projective plane $\mathbb{P^2_C}$.
(1) Show that if $F = GH$ is reducible then $F = 0$ has at least one singular point in $\mathbb{P^2_C}$.
(2) Why is this result false for real curves?
My attempt:
$P=[x_0:y_0:z_0]$ is a singular point of $F=0$ iff $\frac{\partial F}{\partial x}(P)=\frac{\partial F}{\partial y}(P)=\frac{\partial F}{\partial z}(P)=0$.
$\frac{\partial F}{\partial x}=G_xH+GH_x$
$\frac{\partial F}{\partial y}=G_yH+GH_y$
$\frac{\partial F}{\partial z}=G_zH+GH_z$
If $\exists$ a point $P\in F$ such that $G(P)=H(P)=0$,then $\frac{\partial F}{\partial x}(P)=\frac{\partial F}{\partial y}(P)=\frac{\partial F}{\partial z}(P)=0$.
However, I am not sure if such a point $P$ exists.
Are there other methods to prove this proposition?
Bézout's theorem states that if $K$ is an algebraically closed field, then two curves of degree $d$ and $e$ in $\mathbb P^2_{K}$ have $de$ points of intersection, counted with multiplicities. This means, in your case, that $G$ and $H$ have at least a point $P$ in common. Such a point is singular for $F$. If $K=\mathbb R$, you can easily find a counterexample, like $G=x^2+y^2-z^2$ and $H=z$.