Suppose $X = \mathcal{V}(f) \subset \mathbb{P}^2$ is a projective plane curve, for some non-constant, homogenous polynomial $f \in \mathbb{C}[x,y,z]$.
I want to show that $X$ has finitely many singular points. This post answers my question, in the case when $f$ is irreducible.
Suppose $f = f_1\cdots f_n$ is a decomposition of $f$ into irreducible components. Then $$X = \mathcal{V}(f) = \mathcal{V}(f_1) \cup \cdots \cup \mathcal{V}(f_n),$$ is the decomposition of $X$ into irreducible plane curves.
Each $\mathcal{V(f_i)}$ has finitely many singular points.
Furthermore, as long as all of the $f_i$'s are pairwise relatively prime, then by Bezout's theorem, they all intersect in a finite number of points. I conjecture that the singularities of X are exactly the points of intersection of the irreducible components, together with the singularities of the irreducible components. And thus, there are finitely many of them.
Is my above reasoning correct?
It seems to me that my argument fails completely if $f$ is divisible by some $g^k$ where $g$ is irreducible, and $k > 1$. Is there any way to remedy this?
