I am trying to prove the following corollary.
Let $C \subseteq \mathbb{C}^2$ be an affine algebraic curve. If $p \in C$ is a singular point on $C$, such that its order $\operatorname{ord}_p(C) = \deg(C) – 1$, then the curve $C$ has no other singular points.
This is supposed to follow from this next proposition. If points $p_1, \dots, p_k \in C$ lie on the same line and this line isn't a component of the curve $C$, then
$$\operatorname{ord}_{p_1}(C) + \dots + \operatorname{ord}_{p_k}(C) \leq \deg(C)\text{.}$$
I went about proving the corollary as such. Suppose there is another singular point $p' \in C$. Then it is always true that $\operatorname{ord}_{p'}(C) \geq 2$, so if I manage to satisfy the assumptions of the above mentioned proposition, we get that $\operatorname{deg}(C) + 1 \leq \operatorname{ord}_{p}(C) + \operatorname{ord}_{p'}(C) \leq \deg(C)$, which leads us to a contradiction.
Now, the problem I'm facing is showing that the line $L$ connecting the singular points $p$ and $p'$ of the curve $C$ is not a component of $C$.
Definition of an affine algebraic curve I'm working with is the following:
A subset $C \subset \mathbb{C}^2$ is an affine algebraic curve, if there exists a polynomial $f \in \mathbb{C}[x,y]$ of degree $\deg(f) \geq 1$, such that
$$C = \operatorname{V}(f) = \{(a,b) \in \mathbb{C}^2 | f(a,b) = 0\}\text{.}$$
The degree of $C$ is then defined as
$$ \deg(C) = \min\{\deg(f) | \operatorname{V}(f) = C\}\text{,}$$
which I believe is the same as the definition with the minimal polynomial of $C$.
Best Answer
Sorry for the late response - you're having trouble proving this with the definitions in your post because it's false. Here is a counterexample: the curve cut out by $(y-1)(y-x^2)$ has two singular points at $(1,1)$ and $(-1,1)$, and each is a point of order 2 on the curve. In order to prove the statement you want, you need to assume that $C$ has only one component (that is, $C$ is irreducible) in order for your logic to hold. Once you make this extra assumption, things are easy: the line joining the two singular points must be a component of $C$, but this would mean $C$ is just that line. Lines do not have singular points: they're cut out by a nonzero polynomial of degree 1, and thus the derivatives are constants and at least one of them doesn't vanish anywhere.