rational cubic space curves with singularities

Question

Is it true that a rational cubic curve in $\mathbb P^3$ (over an algbraically closed field) with singularities is actually a plane curve? If so, what is the easiest proof/counterexample for this?

Answer 1

Let $C$ be such a curve with a singular point $x_0 \in C$. The restriction to $C$ of the linear projection $\mathbb{P}^3 \dashrightarrow \mathbb{P}^2$ from $x_0$ is a map of degree $3 - 2 = 1$, hence the image of $C$ in $\mathbb{P}^2$ is a line. Therefore, $C$ lies in the preimage of a line, i.e., in a plane.