Singular plane cubic curve birational to $\mathbb{P}^1$

Question

Is it true that every singular plane cubic curve over an algebraically closed field is birationally equivalent to $\mathbb{P}^1$? I know that such a curve has to have only one singular point and that this singular point has multiplicity 2. So I hope that projecting the curve from this singular point to a line that doesn't contain this point I can get a birational equivalence. Is it correct?

Answer 1

Yes, this is true, and follows form Bezout's theorem.

Without loss of generality, we can assume that $\infty = (0:0:1)$ is on the curve (call this the point at infinity). Let the singular point be called $S$.

Now choose a line $L$ in $\mathbb P^2$ that does not pass through $S$. Define a map from this line to the curve as follows: for each point $P \not \in C \cap L$, draw the line between $P$ and $S$. Since $S$ has multiplicity two, this line meets the curve in exactly one other point, by Bezout. Let this be $\pi(P)$.

A little thought shows that this map is generically injective and surjective, hence it is birational.