A rational projective plane curve is a curve $C\subseteq \mathbb{P}^2$ such that there exists a birational map $\mathbb{P}^1\dashrightarrow C$.

My question is whether the following statement is true:

A curve $C\subseteq \mathbb{P}^2$ is rational if and only if there exists a morphism of varieties $f: \mathbb{P}^1\rightarrow \mathbb{P}^2$ with $C = f(\mathbb{P}^1)$.

I've been studying rational curves recently and it seems completely natural to expect this to be the case, but I haven't been able to come up with a proof or counterexamples. The image of $\mathbb{P}^1$ by a nonconstant morphism $\mathbb{P}^1\rightarrow \mathbb{P}^2$ is an irreducible curve, and I was hoping that I could describe every rational plane curve in this way. The only if direction seems to be closely related to the question of whether a birational map $\mathbb{P}^1\dashrightarrow C$ can be extended to a morphism $\mathbb{P}^1\rightarrow C$, but I cannot seem to find a reference for the truth of this statement either.

My primary interest is for when the base field is $\mathbb{C}$, if this can affect the answer. Thanks very much for any help!


1 Answer 1


This is true.

First, a birational morphism between curves $C' \dashrightarrow C$ always extends to a morphism if $C'$ is smooth and $C$ is projective. So being rational is the same as a surjection $f : \Bbb P^1 \to C$ and you are done in one direction.

Conversely, if there is a non-constant morphism $f : \Bbb P^1 \to \Bbb P^2$ , it is a branched cover of its image, and by the Riemann-Hurwitz formula, the image has genus zero. But any genus zero curve is birational to $\Bbb P^1$.

A remark : $C$ will be in general not isomorphic to $\Bbb P^1$, rather it will be a singular curve and $f : \Bbb P^1 \to C$ will be its normalization.

  • $\begingroup$ Excellent, this argument is clear to me. Thanks! $\endgroup$
    – catfish
    Commented Apr 1, 2018 at 16:13
  • $\begingroup$ @catfish : great, glad I could help ! $\endgroup$ Commented Apr 1, 2018 at 19:06

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