Is a smooth cubic in $\mathbb P^2(\mathbb C)$ never birational to a line in $\mathbb P^2(\mathbb C)$?

Question

Let $V$ be a smooth cubic in $\mathbb P^2(\mathbb C)$ ; then is it true that $V$ cannot be bi-rationally equivalent to $\mathbb P^1(\mathbb C)$ ?

We know that $V$ is bi-rationally equivalent to the cubic defined by

$Y^2Z=X^3+aX^2Z+bXZ^2+cZ^3$ . But I'm unable to say anything else (I'm not even sure whether this simplification of $V$ is helpful or not).

Please help. thanks in advance .

EDIT : Please don't use the concept of genus. I think it can be done without using genus.

Answer 1

Let $f : X \to \Bbb P^1$ be a birational morphism. If $X$ is a smooth curve, you can show that $f$ should extend to a regular morphism.

Now, $f$ has degree $1$ so it should be an isomorphism, since it's injective and surjective.

This is a contradiction because a cubic curve is not isomorphic to $\Bbb P^1$ since even topologically they are different.