If $C$ is a plane projective curve which is defined by an irreducible homogeneous cubic polynomial and has a singularity, why must there be a nonconstant morphism $\mathbb{P}^1\rightarrow C$?
(I'm not sure whether $C$ is supposed to have exactly one singularity; is the result still true if it has more than one?)
I think the proof should look something like this:
let $\alpha$ be the projection map from a singularity $P$ of $C$ onto $\mathbb{P}^1$. This is a rational map of degree 1 (why?), so it has an inverse rational map $\alpha^{-1}$ from $\mathbb{P}^1$ to $C$ (why?), which must be a morphism since $\mathbb{P}^1$ is smooth.
But there's two points I can't prove! If we choose coordinates so that $P=[0:0:1]$, then $\alpha([x:y:z])=[x:y]$, so $\alpha$ has domain $C\backslash\{P\}$, and its degree is the sum of ramification indices and therefore at most 3; but why must it be 1? Then given a rational map of degree 1 between two projective curves, why does it need to have a well-defined inverse?
Many thanks for any help with this!
Your rational map in effect projects $C$ to the line at infinity $z=0$ of $\mathbb P^2$.
It has degree one because given a point at infinity $A$ the line $AP$ cuts $C$ in exactly one point distinct from $P$, for all but at most three values of $A$.
The inverse rational map $\alpha^{-1}$ is indeed defined everywhere since the line at infinity is smooth.
So $\alpha^{-1}\colon \mathbb P^1\to C$ is a normalization of $C$.
As to your last question, a birational map does not need to have as inverse a morphism: $\alpha^{-1}$ furnishes a counterexample.
And finally, as an answer to your parenthatical remark, yes a cubic can only have one singularity.
Else the line joining two singularities would cut it in at least four points, contradicting Bézout.