See David Speyer's answer [here](https://mathoverflow.net/a/20115/80013).



> 
> I saw Brian Conrad give an excellent one hour talk to undergraduates where he proved that there do not exist nonconstant, relatively prime, polynomials $a(t)$, $b(t)$ and $c(t) \in \mathbb{C}[t]$ such that
>  $$a(t)^3 + b(t)^3 = c(t)^3.$$
>  He gave an elementary proof, then outlined the better motivated proof where you see that $\mathbb{CP}^1$ can't map holomorphically to a genus $1$ curve.
> 
> 
> 


Could anyone expand a bit on the "outlined the better motivated proof where you see that $\mathbb{CP}^1$ can't map holomorphically to a genus $1$ curve" part? That's not so clear to me. Thanks in advance!



No nonconstant coprime polynomials $a(t)$, $b(t)$, $c(t) \in \mathbb{C}[t]$ where $a(t)^3 + b(t)^3 = c(t)^3$
(With corrections noted by @GH from MO): The map $t\to (a(t),b(t),c(t))$, if at least one of the ratios $a(t)/c(t)$ or $b(t)/c(t)$ is non-constant, would extend to a non-constant map from a lower-genus (compact connected) curve ($\mathbb P^1$) to a higher-genus such, the elliptic curve defined by $a^3+b^3=c^3$. (Maybe open mapping easily shows surjectivity in the complex case, for example.) Impossible, by Riemann-Hurwitz formula. 


Also, in the complex case, since elliptic curves are quotients $\mathbb C/L$ by lattices, and $\mathbb C\mathbb P^1$ is simply-connected, a map to the elliptic curve would lift (by homotopy lifting...) to a map to $\mathbb C$. This would contradict Liouville's theorem.


