TITLE: Showing that $x^3+y^3+z^3=0$ is not rational
QUESTION [5 upvotes]: Is there a short proof that $F:x^3+y^3+z^3=0$ in $\mathbf{P}^2$ is not rational, apart from using the genus? Perhaps this is an  elliptic curve, so every morphism $\mathbf{P}^n\rightarrow F$ is constant. Does this help me somehow?
As I said below, it seems that it is enough to distinguish $\mathrm{Frac} \ \mathbf{C}[x,y]/(x^3+y^3-1)$ and $\mathbf{C}(x)$, how can we do this?

REPLY [4 votes]: Two more ways:

The differential $X^2dX + Y^2dY = -Z^2dZ$ is holomorphic and nowhere vanishing on your curve. But $\mathbf P^1$ has no such differential. 
(This one is silly.) Remark that permutations of $X,Y,Z$ or multiplication of either one of $X,Y$ and $Z$ by a cube root of unity preserve the equation. Hence there is a finite group of order $6\times 3^3$ acting on your curve by automorphisms. Does $\text{PSL}_2(\mathbf C)$ have such a subgroup?