Showing that $x^3+y^3+z^3=0$ is not rational

Question

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?

Answer 1

Suppose that $F$ was birational to $\mathbb{P}^1$, then because they are both smooth curves over $\mathbb{C}$, we'd have that $F\cong \mathbb{P}^1$. But, note that every affine open of $\mathbb{P}^1$ is a localization of $k[x]$, and so a UFD. That said, $\mathbb{C}[x,y]/(x^3+y^3+1)$ is an affine open of $F$, which is not a UFD.

PS: You can show, in fact, that every affine open of the cubic you wrote down has infinite class group.

Answer 2

Two more ways:

1. The differential $X^2dX + Y^2dY = -Z^2dZ$ is holomorphic and nowhere vanishing on your curve. But $\mathbf P^1$ has no such differential.

2. (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?