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Oct 27, 2010 at 17:46 comment added Terry Tao Ah, normalisation was the key ingredient I was missing. Thanks!
Oct 27, 2010 at 17:46 vote accept Terry Tao
Oct 27, 2010 at 16:37 comment added Qing Liu Just to add some details on the genus of C (image of $f: k^2\to k^2$): it is unirational (i.e. dominated by a rational variety), and it is known that unirational curves are rational. The proof is simple: if for all $a\in k$, $x\to f(x, a)$ is constant, then the map $a \to f(0,a)$ is not constant (otherwise $f$ would be constant) and we get a non-constant map from $k$ to $C$. By Luroth, $C$ is rational. If for some $a\in k$, $x\to f(x,a)$ is not constant, then we again have a non-constant map from $k$ to $C$.
Oct 27, 2010 at 8:54 history answered Angelo CC BY-SA 2.5