I'm having difficulties understanding the proof of Theorem XI.3.1 in S. Lang's Algebra.
There is one argument the reader has to fill in himself and I am not sure if I did it correctly:
Suppose $k[a_1, \ldots, a_n]$ is a finitely generated $k$-algebra without zero-divisors ($k$ a perfect field). Suppose also that $K = k(a_1, \ldots, a_n)$ has transcendence degree 1 over $k$. Then $K = k(u)[v]$ and there is an irreducible polynomial $f \in k[X,Y]$ such that $f(u,v) = 0$
If $f$ has infinitely many zeros in $k$ then I can find one zero $(a,b) \in k^2$ such that none of the denominators of the $a_i$ vanish on $(a,b)$. Then I can extend the homomorphism $k[u,v] \to k$ given by $(u,v) \to (a,b)$ to $k[a_1, \ldots, a_n] \to k$. Is this argument correct? In the proof he seems to emphasize that $\partial_Y f(a,b) \ne 0$ but it seems to me, that he does not need it?