Function that is identically zero Is it true that:
Any rational function $f$ on $\mathbb{C}^2$ that vanishes on $S=\{(x,y)\in\mathbb{C}^2 : x=ny  \text{ for some } n \in \mathbb{Z}\}$ must be identically zero.
I have a theorem that says any rational function that vanishes on an open set in Zariski topology must be identically zero, but I can't seem to prove that $S$ is open. Actually, I don't even think $S$ is open.

If the rational function $f = \frac{p}{q}$ vanishes on $S$, then at each point of $S$, so does either the polynomial $p$ or the polynomial $q$. Which means that the polynomial $pq$ vanishes on the whole of $S$. However, if this polynomial is non-zero, this means that $(x-ny)$ is a factor of $pq$ for all $n$, and therefore $pq$ is of infinite degree. This is clearly absurd, so $pq$ must be identically $0$. It cannot be $q$, so therefore it must be $p$ that is identically $0$, and hence also $f$.
