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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.

user44322
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1 Answers1

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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 $f$.

Arthur
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  • Are you implicitly using the fact that if $f(x,y)$ is irreducible, and $f(x,y)\nmid g(x,y)$ then $Z(f)\cap Z(g)$ is finite? – Alex Youcis Aug 15 '13 at 05:55
  • @AlexYoucis I am relying on the fact that the lines $x = ny$ pairwise intersect only at the origin, if that's what you mean. This again means that if you want a polynomial to vanish on $N$ of those lines, then it's either identically zero or at least of degree $N$. – Arthur Aug 15 '13 at 07:15