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Let $X$ be an irreducible, normal variety over an algebraically closed field of characteristic zero. Let $x,y\in X$ be two points such that $f(x)=f(y)$ for every $f\in K(X)$ which is defined at $x$ and at $y$. Can I conclude that $x=y$?

I feel the answer should be affirmative. In fact, the statement can be reduced to the following: Given two effective prime divisors $D_x$ and $D_y$ on $X$, there exists a rational function $f\in K(X)$ with $v_{D_x}(f)\ne 0$ and $v_{D_y}(f)=0$.

If this is true, then assuming $x\ne y$ we could find a divisor $D_x$ containing $x$ but not $y$ and a divisor $D_y$ containing $y$ but not $x$, so a function $f$ as above would yield a contradiction. However, I just can't prove the statement, even though I also think it should be true.

Jesko Hüttenhain
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  • Does variety mean "quasiprojective"? –  Apr 07 '14 at 13:10
  • @AsalBeagDubh: Well. I wanted to avoid it, but if you have a solution for quasi-projective $X$, I'd be curious, too. – Jesko Hüttenhain Apr 07 '14 at 13:24
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    If $X \subset P^n$ is qproj, take hyperplanes $H,K$ such that $y$ lies in $H$, $x$ does not, neither lies in $K$, and both $H$ and $K$ intersect $X$ properly. Then the rational function associated to the principal divisor $H-K$ separates $x$ and $y$. –  Apr 07 '14 at 13:39
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    Your statement holds more generally for any integral separated scheme. Indeed, if $x\in y$, then as subrings of $K(X)$, we have $O_{X,x}\ne O_{X, y}$ by separatedness. Pick an $f\in O_{X,x}$ not in $O_{X,y}$, then $f(x)\ne f(y)$ (if you allow the convention $f(y)=\infty$ when $f\notin O_{X,y}$). – Cantlog Apr 07 '14 at 20:24
  • @Cantlog: That's really great and a clear, elegant proof at that. Why don't you post it as an answer, I'd accept in a heartbeat. – Jesko Hüttenhain Apr 08 '14 at 08:35
  • In the quasi-projective case, one can find an affine open subset $U$ containing $x, y$. There exists obviously $f$ regular on $U$ taking different values at $x, y$ if $x\ne y$. Now $f$ viewed as a rational function on $X$ satisfies the required property. – Cantlog Apr 08 '14 at 11:06

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As mentioned by Cantlog there is true if $X$ is integral and separated. Here is a counterexample showing this is not true in the integral non-separated case.

Take $X$ to be the affine line with doubled origin, $x$ the north pole and $y$ the south pole. Let $f(t) = p(t)/q(t) \in k(t)$ be any rational function on $X$. We see $$f(x) = f(y) = \text{constant coefficient of $p(t)$/ constant coefficient of $q(t)$}$$

but $x \neq y$.