Ellie Benjamin

2022-07-07

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.

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.

Ronald Hickman

Beginner2022-07-08Added 18 answers

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

$f(x)=f(y)=\text{constant coefficient of}p(t)\text{/ constant coefficient of}q(t)$

but $x\ne y$.

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

$f(x)=f(y)=\text{constant coefficient of}p(t)\text{/ constant coefficient of}q(t)$

but $x\ne y$.

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