# Let X be an irreducible, normal variety over an algebraically closed field of characteristic z

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 $f\left(x\right)=f\left(y\right)$ for every $f\in K\left(X\right)$ 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 $f\in K\left(X\right)$ with ${v}_{{D}_{x}}\left(f\right)\ne 0$ and ${v}_{{D}_{y}}\left(f\right)=0$.
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.
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Ronald Hickman
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 $f\left(t\right)=p\left(t\right)/q\left(t\right)\in k\left(t\right)$ be any rational function on $X$. We see

but $x\ne y$.