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

Ellie Benjamin 2022-07-07 Answered
Let X be an irreducible, normal variety over an algebraically closed field of characteristic zero. Let x , y X be two points such that f ( x ) = f ( y ) for every f 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 K ( X ) with v D x ( f ) 0 and v D y ( f ) = 0.
If this is true, then assuming x 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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Answers (1)

Ronald Hickman
Answered 2022-07-08 Author has 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 ) k ( t ) be any rational function on X. We see
f ( x ) = f ( y ) = constant coefficient of  p ( t ) / constant coefficient of  q ( t )
but x y.
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