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.

Question

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&#039;t prove the statement, even though I also think it should be true.

Ronald Hickman · Accepted Answer

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.

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

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