Savanah Boone

2022-07-01

Prove that for any two distinct points of an irreducible curve there exists a rational function that is regular at both, and takes the value 0 at one and 1 at the other.

I think I can construct such a function, for example, $u(x,y)=(x-a{)}^{2}+(y-b{)}^{2}$ for given two points $(a,b)$ and $(c,d)$. However, this doesn't work for general algebraically closed field, for example, the case of $(c,d)=(a+i,b+1)$. Hence now I have no clue. Could you give me a hint for this problem?

I think I can construct such a function, for example, $u(x,y)=(x-a{)}^{2}+(y-b{)}^{2}$ for given two points $(a,b)$ and $(c,d)$. However, this doesn't work for general algebraically closed field, for example, the case of $(c,d)=(a+i,b+1)$. Hence now I have no clue. Could you give me a hint for this problem?

Zachery Conway

Beginner2022-07-02Added 7 answers

Perhaps this is a bit late, but here's what I first thought:

Suppose $a\ne c,b\ne d$, and characteristic is not 2. Then let $u(x,y):=\frac{x-a}{2(c-a)}+\frac{y-b}{2(d-b)}$. Then clearly $u(a,b)=0$ and $u(c,d)=\frac{1}{2}+\frac{1}{2}=1$

Suppose $a\ne c,b\ne d$, and characteristic is not 2. Then let $u(x,y):=\frac{x-a}{2(c-a)}+\frac{y-b}{2(d-b)}$. Then clearly $u(a,b)=0$ and $u(c,d)=\frac{1}{2}+\frac{1}{2}=1$

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