I'm trying to prove the next thm,
(denoted in https://crypto.stanford.edu/pbc/notes/elliptic/divisor.html):
Rational functions with a given divisor are unique up to a constant.
Rational functions with a given divisor are unique up to a constant. Here, divisor of a rational function is formal sum of the order of function at a point times the point.
My attempt: Given that h(x,y),g(x,y) rational functions over eliptic curve, $y^2=f(x)=ax^3+bx+c$
s.t $<h>$=$<g>$, we get that $<\frac{h}{g}>=0$. Denote $\frac{h}{g}=p(x,y)$, which holds canonical represntation as rational function ,say $p(x,y)=a(x)+b(x)y$- with no zeroes and no poles. Similarly, $\overline{p}=a(x)-b(x)y$ has no poles and no roots. We get that $|p|=p*\overline{p}$ , has no root or poles. Now, as a function of x , I would like to say that it must be constant. However, Most of time we work over finite fields, which cannot be algebric closed: (if there are n points, then the polynomial $p(x)=\prod(x-a_i) -1$ has no root)- so I'm stuck about why such function must be constant.
The claim that "Rational functions with a given divisor are unique up to a constant." only makes sense over an algebraically closed field. Otherwise, you could have a rational function such as $f(x) = \frac{x^2+1}{x^2+2}$ over $\mathbb{R}$, which has no zeros or poles over $\mathbb{R}$ but is not constant.
When working over a finite field, one would normally take the algebraic closure of the finite field for divisor calculation purposes. (An alternative but more complicated option is to use scheme theory to allow your "divisors" to have support in the inertial primes.)