I've been very confused about how to define the degree of a rational function over an algebraic curve.
For an Elliptic curve in Weierstrass form $E : Y^2=X^3 - AX -B$ over an algebraically closed field $K$, char $K \neq 2,3$, the set of polynomials on $E$ is given by $$ K[E] := K[X,Y]/\langle Y^2-X^3-AX-B \rangle $$
The set of rational functions is given by $$ K(E):= K[E]^2 / \sim $$
Then, the degree of a polynomial $f \in K[E]$, where $ f(x,y)=v(x)+yw(x)$ is its canonical form is given by $$ \deg(f) := \max \{ 2 \deg_x(v),3+2\deg_x(w) \} $$ Here, $\deg_x(v)$ is the classical degree of a polynomial $v \in K[X].$
Let $C:X^2 + Y^2 = 1+dX^2Y^2$ be an Edwards curve over $K$, how do we define the degree of a polynomial $f \in K[C]$ ?
In particular, for any algebraic curve. How will the result vary when the base field is not algebraically closed (for example, $\mathbb{F}_q$)? From what I've read so far, it looks like it has something to do with the pole divisors of the rational function (x) and (y) in those curves, but this connection is not clear.