Let $p$ be an elliptic curve in $\mathbb{C}[X,Y] $. Consider the quotient ring $A = \mathbb{C}[X,Y]/(p) $ and its field of fractions $F = frac(A) $. For all $f + (p) \in A$, define $deg_A(f + (p))= min\lbrace deg(g) \mid g \in f + (p) \rbrace $.
I want to consider the valuation ring $$\mathcal{O}_{\infty} = \lbrace \frac{f + (p)}{g + (p)} \in F \mid deg_A(f + (p)) \leq deg_A(g + (p)) \rbrace $$ of the algebraic function field in one variable $F/\mathbb{C}$.
The group of units of $\mathcal{O}_{\infty} $ is the set of all rational functions where the degree of the denominator equals the degree of the numerator. Hence the place of $\mathcal{O}_{\infty}$ is the ideal $$P_{\infty} = \lbrace \frac{f + (p)}{g + (p)} \in F \mid deg_A(f + (p)) < deg_A(g + (p)) \rbrace$$
If you accept the fact that every place of a algebraic function field in one variable is principal, then it should be pretty straight forward to find the prime element (generator) of this ideal - but in this case, I'm having trouble finding it explicitly.
In the rational case (when $F = \mathbb{C}(X) $). The prime element of the place $$ P_{\infty} = \lbrace \frac{f}{g} \mid f,g \in \mathbb{C} [X] ,g \neq 0, deg(f) < deg(g) \rbrace $$ is $\frac{1}{X}$ - but if we let $p = Y^2 + X^3 - X$, then both $\frac{1}{X} $ and $\frac{1}{Y}$ do not work here. Any suggestions on a way to do this?