Calculate $\mathcal{v}_P(f)$ on an elliptic curve I asked a more complicated question yesterday that didnt get any hits so let me try something easier that I do not really understand that might put me on the right track.
Take an elliptic curve $E$ over $\mathbb{F}_q$ and a point $P$ on $E$ and a function $f$ in $\mathbb{F}_q(E)$. How does one begin to approach the calculation of $\mathcal{v}_P(f)$?
Can we do an easy case? Lets let $y^2=x^3-x-1$ be the equation for $E$ over $\mathbb{F}_5$. Pick your favorite rational function in two variables $f$ and calculate $\mathcal{v}_P(f)$ for $P=(0,2)$ perhaps?
 A: Let $A$ be the coordinate ring $\mathbb{F}_5[x,y]/(y^2 - (x^3 - x - 1))$ and let $A_P$ be the localization of $A$ at the ideal $\mathfrak{m} = (x, y-2)$.  Given $f \in A_P$, then $v_P(f)$ is the largest positive integer $v$ such that $\DeclareMathOperator{\m}{\mathfrak{m}} f \in \m^v$.  Since your curve is nonsingular, then the ideal $\m$ is principal and $A_P$ is a discrete valuation ring.  Letting $s = y - 2$, then $E$ becomes
$$
x^3 - x - 1 = (s+2)^2 = s^2 + 4s + 4 \implies x + 4s + s^2 - x^3 = 0
$$
and the point $P$ is given by $x = 0, s = 0$.
Since the line $x = 0$ is not tangent to $E$ at $P$, then $x$ is a uniformizing parameter at $P$, i.e., $\m = (x)$.  Then every $f \in A_P$ can be written $f = u x^v$ for some $v \in \mathbb{Z}_{\geq 0}$ (which is $v_P(f)$) and some $u \in (A_P)^\times$ (meaning $u$ doesn't vanish at $P$).
All right, let's do some examples.  Since $x = x^1$, then $v_P(x) = 1$.  To compute $v_P(s) = v_P(y - 2)$, note that
\begin{align*}
s(s+4) = s^2 + 4s = x^3 - x = (x^2-1)x \implies s = \frac{x^2 - 1}{s+4} x \, .
\end{align*}
Since $\frac{x^2 - 1}{s+4}$ doesn't vanish at $P$ then it is a unit, which shows that $v_P(s) = 1$, as well.  (In fact, since $s = 0$ is not tangent to $E$ at $P$, then we could have just as well taken $s$ to be our uniformizer.)  Then $v_P(x/s) = v_P(x) - v_P(s) = 1-1 = 0$.  (And indeed, from the expression for $s$ above we can see that
$$
\frac{x}{s} = \frac{s+4}{x^2-1}
$$
which doesn't vanish at $P$.)  We can also use the fact that $v_P$ is a discrete valuation to compute $v_P(f)$.  For instance, since $x + 4s = s^2 - x^3$ and $s^2$ and $x^3$ have different valuations, then
$$
v_P(x + 4s) = \min\{v_P(s^2), v_P(x^3)\} = \min\{2, 3\} = 2 \, .
$$
These are relatively easy examples, but you can use the same principles to deal with more complicated rational functions.
