I have the following statements:
For $E$, an elliptic curve over $\mathbb{F}_{11}$ given by $y^2=x^3+4x$, I have:
1a. The line through $(0,0)$ and $(2,4)$ as $y-2x=0$ and am told that its order is $2$
For $E$, an elliptic curve over $\mathbb{F}_7$ given by $y^2=x^3+2$, I have:
2a. The line through $(3,6)$ as $4x-y+1=0$ with order $3$
2b. The line through $(6,1)$ as $5x-y-1$ with order $3$
Can someone help me understand how we can work out the order in each of these cases, and the general case?
An elementary way to find the order is via power series. Let $F(x,y) = y^2 - x^3 - Ax - B$ be the equation of the elliptic curve and $(x_0,y_0)$ a point on the curve, and let $\mathbb F$ the field.
If $\frac{\partial}{\partial y }F(x_0,y_0) \neq 0$ then we can find a parametrization $\phi \in T \mathbb F [[T]]$ such that $F(x_0+T,y_0 + \phi(T)) =0 \in \mathbb F[[T]]$. (And similarly if $\frac{\partial}{\partial x} F(x_0,y_0) \neq 0$ then change the roles of the two variables.)
To find the vanishing order of some line $G(x,y) = ux+vy+w$ (or any arbitrary (polynomial) function in the ring of functions over this curve) we plug in our parametrization. The order of that as a power series is defined as the vanishing order of our line $G$ on your curve $F$:
$$ \operatorname{ord} G(x_0+T,y_0 + \phi(T))$$
As an example I use the example 2b: $F(x,y) = y^2 - x^3-2$, $(x_0,y_0) = (6,1)$ and $G(x,y) = 5x-y-1$. In the following all equalities are in $\mathbb F_7$.
$\frac{\partial}{\partial y}F(x_0,y_0) = 2y_0 - 3 x_0 = 5 \neq 0$
So let $\phi(T) = a_1 T + a_2 T^2 + ... \in T\mathbb F[[T]]$. Then
$$0 = F(x_0+T,y_0+\phi(T)) = (y_0+a_1T+a_2T^2+a_3T^3+..)^2-(x_0+T)^3 -2 =:(*)$$ \begin{align}(*) &\equiv (4+2a_1)T \mod T^2 & \implies a_1 =-2 = 5 \\ (*) &\equiv 2a_2T^2 \mod T^3 & \implies a_2 = 0\\ (*) &\equiv (2a_3-1)T^3 \mod T^4 & \implies a_3 = 1/2 = 4\end{align}
e.t.c, so we know $\phi(T) = 5T + 4T^3 + (\text{higher order terms})$. This should suffice to calculate the vanishing order:
$$\begin{align*} G(x_0+T,y_0+\phi(T)) &= G(6+T,1+5T+4T^3+...)\\ & = 5 (6+T)-(1+5T+4T^3 + ...) -1 \\ & = -4T^3 +(\text{higher order terms}) \end{align*}$$
So in this case we see the order is $3$ as the lowest degree monomial with nonzero coefficient is $-4T^3$.
