Order of an equation on an elliptic curve I have the following statements:


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*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?
 A: 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 $-4T^3$. 
