Group Tables of Elliptic Curves over $\mathbb F_5$

Question

I'm trying to construct the group table for the elliptic curve $y^2 = x^3-x-1$ over $\mathbb F_5$.

Using ${\bf o}$ to denote the identity, I got the points on $\mathcal C$ to be: $$\mathcal C/\mathbb F_5 = \{ {\bf o}, (0,2), (0,3), (1,2), (1,3), (2,0), (4,2), (4,3)\}$$

To compose points, say $(0,2)$ and $(1,2)$, I found the equation of the line through $(0,2)$ and $(1,2)$ and then checked for other points of $\mathcal C/\mathbb F_5$ lying on that line. Then I reflected in the $x$-xais and reduced modulo five. For example $(0,2) \oplus (1,2) = (4,3)$

THE PROBLEM I HAVE is that some of these lines do not meet $\mathcal C/\mathbb F_5$ in a third point.

Clearly: if $P \neq Q$ have the same $x$-ordinates, then $P \oplus Q = {\bf o}$

However, if $P$ and $Q$ have different $x$-ordinates then the line through $P$ and $Q$ is tangent to $\mathcal C$ at either $P$ or $Q$ and hence $P \oplus Q \in \{-P,-Q\}$.

I would like to know if there is an easy way of knowing if the line is tangent at $P$ or $Q$ without having to find the equations of the tangent lines at $P$ and $Q$ and then checking.

Below is my group table for $\mathcal C / \mathbb F_5$. The coloured squares on the interior are those for which the line through $P$ and $Q$ only met $\mathcal C$ at $P$ and $Q$, and where $P$ and $Q$ had different $x$-ordinates. I was able to fill these by playing Sudoku. The colour shows which point the line is tangent at, e.g. $(0,2) \oplus (1,3) = (0,3)$, and the line through $(0,2)$ and $(1,3)$ is tangent to $\mathcal C$ at $(0,2)$.

Answer 1

You don't have to find the equation of the tangent line in full. Note that the curve has, by implicit differentiation, $dy/dx = (3x^2-1)/(2y)$. This lets you easily check the derivative at a given point. Compute the slope of the line you're using, $(y_2-y_1)/(x_2-x_1)$; if it matches the derivative at a point, it is tangent.

Answer 2

You can also check the multiplicity of solutions when substituting the linear equation of the line back into cubic. If there is a double root, that is a tangent point. For example, when adding (0,2) and (1,3), you get the linear equation $y=x+2$. When you substitute back into the cubic, you get $$x^2 (x-1) = 0.$$ That gives us a solution at (1,3) and a double solution at (0,2). Since it is a double root at $x=0$, the elliptic curve is tangent to the line at (0,2).