Elliptic curve addition: why does it work in any field?

Solution 1:

When going from continuous fields ($\Bbb R$ and $\Bbb C$, and to a lesser extent $\Bbb Q$) to something like a finite field, you are right that the geometric intuition fails. However, nothing is stopping the algebra from working exactly as before (or, at least, almost exactly like before; as you pointed out, characteristic 2 and 3 mess with things when working with quadratic and cubic polynomials).

Say we have a finite field $K$. Then the line between $(a, b)$ and $(c, d)$ in $K^2$ is simply the set of $(x, y)\in K^2$ which satisfy a linear equation $$ mx + ny + k = 0 $$ where $m, n, k$ are chosen so that the equation is satisfied for $(a, b)$ and $(c, d)$ (if the two points are distinct, then there is exactly one viable choice of $m, n, k$ up to scaling).

A tangent is a bit more tricky, but as we know from the continuous case, a line $mx + ny + k = 0$ which has a point $(a, b)$ in common with $E$ is tangent with $E$ at that point if the intersection at $(a, b)$ has degree greater than $1$. And the degree of the intersection can be found purely algebraically, for instance by looking at $K[x, y]/(mx+ny + k)\cong K[z]$ and considering whether $x^3 + Ax + B - y^2$ has a multiple root at $(a, b)$.

So the moral of the story is that even though the geometric interpretation doesn't make as much sense, we keep the geometric terms and define them algebraically instead, and much of the intuitive results from the continuous case carry right over to the discrete case.