Elliptic curve cryptography is based on finding intersections of lines and elliptic curves:

$$y^2 = x^3 + ax + b ~~\text{and}~~ y = ax + b$$

It make sense when you see it on the graph, but the algorithm itself is using modular arithmetic where those curves are just a messy set of points. Is there any theorem that proves that elliptic curves work the same way on the finite field and the real number field? As per my research it has something to do with modular forms (each EC has a modular form, although that theorem was proven after that cryptography algorithm was introduced)


closed as unclear what you're asking by David Loeffler, Lord Shark the Unknown, user10354138, Mark, Rebellos Nov 7 '18 at 7:40

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  • $\begingroup$ Welcome to stackexchange. That said, I'm voting to close this question because it's much too vague. If you have a particular question about elliptic curves you can ask it instead. $\endgroup$ – Ethan Bolker Oct 31 '18 at 15:26
  • $\begingroup$ The necessary aspects of algebraic geometry are available in the domain of finite fields. Proving that you still get a group out of this may be most conveniently rephrased in the language of divisors of function fields, but it does come out to the same thing. No need for modular forms to do any of that. $\endgroup$ – Jyrki Lahtonen Oct 31 '18 at 15:27
  • $\begingroup$ Are you saying any curve will work the same way as EC (on finite field)? I know it's not true for y=x^2. I was under impression it will only work for EC $\endgroup$ – Mike T Oct 31 '18 at 16:16

The whole story is based on the fact that one can add two points $P$ and $Q$ on an elliptic curve and obtain a third point $P+Q$ on the curve. The points of the curve form an abelian group under this addition. There is a basic access to this result. For this, it is better to work in the projective plane ${\Bbb P}^2({\Bbb K})$ over the field ${\Bbb K}$. The kind of field does not really matter here. A basic fact is that the intersection multiplicity between the curve and a projective line is 0, 1 or 3. Moreover, the only point on the curve not in the affine plane is the base point (i.e., unit element of the corresponding group). All necessary results can be established rather easily. Maybe, the use of modular forms is less boring but I haven't looked into this yet.

  • $\begingroup$ Is there anything special about EC itself, that makes it act the same in both fields? Or it would work the same for many other curves too? I know that trick with finding intersection points will not work for y=x^2 + ax + b . I assumed that this is only works with EC and that's why the algorithm have it in the name. $\endgroup$ – Mike T Oct 31 '18 at 15:49
  • $\begingroup$ There are differences concerning the characteristic $p$ of the underlying field, say when it comes to the ''normal form'' of the Weierstrass equation. The cases $p=2$, $p=3$ and $p\ne 2,3$ are often considered separately. $\endgroup$ – Wuestenfux Oct 31 '18 at 15:53

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