To what extent does the existence of a group-law for adding points depend on the binary operations in the polynomial? For example what if the addition in the polynomial y^2 = ax^3+bx+c is a semigroup/monoid/quasigroup/loop and the multiplication isn't necessarily associative. Will you instead end up with a monoid-law or loop-law ?
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$\begingroup$ What do you mean, binary operations in the polynomial? $\endgroup$– Simon RoseSep 26, 2011 at 17:49
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$\begingroup$ The symbol"+" replaced with a non-group addition and the "." in b.x replaced with non-ring multiplication (but still distributive over "+") $\endgroup$– asriSep 26, 2011 at 17:52
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5$\begingroup$ The group law on the elliptic curve assumes that the underlying algebraic structure is a field. Even just having a commutative ring creates problems and the sum of two points may not always be defined. I can't imagine you will get anything reasonable if the underlying algebraic structure is a semi-ring or worse. $\endgroup$– Felipe VolochSep 26, 2011 at 19:33
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The short answer is that the group law on the set of rational points of an elliptic curve defined by a Weierstrass equation uses all of the structure and properties of a field in an essential way.
To even get a well-defined operation, you really need the fact that if a line intersects the curve in two rational points, then it intersects the curve in exactly one more rational point (counting multiplicity). If you are not working over a field, you may have more or fewer than three intersections. It is not hard to write down counterexamples for any particular weakening of the field axioms, but I do not know of a general counterexample schema that works for every non-field algebraic structure with two binary operations (nor is it clear that such a thing should exist).
For the sake of exercise, let us suppose you were working in some bizarre non-field situation that for miraculous reasons admitted a well-defined operation. You would find that the condition that all three intersection points sum to zero automatically makes the operation commutative, with all elements invertible, so a commutative loop with identity is unavoidable. Associativity is somewhat more complicated to verify in the classical sense, and it is conceivable that in the absence of geometric input of some form (e.g., from theorems like Bezout or Riemann-Roch) it can be violated in your context.
answered Sep 29, 2011 at 18:38