The group structure on a plane cubic curve


The geometric setting.

We are interested to the rational solutions of the equation y^2 = f(x) where f(x) is a cubic polynomial with integral coefficients (for ex. y^2 = x^3 - x) and with 3 distinct roots in C, the field of complex numbers.
By homogeneizing we get F(x,y,z) = 0 where F(x,y,z) = zy^2-f(x, z) (for ex. F(x,y,z) = zy^2 - x^3 + xz^2), which is the equation of a nonsingular cubic curve, C, in the complex projective plane P2.
Now we use an easy version of Bezout's theorem, namely: every line, L, of P2 intersects C in three points, counted with multiplicities
To see it, consider a parametric representation of L: uP+vQ, (u:v) in P1, then the intersection of C with L is given by f(u, v) = 0 where f(u, v) = F(uP+vQ) is a homogeneous polynomial of degree 3 in u, v (f(u,v) is not 0 because C is irreducible hence doesn't contain L); since C is algebraically closed, f(u,v) has 3 roots (counted with multiplicities) in P1.(dehomogeneize to get a degree 3 polynomial in C[X])
Summarizing, the intersection C.L of the plane cubic C with a line L is one of the following:

A) 3 distinct points (this is the general case; f(u,v) has 3 distinct roots)

B) 2 distinct points (one, say P, counted with multiplicity 2): the line L is the tangent to C at P, L intersects C in one further point, Q (f(u,v) has a double root corresponding to P and one simple root corresponding to Q). The intersection multiplicity of C with L in P, i(C, L; p), is 2.

C) 1 point (say P): the line L is the tangent to C at P and P is a flex of C. (f(u,v) has a triple root; i(C, L; p) = 3).
"The third point:"

Let P, Q be two distinct points on C. Consider the line, L = [P, Q], spanned by P and Q. The line L intersects C in a third point denoted P*Q (note that P*Q can be equal to P (or Q) if L is the tangent to C at P (or Q)).
Now if P=Q, instead of [P, Q] take L to be the tangent line to C at P, the third point of intersection of L with C is P*P (which is equal to P if and only if P is a flex of C).

In conclusion to any (P, Q) in CxC we have associated a third point, P*Q, on C.
The group structure (geometric version):

Fix a point, O, of C. Define the addition law as follows: given (P, Q) in CxC, P+Q = (P*Q)*O;
in other words: (i) take the third point P*Q
(ii) then P+Q is the third point of (O, P*Q)

Now we have to verify the axioms of the abelian group structure.

Clearly: P+Q = Q+P (because P*Q = Q*P)
Let's see that, for any P, O+P = P+O = P.
Let Q = P*O. Then P+O is the third point of (P*O, O) which is necessarly P.
The reader will enjoy to verify that -P is obtained as follows: let O' = O*O (the third point on the tangent line to C at O). Then -P = P*O'.
It remains to show that + is associative: for any P, Q, R in C: P+(Q+R) = (P+Q)+R.
This is the difficult part. There are several possibilities:
-using a classical geometrical construction you can prove associativity for a general (P, Q , R) in CxCxC, and then conclude ("passing to the limit") by a topological argument (see for ex. [R]).
-if you know schemes (and "liaison") you will convince yourself that the proof works even if the points envolved are not all distinct.
-take O to be a flex (a smooth plane cubic has always a flex) and work in an affine chart taking as line at infinity the tangent line at O (then the only point of C at infinity is O). Then you have nice analytical formulas for the addition law, and you can check (if you are bold enough!) associativity analytically (cf [S-T])
-using the theory of Weierstrass p-functions you can show that C is the image of an immersion of a 1-dimensional complex torus; the group law on C is nothing else than the (image) of the group structure on the complex torus (see %).
-you may invoke general results from algebraic geometry ("C is isomorphic to its jacobian", see %).
Concluding remarks:

(1) The group structure is algebraic, i.e. the applications: CxC --> C: (P,Q) -->P+Q, and C -->C: P--> -P, are morphisms of algebraic varieties. This gives to C the structure of algebraic group (for instance, since C is a projective algebraic variety, of abelian variety).
(2) A natural question: why only plane smooth cubic curves? In other words: is it possible (for ex. using similar geometrical constructions) to define an (algebraic) group structure on a projective plane curve of degree d?
The answer is NO! (unless C is a smooth plane cubic).
First of all an algebraic group is always a nonsingular variety (if X is an algebraic variety its set of singular points is a proper closed subset, hence X has a smooth point. If X is an algebraic group, P a smooth point of X and Q any point of X, then there is a group automorphism sending Q to P, hence Q has to be a smooth point too).
Secondly, the tangent bundle to an algebraic group is trivial (move a given fiber with an automorphism).
In conclusion if a curve of degree d, C, in P2 is an algebraic group, then C has to be smooth and its tangent bundle, TC, has to be trivial (hence of degree zero). Since TC has degree 2-2g, where g is the genus of C, and since the genus of a smooth plane curve of degree d is g = (d-1)(d-2)/2, you get g = 1 and d = 3: C is a smooth plane cubic!