Saturday, December 10, 2016

Peculiarities of Elliptic Curves And Their Rational Points


Example 1 of an elliptic curve under study for rational points


Example 2 of an elliptic curve. Can you identify its rational points?













In an earlier post from 2010 , e.g.

http://brane-space.blogspot.com/2010/04/looking-at-groups.html


I noted the definition for an Abelian group, in respect of the commutative property i.e. Only if there exist elements a, b Î G such that (a · b) = (b · a), then G is said to be an Abelian Group.

According to Mordell ('On the rational solutions of the indeterminate equation of the third and fourth degrees', Proceedings of the Cambridge Philosophical society, Vol. 21 (1922), 179),

The set E(Q) of rational points of an elliptic curve E defined over Q (the set of rational numbers) forms a finitely generated Abelian group such that:

E(Q) = Zr  ⊕ E(Q)tor

for some non-negative integer r and finite Abelian group E(Q)tor.  Where  'tor' denotes the torsion subgroup.

In general elliptic curves can be considered in long or short Weierstrass form. For the former, we know an elliptic curve over Q is isomorphic to the projective closure of the zero locus of the equation:

y2  + a1 xy  + a3y =  x3      + a2 x2   +  a4 x + a6

But when defining a nonsingular curve the preceding can be transformed over Q to the short form:

y2  =   x3    +  Ax   +  B


for A, B   Î Q   with non-zero discriminant   D = -16 (4 A3  +   27 B2 )

N.B. The non-vanishing of the discriminant ensures the curve is nonsingular.

And we say the elliptic curve given by the short form has height coordinate maximum:

h (E) =   max (4 |A|3 ,   27 B2)

Consider now two examples of elliptic equations with graphs for subsets of the real points shown above.  These are:

1) y2  =   x3     –       x  

And:

2) y2  =   x3     –       x  +  1


For each curve we can apply the short form for the Weierstrass equation. Thus, for (1) we have:

A = -1  and B = 0

Then the discriminant :   D = -16 (4 A3   +  27 B2 ) =   -16( 4 (-1)3    + 0) = 64

The height is:  h (E) =   max (4 |A|3 ,   27 B2) =  4 |(-1)|3 , 0   = 4, 0

Now, for (2) we have: A = -1  and B = 1

Then the discriminant :   D = -16 [4 A3   +  27 (1)2]  =   -16[ 4 (-1)3    + 27]

=   [64 +   27(-16)]  =  [ 64 +   (-432)]  = -368

The height coordinates maximum is:   h (E) =   max (4 |A|3 ,   27 B2) =  4 |(-1)|3     = 4, 27

One peculiarity of such elliptic curves as noted by Ho (Bull. American Math Soc., Jan., 2014, p. 27) is that the elliptic curves occur in two separate domains: the complex points of an elliptic curve make up a one-holed torus, or "genus 1" curve, i.e. like a donut or torus given the genus specifies the number of "holes" or handles. Meanwhile, the real points are smooth curves in R2  with one or two components (compare the graphs shown above)

The "group law" applies, as Ho notes (ibid.) such that the set of solutions in a given field forms a group.

Ho goes on to say that "the group structure on the points of an elliptic curve uses the point 0 at infinity as the identity element and is most easily described geometrically."  Ho gives an example of this which I leave for the energized reader to actually work out using the curve shown in (2). His prescription, which the reader may use as a guide is:

"Construct the line L through any two points P1 and P2 such that they intersect a third point P3, by direct calculation or using Bezout's theorem, e.g.  https://en.wikipedia.org/wiki/B%C3%A9zout's_theorem


The vertical line through P3 then intersects another point on the elliptic curve which is the composition P1 + P2 of P1 and P2"

Further elaborating (ibid.):

"In other words the three intersection points P1, P2, and P3 of any line L with the elliptic curve sum to the identity in the group law.(The identity point 0 may be one of these points,  e.g. a vertical line intersects 0, a point P and its negative. Moreover, if P1 and P2 are rational points then the line L has rational slope, so P1 + P2 is also a rational point."

To get the interested math reader started, you may use as point P1 the vertex of the curve shown in (2)

Other problems for the math enthusiast:

1) Sketch more of the elliptic curve (2) such that the section is shown for x = 4, y = ?

2) Use the short Weierstrass form to generate another elliptic curve and graph it. Then obtain the discriminant and ensure it is non-vanishing. Thence obtain h(E).



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