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) = Z

^{r}⊕ 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

**. For the former, we know an elliptic curve over Q is isomorphic to the projective closure of the zero locus of the equation:**

*Weierstrass form*y

^{2}+ a1 xy + a3y = x

^{3}

^{ }

^{ }+ a2 x

^{2}+ a4 x + a6

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

y

^{2}= x

^{3}

^{ }+ Ax + B

for A, B Î Q with non-zero discriminant D = -16 (4 A

^{3}

^{ }+ 27 B

^{2 })

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 B

^{2})

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

1) y

^{2}= x

^{3}

^{ }–

^{ }x

And:

2) y

^{2}= x^{3}^{ }–^{ }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 A

^{3}^{ }+ 27 B^{2}) = -16( 4 (-1)^{3}^{ }+ 0) = 64
The height is: h (E) = max (4 |A|

^{3}, 27 B^{2}) = 4 |(-1)|^{3}, 0 = 4, 0
Now, for (2) we have: A = -1 and B = 1

Then the discriminant : D = -16 [4 A

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

^{3}^{ }+ 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 B

^{2}) = 4 |(-1)|

^{3}= 4, 27

_{}

^{}

*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

**R**

^{2}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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