Elliptic Curves over C - Elliptic Curves: Number Theory and Cryptography - page 284

n 2 +1 points in E ( C )thatare n -torsion points. This is impossible. Therefore,
E [ n ] is no larger than it should be.
There is also the reverse possibility. How do we know that K is large enough
to account for all the n -torsion points that we found in E ( C )? We need to
show that the n -torsion points in E ( C ) have coordinates that are algebraic
over L (where L is regarded as a subfield of C ). Let P =( x, y )bean n -torsion
point in E ( C ), and suppose that x and y are transcendental over L (since x
and y satisfy the polynomial defining E , they are both algebraic or both
transcendental over K ). Let σ be an automorphism of C such that σ ( x )=
x +1, and such that σ is the identity on L . Such an automorphism exists: take
σ to be the desired automorphism of K ( x ), then use Zorn's Lemma to extend
σ to all of C (see Appendix C). The points σ m ( P )for m =1 , 2 , 3 ,... ,have
distinct x -coordinates x +1 ,x +2 ,x +3 ,... , hence are distinct points. Each
must be an n -torsion point of E in E ( C ). But there are only n 2 such points,
so we have a contradiction. Therefore, the coordinates of the n -torsion points
are algebraic over L , hence are algebraic over K ,since L ⊆ K . Therefore, the

passage from K to C does not affect E [ n ].

Suppose we have an elliptic curve E defined over the real numbers R .
Usually, it is represented by a graph, as in Chapter 2 (see Figure 2.1 on
page 10). It is interesting to see how the torus we obtain relates to this graph.
It can be shown (Exercise 9.5) that the lattice L for E has one of two shapes.
Suppose first that the lattice is rectangular: L = Z ω 1 + Z ω 2 with ω 1 ∈ i R
and ω 2 ∈ R .Then

( ℘ ( z ) ,℘ ( z ))

∈ E ( R )

when

( I ) z = tω 2 with 0 ≤ t< 1 ,

and also when

( II ) z =(1 / 2) ω 1 + tω 2 with 0 ≤ t< 1 .

The first of these is easy to see: if z is real and the lattice L is preserved by
complex conjugation, then conjugating the defining expression for ℘ ( z )leaves
it unchanged, so ℘ maps reals to reals. The second is a little more subtle:
conjugating z =(1 / 2) ω 1 + tω 2 yields z = − (1 / 2) ω 1 + tω 2 , which is equivalent
to z mod L . Therefore, the defining expression for ℘ ( z ) is again unchanged
by complex conjugation, so ℘ maps reals to reals.
Fold the parallelogram into a torus by connecting the right and left sides to
form a tube, then connecting the ends. The paths (I) (see Figure 9.5) starts
and ends at points that differ by ω 2 . Therefore the endpoints are equivalent
mod L , so (I) yields a circle on the torus. Similarly, (I) yields a circle on the
torus.
When the ends of path (I) are disconnected at 0 (which corresponds to ∞
in the Weierstrass form), we obtain a slightly deformed version of the graph
of Figure 2.1(a) on page 10.