What is the proof of - $E[2]$ torsion group is isomophic to $Z_2 \oplus Z_2$?

Question

This is from the book "Elliptic Curves" by Lawrence Washington.

$E$ is an elliptic curve over $K$

$E[n] = \{P \in E(\bar K) \mid nP = \infty\}$

where $\bar K$ is the algebraic closure of $K$

The points of $E[2]$ are ${\infty, (e_1, 0), (e_2, 0), (e_3, 0)}$

I understood how they get up to the above.

But after this the book says

As an abstract group, this is isomorphic to $Z_2 \oplus Z_2$

They haven't expanded further on this - so it looks like the above is something which should be obvious but I don't understand how the above is true.

For it to be isomorphic to $Z_2 \oplus Z_2$, I think the following needs to be shown

$(e_1, 0) + (e_2, 0) = (e_3, 0)$ and so on

How do we prove the above?

Answer 1

I assume that your elliptic curve is given in Weierstrass form $$E:y^2=x^3+c x^2+ax+b$$ with $c=0$. You factorize $$x^3+cx^2+ax+b=(x-e_1)(x-e_2)(x-e_3)$$ So that $e_1+e_2+e_3=c=0$.

For $E$ to be an elliptic curve (not a singular cubic) you need that the $e_j$ are distinct (and $char(K)\ne 2$).

In the group law of $E$ you have that $-(u,v)=(u,-v)$ so the 2-torsion is the points with either $v=0$ or $v=\infty$, which gives $$E[2]=\{ (\infty,\infty),(e_1,0),(e_2,0),(e_3,0)\}$$ $E$ is an abelian group. Whence $E[2]$ is a subgroup.

• $(e_1,0)+(e_2,0)$ can't be $(\infty,\infty)$ as otherwise you'd have $(e_1,0)=-(e_2,0)=(e_2,0)$ so $e_1=e_2$.

• $(e_1,0)+(e_2,0)$ can't be $(e_1,0)=(e_1,0)+(\infty,\infty)$ as otherwise you'd have $(e_2,0)=(\infty,\infty)$.

  Similarly $(e_1,0)+(e_2,0)$ can't be $(e_2,0)$.

• Whence $(e_1,0)+(e_2,0)=(e_3,0)$.

  Similarly $(e_i,0)+(e_j,0)=(e_k,0)$ for $i\ne j\ne k$.

• Otherwise, in general, $E[2]$ is a $\Bbb{F}_2$-vector space so it is automatically $\cong (\Bbb{F}_2)^n$ with $2^n=|E[2]|$ ie. $n=2$.

Answer 2

From @reuns' answer, we get

$e_1 + e_2 + e_3 = 0$

Let $P = (\infty,\infty), Q = (e_1,0), R = (e_2,0),S = (e_3,0)$

Let $N = \{x, y\} = Q + R$

The formula for point addition for elliptic curves is

$m = \frac {y_2 - y_1}{x_2 - x_1}$

$m = \frac {0 - 0} {e_2 - e_1} = 0 $

$x = m^2 - e_1 - e_2 \bmod p$

$x = -e_1 - e_2$

$e_1 + e_2 + e_3 = 0$, so $e_3 = -e_1 - e_2$

So $x = e_3$

$y = m(x_1 - x) - y_1 \bmod p$

$y = m(e_1 - e_3) - 0 \bmod p$

$y = 0$

So $N = Q + R = \{e_3, 0\} = S$

Likewise we can prove that

$R + S = Q$

$S + Q = R$