# 2-Torsion Group Scheme

Consider the elliptic curve $zy^2 + z^2y = x^3.$ I would like to explictly compute the 2-torsion group scheme, $E[2],$ over $\mathbf{Spec}(\mathbb{Z}_2),$ but I'm having a tough time writing down the (co)multiplication law.

The underlying scheme of $E[2]$ is easily computed. The inverse map $[-1]:E \rightarrow E,$ is given by

$$[x,y,z] \mapsto [-x,y+z,-z],$$

and so \begin{align*} E[2] &= \mathbf{Proj}(\mathbb{Z}_2[X,Y,Z]/(ZY^2 + Z^2Y - X^3,2XY + XZ, 2ZY + Z^2)\\&= \mathbf{Spec}(\mathbb{Z}_2[X,Z]/Z + Z^2 -X^3, 2X + XZ, 2Z + Z^2)\\&= \mathbf{Spec}(\mathbb{Z}_2[X,Z]/-Z -X^3, 2X + XZ, 2Z + Z^2) \\ &= \mathbf{Spec}(\mathbb{Z}_2[X]/X^4-2X). \end{align*}.

So, by the general theory, the addition law on $E[2]$ yields a comultiplication law

$$\mu: \mathbb{Z}_2[X]/(X^4-2X) \rightarrow \mathbb{Z}_2[X]/(X^4-2X) \otimes_{\mathbb{Z}_2} \mathbb{Z}_2[X]/(X^4-2X).$$

I would like to compute the homomorphism $\mu,$ but have so far been unable to do so. Any help would be greatly appreciated.

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As you observe, the homogeneous coordinate $Y$ can never be zero on a $2$-torsion point, so we may work in the affine patch with coordinates $x = X/Y$, $z = Z/Y$, where the elliptic curve has equation $z^2 + z = x^3$, and origin (for the group law) equal to the point $(0,0)$.

In these coordinates, the map $[-1]$ is given by $(x,z) \mapsto (-\dfrac{x}{1+z},-\dfrac{z}{1+z})$. (This is not defined when $z = -1$, because $[-1]$ takes $(0,-1)$ to a point which lies at infinity in these coordinates, so we can remove the line $z = -1$ if you like.)

Thus, in these coordinates, $E[2]$ is cut out by the equations $z+z^2 - x^3 = 2x+xz = 2z + z^2 = 0$, just as you indicate, which (as you compute) reduce to the equations $x^4 - 2x = 0$ and $z = -x^3$.

But now you can just use the usual chord tangent law for addition:

Given points $P_1 = (x_1,z_1)$ and $P_2 = (x_2,z_2)$, let $m$ be the gradient of the line joining $P_1$ and $P_2$. The third point of intersection of this line with $E$ then has $x$-coordinate equal to $m^2 - x_1 - x_2$.

The gradient $m$ is given by the formula

$$\dfrac{z_1-z_2}{x_1 - x_2} = \dfrac{(z_1-z_2)(x_1^2 + x_1 x_2 + x_2^2)}{x_1^3 - x_2^3} = \dfrac{(z_1-z_2)(x_1^2 + x_1 x_2 + x_2^2)}{z_1 - z_2 + z_1^2 - z_2^2} = \dfrac{x_1^2 + x_1 x_2 + x_2^2}{1 + z_1 + z_2}.$$

Now we can restrict attention to $E[2]$, where $z = -x^3$, to obtain the addition formula (written in terms of $x$-coordinates alone, because points in $E[2]$ are determined by their $x$-coordinates)

$$(x_1,x_2) \mapsto \dfrac{(x_1^2 + x_1 x_2 + x_2^2)^2}{(1 -x_1^3 - x_2^3)^2} - x_1 - x_2.$$

Of course you can simplify this using the relations $x_1^4 = 2 x_1$ and $x_2^4 = 2 x_2$; I will leave that to you.

Added: In light of a comment below, maybe I should point out that this formula does answer the OP's question!

Namely, $x_1$ and $x_2$ are just more convenient names for the elements $x\otimes 1$ and $1 \otimes x$ of the tensor product. The above formula can then be rephrased by saying that the comultiplication is given by

$$x \mapsto \dfrac{(x_1^2 + x_1 x_2 + x_2^2)^2}{(1 - x_1^3 - x_2^3)^2} - x_1 - x_2.$$

This formula makes sense, because $(1- x_1^3 - x_2^3)$ is invertible in the ring

$$\mathbb Z_2[x_1,x_2]/(x_1^4 - 2 x_1, x_2^4 - 2 x_2) = \mathbb Z_2 [x]/(x^4 - 2 x) \otimes \mathbb Z_2[x]/(x^4 - 2x).$$

One more thing: the elliptic curve $E$ has good reduction over $\mathbb Z[1/3]$, and the above formula actually computes $E[2]$ as a finite flat group scheme over $\mathbb Z[1/3]$. (I.e. we can replace $\mathbb Z_2$ by $\mathbb Z[1/3]$ in the discussion.)

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I thought the OP was looking for an explicit expression of the co-multiplication $\mu$. – Álvaro Lozano-Robledo Dec 18 '11 at 15:26
@Alvaro: Dear Alvaro, This is what the above formula gives: the element $(1-x_1^3 - x_2^3)$ is invertible in the ring $\mathbb Z_2[x_1,x_2]/(x_1^4-2x_1, x_2^4 - 2 x_2)]$, and so the above formula describes this comultiplication. (The element $X$ in the source goes to the element $(x_1^2 + x_1 x_2 + x_2^2)^2/(1-x_1^3 - x_2^3)^2 - x_1 - x_2$ in the target, where I've written $x_1 = x\otimes 1$ and $x_2 = 1 \otimes x$.) Regards, – Matt E Dec 18 '11 at 15:40
Thanks for adding that extra explanation! – Álvaro Lozano-Robledo Dec 18 '11 at 15:51