Let $E$ be an elliptic curve which has CM by $ \Bbb{Z}[ \sqrt{-5}]$ with j invariant $632000+282880\sqrt{5}$.
I want to know whether $E$ has good reduction at $(2,1+\sqrt{―5})$ or not, and to know what is $ \sharp E[ \Bbb{F_2}]$.
But I'm stuck with calculating this by hand.
I couldn't find this elliptic curve in LMGDB.
Thank you for your help.
Best Answer
Consider the following elements in the field $H=\Bbb Q(\sqrt{-1},\sqrt{-5})$: $$ \begin{aligned} k &= \sqrt{-1}\ ,\\ a &= \sqrt{-5}\ ,\\ ak &=\sqrt{5} \text{ (in a manner compatible with the choices for $a,k$)} \\[3mm] s &=\frac 12(\sqrt{-1}+\sqrt{-5})\ . \end{aligned} $$ Then $s$ is an algebraic integer, the minimal polynomial of $s$ is $F=X^4 + 3X^2 + 1$, so $s$ is of norm one, and $1,s,s^2,s^3$ turns out to be an integral basis of $H$. The reduction modulo two of $F$ is $f=x^4 + x^2 + 1=(x^2+x+1)^2$, so it is natural to consider $\Bbb F_4:=\Bbb F_2[u]$, where $u$ has minimal polynomial $(x^2+x+1)$. There is thus a map $\Phi$ from the ring $\mathcal O_H$ to $\Bbb F_4$ defined on the generator $s$ by $s\to u$.
In particular $\Phi(s^2)=u^2=u+1$.
The elements $a,k,ak$ have the following expressions in terms of $s$: $a=s^3+4s$, $k=s^3 +2s$, $ak=-2s^2-3$. We compute $\Phi(2)=0$, $\Phi(1+ak)=0$, so $\Phi$ induces a map from $\mathcal O_H$ modulo $(2,1\pm ak)$ (which corresponds to $(2,1+\sqrt 5)$ in either choice of the square root) to $\Bbb F_4$.
We will use $\Phi$ when passing from an equation defined over $\mathcal O_H$ to one over $\Bbb F_4$. (I.e. we apply the tensor functor $-\otimes_{\mathcal O_H}\Bbb F_4$ with $\Phi$ as structural map.)
Time to get the equation of the curve. Computers are in this century the weapon of choice, when a lookup in the database goes into the void. In the above theoretical setup, we use the code:
And
E
is:(Code was manually rearranged.)
Now the whole discussion about $\Phi$ is computationally useless, (but structurally needed to have a morphism to "something" in characteristic two, thus giving sense to the story,) since over $\Bbb F_2$ most involved coefficients vanish, giving rise to $y^2=x^3$.
$\square$
The answer is finished, but i tried to find other $j$-values (to search for them), those for which we may obtain after reduction regular elliptic curves over $\Bbb F_4$. The list of these possible CM-discriminants, that are not discarded from the start (having $a_4$, $a_6$ not zero modulo $(2)$) was obtained via:
We get:
And over $\Bbb F_4=\Bbb F_2[u]$ we get respectively the curves (recalling $\Phi(s^2)=u+1$) $E_1$, $E_2$, $E_3=E_1$, $E_4=E_2$, with: $$ \begin{aligned} E_1 &:\qquad y^2 = x^3 + ux\ ,\\ E_2 &:\qquad y^2 = x^3 + (u+1)x \ . \end{aligned} $$ They are singular because of $x^3 +ux=x(x+u+1)^2$, and $x^3+(u+1)x=x(x+u)^2$.