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I'm trying to solve the following exercise from Silverman:

Let $E:y^2=x^3+x$. Show that $\# E(\Bbb F_p)$ is divisible by $4$ for prime $p\ge 3$.

My approach is as follows: if $\phi$ denotes the $p$-power Frobenius morphism, then $P\in E(\Bbb F_p)$ if and only if $\phi(P)=P$. Therefore $\# E(\Bbb F_p)=\#\ker(1-\phi)$, which because $1-\phi$ is separable is equal to $\deg(1-\phi)$. I have therefore reduced my problem to showing $\deg(1-\phi)$ is divisible by $4$.

To do this, I would like to use the more general fact that if $\psi$ is an endomorphism of $E$, and we let $\psi_{\ell}$ denote the induced endomorphism on the Tate module $T_{\ell}(E)$ for any prime $\ell$ not equal to the characteristic, then we have

$$\deg(\psi)=\det(\psi_{\ell}).$$

I would like to apply this to our case, when $\psi=1-\phi$ and $\ell=2$. So then I just need to calculate $\det((1-\phi)_2)$. I don't know a good way to do this though, as I'm not sure how to get an explicit $\Bbb Z_{\ell}$-basis of $T_{\ell}(E)\cong\Bbb Z_{\ell}\times\Bbb Z_{\ell}$ (the proof of the latter isomorphism was very non-constructive). I don't even know how to give any explicit points of $E(\Bbb F_p)$ besides $O$, as it relies on $x^3+x$ being a quadratic residue modulo $p$.

Does anybody have any suggestions on how to proceed? Does this approach even seem viable, or is there an easier way to do this?

Alex Mathers
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1 Answers1

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All you have to do is find a subgroup of order $4$ in $E(\Bbb F_p)$. This is easy if $p\equiv1\pmod4$. In this case $(0,0)$, $(i,0)$ and $(-i,0)$ are elements of order $2$, where $i^2\equiv-1\pmod p$.

When $p\equiv 3\pmod 4$ you only have one element of order $2$, viz., $(0,0)$. I suppose in this case you need to find $(x,y)\in E(\Bbb F_p)$ with $[2](x,y)=(0,0)$. Too much like hard work for me at the moment $\smile$.

Angina Seng
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