Lemma 3.2.1 in [Baker, González-Jiménez, González, Poonen, "Finiteness theorems for modular curves of genus at least 2", Amer. J. Math. 127 (2005), 1325–1387.](https://math.mit.edu/%7Epoonen/papers/finiteness.pdf)   

[enter image description here](https://i.stack.imgur.com/ecXLP.png)  

I don't understand why "all supersingular elliptic curves over $\bar{F\_p}$ are isogenous".
Could anyone help me?  

This problem arises from [Supersingular elliptic curves and their "functorial" structure over F\_p^2](https://mathoverflow.net/questions/18982/supersingular-elliptic-curves-and-their-functorial-structure-over-f-p2).  

Moreover, I found some similar question: [Isogenies between elliptic curves and their endomorphism rings](https://math.stackexchange.com/questions/2364125/isogenies-between-elliptic-curves-and-their-endomorphism-rings) and [Isogenies between supersingular elliptic curves](https://mathoverflow.net/questions/86444/isogenies-between-supersingular-elliptic-curves).



Why all supersingular elliptic curves over $\bar{\mathbb{F}_p}$ are isogenous?
This can be proved in several ways (this can be found in the literature as Lemma 42.1.11 in Voight's book on Quaternion algebras, or Proposition 5.2 in <https://arxiv.org/pdf/2005.01537v1.pdf>). Typically the proof relies on Tate's isogeny theorem (1966) which states, among other things, that if two elliptic curves over a finite field have the same number of rational points, then they are isogenous.


$$
\newcommand{\End}{\mathrm{End}}
\newcommand{\Q}{\mathbb{Q}}
\newcommand{\F}{\mathbb{F}}
\newcommand{\Z}{\mathbb{Z}}
\newcommand{\C}{\mathbb{C}}
$$




---


**Proposition.**
Let $p$ be a prime.
Let $E,E'$ be two supersingular elliptic curves over $\F\_{p^2}$. Then $E,E'$ are isogenous over $\F\_{p^{24}}$.


(Note that the $j$-invariant of supersingular elliptic curves over $\overline{\F\_p}$ belongs to $\F\_{p^2}$, we may assume (up to $\overline{\F\_p}$-isomorphism) that $E,E'$ are defined over $\F\_{p^2}$).




---


**Proof.**


Write $\alpha, \beta \in \C$ for the eigenvalues of the $p^2$-Frobenius map on the $\ell$-adic Tate module of $E$ for some $\ell \neq p$, so that $|E(\F\_{p^{2r}})| = p^{2r} + 1 - (\alpha^r + \beta^r)$ for every $r \geq 1$.
Since $E$ is supersingular, its trace $t\_1 := \alpha + \beta$ is $\equiv 0 \pmod p$. By Hasse bound we have $|t\_1| \leq 2 \sqrt{p^2} = 2p$, so we have $t\_1 \in \{ -2p, -p, 0, p, 2p \}$.


In each case, $\alpha, \beta$ can be computed explicitly as roots of a quadratic polynomial (characteristic polynomial of Frobenius):


* If $t\_1 = -2p$, then $X^2 + 2pX + p^2 = (X+p)^2
 = 0$ yields $\alpha = \beta = -p$.
* If $t\_1 = -p$, then $X^2 + pX + p^2 = 0$ yields $\alpha,\beta =
 \frac{-p + \sqrt{p^2 - 4p^2}}{2} = p \cdot \dfrac{-1 \pm i
 \sqrt{3}}{2}$
* If $t\_1 = 0$, then $X^2 + p^2 = 0$ yields $\alpha = - \beta =
 ip$.
* If $t\_1 = p$, then $X^2 - pX + p^2 = 0$ yields $\alpha,\beta =
 p \cdot \dfrac{1 \pm i \sqrt{3}}{2}$
* If $t\_1 = 2p$, then $(X - p)^2 = 0$ yields $\alpha = \beta =
 p$.


In all cases, we observe that $\alpha^{12} = \beta^{12} = p^{12}$ (i.e. $\alpha,\beta$ are equal to $p$ times some $12$-th root of unity [in fact either of order $1,2,3,4$ or $6$]). This proves that
$$
|E(\F\_{p^{24}})| = 
p^{24} + 1 - (\alpha^{12} + \beta^{12})
=
p^{24} + 1 - 2 \cdot p^{12}
=
|E'(\F\_{p^{24}})|
$$
so that $E,E'$ are isogenous over $\F\_{p^{24}}$ by Tate's theorem. This concludes the proof. $\blacksquare$


