Let's take a supersingular elliptic curve $E / \mathbb{F}_q$, where $q = p^2$ for some $p \equiv 1 (\text{mod } 4)$, and $p \geq 5$, let's say. Then $ \# E(\mathbb{F}_q) = (p - 1)^2$, which by Hasse's theorem implies that the $q$-power Frobenius endomorphism $\phi_q$ satisfies the characteristic equation:

$\phi_q^2 - 2p \cdot \phi_q + q = 0 = (\phi_q - p)^2$

in $\text{End}(E)$.

How does this not imply that $\phi_q$ is identically equal to the multiplication-by-p map $[p]$? It's surely not, or else $\phi_q$ would represent an integral element of the quaternion-order $\text{End}(E)$.

EDITED: Looks like I flipped a sign in the trace equation (thanks @reuns). Thus requiring now $p \equiv 1 (\text{mod } 4)$, re: @hutner's comment, at least we have no contradiction in this case, as indeed $E(\mathbb{F}_q)$ is a $p - 1$-group.

The corresponding fact for $p \equiv 3 (\text{mod } 4)$ is that $\phi_q = [-p]$ should act as the identity on $E(\mathbb{F}_q) \cong \left( \mathbb{Z} / (p + 1)\mathbb{Z} \right)^2$. But this in turn is equivalent to $[p + 1]$ being the zero map, so we're again ok.

Thus at least no contradiction follows from @Lord Shark the Unknown's claim! Thanks everyone.

  • 1
    $\begingroup$ What's wrong with $\phi$ equalling $[p]$? It's definitely in the quaternion order, and one expects $\hat\phi\circ\phi=[q]$. $\endgroup$ Nov 9, 2018 at 2:54
  • $\begingroup$ Thanks for your response. But then where are the non-integral endomorphisms going to come from (there should be three which are $\mathbb{Z}$-linearly independent). $\endgroup$
    – BD107
    Nov 9, 2018 at 3:10
  • 1
    $\begingroup$ Alas, in this situation, these non-trivial endomorphisms don't come "for free". $\endgroup$ Nov 9, 2018 at 3:21
  • $\begingroup$ @LordSharktheUnknown I don't think that $\phi$ could equal $[p]$. For example, this would imply that every point in the ground field was annihilated by $[p-1]$, but this contradicts the given count since there are a maximum of $(p-1)^2$ such $\mathbb{F}_q$-points. $\endgroup$
    – hunter
    Nov 9, 2018 at 3:36
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    $\begingroup$ So it would work better if $\phi^2+2p \phi + p^2 = 0$ $\endgroup$
    – reuns
    Nov 9, 2018 at 3:43


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