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After having a look at Silverman's "Arithmetic of Elliptic Curves", I mostly understand the notion of a supersingular Elliptic Curve and its characterizations. However, some subtleties still confuse me. Could someone point out the error in the following, obviously wrong argument?

Let $E$ be a supersingular elliptic curve defined over a $\mathbb{F}_{p^2}$, and assume further that $j(E) \neq 0, 1728$. By one of the characterizations of supersingularity, we see that the multiplication-by-$p$ map is purely inseparable.

By basic algebraic geometry (or e.g. theorem II.2.12 in Silverman's AoEC), it follows that $[p]$ factors as $$ E \ \overset{\pi^2}{\to} \ E^{(p^2)} = E \ \overset{\lambda}{\to} \ E $$ for the $p$-th power Frobenius morphism $\pi$ and a separable isogeny $\lambda$. However, since $[p]$ has separability degree 1, $\lambda$ is already an automorphism. Since $j(E) \neq 0, 1728$ we see already that $\lambda = \pm [1]$. So $\pi^2 = \pm [p]$, and in particular, the trace of the Frobenius endomorphism is $0$, as $\pi^2 \in \mathbb{Z}$.

Since all supersingular j-invariants are defined over $\mathbb{F}_{p^2}$ and the endomorphism ring of $E$ is preserved under isomorphism, this assumption should not make a difference.

However, I am sure that it is not true that the trace of the Frobenius endomorphism is always $0$ in case of a supersingular curve (at least, Silverman and Wikipedia say $E$ supersingular iff the trace is $\equiv 0 \ (\mathrm{mod} \ p)$).

Feanor
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    $\pi=\varphi_p$ is an endomorphism only if $E$ is defined over $\Bbb{F}_p$. The restriction that every endomorphism lie in an imaginary quadratic field gives that either $q=p^{2n}$ is a square and $\varphi_q^2 = \pm q$ (if it is $-$ then $j=1728$, if it is $+$ then $Tr(\varphi_q)=2p^n$) or $q$ is not a square and $\varphi_q^2=-q$. – reuns Feb 27 '22 at 18:04
  • @reuns: Sorry, could you explain the point of your comment? Does it describe where my above argument has a mistake (I do not understand how it does), or is it a general note about which sign we get in the equality $\pi^2 = \pm p$? – Feanor Feb 28 '22 at 14:37
  • I found a possible solution: In Silverman, this statement is Ex. 5.10 and it remarks that the trace of $\pi^2$ is to be computed in the endomorphism ring of the Tate module $\operatorname{End}(T_l(E))$ and not in $\operatorname{End}(E)$. I have not yet thought it through, but maybe this fixes the problem. – Feanor Apr 08 '22 at 13:21

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