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If $E$ is a supersingular elliptic curve over a finite field of characteristic $p$, what is known about its automorphism group (i.e the stabilizer of a point in algebraic curves terminology). Do all the possibilities for the automorphism group $G$ occur for supersingular elliptic curves (if $p\ne 2,3$ this means $|Aut(E)|=2,4,6$)?

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  • $\begingroup$ The two answers skip over $p=2$ and $p=3$, which are the most interesting answers (though Wikipedia describes them in part) $\endgroup$ Commented Feb 4 at 5:22

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I will assume throughout that $p \neq 2,3$.

For $p$ a prime congruent to $3$ mod $4$, and $q$ any power of $p^2$, the curve $y^2 = x^3 - x$ is supersingular over $\mathbb F_q$ and has an automorphism of order $4$. By the upper bound for automorphisms you note, this means its automorphism group has order $4$.

These fields are the only ones that have a supersingular curve of order $4$. If $p $ is congruent to $1$ mod $4$ then because $p$ splits in $\mathbb Q(i)$, $\mathbb Q(i)$ does not embed into any quaternion algebra ramified at $p$, including the automorphism algebra of a supersingular elliptic curve. Otherwise, if $q$ is an odd power of $p$, then supersingular curves must have trace of Frobenius $0$ so their endomorphism field is $\mathbb Q(\sqrt{-p})$, which does not include $i$.

Similarly, for $p$ a prime congruent to $5$ mod $6$, and $q$ any power of $p^2$, the curve $y^2=x^3-1$ is supersingular and has an automorphism of order $6$ over $\mathbb F_q$, and these are the only fields where this is possible.

Now automorphism order $2$ happens if and only if it doesn't have an automorphism of of order $4$ or $6$. Over an algebraically closed field of characteristic $p$, this happens if and only if the $j$ invariant is not $0$ or $1728$. The number of such $j$ invariants is the integer part of $\frac{p-1}{12}$ and so is nonvanishing for $p> 12$. All these $j$ invariants are defined over $\mathbb F_{p^2}$ and so over $\mathbb F_q$ for $q$ an even power of $p$. For $q$ an odd power of $p$ we can take any elliptic curve with characteristic polynomial $T^2 +q$, which exists by Honda's theorem, because its endomorphism field is $\mathbb Q(\sqrt{-q}) = \mathbb Q(\sqrt{-p})$ and does not include $i$. This works even if $p<12$ if we only care about automorphisms defined over the ground field.

Over even powers of $p$ for $p<12$ I think it is not possible to find such a supersingular elliptic curve.

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We only need to look at $j$-invariants $0$ and $1728$, since every other elliptic curve has automorphism group $\{\pm1\}$.

For $p > 3$,

  1. The curve $E_0/\mathbb{F}_p: y^2 = x^3 + 1$ is supersingular if and only if $p \equiv 2 \pmod{3}$, if and only if the automorphism $(x,y)\mapsto(\zeta_3 x, y)$ is not defined over $\mathbb{F}_p$. Over $\mathbb{F}_p$ we have $\#\mathrm{Aut}(E_0) = 2$, but over even-degree extension fields we get $\#\mathrm{Aut}(E_0) = 6$.
  2. Similarly, the curve $E_{1728}/\mathbb{F}_p: y^2 = x^3 + x$ is supersingular if and only if $p \equiv 3 \pmod{4}$, if and only if the automorphism $(x,y)\mapsto(-x,\sqrt{-1}y)$ is not defined over $\mathbb{F}_p$. Over $\mathbb{F}_p$ we have $\#\mathrm{Aut}(E_{1728}) = 2$; over even-degree extensions we get $\#\mathrm{Aut}(E_{1728}) = 4$.

In each case, the irrational automorphism does not commute with the $p$-power Frobenius endomorphism, so the endomorphism ring of the curve (over $\overline{\mathbb{F}}_p$) is non-commutative - and hence the curve is supersingular.

When $p = 2$ and $p = 3$, all supersingular curves have $j$-invariant $0 = 1728$, so they all have extra automorphisms, though again those automorphisms are not necessarily defined over the ground field.

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