# Supersingular elliptic curves and their automorphisms

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 $$|Aut(E)|=2,4,6$$)?

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

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.

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.