[Math] j-invariant of a supersingular elliptic curve

Question

Let E be a supersingular curve over a finite field. Why is the j-invariant always in F_p^2?

Answer 1

(Note: the following argument uses the fact that an isogeny of elliptic curves is inseparable iff it factors through the Frobenius isogeny. This is a result in Silverman's book, for instance.)

Let $E$ be an elliptic curve over an algebraically closed field $k$ of positive characteristic $p$. Recall that $[p]: E \rightarrow E$ is always an inseparable isogeny. Therefore, by the above, it factors through $F: E \rightarrow E^p$. Moreover $E$ is supersingular iff $E[p](k) = 0$ iff $[p]$ is purely inseparable, iff the dual isogeny to Frobenius $V: E^p \rightarrow E$ (the "Verschiebung") is also inseparable. But again, this means that $V$ factors through the Frobenius isogeny for $E^p$ -- i.e., $E^p \rightarrow E^{p^2}$ -- and since both have degree $p$ this means that $E$ is isomorphic to $E^{p^2}$. Thus on $j$-invaraiants we have $j(E)^{p^2} = j(E)$, done.