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.

Answer 2

In characteristic $p$, every map $E_1 \to E_2$ factors as a power of the Frobenius $\varphi_r \colon E_1 \to E_1^{(p^r)}$ followed by a separable morphism $E_1^{(p^r)} \to E_2$, and we find $r$ by looking at the inseparable degree of our map (if the map is separable, then $r=0$, as Pete pointed out).

Now, in the case of interest, if $E$ is supersingular, $\widehat{\varphi}$ is inseparable (as this is equivalent to multiplication by $p$ being purely inseparable). But then $\widehat{\varphi} \colon E^{(p)} \to E$ factors as $E^{(p)} \to E^{(p^2)} \to E$ by comparing degrees, where the first map is the Frobenius and the second is an isomorphism.

It then follows that $j(E) = j(E^{(p^2)}) = j(E)^{p^2}$ so $j(E) \in \mathbb{F}_{p^2}$.