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

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The answers by Pete Clark and Sam Derbyshire below are great, but I'll add here a text reference: Silverman's "The Arithmetic of Elliptic Curves", Chapter V, Section 3, Theorem 3.1. In particular, see the proof of (ii) implies (iii). –  Álvaro Lozano-Robledo Jul 27 '11 at 19:24
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2 Answers

(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.

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Beat me to it. The Frobenius isogeny is always purely inseparable though, right? I think you meant that multiplication by $p$ is purely inseparable iff the dual to the Frobenius is inseparable. –  Sam Derbyshire Mar 19 '10 at 0:52
    
@Sam: yes, you're right. I fixed it accordingly. –  Pete L. Clark Mar 19 '10 at 0:58
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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 $j(E) \in \mathbb{F}_{p^2}$.

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Let me return the favor: every isogeny factors as above with the convention that the power of the Frobenius is allowed to be the zeroth power! By the way, it strikes me that it should be possible to "merge" answers and split the credit. That would be appropriate here... –  Pete L. Clark Mar 19 '10 at 1:00
    
Haha, I guess it was confusing; it's clear that we need the zeroth power whenever our map is separable. As for the rest, I don't see any issues; it's a standard answer anyway! –  Sam Derbyshire Mar 19 '10 at 1:13
    
Thanks for the answers guys –  Josh Mar 19 '10 at 2:35
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