j-invariant of a supersingular elliptic curve - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-19T19:07:49Z http://mathoverflow.net/feeds/question/18688 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/18688/j-invariant-of-a-supersingular-elliptic-curve j-invariant of a supersingular elliptic curve Josh 2010-03-19T00:33:49Z 2010-03-19T01:07:32Z <p>Let E be a supersingular curve over a finite field. Why is the j-invariant always in F_p^2?</p> http://mathoverflow.net/questions/18688/j-invariant-of-a-supersingular-elliptic-curve/18692#18692 Answer by Pete L. Clark for j-invariant of a supersingular elliptic curve Pete L. Clark 2010-03-19T00:49:18Z 2010-03-19T00:57:59Z <p>(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.)</p> <p>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 <code>$E[p](k) = 0$</code> 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 <em>for</em> $E^p$ -- 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. </p> http://mathoverflow.net/questions/18688/j-invariant-of-a-supersingular-elliptic-curve/18693#18693 Answer by Sam Derbyshire for j-invariant of a supersingular elliptic curve Sam Derbyshire 2010-03-19T00:50:35Z 2010-03-19T01:07:32Z <p>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).</p> <p>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.</p> <p>It then follows that $j(E) = j(E^{(p^2)}) = j(E)^{p^2}$ so $j(E) \in \mathbb{F}_{p^2}$. </p>