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Let $p$ be prime and $q = p^n$. Let $E$ be an elliptic curve over $\mathbb{F}_q$, and let $E^{(p)}$ be the pullback of $E$ by the $p$-power Frobenius of $\mathbb{F}_q$. If $E$ is isomorphic (over $\mathbb{F}_q$) to its Galois conjugate $E^{(p)}$, then does it follow that $E$ is the base change of an elliptic curve over $\mathbb{F}_p$? If so, what is the argument?

Note that similar statements are not true over infinite fields: for instance, $\mathbb{Q}$-curves are isomorphic to all their Galois conjugates, yet need not descend to actual elliptic curves over $\mathbb{Q}$.

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  • $\begingroup$ $\mathbb{Q}$-curves are isogenous to all their Galois conjugates, not necessarily isomorphic. $\endgroup$
    – Pete L. Clark
    Oct 23, 2015 at 23:00

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Almost, but not quite. If $j(E)$ is the $j$-invariant of $E$ then, under your hypothesis, $j(E)=j(E^{(p)})=j(E)^p$, so $j(E) \in \mathbb{F}_p$. Hence $E$ is either defined over $\mathbb{F}_p$ or is a twist of such a curve. If you take the quadratic twist in $\mathbb{F}_{p^2}$ of an elliptic curve defined over $\mathbb{F}_p$ you get a counterexample.

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  • $\begingroup$ The quadratic twist in $\mathbb{F}_{p^2}$ of an elliptic curve defined over $\mathbb{F}_p$ is still an elliptic curve defined over $\mathbb{F}_p$, so I don't see how this is a counterexample. I agree with the rest of what you say but don't see how it settles my question. $\endgroup$
    – Lisa S.
    Oct 9, 2015 at 3:24
  • $\begingroup$ If the curve over the prime field is $y^2=f(x)$ take $d \in \mathbb{F}_{p^2}$, not a square in that field and consider $dy^2=f(x)$. That's not the same as twisting over the prime field (this would isomorphic to the first curve over the quadratic equation). $\endgroup$
    – Felipe Voloch
    Oct 9, 2015 at 3:30
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    $\begingroup$ @LisaS.: Let $E_0$ be an elliptic curve over $k_0$ (size $q_0$), and $k/k_0$ finite. For an elliptic curve $E$ over $k$ satisfying $E\simeq E^{(q_0)}$, does $E$ arise from a $k_0$-form of $E_0$; i.e., for $A={\rm{Aut}}_{E_0/k_0}$ the finite etale Aut-scheme, is the image of $f:H^1(k_0,A)\rightarrow H^1(k,A_k)$ characterized by invariance under composing $q_0$-Frobenius $A_k \rightarrow A_k^{(q_0)}$ (isom.!) with scalar extension of the canonical $A^{(q_0)} \simeq A$? This composite is identity if $A=\mu_n$ (constant dual!), so fails when $n|\#k^{\times}$ with $k\supset k_0^{1/n}$. $\endgroup$
    – grghxy
    Oct 9, 2015 at 5:48
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    $\begingroup$ @LisaS.: The same holds for $A = \mathbf{Z}/(2)$ for any $p$ with $[k:k_0]$ is even. But the cases $A = \mu_n$ for some $n$ not divisible by $p$ and $A = \mathbf{Z}/(2)$ for any $p$ cover all cases except for $p=2, 3$ with $j=0=1728$. Hence, we have uniformly handled all cases except supersingular elliptic curves in characteristics 2 and 3. $\endgroup$
    – grghxy
    Oct 9, 2015 at 6:13
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    $\begingroup$ @LisaS.: Perhaps ask your friend to explain their argument or D-R more clearly? Anyway, the D-R argument does not yield a result on your question because at the end of it they introduce a finite extension $k'/k$ of some unknowable degree $m$ and only descend $E_{k'}$ to an elliptic curve $E'_0$ over $k_0$, rather than descend $E$; i.e., $(E'_0)_k$ is merely a $k$-form of $E$ (in my notation, they say nothing about the image on H$^1$'s). Their method exerts no control on the specific field "upstairs" from which they build a descent. $\endgroup$
    – grghxy
    Oct 9, 2015 at 16:44

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