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$\renewcommand{\J}{\mathrm{Jac}} \renewcommand{\F}{\mathbb{F}}$ I am reading this paper by B. Gross, and there is something I don't understand on p. 945. Here is the context: fix a prime $p \equiv 3 \pmod 4$, and define the (hyper)elliptic curves over $\F_p$ given by the (affine) equations $$X_1 : y^2 = x^p - x,\quad X_2 : y^2 = x^{p+1}-1,\quad E_1 : y^2 = x^3-x,\quad E_2 : y^2=x^4-1.$$

I checked (using Tate isogeny theorem) that there is a non-zero isogeny $\alpha : \J(X_1) \to \J(X_2)$ between the jacobian varieties (actually both are isogenous to $E_1^{(p-1)/2}$ over $\F_p$), and there is a non-zero isogeny $\beta : E_1 \to E_2$.

Question: There is a morphism $f_2 : X_2 \to E_2, (x,y) \mapsto (x^{(p+1)/4}, y)$ which has degree $(p+1)/4$. Then it is claimed that we therefore get a morphism $f_1 : X_1 \to E_1$ of degree $(p+1)/4$, but I don't see why/how.


Thoughts: I know that $f_1$ induces a morphism $\phi_2 : \J(X_2) \to E_2$, we get a morphism $\beta \circ \phi_2 \circ \alpha^{\vee} : \J(X_1) \to E_1$, hence a morphism $f_1 : X_1 \to E_1$, but I believe that it has degree at least the degree of $f_2$. Maybe there is a clever way to compose $\phi_2$ with other isogenies to get the equality of degrees?

In general, given a non-constant morphism $f_2 : X_2 \to E_2$, it might not be possible to get a morphism $f_1 : X_1 \to E_1$ of same degree as $f_2$: just take $X_2 = E_2 = X_1, f_2 = \mathrm{id}$ and $E_1$ an elliptic curve isogenous but not isomorphic to $E_2$. I am probably missing something easy, but I prefer asking for clarifications.

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It's easier if we forget about isogenies: $E_1$ and $E_2$ are isomorphic, and $X_1$ and $X_2$ are isomorphic, so the cover $X_2\to E_2$ induces a cover $X_1 \to E_1$ of the same degree.

To make this more explicit: let $d = (p+1)/4$, and rewrite the curve equations in separate coordinate systems: \begin{align*} X_1: Y^2 & = X^p - X \,, & X_2: V^2 & = U^{4d} - 1 \,, \\ E_1: y^2 & = x^3 - x \,, & E_2: v^2 & = u^4 - 1 \,. \end{align*} Let $i$ be a square root of $-1$ in $\mathbb{F}_{p^2}$. Then there is an isomorphism $\psi: X_1 \to X_2$ defined by $$ \psi: (X,Y) \longmapsto (U,V) = \left( \frac{X + i}{X - i} , \frac{(i+1)Y}{(X -i)^{2d}} \right) $$ (here we need $i^p = -i$), the degree-$d$ cover $f_2: X_2 \to E_2$ defined by $$ f_2: (U,V) \longmapsto (u,v) = (U^d, V) \,, $$ and an isomorphism $\phi: E_2 \to E_1$ defined by $$ \phi: (u,v) \longmapsto (x,y) = \left( -i\cdot\frac{u + i}{u - i} , \frac{(i+1)v}{(u-i)^2} \right) \,. $$ Composing, we get a degree-$d$ cover $f_1 = \phi\circ f_2\circ\psi: X_1 \to E_1$, which is what we wanted... Well, almost what we wanted, because we would probably want $f_1$ to be defined over $\mathbb{F}_p$. But expanding, we see that $f_1$ is defined by $$ f_1: (X,Y) \longmapsto (x,y) = \left( -i\cdot\frac{ (X + i)^d + i(X-i)^d }{ (X + i)^d - i(X-i)^d } , \frac{ 2i Y }{ ((X + i)^d - i(X-i)^d)^2 } \right) \,, $$ Both of the rational functions are symmetric with respect to $i \leftrightarrow -i$, so they are defined over $\mathbb{F}_p$, and therefore so is $f_1$.

All four curves have plenty of automorphisms (some over $\mathbb{F}_p$, some over $\mathbb{F}_{p^2}$) which you can compose with these morphisms to produce more solutions.

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  • $\begingroup$ Dear Ben Smith, thank you so much for your helpful answer. But why is $f_1$ defined over $\Bbb F_p$ ? According to my computation, the second component is not invariant under swapping $i$ and $-i$ (set e.g. $X=0, Y=1, p=7$, so $d=2$). $\endgroup$ – Watson Apr 14 at 15:17
  • $\begingroup$ I think that taking $\dfrac{X + i}{X - i}$ in the first coordinate of $\psi$, and by taking $$\phi(u,v) = \left( -i \cdot\frac{u + i}{u - i}, i \cdot\frac{(i-1)v}{(u-i)^2} \right),$$ does give a morphism $f_1$ which is defined over $\Bbb F_p$. $\endgroup$ – Watson Apr 14 at 18:38
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    $\begingroup$ @Watson you're right, I compounded a few typos in my working. I think you might have an $i$ too many on the $y$-coordinate of $\phi$, too. Will update my answer accordingly. $\endgroup$ – Ben Smith Apr 15 at 9:42

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