$\renewcommand{\J}{\mathrm{Jac}} \renewcommand{\F}{\mathbb{F}}$
I am reading [this paper](https://www.ams.org/journals/jams/1990-03-04/S0894-0347-1990-1071117-8/S0894-0347-1990-1071117-8.pdf) 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.



Degree of morphisms and isogenies
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.


