# Degree of morphisms and isogenies


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.

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.
• 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 $d=2$). – Watson Apr 14 at 15:17
• 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 $\Bbb F_p$. – Watson Apr 14 at 18:38
• @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. – Ben Smith Apr 15 at 9:42