Let $(E,\mathcal{O})$ be an elliptic curve over a (perfect) field $K$ and let $\textbf{Cov}(E,\mathcal{O})$ denote the category of finite pointed étale covers of $E$ from smooth varieties where the distinguished point $\mathcal{O}'$ on each cover is a $K$-point. One can show that each of these covering varieties must be an elliptic curve $(E', \mathcal{O}')$ and the covering maps are then isogenies of elliptic curves. The morphisms $(E', \phi)\rightarrow (E'', \psi)$ in this category between two finite étale covers are pointed morphisms of algebraic varieties, and thus become finite étale isogenies in their own right (since the "base change" of an étale morphism is étale).

Let $\textbf{Cov}_{\mathbb{N}} (E,\mathcal{O})\subseteq\textbf{Cov}(E,\mathcal{O})$ denote the full subcategory consisting of multiplication-by-$m$ isogenies $[m]:E\rightarrow E$ for $m\in K^\times$ (see this other question of mine). I'm wondering if there is a functor $\Delta: \textbf{Cov}(E,\mathcal{O})\rightarrow\textbf{Cov}_{\mathbb{N}}(E,\mathcal{O})$ sending each isogeny $\phi: E'\rightarrow E$ to the precomposition with the dual isogeny $[\deg\phi] = \phi\circ\hat{\phi}:E\rightarrow E$?

More precisely, if $f: (E', \phi)\rightarrow(E'', \psi)$ is a morphism in $\textbf{Cov}(E,\mathcal{O})$, does there exist a (functorial) morphism $\Delta(f):E\rightarrow E$ in $\textbf{Cov}_{\mathbb{N}}(E,\mathcal{O})$ such that the following diagram commutes?

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1 Answer 1


After thinking a bit, I think I've solved this: set $\Delta(f) = [\deg f]:E\rightarrow E$. Then it's easy to see that $\zeta: = f\circ \hat{\phi} - \hat{\psi}\circ \Delta(f)$ is a morphism $E\rightarrow E''$ whose image is contained in the kernel of $\psi$. Now $\psi$ is a finite étale morphism so $\ker\psi$ is a finite subgroup of $E''$ of order $\deg\psi$. Since any morphism of smooth curves is either surjective or constant, it follows that $\zeta$ is a constant map. Now in $\textbf{Cov}(E, \mathcal{O})$ the morphism $f$ was pointed (i.e. is an isogeny), and therefore it follows that $\zeta$ is an isogeny. Thus $\zeta$ is identically zero, and therefore the diagram above commutes.

In particular with $\Delta(f) = [\deg f]$ we get $f \circ \hat{\phi} = \hat{\psi}\circ \Delta(f)$. The multiplicative property of degrees of compositions of morphisms implies that $\Delta$ is the functor we wanted.


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