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I have some questions to clarify my understanding about Deuring correspondence between quaternions and isogenies in SQIsign(2D) version 2.0.1 https://sqisign.org/

Let $E_0$ be an elliptic curve with known endomorphism $\operatorname{End}(E_0) \cong O_0$. This means that we know 4 explicit endomorphisms that generate $\operatorname{End}(E_0)$ as rank-4 module over $\mathbb{Z}$. We also know 4 corresponding explicit quaternions that generate the maximal order $O_0$ as module over $\mathbb{Z}$.

Let $I$ be a left ideal of $O_0$. Let $O$ be the right order of $I$. Let $\varphi: E_0 \to E$ be an isogeny that corresponds $I$, i.e. $\ker \varphi = \cap_{\alpha \in I} \ker \alpha$.

  1. What is the naturally induced isomorphism $O \cong \operatorname{End}(E)$?

Let $I_{sk}\cdot I_{chl}$ be ideal that corresponds to the chain of isogenies $\varphi_{chl}\circ \varphi_{sk}: E_0 \to E_{pk} \to E_{chl}$. Also let $I_{com}$ correspond to $\varphi_{com}: E_0 \to E_{com}$.

  1. Why/How does an element $\alpha \in \overline{I_{com}} \cap I_{sk}\cdot I_{chl}$ correspond to an endomorphism of $E_{com}$ that factors through $\overline{\varphi_{com}} \circ \varphi_{sk}\circ\varphi_{chl}$?

  2. Why/How could an $\alpha$ alternatively be seen as an endomorphism of $E_0$?

My general understanding is that a collection of quaternions/endomorphisms (like an ideal) defines a kernel by taking the intersection of the endomorphisms' kernels; and the kernel defines an isogeny (up to ismorphism). But how do you "transport" endomorphisms and kernels between 2 different curves?

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2 Answers 2

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To answer your questions

  1. If $φ:E_0→E$, the map $ω ↦ \hatφωφ$ defines an injection of modules $\mathrm{End}(E) \hookrightarrow \mathrm{End}(E_0)$, and $ω ↦ \frac{\hatφωφ}{\degφ}$ an isomorphism between $\mathrm{End}(E_0)$ and the right order of $I_φ$. You can easily check that these elements stabilise $I_φ$ on the right (or see Voight, Quaternion Algebras, Lemma 42.2.10)

1½. As an aside, observe that if $ω = φψ$ is an endomorphism of $E$, then its image under the isomorphism above is $\hatφφψφ/\degφ = ψφ$, that's "the same" endomorphism with the order of the arrows reversed.

  1. I think it is clear that $α$ corresponds to an endomorphism of $E_0$ that, for want of better terminology, factors through $φ_{\mathrm{sk}}φ_{\mathrm{chl}}$ "at the start" and through $\hatφ_{\mathrm{com}}$ "at the end", pretty much by definition, right? Using point 1½, you see this is precisely the image in $E_0$ of an endomorphism of $E_{\mathrm{com}}$ factoring through $\hatφ_{\mathrm{com}}φ_{\mathrm{sk}}φ_{\mathrm{chl}}$.

  2. The previous point answers this too.

Hope this helps. I always find that the definition of $I_φ$ as "all the endomorphisms that factor through $φ$" is easier to reason about than the (equivalent) definition using kernels, and makes transporting endomorphisms between rings easier.

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So one of my favorite rant on this topic is that the module point of view is often better suited for these kind of questions.

The module Deuring correspondance is that if $E_0/\mathbb{F}_{p^2}$ is a (maximal) supersingular curve and $O_0=\mathrm{End}(E_0)$, then $E \mapsto M_E=Hom(E, E_0)$ is an anti-equivalence of category between (maximal) supersingular elliptic curves and $O_0$-left torsion free modules of rank $1$ (with the obvious morphisms). As a special case, $E_0$ corresponds to $O_0$. [As an aside, this extends naturally to higher dimension by allowing torsion free modules of rank $g$, while of course the ideal point of view can only describe elliptic curves.]

From this point of view, the natural isomorphism $End(E) = End(M_E)$ is given as follows: if $\gamma \in End(E)$ and $\phi: E \to E_0$, then $\phi \cdot \gamma$ is the composition $\phi \circ \gamma$. (Note that $End(E)$ acts on the right of $M$, and $M$ has a natural $(O_0, End(E))$-bimodule structure: if $\gamma_0 \in O_0$, $\gamma_0 \cdot \phi = \phi \circ \gamma_0$.) This answers the first question.

For the second question: if $\phi: E_1 \to E_2$ is an isogeny, then the associated module monomorphism $M_{\phi}: M_2 \to M_1$ is given in the same manner as the construction above: to a morphism $\psi: E_2 \to E_0$ we associate $\psi \circ \phi: E_1 \to E_0$. In particular, the image of $M_2$ in $M_1$ are given by all morphisms $E_1 \to E_0$ that factor through $\phi$.

As a special case, an isogeny $\phi: E_0 \to E$ corresponds to a module monomorphism $M_E \to O_0$, and the image of $M_E$ is the ideal $I_E$ associated to $E$ by the "ideal Deuring correspondance". We recover Luca's point of view that $I_E$ is the set of all endomorphisms of $E_0$ that factor through $\phi$. But note that $I_E$ is not intrinsic (it depends on the choice of $\phi$), while $M_E$ is intrinsic. That's why it is often easier to reason on $M_E$ directly.

So now let's apply this to $\phi_{chl} \circ \phi_{sk}$: we have an associated module morphism $M_{chl} \to O_0$, whose image is, by the discussion above, precisely the ideal of endomorphisms of $E_0$ that factor (at the start) through $\phi_{chl} \circ \phi_{sk}$. Likewise for $I_{com}$, and so its conjugate is the ideal of endomorphism that factor at the end through $\hat{\phi}_{com}$. Assuming that the degree of these isogenies are coprime, it means that $\gamma \in O_0$ is in the intersection if it can be written as $\gamma = \hat{\phi}_{com} \circ \gamma' \circ \phi_{chl} \circ \phi_{sk}$ for some $\gamma': E_{chl} \to E_{com}$.

From the module point of view, the intersection $\overline{I}_{com} \cap I_{sk} \cdot I_{chl}$ is the image of $Hom(M_{com}, M_{chl})$ given by $(\psi: M_{com} \to M_{chl}) \mapsto M_{\phi_{chl} \circ \phi_{sk}} \circ \psi \circ M_{\hat{\phi}_{com}}$ and so the endomorphism $\gamma \in O_0$ is in this intersection if it factors as $\gamma: O_0 \to M_{com} \to M_{chl} \to M_{sk} \to O_0$. (This is just a restatement of the discussion above on isogenies, using contravariance). But such a $\gamma$ also gives an endomorphism of $M_{com}$ (by shifting the individual isogenies), that factor through $\phi_{chl}\circ \phi_{sk} \circ \hat{\phi}_{com}$.

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  • $\begingroup$ Thank you for the insightful answer, but I need help clarifying of the things you wrote. In the 3rd paragraph, where does the order $O$ come in? In the 7th/last paragraph, how do you define $M_f$ where $f$ is an isogeny? You defined $M_E$ above for elliptic curve $E$ but not $f$. $\endgroup$ Commented Sep 29 at 14:37
  • $\begingroup$ $O_0$ is the endomorphism ring of $E_0$. The definition of the module morphism $M_f$ associated to an isogeny $f$ is given in the fourth paragraph. (I added a few precisions in my answer) $\endgroup$ Commented Sep 30 at 19:37

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