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Let $R$ be the ring of integers in a (complete) algebraic closure of $\mathbb Q_p$ with maximal ideal $\mathfrak p$. Suppose I have an Abelian surface $\mathcal A/R$ such that over every $R/\mathfrak p^n$, there exist elliptic curves $E_n, E_n'$ over $R$ with $\mathcal A$ isogenous to $E_n\times E_n'$ over $R/\mathfrak p^n$.

Does this imply that $\mathcal A$ splits over $R$? Note that I do not require any compatibility conditions between the elliptic curves or the isogenies over varying $n$.

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  • $\begingroup$ If you're happy to assume that $\mathcal{A}$ is ordinary, then $E_n,E_n'$ are also ordinary for all $n$ and you can apply Serre--Tate lifting theorem, I believe. $\endgroup$ Commented Dec 6, 2019 at 9:43
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    $\begingroup$ The quotient $R/\mathfrak{p}^n$ should probably be replaced by $R/p^n$ since the ideal $\mathfrak{p}$ is equal to all of its powers: $\mathfrak{p}=\mathfrak{p}^2=\dots$. $\endgroup$
    – SashaP
    Commented Dec 6, 2019 at 16:27

1 Answer 1

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$\newcommand{\cA}{\mathcal{A}}\newcommand{\cB}{\mathcal{B}}\newcommand{\bZ}{\mathbb{Z}}$No, that does not imply that $\cA$ splits over $R$. In fact, if $\cA_1=\cA\times_R R/p$ is isogenous to a product of elliptic curves then all $\cA_n=\cA\times_R R/p^n$ are isogenous to products of elliptic curves:

Lemma. Let $S'\twoheadrightarrow S$ is a surjection of $\bZ/p^n$-algebras with nilpotent kernel. If $\cA,\cB$ are abelian schemes over $S'$ then for any morphism $f:\cA\times_{S'}S\to\cB\times_{S'}S$ of abelian schemes over $S$ there exsists a morphism $g:\cA\to\cB$ over $S'$ reducing to $p^Nf$ for a certain $N$.

Proof. The analogous statement for $p$-divisible groups is a part of Drinfeld's rigidity lemma for quasi-isogenies and the statement for abelian schemes follows from Serre-Tate theorem(Lemma 1.1.3 and Theorem 1.2.1 in Katz's 'Serre-Tate local moduli').

In fact, this particular consequence can be also derived by a more direct argument: we can define the desired $g$ on the level of functors of points. There exists $N$ such that for any $S'$-algebra $T$ the kernel of the surjection $\cB(T)\to \cB_{S}(T\otimes_{S'}S)$ is annihilated by $p^N$. For a section $x\in\cA(T)$ define then $g(x)$ as $p^N\widetilde{f(\overline{x})}$ where $\widetilde{\cdot}$ denotes an arbitrary lifting along the surjection $\cB(T)\to\cB_S(T\otimes_{S'}S)$, the result does not depend on the choice. $\square$

Applying the lemma to the thickening $R/p^n\to R/p$ and the abelian schemes $\cA_n$ and $\widetilde{E}_1\times\widetilde{E'_1}$ where $\widetilde{E}_1,\widetilde{E'_1}$ are arbitrary lifts of the elliptic curves $E_1,E_1'$ gives the claim. The key here is, of course, that we are not bounding the degree of a splitting isogeny.

So, it remains to give an example of a simple abelian surface $\cA$ over $R$ with reduction isogenous to a product of elliptic curves. Here is one way to do this: pick a quaternion algebra $D$ over $\mathbb{Q}$ that is non-split at $p$ but is split over $\mathbb{R}$ and an abelian surface $\cA/R$ with $End_R(\cA)\otimes_{\bZ}\mathbb{Q}\simeq D$ (abelian surfaces with an action of an order in $D$ are parametrized by a Shimura curve and we just need to choose a general enough point on it so that the endomorphism algebra is not larger than $D$). Because of the assumption on $p$ the reduction of $\cA$ modulo $p$ is supersingular and is isogenous to a product of two supersingular elliptic curves(see e.g. Proposition 2 in Chapter III of Boutout-Carayol's 'Uniformisation p-adique des courbes de Shimura').

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