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Let $A/\mathbb{Q}$ be an abelian variety with good reduction at a prime $p$. Assume $\mathcal{A}/\mathbb{Z}_{(p)}$ is an integral model at $p$(hence proper smooth).

For any number field $K$ and any prime ideal $\mathfrak{p}$ of $K$ over $p$ with residue field $\kappa$, by valuation criterion on $\mathcal{O}_{K,\mathfrak{p}}$, any $K$-point of $A$ extends uniquely to an $\kappa$-point of $\mathcal{A}$'s special fiber.

By fixing a prime ideal of $\overline{\mathbb{Z}}$ lying over $p$ and taking injective limit, we get a well defined reduction map\begin{equation} A(\overline{\mathbb{Q}})\rightarrow \mathcal{A}_{\mathbb{F}_p}(\overline{\mathbb{F}_p}). \end{equation}

Does the above map induce isomprhisms on $\ell$-primary parts of both groups? Notice that both of their $\ell$-primary subgroups are isomorphic to $(\mathbb{Q}_{\ell}/\mathbb{Z}_{\ell})^{\oplus 2g}$.

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This is true. One reference for this is Theorem C.1.4 in Hindry-Silverman's "Diophantine Geometry: An Introduction" which says that for an abelian variety $A$, number field $K$ and any prime $\mathfrak p$ above $p$, the reduction map $A(K)\to A(O_K/\mathfrak p)$ is injective on prime-to-$p$-torsion. The result you ask for follows since $\overline{\mathbb Q}$ is a union of number fields.

One way to prove the required injectivity follows from the description of the kernel of the reduction map over the completion $K_{\mathfrak p}$, which can be identified with the formal group associated to $A$. Formal groups over $p$-adic fields have no prime-to-$p$ torsion.

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  • $\begingroup$ How to describe the kernel for the $p$-primary part? $\endgroup$
    – Yuan Yang
    Commented Feb 11, 2023 at 19:36
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    $\begingroup$ @YuanYang It depends on the reduction type of the abelian variety. I believe the reduction map $A(\overline{\mathbb Q})\to A(\overline{\mathbb F_p})$ will always be surjective on torsion, but the $p$-primary part of the latter will always be smaller than of the former. You can describe the structure fully in terms of the height of the associated formal group - the kernel should be isomorphic to $(\mathbb Q_p/\mathbb Z_p)^h$. $\endgroup$
    – Wojowu
    Commented Feb 11, 2023 at 19:44
  • $\begingroup$ Thank you so much!!! $\endgroup$
    – Yuan Yang
    Commented Feb 11, 2023 at 20:28
  • $\begingroup$ Obvious but annoying typo in the third-last word, should be “no”, presumably? $\endgroup$
    – Lubin
    Commented Feb 12, 2023 at 3:26
  • $\begingroup$ @Lubin Correct, thank you. $\endgroup$
    – Wojowu
    Commented Feb 12, 2023 at 3:32

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