Let $K/\mathbb{Q}_p$ be a finite extension with ring of integers $R$ and residue field $k$. Let $A/K$ be an abelian variety with Neron model $\mathcal{A}/R$. We denote by $\tilde{\mathcal{A}}/k$ the special fiber and by $\mathcal{A}^0$ - the connected component of $\mathcal{A}$.

  1. Is it true that $\mathcal{A}^0(R)[n] \hookrightarrow \tilde{\mathcal{A}^0}(k)$ for $p \nmid n$? What is the right reference for this fact?

This is discussed here in case that $A$ has good reduction. Liu's answer in that topic addresses the general case:

(...) when A is not an abelian scheme (for example, when it is the Néron model of an abelian variety over $\mathbb{Q}_p$ with not necessary good reduction), then for any n prime to p, the kernel $A[n]$ is still étale over $\mathbb{Z}_p$ (because the tangent map at 0 of the multiplication by n is just multiplication by n for any commutative algebraic group), but not necessarily finite. There is a biggest closed subscheme H of $A[n]$ which is étale and finite over $\mathbb{Z}_p$. The reduction map on H is injective (see Pete's proof). The generic fiber of H corresponds to the points of the generic fiber of $A[n]$ having specialization mod p. (...)

I don't really understand the last sentence - what does it mean for a point to have a specialization mod $p$? Isn't it true that every point in $A(K)$ corresponds to a point in $\mathcal{A}(R)$ and every point in $\mathcal{A}(R)$ can be reduced to $\mathcal{A}(k)$?

  1. Is it possible to prove the "$(1) \Rightarrow (2)$" implication of Neron-Ogg-Shafarevich criterion for abelian varieties in the language of Silverman's Arithmetic of Elliptic Curves? This would be my attempt:

Theorem If $A/K$ has good reduction, then inertia $I_K$ acts trivially on $A[n]$ for $p \nmid n$.

Proof: Let $L = K(A[n])$ and let $R', k'$ be the ring of integers and residue field of $L$. Let $\mathcal{A}/R'$ be the Neron model of $A_L$. Then any point $P \in A[n]$ corresponds to a point $\mathcal{P} \in \mathcal{A}(R')[n]$. Let $\sigma \in I_K$. Then $\mathcal{P}^{\sigma} - \mathcal{P} \in \mathcal{A}(R')[n]$ and

$$ \widetilde{\mathcal{P}^{\sigma} - \mathcal{P}} = \mathcal{O} $$ in $\tilde{\mathcal{A}}(k')$. Thus (since prime-to-p torsion injects into $\tilde{\mathcal{A}}(k')$ - by [1]) $\mathcal{P}^{\sigma} = \mathcal{P}$ and $P^{\sigma} = P$.


  1. Seems to be true by a formal group argument, just like in the case of good reduction (cf. Proposition 7 in Clark, BOUNDS FOR TORSION ON ABELIAN VARIETIES WITH INTEGRAL MODULI and the preceeding discussion.

  2. I think it is not true in general that reduction commutes with Galois action. Is it true at least on identity component, ie. $$ \widetilde{P^{\sigma}} = (\widetilde{P})^{\sigma} $$ for $P \in \mathcal{A}^0(R_{\overline{K}})$, $\sigma \in Gal(\overline{K}/K)$?

Edit 2:

  1. Ok, I know what is wrong. The Galois group $Gal(L/K)$ doesn't act on $\mathcal{A}$, since it is defined over $R'$ and not $R$. In the case of elliptic curves with good reduction we know that the minimal equation stays the same after base change (since we have a discriminant of valuation 0) and thus the Neron model is basically defined over $R$. In the case of abelian varieties we have no discriminant, that's why it is harder.

I guess we can close this topic.

  • $\begingroup$ Doesn't 1. follow from the infinitesimal lifting criterion for étaleness? $\endgroup$
    – user19475
    Jun 10, 2017 at 15:19
  • $\begingroup$ I believe your proof of 2. is basically correct. $Gal(L/K)$ acts on $\mathcal{A}$ semilinearly, meaning it does not preserve the base ring $\mathcal{O}_L$ (there's the action on $L$ !). If you restrict the action to the inertia, then you obtain an action on the special fiber by endomorphisms because the inertia preserves the residue field. $\endgroup$
    – Lukas
    Jun 12, 2017 at 10:06
  • $\begingroup$ @Lukas - how do you know that $Gal(L/K)$ acts on $\mathcal A$? I mean the generic fiber of $\mathcal{A}$ is defined over $L$ and not over $K$. So if $P \in \mathcal{A}(R')$ then we have no longer guarantee that $P^{\sigma}$ is a well defined point in $\mathcal{A}(R')$, right? $\endgroup$ Jun 12, 2017 at 10:15
  • $\begingroup$ You are completely right. The point $P^\sigma$ is no longer a point of the same $\mathcal{A}(R')$, but is rather a point of $\mathcal{A}_\sigma(R')$, where $\mathcal{A}_\sigma/\mathcal{O}_L$ is the "same" abelian scheme, except the structural morphism is the twisted $\mathcal{A}\to Spec\,\mathcal{O}_L\overset{\sigma}{\to}Spec\,\mathcal{O}_L$. $\endgroup$
    – Lukas
    Jun 12, 2017 at 10:27
  • $\begingroup$ Ok, sounds promising. So we can just conclude that $P^{\sigma} - P \in \mathcal{A}_{\sigma}(R')[n]$ and it must be zero, since it reduces to zero? $\endgroup$ Jun 12, 2017 at 10:37


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.