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We are concerned with slight weakening of a result of the Serre-Tate paper "Good reduction of abelian varieties" google. I think the title of the question conveys what we are in here for. So, I'll move on directly to formulate the question.

Let $K$ be a field, $v$ a discrete valuation of K. Denote by $k$ the residue field of $K$ at $v$. Let $K_s$ be a separable closure of $K$ and $\bar{v}$ an extension of $v$ to $K_s$. Denote $I(\bar{v})$ the inertia subgroup of $\mathrm{Gal}(K_s/K)$. Let $A/K$ be an abelian vareity and for any prime number $\ell$ denote $T_{\ell}(A)$ the Tate-module. Let $$ \rho_{\ell}: \mathrm{Gal}(K_s/K)\rightarrow \mathrm{Aut}(T_{\ell}(A)) $$ be the associated $\ell$-adic representation.

Then corollary 1 to Theorem 2 of the paper says the following :

If $k$ is finite of characteristic $p$ and for some prime number $ \ell \neq p $ , the image $\rho_{\ell}(\mathrm{Gal}(K_s/K))$ in $\mathrm{Aut}(T_{\ell}(A))$ is abelian. Then $A$ has potential good reduction at $v$.

My question is the following:

Suppose we have the weaker condition that the image of the inertia subgroup $\rho_{\ell}(I(\bar{v}))$ in $\mathrm{Aut}(T_{\ell}(A))$ is abelian. Can we still conclude that $A/K$ has potential good reduction at $v$ ?

I was thinking that maybe the same proof as that of the above result might work. If this is the case, then can anyone explain me any arguments of the local Class field theory involved here in this case or suggest appropriate references ?

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  • $\begingroup$ So you want a stronger result, right? It is the hypotheses that are weaker. $\endgroup$ – Qiaochu Yuan Jul 15 '12 at 15:59
  • $\begingroup$ Yes that is right the hypothesis is weaker, but the result is stronger. $\endgroup$ – Bernhard Jul 15 '12 at 16:23
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    $\begingroup$ I don't think so. Consider an elliptic curve with split multiplicative reduction. Inertia acts unipotently through its abelian $\mathbb{Z}_l(1)$ quotient. Indeed, this follows readily from the exact sequence of $G_K$-modules (coming from the theory of Tate curves) $0\to\Z_l(1)\to T_l(E)\to \Z_l\to 0$ and the fact that $\Z_l(1)$ is the maximal pro-l quotient of of the inertia group. $\endgroup$ – user18237 Jul 15 '12 at 17:50
  • $\begingroup$ In fact, the action of inertia will always be "potentially abelian" as $\text{Aut}(T_l(A))$ has a pro-l subgroup of finite index and the maximal pro-l quotient of inertia is abelian. $\endgroup$ – user18237 Jul 15 '12 at 17:58

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