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Jul 15, 2012 at 17:58 comment added user18237 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.
Jul 15, 2012 at 17:50 comment added user18237 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.
Jul 15, 2012 at 16:23 comment added Bernhard Yes that is right the hypothesis is weaker, but the result is stronger.
Jul 15, 2012 at 15:59 comment added Qiaochu Yuan So you want a stronger result, right? It is the hypotheses that are weaker.
Jul 15, 2012 at 14:13 history asked Bernhard CC BY-SA 3.0