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It is a well-known fact that a Tate module $T_p(A)$ of an abelian group (abelian variety or commutative group scheme) $A$ over a field $K$, equipped with a continuous action of the respective absolute Galois group of $K$; but it may not be true for smaller Galois groups than the absolute Galois group.

Let $K$ be a $p$-adic number field with ring of integers $O_K$ and uniformizer $\pi$.

Let $L=K(F[\pi^n])$ be the finite field extension of $K$ by $\pi^n$-torsion points $F[\pi^n]$ of a Lubin-Tate formal group $F$. Consider the Galois group $G:=\text{Gal}(L/K)$ and the Tate module $T_{\pi}F:=\varprojlim_{n} F[\pi^n]$.

But in the Lubin-Tate formal group case, $T_{\pi}F$ admits continuous action by the finite Galois group $G:=\text{Gal}(L/K)$.

  • Why is so?

  • What is the action?


Lubin and Tate proved that the action $G \times T_{\pi}F \to T_{\pi}F$, that is, the map $\chi: G \to \text{Aut}(T_{\pi}F)$ is an isomorphism. Note that $\text{Aut}(T_{\pi}F) \cong O_K^{\times}$.

Thus, for $\sigma \in G$, the Galois action on $T_{\pi}F$ should be $\sigma(\lambda)=[\chi(\sigma)]_F(\lambda)$ for all $\lambda \in T_{\pi}F$, where $\chi(\sigma)$ lands into $O_K^{\times}$.

Why is this action continuous?

And why do other Tate modules of an abelian group $A$ don't admit similar action of smaller Galois group?

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  • $\begingroup$ Your premise is wrong: $T_\pi F$ admits an action of the (infinite) profinite group $\lim_n \mathrm{Gal}(K(F[\pi^n])/K)$. $\endgroup$ Commented Sep 7 at 9:03
  • $\begingroup$ @Satan'sMinion, I found at the bottom of page 3 of this note $\endgroup$
    – Learner
    Commented Sep 7 at 9:18
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    $\begingroup$ $F_{\pi}$ as defined there is the field generated by the $\pi^n$-torsion for ALL $n$. $\endgroup$ Commented Sep 7 at 15:59
  • $\begingroup$ @Satan'sMinion, thank you. I missed that subtle definition "for all $n\geq1$" $\endgroup$
    – Learner
    Commented Sep 7 at 17:19

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