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Let $L/K$ be a normal number field extension with ring of integers $\mathcal O_L/\mathcal O_K$. Let $Q$ be a prime ideal of $\mathcal O_L$ and inertia group $E = \{g \in \operatorname{Gal}(L/K)|\forall\alpha\in\mathcal O_L, g(\alpha) \equiv \alpha \pmod Q \}$.

Let $\pi \in Q-Q^2$. Then show that for all $\sigma \in E, \sigma(\pi) \equiv \alpha \pi \pmod {Q^2}, \alpha \in \mathcal O_L$. This looks like it should be really easy and it probably is but I seem to be missing something. There seems to be very little information to use about $\sigma$...

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  • $\begingroup$ yes but I am interested modulo $Q^2$ and $Q$ might not be principal $\endgroup$ – Asvin May 27 '15 at 10:16
  • $\begingroup$ Principalness of $Q$ is not the issue. See my response. Good question, by the way! $\endgroup$ – Lubin May 28 '15 at 3:06
  • $\begingroup$ Related: i.stack.imgur.com/wLgV5.png $\endgroup$ – Watson Jan 2 '17 at 22:58
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You have, perhaps unwittingly, backed into the (to me) fascinating subarea of number theory called Higher Ramification Theory. I think it causes a lot of people difficulty, but you can get an excellent description of the basics in Serre’s Corps Locaux, translated into English as Local Fields.

Anyway, I hope your proposition wasn’t put to you as a homework assignment, because it’s not true. Here’s your counterexample.

Let $K=\Bbb Q$, $L=\Bbb Q(\omega,\pi)$, where $\omega$ is a primitive cube root of unity and $\pi$ is a cube root of $2$. You let the prime $Q$ upstairs be $(\pi)$, the unique prime of $L$ above the rational prime $2$. The inertia group is the subgroup of the Galois group fixing $\omega$, and its fixed field is $\Bbb Q(\omega)$, over which $L$ is cubic, with Galois group generated by the automorphism that sends $\pi$ to $\omega\pi$. This Galois group is your inertia group, of course. Now, $\omega\pi\not\equiv\pi\pmod{\pi^2}$ because $\omega\not\equiv1\pmod\pi$.

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  • $\begingroup$ If $\sigma(\pi) = \omega\pi$, can't we just take $\alpha =\pi$(as defined in the question?) I don't require the inertia group congruence itself to hold for all powers of $\pi$. $\endgroup$ – Asvin May 28 '15 at 5:58
  • $\begingroup$ I was in fact doing the exercises from Marcus on (what I now realize) are the higher ramification groups. $\endgroup$ – Asvin May 28 '15 at 5:59
  • $\begingroup$ Does this work: $Q = (\pi) + Q^2, \sigma(\pi) \in Q, \sigma(\pi) = \alpha\pi + r (r \in Q^2), \sigma(\pi) \equiv \alpha\pi \pmod Q^2 $. $\endgroup$ – Asvin May 28 '15 at 6:15
  • $\begingroup$ What’s happening, as you can read in your text, is that, locally at $Q$, you have a map from the inertia group into $(\mathscr O_L/Q)^*$, by $\sigma\mapsto(\sigma\pi)/\pi$. A homomorphism, not necessarily one-to-one nor onto. $\endgroup$ – Lubin May 28 '15 at 12:06
  • $\begingroup$ @Lubin: Clearly I am missing something, but the way the original question is framed (and I guess it is ex. 21(a) from Ch4 of Marcus) aren't we just trying to prove that there exists such an $\alpha\in\mathcal{O}_L$, in which case $\alpha=\omega$ works in your example? $\endgroup$ – GaryMak Oct 31 '16 at 20:04

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