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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 $L_E$ be the fixed field of $E$ and $\mathcal O_{L_E}$ it's ring of integers.

Then show that $\mathcal O_L = \mathcal O_{L_E} + Q$(all possible pairs addition). My idea was to write $\alpha = \sum_{g\in E}g(\alpha) - R$ or $\prod_{g \in E}g(\alpha) - R$ and show that $R \in Q$ but I could not quite finish the proof.

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  • $\begingroup$ Not answer but just my thoughts so far. Clearly $\sum\limits_{g \in E} g(\alpha) \in \mathcal O_{L_E}$, and you want to show $\alpha - \sum\limits g(\alpha) \in Q$. If this is true, then letting $g_1, ... , g_s$ be all the nonidentity elements of $E$, you have $$\alpha- \sum\limits_{g \in E} g(\alpha) = -g_1(\alpha) - \cdots - g_s(\alpha) = (\alpha - g_1(\alpha)) + \cdots + (\alpha - g_s(\alpha)) - s \alpha$$ Now $\alpha - g_i(\alpha) \in Q$. If what you want to prove is true, i.e. $\alpha - \sum\limits_{g \in E} g(\alpha) \in Q$, then $ s \alpha$ should be in $Q$ as well. Is this true? $\endgroup$ – D_S May 25 '15 at 15:55
  • $\begingroup$ That's the point at which I got stuck, if you use the product formulation, you need $(\alpha^s - 1)\alpha \in Q$ and this seems easier(maybe?) $\endgroup$ – Asvin May 25 '15 at 16:02
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The subgroup of Galois that you mention is usually called the inertia group. You have $L\supset L^E\supset K$, where the lower extension is unramified, i.e. that’s where the residue field extension happens; and the upper extension is totally ramified, with no residue field extension.

Now, you’re asking whether $\mathscr O_L/Q\cong\mathscr O_{L^E}/Q'$, where $Q'$ is the intersection of $Q$ with $L^E$. This is true: the residue fields of these two primes (one a prime of $L$, the other of $L^E$) are the same.

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  • $\begingroup$ This is a wonderful answer. To my mind at least it would be helpful to mention that you are using the second ring isomorphism theorem (or something equivalent I guess!) to get from the way the OP phrases it, to your succinct statement? $\endgroup$ – GaryMak Oct 31 '16 at 17:22

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