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.

  • $\begingroup$ 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

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.

  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.