7
$\begingroup$

For a Galois extension of local fields $L/K$ with Galois group $G$, define the ramification groups $G_i$ by $$G_i = \{ \sigma \in G : \nu_{L}(\sigma(x) - x) \geq i+1 \text{ } \forall x \in \mathcal{O}_L\}.$$ Here $\nu_{L}$ is the valuation on $L$ such that $\nu_L(\pi_L) = 1$, where $\pi_L$ is a uniformizer of $L$. We would like to show that $G_0$ is the subgroup of $G$ fixing $K^{ur}$, where $K^{ur}$ is the maximal unramified extension of $K$ contained in $L$.

Here is what we have so far. We know that $\mathcal{O}_L$ = $\mathcal{O}_{K^{ur}}[\pi_L]$. Furthermore, elements in $\mathcal{O}_{K^{ur}}$ look like $$\sum_{i \geq 0} a_i \pi_{K^{ur}}^i,$$ where the $a_i$ are units in $\mathcal{O}_{K^{ur}}$. Since $K^{ur}/K$ is unramified, then $\pi_{K^{ur}} = u \cdot \pi_{K}$ for some unit $u$ in $\mathcal{O}_{K^{ur}}$. If we knew that elements in $G_0$ fixed the units in $K^{ur}$, then we could conclude that elements in $G_0$ fixed all sums of the above form in $O_{k^{ur}}$.

Any tips or a full solution would be greatly appreciated. We are just beginning to learn about local fields and are having trouble filling in the details for many of the statements offered without detailed justification, such as this one, in the notes we are reading.

$\endgroup$
  • $\begingroup$ Well, we have $[L:K]=e(L/K)f(L/K)$. Since both $e$ and $f$ are multiplicative for tower of extension, it is enough to show that $L/K^{\mathrm{ur}}$ is totally ramified of degree $e(L/K)$, which follows from the fact that ramification degree coincides with the size of the inertia group, which is $G_0$ (for both $L/K$ and $L/K^\mathrm{ur}$). $\endgroup$ – Jiangwei Xue Jun 5 '13 at 13:00
  • $\begingroup$ By "to show $L/K^{\mathrm{ur}}$ is totally ramified..", I meant to take $K'$ to be the subfield fixed by $G_0$, and show $L/K'$ to be totally ramified of degree $e(L/K)$. Then it follows that $K'/K$ is unramified of degree $f(L/K)$. $\endgroup$ – Jiangwei Xue Jun 5 '13 at 13:46

Your Answer

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

Browse other questions tagged or ask your own question.