1
$\begingroup$

I am assuming the usual framework and notation of ramification theory. Let $G=\operatorname{Gal}(L/K)$.

We define the decomposition group of a prime ideal $\mathfrak{q}$ above $\mathfrak{p}$ as \begin{equation} G^{Z}(\mathfrak{q}\:|\:\mathfrak{p})=\{\sigma\in G\:|\:\sigma(\mathfrak{q})=\mathfrak{q}\}. \end{equation}

Now, from the valuation theory point of view, we define, for extensions of valuations $w\:|\:v$, \begin{equation} G_{w}=\{\sigma\in G\:|\:w\circ\sigma=w\}. \end{equation}

Are these the same groups? I guess they are, but I am not sure how to justify this:

If $v$ is a valuation that comes from $\mathfrak{p}$, does every extension of $w$ to $v$ come from an ideal $\mathfrak{q}$ over $\mathfrak{p}$?

Thanks.

$\endgroup$
1
$\begingroup$

Yes, they are the same. Let me denote $w_\mathfrak{q}$ the valuation of $\mathfrak{q}$ and $v_\mathfrak{p}$ the valuation for $\mathfrak{p}$.

We then have a map

$$\mathrm{Gal}(L_{w_\mathfrak{q}}/K_{v_\mathfrak{p}})\to D(\mathfrak{q}\mid \mathfrak{p})$$

(where I've used $D$ for the decomposition) group which maps

$$\sigma\mapsto \sigma\mid_L$$

where we have the canonical inclusion $L\hookrightarrow L_{w_\mathfrak{q}}$. To see that $\sigma\mid_L$ is actually in $D(\mathfrak{q}\mid\mathfrak{p})$ note that since $\sigma$ is continuous with respect to the $w_\mathfrak{q}$-topology that $\sigma\mid_L$ satisfies the same property. But, that means that $\sigma\mid_L(\mathfrak{q})=\mathfrak{q}$ (why?). This map is injective since $L$ is dense in $L_{w_\mathfrak{q}}$.

To see the inverse note that $\sigma\in D(\mathfrak{q}\mid\mathfrak{p})$ is uniformly continuous for the $w_\mathfrak{q}$-adic topology and so extends uniquely to a map $\widetilde{\sigma}:L_{w_\mathfrak{q}}\to L_{w_\mathfrak{q}}$ which is a continuous ring isomorphism since one can check this on dense subsets (why?). Thus, it suffices to see that $\widetilde{\sigma}$ actually fixes $K$. But, note that $\widetilde{\sigma}$ fixes $K$ and so fixes its closure, which is $K_{v_\mathfrak{p}}$.

It's easy to see that these maps are mutually inverse.

$\endgroup$

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.