Two decomposition groups. Are they the same?

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.

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.