September 28th
Today I learned the definition of ramification groups. Let $L/K$ be a normal extesion of number fields. I think it's natural to introduce these as generalizations of the inertia subgroup, noting that we can generalize\[E(\mf q/\mf p)=\{\sigma\in\op{Gal}(L/K):\sigma(\alpha)\equiv\alpha\pmod{\mf q}~~\forall\alpha\}\]to\[V_m=\left\{\sigma\in\op{Gal}(L/K):\sigma(\alpha)\equiv\alpha\pmod{\mf q^{m+1}}\right\}.\]In particular, $V_{-1}=\op{Gal}(L/K)$ and $V_0=E.$ Composing automorphisms shows that these are indeed subgroups; in particular, $\op{id}$ is in every $V_m$ because it's the identity, and we also have $\op{id}(\alpha)=\alpha\equiv\alpha\pmod{\mf q^\bullet}$ always.
What follow are some basic properties. For example, we can quickly show that the $V_m$ are normal subgroups of $D(\mf q/\mf p)$ because for $\sigma\in D(\mf q/\mf p),$ we see\[\sigma V_m\sigma^{-1}\left(\alpha+\mf q^{m+1}\right)=\sigma V_m\left(\sigma^{-1}(\alpha)+\mf q^{m+1}\right)=\sigma\left(\sigma^{-1}(\alpha)+\mf q^{m+1}\right)=\alpha+\mf q^{m+1}.\]This implies $\sigma V_m\sigma^{-1}\subseteq V_m,$ which is what we wanted.
Because $\mf q^{m+1}\subsetneq\mf q^m,$ we see that $\sigma(\alpha)\equiv\alpha\pmod{q^{m+1}}$ implies $\sigma(\alpha)\equiv\alpha\pmod{\mf q^m},$ so the $V_m$ are a (not strictly) decreasing sequence of subgroups. And in fact, we claim that there is an $M$ for which $V_m=\{\op{id}\}$ for any integer $m\ge M.$ In fact, we can construct that $M$ explicitly: fix $L=K[\alpha]$ where $\alpha\in\mathcal O_L.$ (Certainly an $\alpha$ exists; multiplying it by a sufficient constant will make its polynomial monic.) Then we claim/I think\[M=\max\left\{\nu_\mf q\left((\sigma(\alpha)-\alpha)\mathcal O_L\right):\sigma\in\op{Gal}(L/K)\setminus\{\op{id}\}\right\}\]works and is in fact sharp. Quickly, note that $M$ is well-defined because for $\alpha\ne\op{id},$ we have $\sigma(\alpha)\ne\alpha.$ Indeed, $\sigma(\alpha)=\alpha$ forces all polynomials and ratios of polynomials of $\alpha$ to be fixed by $\sigma,$ which is all of $L.$
Now we show that this $M$ works. Suppose $m\ge M$ and $\sigma\ne\op{id}.$ We want to show that $\sigma\notin V_m,$ from which $V_m=\{\op{id}\}$ will follow. Note that $\sigma(\alpha)\ne\alpha$ as before, so we may talk about $(\sigma(\alpha)-\alpha)\mathcal O_L.$ Now, $m\ge M\ge\nu_\mf q((\sigma(\alpha)-\alpha)\mathcal O_L)$ if and only if\[\mf q^{m+1}\nmid(\sigma(\alpha)-\alpha)\mathcal O_L.\]But then the definition of divides means that this holds if and only if $(\sigma(\alpha)-\alpha)\mathcal O_L\not\subseteq\mf q^{m+1},$ which is true if and only if\[\sigma(\alpha)-\alpha\notin\mf q^{m+1}\]because $\mf q^{m+1}$ is an ideal. But this is equivalent to $\sigma(\alpha)\not\equiv\alpha\pmod{\mf q^{m+1}},$ which translates into $\sigma\notin V_m$ because $\alpha$ generates all of $L.$ We get sharpness of our bound $M$ by taking this over each $\sigma$ and noting that all manipulations were equivalences.