Let $F$ be a finite field extension of the $p$-adic numbers $\mathbb{Q}_p$, whose residue field has $q$ elements. Let $\mathfrak{p}$ be the prime ideal of $F$. Given a finite field extension $K/F$, write $\operatorname{Disc}_\mathfrak{p}(K)$ for $q^{v_\mathfrak{p}(\operatorname{Disc}(K/F))}$ and define the mass of $K$ to be $$ m(K) := \frac{1}{\#\operatorname{Aut}(K/F)}\cdot\frac{1}{\operatorname{\operatorname{Disc}_{\mathfrak{p}}(K)}}. $$ Then Serre's mass formula states that for all $n\geq 1$, we have $$ \sum_{[K:F] = n \text{ totally ramified}} m(K) = \frac{1}{q^{n-1}}. $$
I am trying to answer the following question:
Question 1: For $k = 1, 2, 4$, what is the value of $$ \sum_{\substack{[K:F] = 4 \text{ totally ramified} \\ \#\operatorname{Aut}(K/F) = k}} m(K)? $$
So Question 1 is basically the same as Serre's mass formula with $n=4$, but where we restrict to extensions with a given number of automorphisms (e.g. Galois extensions).
The tamely ramified case is easy, but I haven't been able to find a nice formula for residue characteristic $2$.
The obvious more general question is:
Question 2: Given a positive integer $n$ and $k \mid n$, what is the value of $$ \sum_{\substack{[K:F] = n \text{ totally ramified} \\ \#\operatorname{Aut}(K/F) = k}} m(K)? $$
It seems like the paper Serre's "formule de masse" in prime degree, by Dalawat, proves this sort of refinement in the case where $n$ is prime. Since Question 2 is easier for $n$ prime, maybe the next simplest case is $n = p^2$ for some prime $p$. This would actually suffice for my purposes, since I am interested in $n=4$. Does anybody know if this has been done? Alternatively, does anyone have any ideas for approaching this question?
My ideas
I have two vague ideas. The first is to try to generalise the methods of Dalawat's paper linked above. I haven't yet understood the details of the paper (I thought I'd ask here first in case the solution I'm looking for already exists), and I suspect that it would probably break down for non-prime $n$, since it uses a lot of representation theory over $\mathbb{F}_p$. It seems likely that switching to general $n$ would replace $\mathbb{F}_p$ with $\mathbb{Z}/n\mathbb{Z}$, and in that case much of the representation theory would fail. I'm not actually sure if we do land up with $\mathbb{Z}/n\mathbb{Z}$, since I'm not sure exactly where the $\mathbb{F}_p$ comes from in the proof. I'm sure I'll be able to understand these details once I've put some time into them - as I said, my objective in asking here is mainly to find out if this has already been done or is somehow easy.
My second idea is to try to adapt Serre's original proof of the mass formula (specifically, the first proof he gives in his paper). Serre starts by defining $P_n$ to be the set of degree $n$ monic polynomials over $F$ that are Eisenstein, viewed naturally as a subspace of $F^n$. It is easy to see that $$ \mu(P_n) = q^{-n}(1-q^{-1}), $$ where $\mu$ is the natural Haar measure on $F^n$. He then defines $P_n'$ to be the subspace consisting of polynomials with nonzero discriminant (in my case $P_n' = P_n$, since we are in characteristic $0$), so that $$ \mu(P_n') = \mu(P_n) = q^{-n}(1 - q^{-1}). $$ Given a degree $n$ totally ramified extension $K$ of $F$, define $P_n^K$ to be the set of $f \in P_n'$ such that $F[X]/(f(X)) \cong K$. Then $P_n'$ is the disjoint union of the $P_n^K$, so we have $$ q^{-n}(1-q^{-1}) = \mu(P_n') = \sum_L \mu(P_n^K). $$ Finally (and quite lengthily), Serre shows that $$ \mu(P_n^K) = \frac{q^{-1 - v_\mathfrak{p}(\operatorname{Disc}(K/F))}(1-q^{-1})}{\#\operatorname{Aut}(K/F)}, $$ and the mass formula follows. In order to adapt this proof, I would need to know the Haar measure of the set of Eisenstein polynomials $f$ such that the number of $F$-automorphisms of $F[X]/(f(X))$ is $k$. This seems to me quite hard because, as far as I know, there is no neat way of reading off the automorphism count from the coefficients.
