Are there explicit formulas for traces and norms of $p$-adic fields with given ramification index? I met the following "elementary fact" on Henniart-Bushnell's text Local Langlands Conjectures for $GL(2)$ (which they didn't include a proof):
Let $F$ be a $p$-adic field and $E/F$ a finite extension. Then $E/F$ is tamely ramified if and only if $Tr_{E/F}(\mathfrak{o}_E)=\mathfrak{o}_F$. And suppose $E/F$ is tamely ramified with ramification index $e=e(L/K)$. Then we have:
$$Tr_{E/F}(\mathfrak{p}_E^{1+r})=\mathfrak{p}_F^{1+[r/e]}=\mathfrak{p}_F^{1+r}\cap F.$$
I didn't know how to prove this, though I realized this has something to do with the notion of the different for an extension of Dedekind domains. (I think I didn't have enough experience about how to deal with such concrete properties.) Can anyone give me some hint/advice? And I want to ask: suppose we are given an extension of $p$-adic fields $E/F$ with ramification index $e$ and degree $n=ef$. Is there any explicit formula of the trace of $\mathfrak{p}_E^{r}$ as above? If there's no such a formula for wildly ramified extensions, then what's the essential difficulty? (or essential difference between the tame and wild cases?)
Thanks a lot in advance!
 A: I think Serre’s Local fields book is a good reference for questions of this kind.
Anyway, assume first that $E/F$ is totally ramified with degree $e$, let $\varpi$ be a uniformizer of $f$.
$M=\mathrm{Tr}_{E/F}(O_E)$ is a submodule of $O_F$. So it is either $O_F$, or is contained in $\varpi O_F$.
Thus, $M \neq O_F$ iff $M \subset \varpi O_F$ iff $ \mathrm{Tr}_{E/F}(O_E/\varpi) \subset O_F$ iff the exponent of the different of $E/F$ is at least $e$. By Serre, Chapter III, Prop. 13, this holds iff $E/F$ is wildly ramified. (*)
In general, let $F \subset F’ \subset E$ be the maximal ramified subextension, and $\varpi \in F$ a uniformizer. Let $T=\mathrm{Tr}_{E/F’}(O_E)$, it is a submodule of $O_{F’}$ so of the form $\varpi^s O_{F’}$.
Now, $\mathrm{Tr}_{E/F}(O_E)=O_F$ iff $\mathrm{Tr}_{F’/F}(T) =O_F$, iff $\mathrm{Tr}_{F’/F}{\varpi^s O_{F’}}=O_F$, iff $s=0$ (the trace is surjective on integer rings for unramified extensions, that’s more or less Serre, Chapter III, Th. 1), iff $\mathrm{Tr}:O_E \rightarrow O_{F’}$ is onto, iff $E/F’$ (hence $E/F$) is tamely ramified.

Here’s an elementary, self-contained proof for (*). If $E/F$ is tamely ramified, $\mathrm{Tr}_{E/F}(1)=e \in O_F^{\times}$, so $\mathrm{Tr}_{E/F}: O_E \rightarrow O_F$ is onto.
Conversely, if $u \in O_E$ has invertible trace, then (because the extension is totally ramified) we can assume that $u=1+w$ with $w$ non-invertible in $O_E$. Then the trace of $w$ is non-invertible in $O_F$ (it has positive valuation), so it means that the trace of $1$ is invertible in $O_F$ – but this trace is $e$, so $E/F$ is tamely ramified.
