Comparison of trace maps of extensions of local fields and their residue fields Let $K/\mathbb{Q}_p$ be a finite extension of local fields and we denote by $k/\mathbb{F}_p$ the extension of their residue fields. Let $\mathrm{Tr}_K:K\to \mathbb{Q}_p$ be the trace map. 
The kernel of the composition of $\mathrm{Tr}_{K}$ with the quotient $\mathbb{Q}_p\to\mathbb{Q}_p/\mathbb{Z}_p$ is the fractional ideal $\mathfrak{m}_K^{-n}$ for some $n\geqslant 1$. Therefore, choosing an uniformiser $\varpi_K$ of $K$ we obtain an additive character $$\psi_k:k\xrightarrow{\times\,\varpi_K^{-n-1}}\mathfrak{m}_K^{-n-1}/\mathfrak{m}_K^{-n}\xrightarrow{\mathrm{Tr}_K}p^{-1}\mathbb{Z}_p/\mathbb{Z}_p\xrightarrow{\frac{z}{p}+\mathbb{Z}_p\,\mapsto z+p\mathbb{Z}_p} \mathbb{F}_p.$$ The "canonical" additive character of $k$ is the trace map $$\mathrm{Tr}_k:k\to\mathbb{F}_p.$$ The general theory of additive characters tells that there exists $a\in k$ such that for all $x\in k$ we have $$\psi_k(x)=\mathrm{Tr}_k(ax).$$ Can we determine more precisely this $a$? A priori it depends on the choice of $\varpi_K$, but there's no canonical choice for the uniformizer. Is it too much to expect $a=1$ ?
 A: Perhaps certain properties of « dual » lattices w.r.t. the bilinear form $Tr_{K/F}(xy)$ defined by the trace map of a finite extension $K/F$ of local fields, with rings of integers $O_K$ and $O_F$, need to be recalled first. The set of $y\in K$ s.t. $Tr_{K/F}(xy)\in O_F$ for all $x\in O_K$, called the codifferent (or inverse different) and denoted $D^{-1}(K/F)$, is a fractional ideal of $K$, the largest $O_K$-submodule $M$ of $K$ s.t. $Tr_{K/F}(M)\subset O_F$. It is also characterized by the following criterion: (*) For a fractional ideal $A$ (resp. $B$) of $F$ (resp. $K$), $Tr_{K/F}B\subset A$ iff $B\subset A.D^{-1}(K/F)$ (see e.g. Serre's "Local Fields", chap.3, §3). When applying this to your situation (and keeping your notations), we get that the kernel of the composition of $Tr_{K}$ with the quotient $\mathbf Q_p \to \mathbf Q_p/\mathbf Z_p$ is the fractional ideal $\mathfrak m_K^{-n}$, where $-n$ is the valuation of the codifferent $D^{-1}(K/\mathbf Q_p)$. A precise formula giving $n$ in terms of ramification subgroups is known (op. cit., chap.4, propos.4). In any case, $n\ge e-1$, where $e=v_K (p)$ is the ramification index of $K$ (chap.3, propos.13), with equality iff $p$ does not divide $e$ (tame ramification) .
Now I have 3 questions : 1) You assert that the two « additive characters » $Tr_k$ and $\psi_k$ should be proportional, but I don’t see why. Considered as linear forms on the $\mathbf F_p$- vector space $k$, they are proportional iff they have the same kernel (which is a hyperplane). 2) In the last part of your definition of $\psi_k$, the notation $p^{-1} \mathbf Z_p/\mathbf Z_p$ usually means the subgroup of order $p$ of $\mathbf Q_p/\mathbf Z_p$, so that multiplication by $p$ will give zero ! I think one should not multiply by $p$ and stay inside $\mathbf Q_p/\mathbf Z_p$, identifying $\mathbf Z_p/p\mathbf Z_p$ with $p^{-1} \mathbf Z_p/\mathbf Z_p$ by mapping $a$ mod $p\mathbf Z_p$ to $\frac ap$ mod $\mathbf Z_p$ .  3) At the beginning of your definition of $\psi_k$, you make $Tr_K$ send $\mathfrak m_K^{-n-1}$ to $p^{-1}\mathbf Z_p$, but the choice of $\mathfrak m_K^{-n-1,}$ is not the optimal one, see below. Actually, applying the characterization (*) of the codifferent, we have $Tr_{K}(u\pi^{-m}) \in p^{-1}\mathbf Z_p$ iff $m\le e+n$, which still leaves much choice to define $\psi_{k,m}$ analogously to $\psi_k$ .
Let us look for a general formula relating $Tr_k$ and $\psi_{k,m}$. Denote by $x \to \bar x$ the residue class map and, for a unit $u$ of $K$, let us compute $\psi_k (\bar u)$. There are many different characterizations of the residual field $k$, but let us simply stick to the definition $k= O_K /\mathfrak m_K$. With an appropriate choice of $m$, we have $Tr_{K}(u\pi^{-m}) \in p^{-1}\mathbf Z_p$, or  $pTr_{K}(u\pi^{-m})= Tr_{K}(p\pi^{-m}u)\in \mathbf Z_p$ . Impose further that the element $x=p\pi^{-m}u$ actually lies in $O_K$, i.e. $m\le e$ and so $\psi_k (\bar u)=\bar (Tr_{K}(x))$ (there is a bar over all the RHS).
But it is classically known that $\bar (Tr_{K}(x))=e.Tr_k(\bar x)$ (Cassels-Fröhlich, chap.1, §5, lemma 2). With our extra condition on $m$, we thus get $\psi_{k,m} (\bar u)=e.Tr_k(\bar x)$ . Note that this extra condition is verified by $n+1$ iff $n=e-1$, i.e. in the tamely ramified case.
