Extensions of Local Fields and the Trace Map I am now having some trouble dealing with the trace map in the extension $K/F$ of local fields. Here are some questions.


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*To introduce the discriminant, we may have two ways:


Consider the valuation ring $\mathcal{o}_K$ and $\mathcal{o}_F$, suppose we have a basis $\{e_1,\cdots,e_n\}$ of $\mathcal{o}_K$ over $\mathcal{o}_F$, then we may define $\Delta_{K/F}=det(tr(e_ie_j))$
Consider the codifferent of the extension $\mathcal{D}^{-1}_{K/F}$ defined to be $\pi_K^{-d}\mathcal{o}_K$, where $d$ is the largest integer s.t. $tr_{K/F}(\pi_K^{-d}\mathcal{o}_K)\subseteq\mathcal{o}_F$, then define $\Delta_{K/F}=N_{K/F}(\pi^d_K\mathcal{o}_K)$
so how to show that these definitions are the same? Well, I tried to approach this question with the view of dual basis but failed. I found it hard to study the valuation of sum of elements in a local field.


*How to show that $K/F$ is tamely ramified iff $v_K(\mathcal{D}_{K/F})=e-1$
 A: Perhaps things will get clearer if you adopt a functorial point of view, avoiding as much  as possible the recourse to computational (as opposed to functorial) arguments. 
0) So let $A\subset B$ be two principal domains, with respective fields of fractions $F\subset K$.  Two $A$-free submodule $M,N$  of $V$ of rank $n$ are obviously isomorphic via a linear isomorphism $\lambda$ whose determinant is non zero and defined by $M,N$ up to a unit of $A$, so we can define a "generalized index", which is the fractional ideal $[M:N]_A:= det(\lambda)A$. Let moreover $V$ be an $F$-vector space containing $M$, equipped with a non degenerate, symmetric, $F$- bilinear form $b$, and introduce the dual module $M^*$ = the  $A$-submodule of $V$ consisting of all the vectors $v$ of $V$ s.t. $b(v, M)\subset A$. Clearly $M^*$ is also $A$-free of rank $n$, and we can define its discriminant $\Delta_A(M):=[M^*:M]_A$. 
1) In your concrete case here, $b(u,v)=Tr_{K/F}(uv)$ (= the Gram determinant), $[B:I]_A=N_{K/F} (I)$ for any non null fractional ideal $I$ of $K$ (because $I$ spans $K$ over $F$), and the link with the usual norm is $N_{K/F}(aB)=N_{K/F}(a)A$ (because $a$ is the determinant of the linear map $x\to ax$). In particular, by definition of the generalized index,  $\Delta_{K/F}:=\Delta_A (B)$ is the codifferent ${\frak D}^{-1}$ of $K/F$ (hence an integral ideal since $B^*$ contains $B$). Moreover $\Delta_A (B)= [B^*:B]=[B:B^*]^{-1}=N_{K/F}({\frak D}^{-1})^{-1}=N_{K/F}({\frak D})$. This answers your first question because $A$ is a discrete valuation ring.  NB: a large part of the above remains valid when $A,B$ are Dedekind rings. 
2) As for your second question,  note that for any $b\in B$, the residual class of $Tr_{K/F}(b)$ is $e$ times the residual trace of $\bar b$ (where $\bar b$ is the residual class of $b$), which shows immediately that $K/F$ is tamely ramified iff $Tr_{K/F}(B)=A$. Let us prove then that this surjection is equivalent to $v({\frak D)}= e-1$. For any $a\in F,Tr_{K/F}(aB)=aTr_{K/F}(B)$, hence $Tr_{K/F}(B)={\frak D}^{-1}\cap K$. Putting $\alpha=v_F(Tr_{K/F}(B))$ and $\beta=v_K({\frak D)}$, this means that $\beta\le \alpha/e<\beta +1$. If $\alpha=e-1$, then $\beta=0$. If $\alpha =0$, then $\beta <e$.
It is classically known (but not obvious, this must be a theorem in your course) that $\beta\ge e-1$, so the latter case is equivalent to $\beta = e-1$. 
