Norm map of an extension of nonarchimedean local fields Let $k_1/k$ be a finite extension of nonarchimedean local fields. Let $A_1 = \{x \in k_1\,: \, \text{mod}_{k_1}(x) \leq 1\}$ denote the ring of integers of $k_1$ and let $A$ be the ring of integers of $k$. Here $\text{mod}_{k_1}(x)$ denotes the module of the automorphism $y \mapsto xy$ on $k_1$, following Weil.
Is it true that $N_{k_1/k}(A_1) \subset A$, where $N_{k_1/k}$ denotes the norm map of the extension $k_1/k$? 
 A: Yes, if $|k_1:k|=n$ then $A_1$ is a free $A$-module of rank $n$.
For $y\in A_1$, $N(y)$ is the determinant of the $A$-endomorphism
of $A_1$ given by multiplication by $y$. So $N(y)$ is the determinant
of some $n$-by-$n$ matrix over $A$, and so is an element of $A$.
A: In your situation ($k_1 /k, A_1 /A$, etc.), your question amounts to the equality of the ring of integers (in the usual sense) of $k_1$ and its valuation ring which you denote by $A_1$. There are many different proofs, but there are all intertwined, depending on the chosen definitions. It seems to me that the most convenient answer (and which brings enlightening extra information) to your question is the following general property: for a field $K$ which is complete w.r.t. a discrete valuation $v$ and $L/K$ a finite extension, there exists a unique valuation $w$ of $L$ which extends $v$ (and $L$ is automatically complete w.r.t. $w$), and $w$ is given by the formula $w(x)=v(N_{L/K} (x))/f$, where $f$ is the inertia index of $L/K$. See e.g. Serre's "Local Fields", chap. II, §2.
A: The norm is continuous and multiplicative. Since $A_1$ is a compact subring of $k_1$, the set $N_{k_1/k}(A_1)$ is a multiplicatively closed and compact subset of $k$. Let $S \subset k$ be any multiplicatively closed and compact subset of $k$. Then $S \subset A$: Let $s \in S$, then all powers $s^{n}$ belong to $S$ as well. Since $S$ is compact and $\text{mod}_{k}: k \rightarrow [0, \infty)$ is continous, there exits a constant $C \geq 0$ such that for all $t\in S$ we have $\text{mod}_k{(t)} \leq C$. In particular $(\text{mod}_{k}(s))^n \leq C$ for all $n \geq 1$ and hence $\text{mod}_{k}(s) \leq 1$, that is $s \in A$.
