Is the Field Norm of a unit a unit?

Question

I am currently studying various properties of norms and was curious if the following result is true, or if there is something close to it: Given a Galois extension $K/F$ and an intermediate field $L$ with $[K:L]=p$ (where $p$ is prime), if $A\subset K$ is an order of $K$ and if $t\in A^{\times}$ (unit group of $A$), is $$Nm_{K/L}(t)\in A^{\times} \cap L?$$

I was able to show $Nm_{K/L}(t)\in L$ since $[K:L]$ is prime, but I am not sure if $$Nm_{K/L}(t)\in A^{\times}.$$

Answer 1

Ok so it turns out that this is not always true. I found a counter example with $K=\mathbb{Q}(i,\sqrt{2})$, $\alpha=1+\sqrt{2}+i$, and $A=\mathbb{Z}[\alpha]$. $A$ is indeed an order of $K$. If $L=\mathbb{Q}(\sqrt{2})$, $Nm_{K/L}(\alpha)=4+2\sqrt{2}$ and it is not too hard to show that $4+2\sqrt{2}\notin A$.

Of course, it easy to show that if $K$ is a number field that is also a Galois extension of $\mathbb{Q}$ and if $A=\mathcal{O}_K$, then for any intermediate field $L$, $Nm_{K/L}(\alpha)\in A$, $\forall \alpha\in A$.

Answer 2

Given a finite extension $K/L$ and a subring $A \subset K, Frac(A)=K$, $B = A \cap L$,

For $a \in A$, let $R_1 = B[a]$, and pick iteratively some $a_{m+1} \in A$ such that $R_{m+1}= R_m[a_{m+1}]$ is a free $R_m$-module, until $R_M$ is of finite index in $A$. So $R_M$ is also a free $B$-module.

The field norm $N_{K/L}(a)$ is the determinant of the $B$-linear map $x \mapsto a x , R_M \to R_M$. So $N_{K/L}(a) \in B$.

If $a \in A^\times$ then $1=N_{K/L}(aa^{-1})=N_{K/L}(a)N_{K/L}(a^{-1})$ and hence $N_{K/L}(a) \in B^\times$.