On conductors, levels and traces on quaternion algebras I am currently working on level issues in the division central simple algebra case, say $D$ over a local non-archimedean field $F$ (e.g. $\mathbf{Q}_p$). Let say that $\mathcal{O}_D$ and $\mathcal{O}_F$ are the maximal orders of $D$ and $F$, with uniformizers $P_D$ and $P_F = P_D^d$. If $I_D$ is the unique maximal ideal of $\mathcal{O}_D$, which is a valuation ring, then the uniformizer $P_D$ is defined as a generator (minimal valuation element) of $I_D$.
Any conductor $\chi$ of $D^\times$ can be written $\chi = \chi_F \circ n$ where $\chi_F$ is a character of $F^\times$ and $n$ the reduced norm on $D$. I would like to verify some well-known relations between the level of $\chi = \chi_F \circ n$ character of $D^\times$, and the the level of $\chi_F$. 
Because of the fact that $\chi(1+x) = \chi_F(1+tr(x)+n(x))$, it seems sufficient to know, for $x \in P_D^k$, where lies its trace and norm. Some answers to those -- probably naive -- questions would be of great help :


*

*The trace is surjective from $\mathcal{O}_D$ onto $\mathcal{O}_F$. Could we say that it goes from $P_D$ to $P_F$, $P_D^2$ to $P_F$, ... $P_D^{d-1}$ to $P_F$, and then $P_D^d$ to $P_F^2$, etc. ? 
More generaly, do we have $Tr : P_D^k \to P_F^{[k/d]}$ ? And is there surjectivity ?

*Similarly, what do we know about the norm ? It seems plausible that the $P_F$-valuation of the norm of $x$ is always greater than the valuation of its trace : is it always true ?

*Could we say anything precise about the ramification index $d$ ? For instance in the case of quaternion algebras, there is only one such division algebra $D$, and a unique maximal order $\mathcal{O}_D$ : is $d$ always 2 ?
I am quite not at ease with the effective computations and behaviors of traces and norms in those cases : any idea on one of those questions, or any reference, would be very helpful !
Best regards.
 A: I think I can give you at least a sketch of answer now, but I have not thought carefully about details, so you should check though things carefully.  
Let $K$ be a cyclic extension of $F$ of degree $n$, and $\sigma$ a generator of the Galois group Gal($K/F$).  Then we can write $D$ as a cyclic algebra $(K/F, \sigma, a)$ where $a \in F^\times$ has order $n$ modulo $N_{K/F}(K^\times)$.  This is a subalgebra of $M_n(K)$ generated by elements of the form
$$ \iota(\alpha) = diag(\alpha, \alpha^\sigma, \ldots, \alpha^{\sigma^{n-1}}), \quad \alpha \in K$$
and some element $y$ such that 
$$ y^n = a, \quad \iota(\alpha)y = y\iota(\alpha)^\sigma \quad (\alpha \in K).$$
Then you can consider the order $\mathcal O_D$ generated by $y$ and $\alpha \in \mathcal O_K$.  I think this should usually (always?) be a maximal order with maximal ideal $I_D$ generated by $y$ and $\alpha \in \varpi_K \mathcal O_K$.  (Here you may need $K/F$ is unramified and some assumption on residual/characteristic, but I'm not sure.) When this is the case, you should be able to take for $P_D$ simply $\iota(\varpi_K)$.  


*

*If $K/F$ is unramified, the answer to your 3rd question is that
$d$ should simply be $n$.

*For your first question, the trace of any element will be the trace
of some $\iota(\alpha)$, with $\alpha$ lying in an appropriate ideal
of $K$.  Thus the image of the trace on $I_D$ is simply the image of
trace to $\mathcal O_F$ on $\varpi_K \mathcal O_K$.  Probably you can work out an example to see the image is not in $\varpi_F \mathcal O_F$.

*For the second question, the reduced norm of an element in $I_D$ will lie in $\varpi_F \mathcal O_F$.  However, it is not true that the norm always has greater valuation than the trace.  For example, for a quaternion algebra with $F=\mathbb Q_p$ and $K=F(\delta)$ such that $\sigma(\delta) = -\delta$, consider $x=\iota(p^m+p\delta)$.  Generically (avoiding special cases), the norm has valuation 2, but the trace has valuation $m$.

