# 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.

• Can you slow down to explain your question clearly from the beginning? There are too many things undefined, for one. E.g., what are $F$, $\chi_F$, $n$? (The latter 2 can be easily guessed, but your assumptions on $F$ are somewhat ambiguous.) More crucially, how are you defining $P_D$ and the conductor of $\chi$? Mar 28, 2015 at 15:09
• @Kimball Sorry for so many rush in my explanations, I will edit the original post in order to be more explicite. Here are the details : $F$ is a local field, let's say $\mathbf{Q}_p$. 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 $D$. If $I_D$ is the unique maximal ideal of $\mathcal{O}_D$, the integer (valuation) ring of $D$, then the uniformizer $P_D$ is defined as a generator (minimal valuation element) of $I_D$. Let's forget the conductors, and say I am only talking about levels ;) Mar 29, 2015 at 16:19
• @Kimball Maybe should I delete my post and split it more independent and precise questions, in order to be more accessible and swiftly readable ? Mar 29, 2015 at 16:25
• I think your definition of $P_D$, $P_F$ and $d$ still has issues. I suppose it may be that no nontrivial power of $P_D$ lies in $F$, so $P_F = P_D^d$ is impossible. And do you know that $d$ does not depend on the choice of $P_D$? If so, please say why. Then I think I can probably answer at least 2 of the questions. Mar 29, 2015 at 17:17
• One thing that was confusing me was that you define $P_F$ and $P_D$ as uniformizers (elements), then seem to use them to mean ideals later in your question. If you mean ideals, I think $d$ is well-defined (if you write $P_F \mathcal O_D = P_D^d$), and you should be able to determine it using ideal norms. Mar 31, 2015 at 1:50

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 $m$.