# What is the quotient group $D^*/{F^*(1+P_D)}$ for a quaternion division algebra $D$ over a local field $F$?

Let $$F$$ be the non-archimedean local field $$\mathbb{Q}_p$$ for some prime $$p$$ and $$D$$ be a quaternion division algebra over $$F$$. Let $$\mathcal{O}_D$$ and $$\mathcal{P}_D$$ denote the ring of integers of $$D$$ and its unique maximal ideal (respectively). Then, what is the finite group $$\frac{D^*}{F^*(1+ \mathcal{P}_D)} = {?}$$ where $$D^*=D-\{0\}$$ and $$F^*=F-\{0\}$$ are multiplicative groups.

Consider the reduced norm map $$N_\text{rd}:D \rightarrow F$$, then $$N_\text{rd}(D^*)=F^*$$ and if $$D^1$$ denotes the reduced norm one elements of $$D$$, then we have an exact sequence $$1 \rightarrow D^1 \rightarrow D^* \rightarrow F^* \rightarrow 1$$ but we have $$D^1 \cap F^*=\{\pm 1\}$$. We know from Carl Riehm's article that $$\frac{D^1}{(1+ \mathcal{P}_D)} \cong {_N}(\mathbb{F}_{p^2})= \text{Finite cyclic group of order } (p+1).$$ Here $${_N}(\mathbb{F}_{p^2})$$ is the subgroup of $$\mathbb{F}_{p^2}$$ consisting of norm 1 elements.

Question: Similarly, can we write $$\frac{D^*}{F^*(1+ \mathcal{P}_D)}$$ in terms of finite fields?

• What is Carl Riehm's article? May 16, 2022 at 21:11
• @LSpice, "The Norm 1 Group of p-Adic Division Algebra" was Riehm's article. Thanks!
– BPK
May 17, 2022 at 7:51

Yes, we can.$$\newcommand{\order}{\mathcal{O}}$$ $$\newcommand{\Z}{\mathbb{Z}}$$ $$\newcommand{\prim}{\mathcal{P}}$$ $$\newcommand{\F}{\mathbb{F}}$$

First, let me remind you of the following explicit description of $$\order_D$$. I won't use it explicitly but it is convenient to check some of my claims below. Let $$\pi$$ be a uniformiser of $$F$$ and $$L/F$$ the unique unramified quadratic extension with Galois group generated by $$\sigma$$. We have $$\order_D = \order_L + \order_L \Pi$$ where $$\Pi^2 = \pi$$ and $$\Pi \lambda = \sigma(\lambda) \Pi$$ for $$\lambda\in L$$ (you also trivially get $$D$$ by extension of scalars). Let $$v$$ denote the normalised valuation on $$D$$, so that $$v(\Pi) = 1$$ (and $$v(\pi) = 2$$).

The valuation induces an isomorphism $$D^\times / \order_D^\times \cong \Z$$, and therefore $$D^\times / F^\times\order_D^\times \cong \Z/2\Z$$. Since $$\order_D^\times \cap F^\times = \order_F^\times$$ we get an exact sequence $$1 \to \order_D^\times/\order_F^\times(1+\prim_D) \to D^\times/F^\times(1+\prim_D) \to \Z/2\Z \to 1$$ and this sequence is split by the existence of the element $$\Pi$$ since $$\Pi^2 = \pi \in F^\times$$.

Since $$\order_D^\times/(1+\prim_D) \cong \F_{q^2}^\times$$ and the image of $$\order_F^\times$$ is $$\F_q^\times$$, we obtain $$\order_D^\times/\order_F^\times(1+\prim_D) \cong \F_{q^2}^\times/\F_q^\times.$$ Moreover, the action of the nontrivial element of $$\Z/2\Z$$ is via conjugation by $$\Pi$$, which is the same as the action of the Frobenius automorphism $$x\mapsto x^q$$.

We therefore obtain $$D^\times/F^\times(1+\prim_D) \cong \F_{q^2}^\times/\F_q^\times \rtimes \mathrm{Gal}(\F_{q^2}/\F_q) \cong C_{q+1} \rtimes C_2,$$ where in the first semidirect product the action is the natural one, and in the second one the nontrivial element of $$C_2$$ acts by inversion on $$C_{q+1}$$.

• I always forget. Does this also work when $p = 2$? May 16, 2022 at 20:56
• @LSpice Yes, there is nothing special for $p=2$, for once. May 16, 2022 at 21:07
• Thanks! But surely you meant to say "it works even for $p = 2$; for once, $p = 2$ is not odd." May 16, 2022 at 21:10
• @LSpice Exactly! :-) May 16, 2022 at 21:12
• I don't have much to add to @LSpice's answer: the next steps of the filtrations are $p$-groups, and they don't have such a simple structure. I don't know of an explicit presentation for the larger quotient. May 17, 2022 at 12:13

To give an alternative answer, let us first recall the article "Construction of Locally Compact Near-Fields from $$p$$-Adic Division Algebras" by Detlef Groger:

Fix a prime element $$\pi_F$$ of $$F$$. Then $$D$$ is generated as a non-commutative $$F$$-algebra by an unramified extension $$E/F$$ of degree 2 and an element $$\pi$$ with $$\pi^2=\pi_F$$. we consider $$\varpi$$ for a (fixed) primitive $$(p^2 -1)^\text{th}$$ root of unity in $$E$$ and $$\varpi_F=\varpi^\frac{p^2 -1}{p-1} \in F$$.

Suppose $$U_F$$, $$U_D$$ denote the groups of units in $$F$$ and $$D$$ with $$U^1_F := 1 + P_F$$ and $$U^1_D := 1 + P_D$$. Put $$C = \langle \varpi, \pi \rangle$$ and $$C_F=C \cap F^* = \langle \varpi_F, \pi_F \rangle$$. Then, $$C$$ is a complement of $$U^1_D$$ in the semidirect product $$D^*=U^1_D \rtimes C$$, whereas the product $$F^*=U^1_F \times C_F$$ is direct product. Therefore, $$F^* U^1_D= U^1_D \times C_F$$ and, $$\frac{D^*}{F^*(1+P_D)} \cong C/C_F \cong \langle\overline{\varpi}\rangle \rtimes \langle\overline{\pi}\rangle \cong Z_{p+1} \rtimes Z_2 .$$

• TeX note: Please use $\langle\overline{\varpi}\rangle$ \langle\overline{\varpi}\rangle, not $<\overline{\varpi}>$ <\overline{\varpi}>. I have edited accordingly. May 17, 2022 at 10:23