Suppose $p$ is a prime, that $F$ is a finite extension of the field $\mathbb{Q}\_p$, $D$ is the division quaternion algebra over $F$ and $\mathcal{O}\_D$ is the valuation ring of $D$. What is the abelianisation of the group of units $\mathcal{O}\_D^\times$? I'd also appreciate a reference.


Apologies that this may look like a homework problem. It isn't, for me at least. It is clear to me that there is a surjective group homomorphism from $\mathcal{O}\_D^\times$ to $\mathcal{O}\_F^\times$ given by reduced norm and a surjective homomorphism from $\mathcal{O}\_D^\times$ to the group of units in the quadratic field extension of the residue field of $F$ with kernel $1+P\_D$ where $P\_D$ is the unique maximal ideal in $\mathcal{O}\_D$. What isn't clear to me is whether the derived subgroup is precisely the intersection of the kernels of these two homomorphisms or it is smaller than that.



Abelianization of unit quaternions over a p-adic field
The answer is yes: it is a result of Riehm, Corollary to Theorem 21 in The norm 1 group of $\mathfrak{p}$-adic division algebras *Amer. J. Math.* **92** 2 (1970), 499--523, see also Theorem 1.9 p.33 and the following Remark in Platonov and Rapinchuk, Algebraic groups and number theory *Pure and Applied Math.* **139** (1994).


The precise statement is as follows: define $C\_i = \ker(\mathrm{nrd}\colon U\_i \to \mathcal{O}\_F^\times)$ ($H\_r$ in Riehm's notation), where $U\_0 = \mathcal{O}\_D^\times$ and $U\_i = 1+P\_D^i$ is the usual filtration. Then $C\_1 = [C\_0,C\_0]$.


