Abelianization of unit quaternions over a p-adic field

Question

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.

Answer 1

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 $C_1 = [C_0,C_0]$.