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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.

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  • $\begingroup$ "of the field": you mean "of the field $\mathbf{Q}_p$"? $\endgroup$ Commented Oct 26, 2022 at 16:52
  • $\begingroup$ Sorry. Yes, I do, of course. I'll edit. $\endgroup$ Commented Oct 26, 2022 at 16:55
  • $\begingroup$ A point of terminology: although you say clearly what you mean, without context I would expect "unit quaternions" to refer to the group of norm-$1$ quaternions, not the unit group of the ring of integers in the quaternion division algebra. $\endgroup$ Commented Oct 26, 2022 at 19:34
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    $\begingroup$ Thanks for the terminology comment. The tension was between brevity and accuracy. $\endgroup$ Commented Oct 27, 2022 at 11:20

1 Answer 1

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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 $C_1 = [C_0,C_0]$.

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    $\begingroup$ I knew I'd seen someone discussing exactly this article recently! $\endgroup$ Commented Oct 27, 2022 at 2:16
  • $\begingroup$ Amazing. Thank you! And I'm pleased to see that the proof sufficiently involved to justify my question. $\endgroup$ Commented Oct 27, 2022 at 11:25
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    $\begingroup$ @LSpice Indeed, I had almost forgotten about that previous question! $\endgroup$ Commented Oct 27, 2022 at 13:43
  • $\begingroup$ @SimonWadsley You are welcome. It is not trivial indeed! $\endgroup$ Commented Oct 27, 2022 at 13:43

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