Quaternion algebras over local fields

Abstract

In this chapter, we approach the classification of quaternion algebras over local fields in a second way, using valuations.

In this chapter, we approach the classification of quaternion algebras over local fields in a second way, using valuations.

1 Extending the valuation

Recall (section 12.1) the valuation $v=v_p$ on $\mathbb Q _p$, measuring divisibility by p. We have

$$\begin{aligned} \begin{aligned} \mathbb Z _p&= \{x \in \mathbb Q _p : v(x) \ge 0\}, \text { and} \\ p\mathbb Z _p&= \{x \in \mathbb Q _p : v(x) > 0\}; \end{aligned} \end{aligned}$$

(13.1.1)

Indeed, these can profitably be taken as their definition.

For any finite extension $K \supseteq \mathbb Q _p$ of fields, there is a unique valuation w on K such that $w|_\mathbb{Q _p}=v$ (so w extends v), defined by

$$\begin{aligned} w(x) :=\frac{v({{\,\mathrm{Nm}\,}}_{K|\mathbb Q _p}(x))}{[K:\mathbb Q _p]}. \end{aligned}$$

(13.1.2)

The integral closure of $\mathbb Z _p$ in K is the valuation ring $\{x \in K : w(x) \ge 0\} \supseteq \mathbb Z _p$, and its unique maximal ideal is $\{x \in K : w(x) > 0\}$, as in (13.1.1).

For example, there is a unique unramified quadratic extension K of $\mathbb Q _p$: we have $K=\mathbb Q _p(\sqrt{e})$, where $e=-3$ for $p=2$ and otherwise $e \in \mathbb Z$ is a quadratic nonresidue modulo p for p odd. It is common to write $K=\mathbb Q _{p^2}$ for this field and $\mathbb Z _{p^2}$ for its valuation ring, since the residue field of K is $\mathbb F _{p^2}$.

In a completely parallel fashion, let B be a division quaternion algebra over $\mathbb Q _p$. Then there is again a unique valuation w extending v, defined by

$$\begin{aligned} \begin{aligned} w :B&\rightarrow \mathbb R \cup \{\infty \} \\ \alpha&\mapsto \frac{v({{\,\mathrm{nrd}\,}}(\alpha ))}{2}. \end{aligned} \end{aligned}$$

(13.1.3)

The valuation ring

$$\begin{aligned} \mathcal {O}:=\{\alpha \in B : w(\alpha ) \ge 0\} \end{aligned}$$

(13.1.4)

is the unique (!) maximal R-order in B, consisting of all elements of B that are integral over $\mathbb Z _p$. The set

$$\begin{aligned} P :=\{\alpha \in B : w(\alpha )>0\} \end{aligned}$$

(13.1.5)

is the unique maximal two-sided (bilateral) ideal of $\mathcal {O}$.

Using the unique extension of the valuation, we obtain the following main result of this chapter (a special case of Theorem 13.3.11).

Theorem 13.1.6

Let $q :=p^2$. Then the following statements hold.

• (a) There is a unique division quaternion algebra B over $\mathbb Q _p$, up to isomorphism given by .

• (b) The valuation ring of B is $\mathcal {O}\simeq \mathbb Z _{q} \oplus \mathbb Z _{q} j$.

• (c) The maximal ideal $P=\mathcal {O}j$ has $P^2=p\mathcal {O}$ and $\mathcal {O}/P \simeq \mathbb Z _{q}/p\mathbb Z _{q} \simeq \mathbb F _{q}$.

The method of proof used in this classification can also be used to classify central division algebras over local fields in much the same manner.

2 Valuations

To begin, we briefly review extensions of valuations; for further reading, see the references given in section 12.2.

Let R be a complete DVR with valuation $v :R \rightarrow \mathbb Z _{\ge 0} \cup \{\infty \}$, field of fractions F, maximal ideal $\mathfrak p$ generated by a uniformizer $\pi$ (with $v(\pi )=1$), and residue field $k :=R/\mathfrak p$. Then R is an integrally closed PID (every ideal is a power of the maximal ideal $\mathfrak p$), and $R=\{x \in F : v(x) \ge 0\}$. Let $|\, \,|_v$ be an absolute value attached to v, as in (12.2.11).

Let $K \supseteq F$ be a finite separable extension of degree $n :=[K:F]$. Then in fact K is also a nonarchimedean local field; more precisely, we have the following lemma.

Lemma 13.2.1

There exists a unique valuation w on K such that $w|_F=v$, defined by

$$\begin{aligned} w(x) :=\frac{v({{\,\mathrm{Nm}\,}}_{K|F}(x))}{[K:F]}. \end{aligned}$$

(13.2.2)

The integral closure of R in K is the valuation ring

$$\begin{aligned} S :=\{x \in K : w(x) \ge 0\}. \end{aligned}$$

When $w|_F=v$, we say that w extends v.

Proof. See e.g. Neukrich [Neu99, Chapter II, Theorem (4.8)], Cassels [Cas86, Chapter 7, Theorem 1.1], or Serre [Ser79, Chapter II, §2, Proposition 3]. $\square$

By the same token using (13.2.2), there exists a unique absolute value $|\, \,|_w$ on K which restricts to $|\, \,|_v$ on F; we pass freely between these two formulations.

13.2.3

We say $K \supseteq F$ is unramified if a uniformizer $\pi$ for F is also a uniformizer for K. We say $K \supseteq F$ is totally ramified if a uniformizer $\pi _K$ has the property that $\pi _K^n$ is a uniformizer for F.

In general, there is a (unique) maximal unramified subextension $K_{un } \subseteq K$, and the extension $K \supseteq K_{un }$ is totally ramified.

We say that $e=[K:K_{un }]$ is the ramification degree and $f=[K_{un }:F]$ the inertial degree, and the fundamental equality

$$\begin{aligned} n=[K:F]=ef \end{aligned}$$

(13.2.4)

holds.

13.2.5

Suppose that F is a local field (equivalently, the residue field k is a finite field). Then for all $f \in \mathbb Z _{\ge 1}$, there is a unique unramified extension of F of degree f and such a field corresponds to the unique extension of the residue field k of degree f. In an unramified extension $K \supseteq F$ of degree $[K:F]=f$, we have ${{\,\mathrm{Nm}\,}}_{K|F}(K^\times )=R^\times \pi ^{f\mathbb Z }$, so $b \in {{\,\mathrm{Nm}\,}}_{K|F}(K^\times )$ if and only if $f \mid v(b)$.

If ${{\,\mathrm{char}\,}}k \ne 2$, then by Hensel’s lemma, the unramified extension of degree 2 is given by adjoining a square root of an element of R which reduces to the unique nontrivial class in $k^\times /k^{\times 2}$; if ${{\,\mathrm{char}\,}}k=2$, then the unramified extension of degree 2 is given by adjoining a root of the polynomial $x^2+x+t$ where $t \in R$ reduces to an element which is nontrivial in the Artin-Schreier group $k/\wp (k)$ (recalling 12.3.11).

Before proceeding further, we describe local fields by their defining polynomials—we will need this later in the study of norms and strong approximation.

Lemma 13.2.6

(Krasner’s lemma). Let $K \supseteq F$ be a finite, Galois extension with absolute value $|\, \,|_w$. Let $\alpha ,\beta \in K$, and suppose that for all $\sigma \in {{\,\mathrm{Gal}\,}}(K\,|\,F)$ with $\sigma (\alpha ) \ne \alpha$, we have

$$\begin{aligned} |\alpha -\beta \,|_w < |\alpha -\sigma (\alpha ) \,|_w. \end{aligned}$$

(13.2.7)

Then $F(\alpha ) \subseteq F(\beta )$.

Intuitively, we can think of Krasner’s lemma as telling us when $\beta$ is closer to $\alpha$ than any of its conjugates, then $F(\beta )$ contains $\alpha$. It is for this reason that we state the lemma in terms of absolute values (instead of valuations).

Proof. Let $\sigma \in {{\,\mathrm{Gal}\,}}(K\,|\,F(\beta ))$ have $\sigma (\alpha ) \ne \alpha$. Then by the ultrametric inequality,

$$\begin{aligned} \begin{aligned} |\sigma (\alpha )-\alpha \,|_w&= |\sigma (\alpha )-\beta +\beta -\alpha \,|_w \le \max (|\sigma (\alpha )-\beta \,|_w,|\beta -\alpha \,|_w) \\&= \max (|\sigma (\alpha -\beta ) \,|_w,|\beta -\alpha \,|_w)=|\alpha -\beta \,|_w, \end{aligned} \end{aligned}$$

(13.2.8)

the final equality a consequence of (13.2.2) and the fact that Galois conjugates have the same norm. This contradicts the existence of $\sigma$, so $\sigma (\alpha )=\alpha$ for all $\sigma \in {{\,\mathrm{Gal}\,}}(K\,|\,F(\beta ))$. By Galois theory, we conclude that $F(\alpha ) \subseteq F(\beta )$. $\square$

Corollary 13.2.9

Let $f(x)=x^n+a_{n-1}x^{n-1}+\dots +a_0 \in F[x]$ be a separable, monic polynomial. Then there exists $\delta > 0$ such that whenever $g(x)=x^n+b_{n-1}x^{n-1}+\dots + b_0 \in F[x]$ has $|a_i-b_i \,|_v < \delta$, then

$$\begin{aligned} F[x]/(f(x)) \simeq F[x]/(g(x)). \end{aligned}$$

In particular, if f(x) is irreducible then g(x) is irreducible.

Proof. Since f(x) is separable, its discriminant ${{\,\mathrm{disc}\,}}(f)$ is nonzero. The discriminant is a polynomial function in the coefficients, so by continuity (multivariate Taylor expansion), there exists $\delta _1 >0$ such that if $g(x)=x^n+\dots +b_0 \in F[x]$ has $|a_i-b_i \,|_v<\delta _1$ for all i, then $|{{\,\mathrm{disc}\,}}(g)-{{\,\mathrm{disc}\,}}(f) \,|_v<|{{\,\mathrm{disc}\,}}(f) \,|_v$; by the ultrametric inequality, we conclude that for such g(x) we have $|{{\,\mathrm{disc}\,}}(g) \,|_v=|{{\,\mathrm{disc}\,}}(f) \,|_v$ so in particular ${{\,\mathrm{disc}\,}}(g) \ne 0$.

Let g(x) be as in the previous paragraph; then g(x) is separable. We first consider the case where f(x) is irreducible. Let $K \supseteq F$ be a splitting field for the polynomials $f(x)=\prod _{i=1}^n (x-\alpha _i) \in K[x]$ and $g(x)=\prod _{i=1}^n (x-\beta _i)$. Let $|\, \,|_w$ on K extend $|\, \,|_v$. Let

$$\begin{aligned} \epsilon :=\min _{i \ne j} |\alpha _i-\alpha _j \,|_w. \end{aligned}$$

(13.2.10)

Finally, let

$$\begin{aligned} \rho (g)=\rho (b_0,\dots ,b_{n-1}) :=\prod _{i=1}^n g(\alpha _i) = \prod _{i,j=1}^n (\alpha _i-\beta _j). \end{aligned}$$

(13.2.11)

The map $g \mapsto \rho (g)$ is again a polynomial in the coefficients $b_0,\dots ,b_{n-1}$ (indeed, it is a polynomial resultant). Therefore there exists $\delta >0$, with $\delta < \delta _1$, such that if $|a_i-b_i \,|_v < \delta$, then $|\rho (g) \,|_w < \epsilon ^{n^2}$. Therefore in (13.2.11), there exists i, j such that $|\alpha _i-\beta _j \,|_w < \epsilon$. Together with (13.2.10), we have

$$\begin{aligned} |\alpha _i-\beta _j \,|_w < \epsilon \le |\alpha _i-\alpha _k \,|_w \end{aligned}$$

for all $k \ne i$. By Krasner’s lemma (Lemma 13.2.6), we conclude that $F(\alpha _i) \subseteq F(\beta _j)$. Since $f(\alpha _i)=g(\beta _j)=0$ and f is irreducible, we have

$$\begin{aligned}{}[F(\alpha _i):F]=n \le [F(\beta _j):F] \le n \end{aligned}$$

so in fact $[F(\beta _j):F]=n$ and $F(\alpha _i)=F(\beta _j)$. Finally,

$$\begin{aligned} F[x]/(f(x)) \simeq F(\alpha _i)=F(\beta _j) \simeq F[x]/(g(x)) \end{aligned}$$

as desired.

The case when $f(x)=f_1(x)\cdots f_r(x)$ is reducible follows by repeating the above argument on each factor, and finishing using the continuity of multiplication among the coefficients: the details are requested in Exercise 13.16. $\square$

3 Classification via extensions of valuations

We now seek to generalize this setup to the noncommutative case; we retain the notation from the previous section. Let D be a central (simple) division algebra over F with $\dim _F D = [D:F]=n^2$. We extend the valuation v to a map

$$\begin{aligned} \begin{aligned} w:D&\rightarrow \mathbb R \cup \{\infty \} \\ \alpha&\mapsto \frac{v({{\,\mathrm{Nm}\,}}_{D|F}(\alpha ))}{[D:F]} = \frac{v({{\,\mathrm{nrd}\,}}(\alpha ))}{n}, \end{aligned} \end{aligned}$$

(13.3.1)

where the equality follows from the fact that ${{\,\mathrm{Nm}\,}}_{D|F}(\alpha )={{\,\mathrm{nrd}\,}}(\alpha )^n$ (see section 7.8).

Lemma 13.3.2

The map w is the unique valuation on D extending v, i.e., the following hold:

• (i) $w(\alpha )=\infty$ if and only if $\alpha =0$.

• (ii) $w(\alpha \beta )=w(\alpha ) + w(\beta )=w(\beta \alpha )$ for all $\alpha ,\beta \in D$.

• (iii) $w(\alpha +\beta ) \ge \min (w(\alpha ),w(\beta ))$ for all $\alpha ,\beta \in D$.

• (iv) $w(D^\times )$ is discrete in $\mathbb R$.

Proof. Since D is a division ring, statement (i) is immediate. Statement (ii) follows from the multiplicativity of ${{\,\mathrm{nrd}\,}}$ and v. To prove (iii), we may suppose $\beta \ne 0$ and so $\beta \in D^\times$. We have

$$\begin{aligned} w(\alpha +\beta )=w((\alpha \beta ^{-1}+1)\beta )=w(\alpha \beta ^{-1}+1)+w(\beta ). \end{aligned}$$

But the restriction of w to $F(\alpha \beta ^{-1})$ is a discrete valuation, thus $w(\alpha \beta ^{-1}+1) \ge \min (w(\alpha \beta ^{-1}),w(1))$ and by (ii) $w(\alpha +\beta ) \ge \min (w(\alpha ),w(\beta ))$, as desired. Finally, (iv) holds since $w(D^\times ) \subseteq v(F^\times )/n$ and the latter is discrete. The valuation is unique because it is unique whenever it is restricted to a subfield. $\square$

13.3.3

From Lemma 13.3.2, we say that w is a discrete valuation on D since it satisfies the same axioms as for a field. It follows from Lemma 13.3.2 that the set

$$\begin{aligned} \mathcal {O}:=\{\alpha \in D : w(\alpha ) \ge 0\} \end{aligned}$$

is a ring, called the valuation ring of D.

Proposition 13.3.4

The ring $\mathcal {O}$ is the unique maximal R-order in D, consisting of all elements of D that are integral over R.

Proof. First, we prove that

$$\begin{aligned} \mathcal {O}=\{\alpha \in D : \alpha \text { is integral over }R\}. \end{aligned}$$

(13.3.5)

We first show the inclusion $(\supseteq )$ of (13.3.5), and suppose $\alpha \in D$ is integral over R. Since R is integrally closed, by Lemma 10.3.5 the coefficients of the minimal polynomial $f(x) \in F[x]$ of $\alpha$ belong to R. Since D is a division ring, f(x) is irreducible and hence the reduced characteristic polynomial g(x) is a power of f(x) and thus has coefficients in R. Up to sign, the constant coefficient of g(x) is ${{\,\mathrm{nrd}\,}}(\alpha )$, so $w(\alpha )=v({{\,\mathrm{nrd}\,}}(\alpha )) \ge 0$, hence $\alpha \in \mathcal {O}$.

Next we prove $(\subseteq )$ in (13.3.5). Suppose $\alpha \in \mathcal {O}$, so that $w(\alpha ) \ge 0$, and let $K=F(\alpha )$. Let $f(x) \in F[x]$ be the minimal polynomial of $\alpha$. We want to conclude that $f(x) \in R[x]$ knowing that $w(\alpha ) \ge 0$. But the restriction of w to K is the unique extension of v to K, and this is a statement about the extension $K \supseteq F$ of fields and therefore follows from the commutative case, Lemma 13.2.1.

We can now prove that $\mathcal {O}$ is an R-order. Scaling an element of $D^\times$ by an appropriate power of $\pi$ gives it positive valuation, so $\mathcal {O}{}F=D$. To conclude, we must show that $\mathcal {O}$ is finitely generated as an R-module. Recall that D is a central division algebra over F, hence a separable F-algebra, so we may apply Lemma 10.3.7: every $\alpha \in \mathcal {O}$ is integral over R and $\mathcal {O}$ is a ring, and the lemma implies that $\mathcal {O}$ is an R-order.

Finally, it follows immediately that $\mathcal {O}$ is a maximal R-order: by Corollary 10.3.3, every element of an R-order is integral over R, and $\mathcal {O}$ contains all such elements. $\square$

Remark 13.3.6. For a quaternion division algebra D, we can argue more directly in the proof of Proposition 13.3.4 using the reduced norm: see Exercise 13.4.

13.3.7

It follows from Proposition 13.3.4 that $\mathcal {O}$ is a finitely generated R-submodule of D. But R is a PID so in fact $\mathcal {O}$ is free of rank [D : F] as an R-module. We have

$$\begin{aligned} \mathcal {O}^\times = \{\alpha \in D : w(\alpha )=0\} \end{aligned}$$

(13.3.8)

since $w(\alpha ^{-1})=-w(\alpha )$, and in particular $\alpha \in \mathcal {O}^\times$ if and only if ${{\,\mathrm{nrd}\,}}(\alpha )\in R^\times$. Consequently,

$$\begin{aligned} P :=\{\alpha \in D : w(\alpha ) > 0\} = \mathcal {O}\smallsetminus \mathcal {O}^\times \end{aligned}$$

(13.3.9)

is the unique maximal two-sided (bilateral) ideal of $\mathcal {O}$, as well as the unique left or right ideal of $\mathcal {O}$. Therefore $\mathcal {O}$ is a noncommutative local ring, a noncommutative ring with a unique maximal left (equivalently, right) ideal.

13.3.10

Let $\beta \in P$ have minimal (positive) valuation $w(\beta )>0$. Then for all $0 \ne \alpha \in P$ we have $w(\alpha \beta ^{-1})=w(\alpha )-w(\beta ) \ge 0$ so $\alpha \beta ^{-1} \in \mathcal {O}$ and $\alpha \in \mathcal {O}\beta$. Arguing on the other side, we have also $\alpha \in \beta \mathcal {O}$. Thus $P=\mathcal {O}\beta =\beta \mathcal {O}= \mathcal {O}\beta \mathcal {O}$.

Arguing in the same way, we see that every one-sided ideal of $\mathcal {O}$ is in fact two-sided, and every two-sided ideal of $\mathcal {O}$ is principally generated by any element with minimal valuation hence of the form $P^r$ for some $r \in \mathbb Z _{\ge 0}$.

We are now prepared to give the second proof of the main result in this chapter (Main Theorem 12.3.2). We now add the hypothesis that F is a local field, so that k is a finite field.

Theorem 13.3.11

Let F be a nonarchimedean local field. Then the following statements hold.

• (a) There is a unique division quaternion algebra B over F, up to F-algebra isomorphism given by

  

  where K is the unique quadratic unramified (separable) extension of F.

• (b) Let B be as in (a). Then the valuation ring of B is $\mathcal {O}\simeq S \oplus S j$, where S is the integral closure of R in K. Moreover, the ideal $P=\mathcal {O}j$ is the unique maximal ideal; we have $P^2=\pi \mathcal {O}$, and $\mathcal {O}/P \supseteq R/\mathfrak p$ is a quadratic extension of finite fields.

Proof. We begin with existence in part (a), and existence: we prove that $B=({K,\pi } \mid {F})$ is a division algebra. We recall that B is a division ring if and only if $\pi \not \in {{\,\mathrm{Nm}\,}}_{K|F}(K^\times )$ by Main Theorem 5.4.4 and Theorem 6.4.11. Since $K \supseteq F$ is unramified, we have ${{\,\mathrm{Nm}\,}}_{K|F}(K^\times ) = R^\times \pi ^{2\mathbb Z }$ by 13.2.5. Putting these together gives the result.

Continuing with (a), we now show uniqueness. Let B be a division quaternion algebra over F. We refer to 13.3.10, and let $P=\mathcal {O}\beta$. Then $w(\beta ) \in \frac{1}{2} \mathbb Z _{>0}$, so

$$\begin{aligned} w(\beta ) \le w(\pi )=v(\pi )=1 \le 2w(\beta )=w(\beta ^2); \end{aligned}$$

(13.3.12)

we conclude that $\beta \mathcal {O}= P \supseteq \pi \mathcal {O}\supseteq P^2 = \beta ^2 \mathcal {O}$. The map $\alpha \mapsto \alpha \beta$ yields an isomorphism $\mathcal {O}/P \xrightarrow {\smash {{\sim }}}P/P^2$ of k-vector spaces, so

$$\begin{aligned} 4 = \dim _k(\mathcal {O}/\pi \mathcal {O}) \le \dim _k(\mathcal {O}/P^2) = \dim _k(\mathcal {O}/P) + \dim _k(P/P^2) = 2 \dim _k(\mathcal {O}/P) \end{aligned}$$

(13.3.13)

and thus $\dim _k(\mathcal {O}/P) \ge 2$, with equality if and only if $\pi \mathcal {O}=P^2$.

As in (13.3.9), we have $\mathcal {O}\smallsetminus P = \mathcal {O}^\times$, so the ring $\mathcal {O}/P$ is a division algebra over k and hence a finite division ring. By Wedderburn’s little theorem (Exercise 7.30), we conclude that $\mathcal {O}/P$ is a finite field! So there exists $i \in \mathcal {O}$ such that its reduction generates $\mathcal {O}/P$ as a finite extension of k. But i satisfies its reduced characteristic polynomial, a monic polynomial of degree 2 with coefficients in R, so its reduction satisfies a polynomial of degree 2 with coefficients in k. Since i is a generator, we conclude $[\mathcal {O}/P:k] \le 2$. Together with the conclusion of the previous paragraph, we conclude that $[\mathcal {O}/P:k]=\dim _k(\mathcal {O}/P)=2$, in other words $\mathcal {O}/P$ is a (separable) quadratic field extension of k. It then follows from 13.2.5 that $K :=F(i)$ is the unique, unramified (separable) quadratic extension of F. Therefore equality holds in (13.3.13) and $P^2=\pi \mathcal {O}$. Since $\beta ^2\mathcal {O}= P^2 = \pi \mathcal {O}$, we have $w(\beta )=1/2$.

By Exercise 6.2 or 7.27, there exists $b \in F^\times$ such that $B \simeq ({K,b} \mid {F})$. Recalling the first paragraph above, since B is a division algebra, we have $b \not \in {{\,\mathrm{Nm}\,}}_{K|F}(K^\times )=R^\times \pi ^{2\mathbb Z }$. Applying Exercise 6.4, we may multiply b by a norm from $K^\times$, so we may suppose $b=\pi$, and therefore $B \simeq ({K,\pi } \mid {F})$. This concludes the proof of (a).

We turn now to (b), with $B=({K,\pi } \mid {F})=K + Kj$ with $j^2=\pi$. Let $\alpha =u+vj \in B$ with $u,v \in K$. Then ${{\,\mathrm{nrd}\,}}(\alpha )={{\,\mathrm{Nm}\,}}_{K|F}(u)-\pi {{\,\mathrm{Nm}\,}}_{K|F}(v)=x-\pi y$ with $x,y \in F$ and v(x) even and $v(\pi y)$ odd (as norms from K). By the ultrametric inequality, we have $w(\alpha )=v({{\,\mathrm{nrd}\,}}(\alpha )) \ge 0$ if and only if $v(x),v(y) \ge 0$ if and only if $u,v \in S$ (as S is the valuation ring of K). Therefore $\mathcal {O}=S+Sj$. Since $j^2=\pi$, we have $w(j)=1/2$, so $j\mathcal {O}=P$. The remaining statements were proven in the course of proving (a). $\square$

Corollary 13.3.14

Let $F \ne \mathbb C$ be a local field. Let K be the unramified quadratic extension of F, with $\langle \sigma \rangle = {{\,\mathrm{Gal}\,}}(K\,|\,F)$. Then the F-subalgebra

$$\begin{aligned} B = \left\{ \begin{pmatrix} u &{} \pi v \\ \sigma (v) &{} \sigma (u) \end{pmatrix} : u,v \in K\right\} \subset {{\,\mathrm{M}\,}}_2(K) \end{aligned}$$

is the unique division quaternion algebra over F (up to isomorphism).

Proof. Using Theorem 13.3.11(a), we split B over K as in 2.3.4. (We may also put $\pi$ below the diagonal as in 2.3.12.) $\square$

4 Consequences

We now observe a few consequences of Theorem 13.3.11.

Corollary 13.4.1

Let F be a nonarchimedean local field with valuation v, let K be a separable, unramified quadratic F-algebra, and let $B=({K,b} \mid {F})$ with $b \in F^\times$. If $v(b)=0$, then $B \simeq {{\,\mathrm{M}\,}}_2(F)$.

Proof. Either $K \simeq F \times F$ or $K \supseteq F$ is the unique unramified quadratic field extension. In the first case, K has a zerodivisor so $B \simeq {{\,\mathrm{M}\,}}_2(F)$. In the second case, we conclude as in the first paragraph of the proof of Theorem 13.3.11, since $b \in R^\times \le {{\,\mathrm{Nm}\,}}_{K|F}(K^\times )$. $\square$

13.4.2

Let B be a division quaternion algebra over F. In analogy with the case of field extensions (13.2.4), we define the ramification index of B over F as $e(B|F)=2$ since $P^2=\pi \mathcal {O}$, and the inertial degree of B over F as $f(B|F)=2$ since B contains the unramified quadratic extension K of F, and note the equality

$$\begin{aligned} e(B|F)f(B|F)=4=[B:F], \end{aligned}$$

as in the commutative case. (Viewed in this way, B is obtained from first an unramified extension and then a “noncommutative” ramified extension.)

Remark 13.4.3. Theorem 13.3.11, the fundamental result describing division quaternion algebras over a local field, is a special case of a more general result as follows. Let R be a complete DVR with maximal ideal $\mathfrak p =\pi R$ and $F :={{\,\mathrm{Frac}\,}}(R)$.

Let D be a (finite-dimensional) division algebra over F, and let $\mathcal {O}\subseteq D$ be the valuation ring and $P \subset \mathcal {O}$ the maximal ideal. Then $P^e=\mathfrak p \mathcal {O}$ for some $e \ge 1$, called the ramification index; the quotient $\mathcal {O}/P$ is a division algebra over the field $k=R/\mathfrak p$, and we let the inertial degree be $f=\dim _k(\mathcal {O}/P)$. Then $ef=\dim _F D=n^2$; moreover, if k is finite (F is a local field), then $e=f=n$. For a proof, see Exercise 13.10; or consult Reiner [Rei2003, Theorems 12.8, 13.3, 14.3]. However, the uniqueness of D up to F-algebra isomorphism no longer holds. If F is a local field, then the possibilities for D are classified up to isomorphism by a local invariant ${{\,\mathrm{inv}\,}}D \in (\frac{1}{n} \mathbb Z )/\mathbb Z \simeq \mathbb Z /n\mathbb Z$. These patch together to give a global result: see Remark 14.6.10.

This classification can be further extended to an arbitrary central simple algebra $B \simeq {{\,\mathrm{M}\,}}_n(D)$ over F: see Reiner [Rei2003, §17–18].

Splitting of local division quaternion algebras over extension fields is given by the following simple criterion.

Proposition 13.4.4

Let B be a division quaternion algebra over a local field F, and let L be a separable field extension of F of finite degree. Then L is a splitting field for B if and only if [L : F] is even.

Proof. If F is archimedean, then either $F=\mathbb C$ and there is no such L, or $F=\mathbb R$ and $B=\mathbb H$ and $L=\mathbb C$, and the result holds. So suppose F is nonarchimedean. We have $B \simeq ({K,\pi } \mid {F})$ where K is the unramified quadratic extension of F. Let e, f be the ramification index and inertial degree of L, respectively. Then $[L:F]=n=ef$, and n is even if and only if e is even or f is even. But f is even if and only if L contains an unramified quadratic subextension, necessarily isomorphic to K; but then K splits B so L splits B.

Having established the claim when f is even, suppose that f is odd. Then L is linearly disjoint from K and $K \otimes _F L=KL$ is the unramified quadratic extension of L. Therefore $B \otimes _F L \simeq ({KL,\pi } \mid {L})$. Let $R_L$ be the valuation ring of L and let $\pi _L$ be a uniformizer for L. Then ${{\,\mathrm{Nm}\,}}_{KL/L}(KL^\times ) = R_L^\times \pi _L^{2\mathbb Z }$. We have $\pi =u\pi _L^e$ for some $u \in R_L^\times$. Putting these together, we see that $B \otimes _F L$ is split if and only if $\pi$ is a norm from KL if and only if e is even. $\square$

As a consequence, B contains every separable quadratic extension of F.

Corollary 13.4.5

If B is a division quaternion algebra over a local field F and $K \supseteq F$ is a separable quadratic field extension, then $K \hookrightarrow B$.

Proof. Combine Proposition 13.4.4 with Lemmas 5.4.7 and 6.4.12. $\square$

We repeat now Lemma 12.4.6, giving a proof that works without restriction on characteristic.

Corollary 13.4.6

If ${{\,\mathrm{char}\,}}F \ne 2$, then the Hilbert symbol defines a symmetric, nondegenerate bilinear form on $F^\times /F^{\times 2}$.

Proof. Let $K :=F[x]/(x^2-a)$. The Hilbert symbol gives a well-defined map of sets

$$\begin{aligned} F^\times /F^{\times 2}&\rightarrow \{\pm 1\} \\ b&\mapsto (a,b)_F \end{aligned}$$

and we may conclude as in Lemma 12.4.6 if we show that this is a nontrivial group homomorphism.

First we show it is nontrivial. By Corollary 13.4.5, the field K embeds in the division quaternion algebra B, so by Exercise 2.5, there exists b such that $B \simeq ({a,b} \mid {F})$, whence $(a,b)_F = -1$.

Next, we show it is a homomorphism. We appeal to Main Theorem 5.4.4. We have $(a,b)_F=1$ if and only if $b \in {{\,\mathrm{Nm}\,}}_{K|F}(K^\times )$. So we reduce to showing that if $(a,b)_F=(a,b')_F=-1$ for $b,b'\in F^\times$, then $(a,bb')_F=1$. But by Corollary 7.7.6, since there is a unique division quaternion algebra, we conclude that $b/b' \in {{\,\mathrm{Nm}\,}}_{K|F}(K^\times )$; thus $bb'=(b')^2(b/b') \in {{\,\mathrm{Nm}\,}}_{K|F}(K^\times )$ and $({a,bb'} \mid {F}) \simeq {{\,\mathrm{M}\,}}_2(F)$ so $(a,bb')_F=1$, as claimed. $\square$

Remark 13.4.7. The proof of Corollary 13.4.6 (pairing with any $F^\times /F^{\times 2}$) shows that

$$\begin{aligned} F^\times /{{\,\mathrm{Nm}\,}}_{K|F}(K^\times ) \simeq \mathbb Z /2\mathbb Z . \end{aligned}$$

(13.4.8)

Conversely, if we know (13.4.8) then the properties of the Hilbert symbol are immediate. Although it was not hard to prove (13.4.8) when ${{\,\mathrm{char}\,}}k \ne 2$, to establish its truth when ${{\,\mathrm{char}\,}}k = 2$, one is led to study higher ramification groups (e.g. Serre [Ser79, Chapter XV]) eventually leading to local class field theory.

The norm groups played an important role in the proof above, so we conclude by recording the image of the reduced norm ${{\,\mathrm{nrd}\,}}(B_v^\times ) \subseteq F_v^\times$.

Lemma 13.4.9

We have

$$ {{\,\mathrm{nrd}\,}}(B^\times )= {\left\{ \begin{array}{ll} \mathbb R ^\times _{>0}, &{}\text { if }B \simeq \mathbb H ; \\ F^\times , &{} \text { otherwise.} \end{array}\right. } $$

Moreover, if F is nonarchimedean and $\mathcal {O}\subseteq B$ is a maximal R-order, then ${{\,\mathrm{nrd}\,}}(\mathcal {O}^\times )=R^\times$.

Proof. If $B \simeq {{\,\mathrm{M}\,}}_2(F)$ is split, then ${{\,\mathrm{nrd}\,}}(B^\times ) = \det ({{\,\mathrm{GL}\,}}_2(F))=F^\times$. So suppose B is a division algebra. If $B \simeq \mathbb H$ then ${{\,\mathrm{nrd}\,}}(B^\times ) = \mathbb R _{>0}^{\times }$, so we suppose F is nonarchimedean. Then $B \simeq ({K,\pi } \mid {F})$ where as above K is the unramified quadratic extension of F and $\pi$ is a uniformizer. But $F^\times = R^\times \times \langle \pi \rangle$, and ${{\,\mathrm{nrd}\,}}(K^\times )={{\,\mathrm{Nm}\,}}_{K|F}(K^\times )=R^\times \pi ^{2\mathbb Z }$ and ${{\,\mathrm{nrd}\,}}(j)=\pi$. The result then follows by multiplicativity of the norm.

The second statement follows similarly: if $B \simeq {{\,\mathrm{M}\,}}_2(F)$ then $\mathcal {O}\simeq {{\,\mathrm{M}\,}}_2(R)$ and ${{\,\mathrm{nrd}\,}}(\mathcal {O}^\times )=\det ({{\,\mathrm{GL}\,}}_2(R))=R^\times$; otherwise $\mathcal {O}\simeq ({S,\pi } \mid {R})$ where S is the ring of integers of K, and ${{\,\mathrm{nrd}\,}}(S^\times )={{\,\mathrm{Nm}\,}}_{K|F}(S^\times ) \supseteq R^\times$ and again ${{\,\mathrm{nrd}\,}}(j)=\pi$. $\square$

5 Some topology

In this section, we dive into the basic topological adjectives relevant to the objects we have seen and that will continue to play an important role. Throughout, let F be a local field.

13.5.1

F is a locally compact topological field (by definition) but F is not itself compact. The subgroup $F^\times = F \smallsetminus \{0\}$ is given the topology induced from the embedding

$$\begin{aligned} F^\times&\hookrightarrow F \times F \\ x&\mapsto (x,x^{-1}); \end{aligned}$$

it turns out here that this coincides with the subspace topology $F^\times \subseteq F$ (see Exercise 13.14(a)). Visibly, $F^\times$ is open in F so $F^\times$ is locally compact.

If F is nonarchimedean, with valuation ring R and valuation v, then $F^\times$ is totally disconnected and further

$$\begin{aligned} R^\times =\{x \in R : v(x)=0\} \subset R \end{aligned}$$

is closed so is a topological abelian group that is compact (and totally disconnected).

Now let B be a finite-dimensional F-algebra.

13.5.2

As an F-vector space, B has a unique topology compatible with the topology on F as all norms on a topological vector F-space extending the norm on F are equivalent (the sup norm is equivalent to the sum of squares norm, etc.): see Exercise 13.12. In particular, two elements are close in the topology on B if and only if their coefficients are close with respect to a (fixed) basis: for example, two matrices in ${{\,\mathrm{M}\,}}_n(F)$ are close if and only if all of their coordinate entries are close. (Of course, the precise notion of “close” depends on the choice of norm.) Consequently, B is a complete, locally compact topological ring, taking a compact neighborhood in each coordinate.

13.5.3

The group $B^\times$ is a topological group, with the topology given by the embedding $B^\times \ni \alpha \mapsto (\alpha ,\alpha ^{-1}) \in B \times B$. This topology coincides with the subspace topology (see Exercise 13.14(b)). From this, we can see that $B^\times$ is locally compact: the norm ${{\,\mathrm{Nm}\,}}_{B|F}:B^\times \rightarrow F^\times$ is a continuous map, so $B^\times = {{\,\mathrm{Nm}\,}}_{B|F}^{-1}(F^\times )$ is open in B, and an open subset of a Hausdorff, locally compact space is locally compact in the subspace topology (Exercise 13.14(c)).

Example 13.5.4

If $B={{\,\mathrm{M}\,}}_n(F)$, then $B^\times ={{\,\mathrm{GL}\,}}_n(F)$ is locally compact: a closed, bounded neighborhood that avoids the locus of matrices with determinant 0 is a compact neighborhood. When F is archimedean, this is quite visual: a matrix of nonzero determinant is at some finite distance away from the determinant zero locus! Note however that ${{\,\mathrm{GL}\,}}_n(F)$ is not itself compact: already $F^\times ={{\,\mathrm{GL}\,}}_1(F)$ is not compact.

Now suppose F is nonarchimedean with valuation v and valuation ring R.

13.5.5

We claim that R is the maximal compact subring of F. Indeed, $x \in F$ lies in a compact subring if and only if $v(x) \ge 0$ if and only if x is integral over R. The only new implication here is the statement that if $v(x)<0$ then x does not lie in a compact subring, and that is because the sequence $x_n=x^n$ does not have a convergent subsequence as $|x_n \,| \rightarrow \infty$.

Next, let $\mathcal {O}$ be an R-order in B.

13.5.6

Choosing an R-basis, we have an isomorphism $\mathcal {O}\simeq R^n$, and this isomorphism is also a homeomorphism. Therefore, $\mathcal {O}$ is compact as the Cartesian power of a compact set. The group $\mathcal {O}^\times$ is therefore also compact because it is closed: for $\gamma \in \mathcal {O}$, we have $\gamma \in \mathcal {O}^\times$ if and only if ${{\,\mathrm{Nm}\,}}_{B|F}(\gamma ) \in R^\times$, the norm map is continuous, and $R^\times =\{x \in R : v(x)=0\} \subseteq R$ is closed.

Example 13.5.7

For $R=\mathbb Z _p \subseteq F=\mathbb Q _p$ and $B={{\,\mathrm{M}\,}}_n(\mathbb Q _p)$, the order $\mathcal {O}={{\,\mathrm{M}\,}}_n(\mathbb Z _p)$ is compact (neighborhoods of a matrix can be taken as neighbhoods in each coordinate) and the subgroup $\mathcal {O}^\times ={{\,\mathrm{GL}\,}}_n(\mathbb Z _p)$ is compact: there is no way to “run off to infinity”, either in a single coordinate or via the determinant.

13.5.8

Suppose $B=D$ is a division algebra. Then the valuation ring $\mathcal {O}$ is the maximal compact subring of B, for the same reason as in the commutative case (see 13.5.5, details requested in Exercise 13.17(a)). There is a filtration

$$\begin{aligned} \mathcal {O}\supset P \supset P^2 \supset \dots \end{aligned}$$

giving rise to a filtration

$$\begin{aligned} \mathcal {O}^\times \supset 1+P \supset 1+P^2 \supset \dots . \end{aligned}$$

(13.5.9)

As in the second proof of Main Theorem 12.3.2, the quotient $\mathcal {O}/P$ is a finite extension of the finite residue field k, so $(\mathcal {O}/P)^\times$ is a finite cyclic group. The maximal two-sided ideal P is principal, generated by an element j of minimal valuation, and multiplication by $j^n$ gives an isomorphism $\mathcal {O}/P \xrightarrow {\smash {{\sim }}}P^n/P^{n+1}$ of k-vector spaces (or abelian groups) for all $n \ge 1$.

Furthermore, for each $n \ge 1$, there is an isomorphism of groups

$$\begin{aligned} \begin{aligned} \mathcal {O}/P \simeq P^n/P^{n+1}&\xrightarrow {\smash {{\sim }}}(1+P^n)/(1+P^{n+1}) \\ \alpha&\mapsto 1+\alpha . \end{aligned} \end{aligned}$$

(13.5.10)

Therefore, $\mathcal {O}^\times = \varprojlim _n (\mathcal {O}/P^n)^\times$ is a projective limit of solvable groups, also called a prosolvable group.

Example 13.5.11

If B is a division quaternion algebra over $\mathbb Q _p$, with valuation ring $\mathcal {O}$ and maximal ideal P, then the filtration (13.5.9) has quotients isomorphic to $\mathcal {O}/P \simeq \mathbb F _{p^2}$.

13.5.12

We will also want to consider norm 1 groups; for this, we suppose that B is a semisimple algebra. Let

$$\begin{aligned} B^1 :=\{\alpha \in B : {{\,\mathrm{nrd}\,}}(\alpha )=1\}; \end{aligned}$$

some authors also write ${{\,\mathrm{SL}\,}}_1(B) :=B^1$. Then $B^1$ is a closed subgroup of $B^\times$, since the reduced norm is continuous.

If B is a division ring and F is archimedean, then $B \simeq \mathbb H$ and $B^1 \simeq \mathbb H ^1 \simeq {{\,\mathrm{SU}\,}}(2)$ is compact (it is identified with the 3-sphere in $\mathbb R ^4$). In a similar way, if B is a division ring and F is nonarchimedean, then $B^1$ is compact: for B has a valuation v and valuation ring $\mathcal {O}$, and if $\alpha \in B$ has ${{\,\mathrm{nrd}\,}}(\alpha )=1$ then $v(\alpha )=0$ and $\alpha \in \mathcal {O}$, and consequently $B^1 \subseteq \mathcal {O}^\times$ is closed in a compact set so compact.

If B is not a division ring, then either B is the product of two algebras or B is a matrix ring over a division ring, and in either case $B^1$ is not compact.

Remark 13.5.13. The locally compact division algebras over a nonarchimedean field are necessarily totally disconnected. On the other hand, it is a theorem of Pontryagin [War89, Theorem 27.2] that if A is a connected locally compact division ring, then A is isomorphic as a topological ring to either $\mathbb R$, $\mathbb C$, or $\mathbb H$.

Exercises

• 1. Let .

  • (a) Show that B is a division ring that is complete with respect to the discrete valuation w defined by $w(t+xi+yj+zij) = v(t^2+x^2+y^2+z^2)$ for $t,x,y,z \in \mathbb Q _2$.

  • (b) Prove that

    $$\begin{aligned} \mathcal {O}:=\mathbb Z _2 \oplus \mathbb Z _2 i \oplus \mathbb Z _2 j \oplus \mathbb Z _2 \frac{1+i+j+ij}{2} \subset B \end{aligned}$$

    is the valuation ring of B.

• 2. Let B be a division quaternion algebra over a nonarchimedean local field F. Give another proof that the unramified quadratic extension K of F embeds in B as follows.

  Suppose it does not: then for all $\alpha \in \mathcal {O}$, the extension $F(\alpha ) \supseteq F$ is ramified, so there exists $a \in R$ such that $\alpha -a \in P \cap K(\alpha )$; let $P=j\mathcal {O}$ and write $\alpha =\alpha _0=a+j\alpha _1$, and iterate to conclude that $\alpha =\sum _{n=0}^{\infty } a_n j^n$ with $a_n \in R$. But F(j) is complete so $\mathcal {O}\subseteq F(j)$, a contradiction.

• 3. Let F be a local field with $F \not \simeq \mathbb C$, let K the unramified (separable) quadratic extension of F (take $K=\mathbb C$ if $F \simeq \mathbb R$), and let $\langle \sigma \rangle = {{\,\mathrm{Gal}\,}}(K\,|\,F)$, so that $\sigma$ is the standard involution on K. Let B be a division quaternion algebra B over F.

  Show that

  $$\begin{aligned} B \simeq \left\{ \begin{pmatrix} a &{} b \\ \pi \sigma (b) &{} \sigma (a) \end{pmatrix} : a,b \in K \right\} \subseteq {{\,\mathrm{M}\,}}_2(K). \end{aligned}$$

  [Hint: Compute the regular representation 2.3.8.] Identify the maximal order $\mathcal {O}$ its maximal ideal J under this identification.

• 4. Let B be a division quaternion algebra over F. Show that $\alpha \in B$ is integral over R if and only if ${{\,\mathrm{nrd}\,}}(\alpha ),{{\,\mathrm{nrd}\,}}(\alpha +1) \in R$ if and only if $w(\alpha ),w(\alpha +1) \ge 0$, where w is the valuation on B.

• 5. Extend Theorem 13.3.11 as follows. Let R be a complete DVR with field of fractions F, and let B be a quaternion division algebra over F. Show that where $K \supseteq F$ is an unramified separable quadratic extension of F and $b \not \in {{\,\mathrm{Nm}\,}}_{K|F}(K^\times )$.

• 6. Let F be a nonarchimedean local field with residue field k having ${{\,\mathrm{char}\,}}k \ne 2$, and let $K \supseteq F$ be a separable quadratic field extension.

  • 1. Let $b \in F^\times$. Show that if K is unramified then $b \in {{\,\mathrm{Nm}\,}}_{K|F}(K^\times )$ if and only if v(b) is even; and if $K=F(\sqrt{a})$ is ramified, then $b \in {{\,\mathrm{Nm}\,}}_{K|F}(K^\times )$ if and only if $b=c^2$ or $b=-ac^2$ for some $c \in F^{\times 2}$.

  • 2. Deduce $[F^\times : {{\,\mathrm{Nm}\,}}_{K|F}(K^\times )]=2$ and Corollary 13.4.6 in this case.

• 7. Let $R :=\mathbb Q [[t]]$ be the ring of formal power series over $\mathbb Q$; then R is a complete DVR with fraction field $F=\mathbb Q ((t))$, the Laurent series over $\mathbb Q$. Let $B_0 :=({a,b} \mid \mathbb{Q })$ be a division quaternion algebra over $\mathbb Q$, and let $B :=B_0 \otimes _\mathbb{Q } F = ({a,b} \mid {F})$. Show that B is a division quaternion algebra over F, with valuation ring $\mathcal {O}:=R + Ri + Rj + Rij$.

• 8. Let B be a division quaternion algebra over a nonarchimedean local field F, and let $\mathcal {O}$ be the valuation ring.

  • (a) Show that every one-sided (left or right) ideal of $\mathcal {O}$ is a power of the maximal ideal P and hence is two-sided.

  • (b) Let

    $$\begin{aligned}{}[\mathcal {O},\mathcal {O}] :=\langle \alpha \beta -\beta \alpha : \alpha ,\beta \in \mathcal {O}\rangle \end{aligned}$$

    be the commutator ideal $[\mathcal {O},\mathcal {O}]$ of $\mathcal {O}$, the two-sided ideal generated by commutators of elements of $\mathcal {O}$. Show that $P=[\mathcal {O},\mathcal {O}]$.

• 9. Let F be a nonarchimedean local field, let $B={{\,\mathrm{M}\,}}_2(F)$ and $\mathcal {O}={{\,\mathrm{M}\,}}_2(R)$. Show that there are $q+1$ right $\mathcal {O}$-ideals of norm $\mathfrak p$ corresponding to the elements of $\mathbb P ^1(k)$ or equivalently the lines in $k^2$.

• 10. Give another proof of Lemma 13.4.9 using quadratic forms.

• 11. Let F be a nonarchimedean local field with valuation ring R, maximal ideal $\mathfrak p$, and residue field k. Let D be a division algebra over F with $\dim _F D = n^2$, with valuation ring $\mathcal {O}$ and maximal two-sided ideal P. Show that $\mathcal {O}/P$ is finite extension of k of degree n, and $P^n=\mathfrak p \mathcal {O}$ (cf. Remark 13.4.3).

• 12. Show that (13.5.10) is an isomorphism of (abelian) groups.

• 13. Let F be a field with absolute value $|\, \,|$ and V a finite-dimensional F-vector space.

  • (a) Let $x_1,\dots ,x_n$ be a basis for V, and define

    $$\begin{aligned} \Vert a_1x_1+\dots +a_nx_n \Vert :=\max (|a_1 \,|,\dots ,|a_n \,|) \end{aligned}$$

    for $a_i \in F$. Show that V is a metric space with distance $d(x,y)=\Vert x-y\Vert$.

  • (b) Show that the topology on V is independent of the choice of basis in (a).

  • (c) Finally, show that if F is complete with respect to $|\, \,|$, then V is also complete.

• 14. Let F be a topological field. Show that the coarsest topology (fewest open sets) in which multiplication on ${{\,\mathrm{M}\,}}_n(F)$ is continuous is given by the coordinate topology.

• 15. Let F be a local field.

  • (a) The group $F^\times$ has the structure of topological group under the embedding $x \mapsto (x,x^{-1}) \in F \times F$ (under the subspace topology in $F \times F$). Show that this topology coincides with the subspace topology $F^\times \subseteq F$.

  • (b) More generally, let B be a finite-dimensional F-algebra. Show that the the topology on $B^\times$ induced by $\alpha \mapsto (\alpha ,\alpha ^{-1}) \in B \times B$ coincides with the subspace topology $B^\times \subseteq B$.

  • (c) Show that an open subset of a Hausdorff, locally compact space is locally compact in the subspace topology.

• 16. Let F be a finite extension of $\mathbb Q _2$. Show that $(-1,-1)_F=(-1)^{[F:\mathbb Q _2]}$.

• 17. Finish the proof of Lemma 13.2.9.

• 18. Let D be a division algebra over a nonarchimedean local field F. We recall (see 13.5.2) that D is a complete, locally compact topological ring.

  • (a) Verify (as in 13.5.5) that $\mathcal {O}$ is the maximal compact subring of B.

  • (b) Show that $B^\times /F^\times$ is a compact topological group.