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.

  1. (a)

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

  2. (b)

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

  3. (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 ij 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:

  1. (i)

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

  2. (ii)

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

  3. (iii)

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

  4. (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.

  1. (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.

  2. (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 ef 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. 1.

    Let .

    1. (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 \(t,x,y,z \in \mathbb Q _2\).

    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. 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. 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. 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. 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. 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. 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. 2.

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

  7. 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. 8.

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

    1. (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.

    2. (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. 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. 10.

    Give another proof of Lemma 13.4.9 using quadratic forms.

  11. 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. 12.

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

  13. 13.

    Let F be a field with absolute value \(|\, \,|\) and V a finite-dimensional F-vector space.

    1. (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 \(d(x,y)=\Vert x-y\Vert \).

    2. (b)

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

    3. (c)

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

  14. 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. 15.

    Let F be a local field.

    1. (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 \(F^\times \subseteq F\).

    2. (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 \(B^\times \subseteq B\).

    3. (c)

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

  16. 16.

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

  17. 17.

    Finish the proof of Lemma 13.2.9.

  18. 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.

    1. (a)

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

    2. (b)

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