Ternary quadratic forms over local fields

Abstract

In this chapter, we classify quaternion algebras over local fields using quadratic forms; this generalizes the classification of quaternion algebras over $\mathbb R$.

In this chapter, we classify quaternion algebras over local fields using quadratic forms; this generalizes the classification of quaternion algebras over $\mathbb R$.

1 $\triangleright$ The p-adic numbers and local quaternion algebras

Before beginning, we briefly remind the reader about the structure of the p-adic numbers. The p-adics were developed by Hensel, who wanted a uniform way to say that a Diophantine equation has a (consistent) solution modulo $p^n$ for all n. In the early 1920s, Hasse used them in the study of quadratic forms and algebras over number fields. At the time, what is now called the local-global principle then was called the p-adic transfer from the “small” to the “large”. As references on p-adic numbers, see for example Gouvêa [Gou97], Katok [Kat2007], or Koblitz [Kob84].

Just as elements of $\mathbb R$ can be thought of infinite decimals, an element of $\mathbb Q _p$ can be thought of in its p-adic expansion

$$\begin{aligned} a = (\dots a_3a_2a_1a_0.a_{-1}a_{-2}\cdots a_{-k})_p = \sum _{n=-k}^{\infty } a_n p^n \end{aligned}$$

(12.1.1)

where each $a_i \in \{0,\dots ,p-1\}$ are the digits of a. We continue “to the left” because a decimal expansion is a series in the base $1/10<1$ and instead we have a base $p>1$.

Put a bit more precisely, we define the p -adic absolute value on $\mathbb Q$ by defined by $|0 \,|_p :=0$ and

$$\begin{aligned} |c \,|_p :=p^{-v_p(c)} \quad \text {for }c \in \mathbb Q ^\times , \end{aligned}$$

(12.1.2)

where $v_p(c)$ is the power of p occurring in c in its unique factorization (taken to be negative if p divides the denominator of c written in lowest terms). Then the field $\mathbb Q _p$ is the completion of $\mathbb Q$ with respect to $|\,|_p$, that is to say, $\mathbb Q _p$ is the set of equivalence classes of Cauchy sequences of rational numbers, and it obtains a topology induced by the metric $d_p(x,y)=|x-y \,|_p$. We have $|a \,|_p=p^{k}$ for a as in (12.1.1) with $a_{-k} \ne 0$. Of course, all of the information in the p-adic absolute value is encoded in the p -adic valuation $v_p :\mathbb Q \rightarrow \mathbb R \cup \{\infty \}$.

Inside $\mathbb Q _p$ is the ring $\mathbb Z _p$ of p-adic integers, the completion of $\mathbb Z$ with respect to $|\,|_p$: the ring $\mathbb Z _p$ consists of those elements of $\mathbb Q _p$ with $a_n=0$ for $n<0$. (The ring $\mathbb Z _p$ might be thought of intuitively as $\mathbb Z /p^{\infty } \mathbb Z$, if this made sense.)

Equipped with their topologies, the ring $\mathbb Z _p$ is compact and the field $\mathbb Q _p$ is locally compact. These statements can be understood quite easily by viewing $\mathbb Z _p$ in a slightly different way, as a projective limit with respect to the natural projection maps $\mathbb Z /p^{n+1} \mathbb Z \rightarrow \mathbb Z /p^n \mathbb Z$:

$$\begin{aligned} \begin{aligned} \mathbb Z _p&= \varprojlim _n \mathbb Z /p^n \mathbb Z \\&= \left\{ x=(x_n)_n \in \prod _{n=1}^{\infty } \mathbb Z /p^n \mathbb Z : x_{n+1} \equiv x_n ~(\text{ mod } ~{p^n}) \text { for all }n \ge 1\right\} . \end{aligned} \end{aligned}$$

(12.1.3)

In other words, each element of $\mathbb Z _p$ is a compatible sequence of elements in $\mathbb Z /p^n \mathbb Z$ for each n. The equality (12.1.3) is just a reformulation of the notion of Cauchy sequence for $\mathbb Z$, and so for the purposes of this introduction it can equally well be taken as a definition.

As for the topology in (12.1.3), each factor $\mathbb Z /p^n \mathbb Z$ is given the discrete topology, the product $\prod _{n=0}^{\infty } \mathbb Z /p^n \mathbb Z$ is given the product topology, and $\mathbb Z _p$ is given the subspace topology. Since each $\mathbb Z /p^n \mathbb Z$ is compact (it is a finite set!), by Tychonoff’s theorem the product $\prod _{n=0}^{\infty } \mathbb Z /p^n \mathbb Z$ is compact; and $\mathbb Z _p$ is closed inside this product (a convergent limit of Cauchy sequences is a Cauchy sequence), so $\mathbb Z _p$ is compact and still Hausdorff. The topology on $\mathbb Z _p$ is a bit strange though, as $\mathbb Z _p$ is totally disconnected: every nonempty connected subset is a single point. In fact, $\mathbb Z _p$ is homeomorphic to the Cantor set, which is itself homeomorphic to the product of countably many copies of $\{0,1\}$. (More generally, every nonempty totally disconnected compact metric space with no isolated points is homeomorphic to the Cantor set.)

The set $\mathbb Z _p$ is a compact neighborhood of 0, as it is the closed ball of radius 1 around 0:

$$\begin{aligned} \mathbb Z _p = \{x \in \mathbb Q _p : |x \,|_p \le 1 \} = \{x \in \mathbb Q _p : v_p(x) \ge 0\}. \end{aligned}$$

(12.1.4)

In a similar way, the disc of radius 1 around $a \in \mathbb Q _p$ is a compact neighborhood of a homeomorphic to $\mathbb Z _p$, so $\mathbb Q _p$ is locally compact. Being able to make topological arguments like the one above is the whole point of looking at fields like $\mathbb Q _p$: our understanding of infinite algebraic objects is informed by topology.

With this review, and topological arguments now at our disposal, we consider quaternion algebras over $\mathbb Q _p$. The ‘original’ quaternion algebra, of course, was the division ring $\mathbb H$ of Hamiltonians over the real numbers (the ‘original’ field with a topology), and indeed $\mathbb H$ is the unique division quaternion algebra over $\mathbb R$ (Corollary 3.5.8). We find a similar result over $\mathbb Q _p$ (a special case of Theorem 12.3.2), as follows.

Theorem 12.1.5

There is a unique division quaternion algebra B over $\mathbb Q _p$, up to isomorphism; if $p \ne 2$, then

where $e \in \mathbb Z$ is a quadratic nonresidue modulo p.

For example, if $p \equiv 3 \pmod {4}$ then $-1$ is quadratic nonresidue and $({-1,p} \mid \mathbb{Q _p})$ is the unique division quaternion algebra over $\mathbb Q _p$.

Because we have exactly two such possibilities, we define the Hilbert symbol: for $a,b \in \mathbb Q _p^\times$, we have $(a,b)_\mathbb{Q _p}=1,-1$ according as the quaternion algebra $({a,b} \mid \mathbb{Q _p}) \simeq {{\,\mathrm{M}\,}}_2(\mathbb Q _p)$ is split or not. According to Theorem 12.1.5, the Hilbert symbol over $\mathbb Q _p$ uniquely identifies the two possible isomorphism classes of quaternion algebras over $\mathbb Q _p$—just like it does over $\mathbb R$.

Our approach to Theorem 12.1.5 uses quadratic forms: we use the classification of isomorphism classes of quaternion algebras given in terms of similarity classes of ternary quadratic forms (Theorem 5.1.1). The following proposition then implies Theorem 12.1.5.

Proposition 12.1.6

There is a unique ternary anisotropic quadratic form Q over $\mathbb Q _p$, up to similarity; if $p \ne 2$, then $Q \sim \langle 1, -e, -p \rangle$ where e is a quadratic nonresidue modulo p.

Happily, this proposition can be proved using some rather direct manipulations with quadratic forms and gives a very “hands on” feel; it is also suggests the arguments we use for a more general result. The main input we need is a quadratic Hensel’s lemma, or more precisely, the following consequence.

Lemma 12.1.7

For $p \ne 2$, the classes in $\mathbb Q _p^\times /\mathbb Q _p^{\times 2}$ are represented by 1, e, p, ep where e is a quadratic nonresidue modulo p.

Proof. Let $a \in \mathbb Q _p^\times$ and let $m :=v_p(a)$. Then $a=bp^m$ with $b :=a/p^m \in \mathbb Z _p^\times$, and by squaring $a \in \mathbb Q _p^{\times 2}$ if and only if $b \in \mathbb Z _p^{\times 2}$ and m is even. We claim that $b \in \mathbb Z _p^{\times 2}$ if and only if its reduction b modulo p is a square in $(\mathbb Z /p\mathbb Z )^\times$. With the forward implication immediate, suppose $b \equiv c^2 ~(\text{ mod } ~{p})$ with $c \in \mathbb Z _p^\times$, then $b/c^2 \in 1 + p\mathbb Z _p$. But squaring is a bijection on $1+p\mathbb Z _p$, by expanding the square root as a convergent series (see Exercise 12.1) and using that $p \ne 2$. Thus $b/c^2 \in \mathbb Z _p^{\times 2}$, and the result follows. $\square$

We now proceed with the proof when $p \ne 2$.

Proof of Proposition 12.1.6, $p \ne 2$.  We start by showing that $Q(x,y,z)=x^2-ey^2-pz^2$ is anisotropic. Suppose $Q(x,y,z)=0$ with not all $x,y,z \in \mathbb Q _p$ zero. Rescaling by p, we may assume that $x,y,z \in \mathbb Z _p$ and not all $x,y,z \in p\mathbb Z _p$. We then reduce modulo p to find that $x^2 \equiv ey^2 \pmod {p}$. If $p \not \mid y$, then $(x/y)^2 \equiv e \pmod {p}$; but e is a quadratic nonresidue modulo p, a contradiction. So $p \mid y$; thus $p \mid (ey^2+pz^2)=x^2$, so $p \mid x$; thus $p^2 \mid (x^2-ey^2)=pz^2$, so $p \mid z$, a contradiction.

To show uniqueness, let Q be a ternary anisotropic form over $\mathbb Q _p$. Since $p \ne 2$, we may diagonalize. In such a diagonal form taken up to similarity, we may also rescale each coordinate up to squares as well as rescale the entire quadratic form. Putting this together with Lemma 12.1.7, without loss of generality we may suppose $Q(x,y,z)=\langle 1, -b, -c \rangle = x^2-by^2-cz^2$ with $b,c \in \{1,e,p,ep\}$, the signs chosen for convenience. If $b=1$ or $c=1$, the form is isotropic by inspection. So we are left to consider the cases $(b,c)=(e,e),(e,p),(e,ep),(p,p),(p,ep),(ep,ep)$.

• When $(b,c)=(e,e)$, we have after rescaling $x^2+y^2-ez^2$. We claim this form is always isotropic. Indeed, the form reduces to a nondegenerate ternary quadratic form over $\mathbb F _p$. Such a form is always isotropic by a delightful counting argument (Exercise 5.5(b), or a second chance in Exercise 12.6!). Lifting, there exist $x,y,z \in \mathbb Z _p$, not all zero modulo p, such that $x^2 \equiv -y^2+ez^2 \pmod {p}$. Since e is a nonsquare, we have $p \not \mid x$ (arguing similarly as in the first paragraph). Let $d :=-y^2+ez^2 \in \mathbb Q _p^\times$. By the possibilities in Lemma 12.1.7, we must have $d \in \mathbb Q _p^{\times 2}$; solving $x^2 = d$ for $x \in \mathbb Q _p$ then shows that Q is isotropic.

• The case (e, p) is our desired form.

• In the third case (eep!), we substitute $x \leftarrow ex$ and divide by e to obtain the form $-y^2+ex^2-pz^2$. We claim that there is an isometry $\langle -1,e \rangle \simeq \langle 1,-e \rangle$: indeed, in the first bullet we showed that the quadratic form $\langle -1,e,-1 \rangle$ is isotropic, so $-x^2+ey^2$ represents 1; using this representation as the first basis vector, extending to a basis, and diagonalizing, we conclude that $\langle -1,e \rangle \simeq \langle 1,b \rangle$. By discriminants, we have $-e=b$ up to squares. This brings us back to the first case.

• In cases (p, p) or (ep, ep), replacing $x \leftarrow px$ and dividing gives the quadratic forms $x^2+y^2-pz^2$ and $x^2+y^2-epz^2$. If $-1 \in \mathbb Z _p^{\times 2}$, then the form is isotropic; otherwise, we may take $e=-1$ and we are back to cases (e, p), (e, ep).

• In the final case (keeping pep!), we substitute $x \leftarrow px$ and divide by $-p$ to get $y^2+ez^2-px^2$. If $-1 \in \mathbb Z _p^{\times 2}$, then by substitution we change the middle sign to return to the first case. Otherwise, we may take $e=-1$, and the form is isotropic, a contradiction.

This consideration of cases completes the proof. $\square$

Proof of Proposition 12.1.6, $p \ne 2$.  We start by showing that $Q(x,y,z)=x^2-ey^2-pz^2$ is anisotropic. Suppose $Q(x,y,z)=0$ with not all $x,y,z \in \mathbb Q _p$ zero. Rescaling by p, we may assume that $x,y,z \in \mathbb Z _p$ and not all $x,y,z \in p\mathbb Z _p$. We then reduce modulo p to find that $x^2 \equiv ey^2 \pmod {p}$. If $p \not \mid y$, then $(x/y)^2 \equiv e \pmod {p}$; but e is a quadratic nonresidue modulo p, a contradiction. So $p \mid y$; thus $p \mid (ey^2+pz^2)=x^2$, so $p \mid x$; thus $p^2 \mid (x^2-ey^2)=pz^2$, so $p \mid z$, a contradiction.

To show uniqueness, let Q be a ternary anisotropic form over $\mathbb Q _p$. Since $p \ne 2$, we may diagonalize. In such a diagonal form taken up to similarity, we may also rescale each coordinate up to squares as well as rescale the entire quadratic form. Putting this together with Lemma 12.1.7, without loss of generality we may suppose $Q(x,y,z)=\langle 1, -b, -c \rangle = x^2-by^2-cz^2$ with $b,c \in \{1,e,p,ep\}$, the signs chosen for convenience. If $b=1$ or $c=1$, the form is isotropic by inspection. So we are left to consider the cases $(b,c)=(e,e),(e,p),(e,ep),(p,p),(p,ep),(ep,ep)$.

• When $(b,c)=(e,e)$, we have after rescaling $x^2+y^2-ez^2$. We claim this form is always isotropic. Indeed, the form reduces to a nondegenerate ternary quadratic form over $\mathbb F _p$. Such a form is always isotropic by a delightful counting argument (Exercise 5.5(b), or a second chance in Exercise 12.6!). Lifting, there exist $x,y,z \in \mathbb Z _p$, not all zero modulo p, such that $x^2 \equiv -y^2+ez^2 \pmod {p}$. Since e is a nonsquare, we have $p \not \mid x$ (arguing similarly as in the first paragraph). Let $d :=-y^2+ez^2 \in \mathbb Q _p^\times$. By the possibilities in Lemma 12.1.7, we must have $d \in \mathbb Q _p^{\times 2}$; solving $x^2 = d$ for $x \in \mathbb Q _p$ then shows that Q is isotropic.

• The case (e, p) is our desired form.

• In the third case (eep!), we substitute $x \leftarrow ex$ and divide by e to obtain the form $-y^2+ex^2-pz^2$. We claim that there is an isometry $\langle -1,e \rangle \simeq \langle 1,-e \rangle$: indeed, in the first bullet we showed that the quadratic form $\langle -1,e,-1 \rangle$ is isotropic, so $-x^2+ey^2$ represents 1; using this representation as the first basis vector, extending to a basis, and diagonalizing, we conclude that $\langle -1,e \rangle \simeq \langle 1,b \rangle$. By discriminants, we have $-e=b$ up to squares. This brings us back to the first case.

• In cases (p, p) or (ep, ep), replacing $x \leftarrow px$ and dividing gives the quadratic forms $x^2+y^2-pz^2$ and $x^2+y^2-epz^2$. If $-1 \in \mathbb Z _p^{\times 2}$, then the form is isotropic; otherwise, we may take $e=-1$ and we are back to cases (e, p), (e, ep).

• In the final case (keeping pep!), we substitute $x \leftarrow px$ and divide by $-p$ to get $y^2+ez^2-px^2$. If $-1 \in \mathbb Z _p^{\times 2}$, then by substitution we change the middle sign to return to the first case. Otherwise, we may take $e=-1$, and the form is isotropic, a contradiction.

This consideration of cases completes the proof. $\square$

Although direct, the proof we just gave has the defect that quadratic forms behave differently in characteristic 2, and so one may ask for a proof that works uniformly in all characteristics: we give such a proof in the next chapter by extending valuations.

One of the nice applications of this classification is that it gives a necessary condition for two quaternion algebras to be isomorphic. Let $B=({a,b} \mid \mathbb{Q })$ be a quaternion algebra over $\mathbb Q$ and consider its scalar extension $B_p=B \otimes _\mathbb Q \mathbb Q _p \simeq ({a,b} \mid \mathbb{Q _p})$. If $B'$ is another quaternion algebra over $\mathbb Q$ and $B \simeq B'$, then this implies $B_p \simeq B_p'$ for all primes p, and of course the same is true over $\mathbb R$. Perhaps surprisingly, it turns out that the collection of all of these tests is also sufficient: if $B,B'$ become isomorphic over $\mathbb R$ and over $\mathbb Q _p$ for all primes p, then in fact $B \simeq B'$ are isomorphic over $\mathbb Q$! This profound and powerful principle—detecting global isomorphism from local isomorphisms, a local-global principle—will be examined in chapter 14.

2 Local fields

In this section, we set up notation and basic results from the theory of local fields. The theory of local fields is described in many places, including Neukirch [Neu99, Chapters II, V], the classic texts by Cassels [Cas86] and Serre [Ser79]. Weil [Weil74] approaches number theory from the ground up in the language of local fields, building up the theory of local division rings.

Our motivation for local fields is as follows: we want a topology compatible with the field operations in which the field is Hausdorff and locally compact (every element has a compact neighborhood), analogous to what holds over the real and complex numbers. And to avoid trivialities, we will insist that this topology is not the discrete topology (where every subset of F is open). To carry this out, we begin with some basic definitions.

Definition 12.2.1

A topological group is a group equipped with a topology such that the group operation and inversion are continuous. A homomorphism of topological groups is a group homomorphism that is continuous.

A topological ring is a ring A equipped with a topology such that the ring operations (addition, negation, and multiplication) are continuous. A homomorphism of topological rings is a ring homomorphism that is continuous. A topological field is a field that is also a topological ring in such a way that division by a nonzero element is continuous.

One natural way to equip a ring with a topology is by way of an absolute value. To get started, we consider such notions first for fields. Throughout this section, let F be a field.

Definition 12.2.2

An absolute value on F is a map

$$\begin{aligned} |\,|:F \rightarrow \mathbb R _{\ge 0} \end{aligned}$$

such that:

• (i) $|x \,|=0$ if and only if $x=0$;

• (ii) $|xy \,| = |x \,||y \,|$ for all $x,y \in F$; and

• (iii) $|x+y \,| \le |x \,|+|y \,|$ for all $x,y \in F$ (triangle inequality).

An absolute value $|\,|$ on F gives F the structure of a topological field by the metric $d(x,y)=|x-y \,|$. Two absolute values $|\,|_1,|\,|_2$ on F are (strictly) equivalent  if there exists $c>0$ such that $|x \,|_1=|x \,|_2^c$ for all $x \in F$; equivalent absolute values induce the same topology on F.

Remark 12.2.3. If $|\,|$ is an absolute value on F, then it need not be the case that $x \mapsto |x|^c$ for $c>0$ is again absolute value, because it need not satisfy the triangle inequality. In particular, we will find it convenient to consider the square of the usual absolute value on $F=\mathbb C$, which suffers from this deficiency. There are various ways around this problem; perhaps the simplest is just to ignore it.

Definition 12.2.4

An absolute value is nonarchimedean  if the ultrametric inequality

$$\begin{aligned} |x+y \,|\le \sup \{|x \,|,|y \,|\} \end{aligned}$$

is satisfied for all $x,y \in F$, and archimedean  otherwise.

Example 12.2.5

The fields $\mathbb R$ and $\mathbb C$ are topological fields with respect to the usual archimedean absolute value.

Remark 12.2.6. A field with absolute value is archimedean if and only if it satisfies the archimedean property: for all $x \in F^\times$, there exists $n \in \mathbb Z$ such that $|n x \,|>1$. In particular, a field F equipped with an archimedean absolute value has ${{\,\mathrm{char}\,}}F = 0$.

Example 12.2.7

Every field has the trivial  (nonarchimedean) absolute value, defined by $|0 \,|=0$ and $|x \,|=1$ for all $x \in F^\times$; the trivial absolute value induces the discrete topology on F.

A nonarchimedean absolute value on a field F arises naturally by way of a valuation, as follows.

Definition 12.2.8

A valuation of a field F is a map $v :F \rightarrow \mathbb R \cup \{\infty \}$ such that:

• (i) $v(x)=\infty$ if and only if $x=0$;

• (ii) $v(xy) = v(x) + v(y)$ for all $x,y \in F$; and

• (iii) $v(x+y) \ge \min (v(x),v(y))$ for all $x,y \in F$.

A valuation is discrete  if the value group  $v(F^\times )$ is discrete in $\mathbb R$ (has no accumulation points).

Here, we set the convention that $x+\infty =\infty +x=\infty$ for all $x \in \mathbb R \cup \{\infty \}$. By (ii), the value group $v(F^\times )$ is a subgroup of the additive group $\mathbb R$, and so whereas an absolute value is multiplicative, a valuation is additive.

Example 12.2.9

Each $x \in \mathbb Q ^\times$ can be written $x=p^r a/b$ with $a,b \in \mathbb Z$ relatively prime and $p \not \mid ab$; the map $v_p(x)=r$ defines the p-adic valuation on $\mathbb Q$.

Example 12.2.10

Let k be a field and $F=k(t)$ the field of rational functions over k. For $f(t)=g(t)/h(t) \in k(t) \smallsetminus \{0\}$ with $g(t),h(t) \in k[t]$, define $v(f(t)) :=\deg h(t) - \deg g(t)$. Then v is a discrete valuation on F.

Given the parallels between them, it should come as no surprise that a valuation gives rise to an absolute value on F by defining

$$\begin{aligned} |x \,|=c^{-v(x)} \end{aligned}$$

(12.2.11)

for a fixed $c>1$; the induced topology on F is independent of the choice of c. By condition (iii), the absolute value associated to a valuation is nonarchimedean.

Example 12.2.12

The trivial valuation is the valuation v satisfying $v(0)=\infty$ and $v(x)=0$ for all $x \in F^\times$. The trivial valuation gives the trivial absolute value on F.

Two valuations v, w are equivalent if there exists $a \in \mathbb R _{>0}$ such that $v(x)=aw(x)$ for all $x \in F$; equivalent valuations give the same topology on a field. A nontrivial discrete valuation is equivalent after rescaling (by the minimal positive element in the value group) to one with value group $\mathbb Z$, since a nontrivial discrete subgroup of $\mathbb R$ is cyclic; we call such a discrete valuation normalized.

12.2.13

Given a field F with a nontrivial discrete valuation v, the valuation ring is $R :=\{x \in F : v(x) \ge 0\}$. We have $R^\times =\{x \in F : v(x)=0\}$ since

$$\begin{aligned} v(x)+v(x^{-1})=v(xx^{-1})=v(1)=0 \end{aligned}$$

for all $x \in F^\times$. The valuation ring is a local domain with unique maximal ideal

$$\begin{aligned} \mathfrak p :=\{x \in F : v(x) > 0\} = R \smallsetminus R^\times . \end{aligned}$$

An element $\pi \in \mathfrak p$ with smallest valuation is called a uniformizer, and comparing valuations we see that $\pi R=(\pi )=\mathfrak p$. Since $\mathfrak p \subsetneq R$ is maximal, the quotient $k :=R/\mathfrak p$ is a field, called the residue field of R (or of F).

Recall that a topological space is locally compact if each point has a compact neighborhood (every point is contained in a compact set containing an open set).

Definition 12.2.14

A local field is a Hausdorff, locally compact topological field with a nondiscrete topology.

In a local field, we can hope to understand its structure by local considerations in a compact neighborhood, hence the name. Local fields have a very simple classification as follows.

Theorem 12.2.15

A field F with absolute value is a local field if and only if F is one of the following:

• (i) F is archimedean, and $F \simeq \mathbb R$ or $F \simeq \mathbb C$;

• (ii) F is nonarchimedean with ${{\,\mathrm{char}\,}}F=0$, and F is a finite extension of $\mathbb Q _p$ for some prime p; or

• (iii) F is nonarchimedean with ${{\,\mathrm{char}\,}}F=p$, and F is a finite extension of the Laurent series field $\mathbb F _p((t))$ for some prime p; in this case, there is a (non-canonical) isomorphism $F \simeq \mathbb F _q((t))$ where q is a power of p.

A field F with absolute value $|\ \,|$ is a nonarchimedean local field if and only if F is complete with respect to $|\ \,|$, and $|\ \,|$ is equivalent to the absolute value associated to a nontrivial discrete valuation $v :F \rightarrow \mathbb R \cup \{\infty \}$ with finite residue field.

Proof. See Neukirch [Neu99, Chapter II, §5], Cassels [Cas86, Chapter 4, §1], or Serre [Ser79, Chapter II, §1]. $\square$

Although a local field is only locally compact, the valuation ring is itself compact, as follows.

Lemma 12.2.16

Suppose F is nonarchimedean. Then F is totally disconnected and the valuation ring $R \subset F$ is a compact, totally disconnected topological ring.

Proof. To see that F is totally disconnected (whence R too is totally disconnected), by translation it suffices to show that the only connected set containing 0 is $\{0\}$. Let $x \in F^\times$ with $|x \,|=\delta >0$. The image $|F^\times \,| \subseteq \mathbb R _{>0}$ is discrete, so there exists $0<\epsilon <\delta$ such that $|y \,|<\delta$ implies $|y \,| \le \delta -\epsilon$ for all $y \in F$. Thus an open ball is a closed ball

$$\begin{aligned} D(0,\delta )=\{y \in F : |y \,| < \delta \}=\{y \in F : |y \,| \le \delta -\epsilon \}=D[0,\delta -\epsilon ]; \end{aligned}$$

since $x \in F^\times$ and $\delta >0$ were arbitrary, the only connected subset containing 0 is $\{0\}$.

Next, we show R is compact. There is a natural continuous ring homomorphism

$$\begin{aligned} \phi :R \rightarrow \prod _{n=1}^{\infty } R/\mathfrak p ^n \end{aligned}$$

where each factor $R/\mathfrak p ^n$ is equipped with the discrete topology and the product is given the product topology. The map $\phi$ is injective, since $\bigcap _{n=1}^{\infty } \mathfrak p ^n =\{0\}$ (every nonzero element has finite valuation). The image of $\phi$ is obviously closed. Therefore R is homeomorphic onto its closed image. But by Tychonoff’s theorem, the product $\prod _{n=1}^{\infty } R/\mathfrak p ^n$ of compact sets is compact, and a closed subset of a compact set is compact, thus R is compact. $\square$

One key property of local fields we will use is Hensel’s lemma.

Lemma 12.2.17

(Hensel’s lemma, univariate). Let F be a nonarchimedean local field with valuation v and valuation ring R, and let $f(x) \in R[x]$. Let $a \in R$ satisfy $m :=v(f(a)) > 2 v(f'(a))$. Then there exists $\widetilde{a} \in R$ such that $f(\widetilde{a})=0$ and $\widetilde{a} \equiv a \pmod \mathfrak{p ^m}$.

Proof. The result is straightforward to prove using Taylor expansion or the same formulas as in Newton’s method. $\square$

Perhaps less well-known is the multivariate version.

Lemma 12.2.18

(Hensel’s lemma). Let F be a nonarchimedean local field with valuation v and valuation ring R, and let $f(x_1,\dots ,x_n) \in R[x_1,\dots ,x_n]$ with $n \ge 1$.

Let $a \in R^n$ have $m :=v(f(a))$ and suppose that

$$\begin{aligned} m > 2\min _i v\left( \frac{\partial f}{\partial x_i}(a)\right) \ge 0. \end{aligned}$$

Then there exists $\widetilde{a} \in R^n$ such that $f(\widetilde{a})=0$ and

$$\begin{aligned} \widetilde{a} \equiv a \pmod \mathfrak{p ^m}. \end{aligned}$$

Proof. One can reduce from several variables to the one variable version of Hensel’s lemma (Lemma 12.2.17): see Exercise 12.11. $\square$

Remark 12.2.19. With essentially the same proof, Hensel’s lemma holds more generally for R a complete DVR (without the condition on the residue field) and becomes axiomatically the property of Henselian rings.

3 Classification via quadratic forms

We now seek to classify quaternion algebras over local fields.

12.3.1

First, suppose F is archimedean. When $F=\mathbb C$, the only quaternion algebra over $\mathbb C$ up to isomorphism is $B \simeq {{\,\mathrm{M}\,}}_2(\mathbb C )$. When $F=\mathbb R$, by the theorem of Frobenius (Corollary 3.5.8), there is a unique quaternion division algebra over $\mathbb R$.

The classification of quaternion algebras over nonarchimedean local fields is quite analogous to the classification over $\mathbb R$, as follows.

Main Theorem 12.3.2

Let $F \ne \mathbb C$ be a local field. Then there is a unique division quaternion algebra B over F up to F-algebra isomorphism.

We approach the proof of Main Theorem 12.3.2 from two vantage points. In this section, we give a proof using quadratic forms; in the next section, we give another proof by extending the valuation (valid in all characteristics).

To prove this theorem, having dispatched the cases $F=\mathbb R ,\mathbb C$ in 12.3.1 above, from the previous section we may suppose F is a nonarchimedean local field with discrete valuation v, valuation ring R, maximal ideal $\mathfrak p =\pi R$ with uniformizer $\pi$, and residue field $R/\mathfrak p =k$.

12.3.3

Since R is a DVR, all R-lattices M are free (and we only consider those of finite rank): i.e., $M \simeq R^n$ for some $n \in \mathbb Z _{\ge 0}$. Given such an R-lattice M, we can reduce modulo $\mathfrak p$ to get $M/\mathfrak p M \simeq M \otimes _R k \simeq k^n$; conversely, any lift to M of any k-basis of $M/\mathfrak p M$ is an R-basis for M, by Nakayama’s lemma.

We recall Main Theorem 5.2.5, Corollary 5.2.6, and Main Theorem 5.4.4: isomorphism classes of quaternion algebras over a field F are in natural bijection with nondegenerate ternary quadratic forms up to similarity, and the matrix algebra corresponds to any isotropic form. So to prove Main Theorem 12.3.2, it is equivalent to prove the following statement.

Theorem 12.3.4

Let $F \ne \mathbb C$ be a local field. Then there is a unique anisotropic ternary quadratic form over F up to similarity.

Rescaling shows equivalently that there is a unique anisotropic ternary quadratic form over a local field $F \ne \mathbb C$ of discriminant 1 up to isometry. So our task becomes a hands-on investigation of ternary quadratic forms over F. The theory of quadratic forms over F is linked to that over its residue field k, so we first need to examine isotropy of quadratic forms over a finite field.

Lemma 12.3.5

A quadratic form $Q :V \rightarrow k$ over a finite field k with $\dim _k V \ge 3$ is isotropic.

Proof. The proof is a delightful elementary exercise (Exercise 12.6). $\square$

Recall definitions and notation for quadratic forms over R provided in section 9.7, we embark on a proof in the case where ${{\,\mathrm{char}\,}}k \ne 2$, beginning with the following lemma.

Lemma 12.3.6

Suppose ${{\,\mathrm{char}\,}}k \ne 2$. Let $Q:M \rightarrow R$ be a nonsingular quadratic form over R. Then the reduction $Q_k :M \otimes _R k \rightarrow k$ of Q modulo $\mathfrak p$ is nonsingular (equivalently, nondegenerate) over k; moreover, Q is isotropic over R if and only if is isotropic.

Proof. For the first statement, by definition we have ${{\,\mathrm{disc}\,}}Q \in R^\times$, so ${{\,\mathrm{disc}\,}}Q_k \in k^\times$ by reduction.

For the second, we first prove ($\Rightarrow$), let $x \in M \smallsetminus \{0\}$ have $Q(x)=0$. Since Q is homogeneous, we may suppose that $x \not \in \mathfrak p M$ (divide by powers of $\pi$ as necessary), so its image in $M \otimes _r k$ is nonzero and thereby shows that $Q_k$ is isotropic. For ($\Leftarrow$), let $a \in M$ be such that $Q_k(a) =0 \in k$ and a has nonzero reduction. Choose a basis $M \simeq R^n$ and write $Q(x_1,\dots ,x_n)=Q(x_1e_1+\dots +x_ne_n) \in R[x_1,\dots ,x_n]$ in the standard basis as a homogeneous polynomial of degree 2, let T be the associated symmetric bilinear form and $[T]=(T(e_i,e_j))_{i,j}$ the Gram matrix. We are almost ready to apply Hensel’s lemma (Lemma 12.2.18), but need to ensure convergence. We observe that

$$\begin{aligned} \frac{\partial Q}{\partial x_i}(x_1,\dots ,x_n) = \sum _{j=1}^n T(x_i,x_j)x_j \end{aligned}$$

(12.3.7)

so the vector of partial derivatives $((\partial Q/\partial x_i)(a))_i=[T]a$ is just the matrix product of the Gram matrix with the vector $a=(a_i)_i$. Working modulo $\mathfrak p$, we have ${{\,\mathrm{disc}\,}}Q_k = 2^{-n} \det [T] \in k^\times$, using that $2 \in k^\times$, so the kernel of is zero. Since a has nonzero reduction, we conclude that [T]a also has nonzero reduction, which means that $\min _i v((\partial Q/\partial x_i)(a))=0$. Therefore the hypotheses of Hensel’s lemma are satisfied with $m=1$, and we conclude there exists a nonzero $\widetilde{a} \in M$ such that $Q(\widetilde{a})=0$ and so Q is isotropic. $\square$

From Lemma 12.3.6, we obtain the following.

Proposition 12.3.8

Suppose ${{\,\mathrm{char}\,}}k \ne 2$. Let $Q:M \rightarrow R$ be a nonsingular quadratic form over R with M free of rank at least 3. Then Q is isotropic.

Proof. Combine Lemmas 12.3.5 and 12.3.6. $\square$

Considering valuations, we also deduce the following from Lemma 12.3.6.

Lemma 12.3.9

Suppose ${{\,\mathrm{char}\,}}k \ne 2$. Then $F^{\times }/F^{\times 2} \simeq (\mathbb Z /2\mathbb Z )^2$ and is represented by the classes of $1,e,\pi ,e\pi$ where $e \in R^\times$ is an element which reduces modulo $\mathfrak p$ to a nonsquare in k.

We now turn to the proof of our theorem in the case ${{\,\mathrm{char}\,}}k \ne 2$.

Proof of Theorem 12.3.4 (${{\,\mathrm{char}\,}}k \ne 2$).  Let $Q \simeq \langle a, -b, -c \rangle$ be an anisotropic ternary quadratic form over F. Then Q is nondegenerate. After rescaling and a change of basis (Exercise 12.8), we may suppose that $a=1$ and $0=v(b) \le v(c)$. If $v(b)=v(c)=0$ then the quadratic form modulo $\mathfrak p$ is nonsingular, so by Lemma 12.3.5 it is isotropic and by Lemma 12.3.6 we conclude Q is isotropic, a contradiction.

We are left with the case $v(b)=0$ and $v(c)=1$. By Lemma 12.3.9, we may suppose $b=1$ or $b=e$ where e is a nonsquare in k. If $b=1$, then the form is obviously isotropic, so $b=e$. Similarly, $c=\pi$ or $c=e\pi$. In fact, the latter case is similar to the former: dividing by e, we have

$$\begin{aligned} \langle 1,-e,-e\pi \rangle \sim \langle e^{-1}, -1, -\pi \rangle \simeq \langle -1,e,-\pi \rangle \end{aligned}$$

and since $\langle -1, e \rangle \simeq \langle 1, -e \rangle$ (Exercise 12.7), we conclude $Q \sim \langle 1, -e, -\pi \rangle$.

To finish, we show that the form $\langle 1,-e,-\pi \rangle$ is anisotropic. Suppose that $x^2-ey^2=\pi z^2$ with $x,y,z \in F^3$ not all zero. By homogeneity, rescaling by a power of $\pi$ if necessary, we may suppose $x,y,z \in R$ and at least one of $x,y,z \in R^\times$. Reducing modulo $\mathfrak p$ we have $x^2 \equiv ey^2 \pmod \mathfrak{p }$; since e is a nonsquare in k, we must have $v(x),v(y) \ge 1$. But this implies that $v(z)=0$ and so $v(\pi z^2)=1=v(x^2-ey^2) \ge 2$, a contradiction. $\square$

Predictably, the proof when ${{\,\mathrm{char}\,}}k \ne 2$ involving quadratic forms does not generalize in a simple way. However, armed with the above outline and acknowledging these complications, we now pursue the case ${{\,\mathrm{char}\,}}k = 2$. A key ingredient will understanding certain binary quadratic forms, as follows.

Lemma 12.3.10

There is a unique anisotropic binary quadratic form over k, up to isometry. Moreover, there is a unique (anisotropic) binary quadratic form over R whose reduction modulo $\mathfrak p$ is anisotropic, up to isometry (over R).

When ${{\,\mathrm{char}\,}}k \ne 2$, this unique class of binary forms over R is $\langle 1, e \rangle$ in the notation above. We will want a similar bit of notation in the case ${{\,\mathrm{char}\,}}k = 2$.

12.3.11

Recall the issues (6.1.4) with inseparability in characteristic 2. Let $\wp (k)=\{ z+z^2 : z \in k\}$ be the Artin–Schreier group of k. The polynomial $x^2+x+a \in k[x]$ is reducible if and only if $a \in \wp (k)$, and since k is finite, $k/\wp (k) \simeq \mathbb Z /2\mathbb Z$ (Exercise 12.9).

Let $t \in R$ be such that its reduction to k represents the nontrivial class in $k/\wp (k)$.

Proof of Lemma 12.3.10

We begin with the first part of the statement. Let Q be an anisotropic binary quadratic form over k, with T the associated bilinear form. The quadratic form $Q \boxplus \langle -1 \rangle$ is isotropic by Lemma 12.3.5, and Q is anisotropic, so Q represents 1, say $Q(x)=1$. Extending to a basis, we may rescale the second basis element y so that $T(x,y)=0,1$.

• Suppose $T(x,y)=0$, so $Q=\langle 1,-b \rangle$ for $b \in k^\times$. Since Q is isotropic, $b \not \in k^{\times 2}$, so ${{\,\mathrm{char}\,}}k \ne 2$, the class $b \in k^{\times }/k^{\times 2}$ is unique, and indeed Q is anisotropic.

• Suppose $T(x,y)=1$. If ${{\,\mathrm{char}\,}}k \ne 2$, we may complete the square and reduce to the previous case, so we suppose ${{\,\mathrm{char}\,}}k = 2$ and $Q(x,y)=x^2+xy+cy^2$ with $c \in k$. Working now with the Artin–Schreier group, since Q is anisotropic we must have $c + \wp (k)=t + \wp (k) \in k$, giving uniqueness.

For the second statement, if Q is a binary quadratic form over R whose reduction modulo $\mathfrak p$ is anisotropic, then we can find a change of basis over k to put $Q_k$ in the unique isometry class found above; lifting this basis, we may suppose that $Q_k$ is equal to this fixed form. The statement then follows by using Hensel’s lemma to lift the identity between any two such lifts, and it is a nice application of Hensel’s lemma in two (several) variables: see exercise (Exercise 12.12). $\square$

We now return to the proof of our theorem.

Proof of Theorem 12.3.4 (${{\,\mathrm{char}\,}}k = 2$).  We first claim that the form $[1,1,t] \boxplus \langle \pi \rangle$ is anisotropic; this follows from a straightforward modification of the argument as in the proof when ${{\,\mathrm{char}\,}}k \ne 2$ above.

We now show this form is the unique one up to similarity. Suppose that Q is a ternary anisotropic form over R. Let $x \in V$ be nonzero; since Q is anisotropic, we may scale x so that $a :=Q(x) \in R$. Since $\dim V \ge 3$, there exists nonzero $y' \in V$ such that $T(x,y')=0$; rescale $y'$ so that $Q(y') \in R$. Let $y :=x+y'$. Then $T(x,y)=T(x,x+y')=a$ and $b :=Q(y) \in R$, so Q on this basis is $ax^2+axy+b^2 \simeq [a,a,b]$. We compute that ${{\,\mathrm{disc}\,}}\,[a,a,b]=a(a-4b) \equiv a^2 \pmod \mathfrak{p }$, so ${{\,\mathrm{disc}\,}}\, [a,a,b] \in R^\times$ and [a, a, b] is nonsingular; completing to a basis with a nonzero element in the orthogonal complement and rescaling, we may suppose without loss of generality that $Q=[a,a,b] \boxplus \langle c \rangle$ with $a,b,c \in R$ and both $v(a)=0,1$ and $v(c)=0,1$, with ${{\,\mathrm{disc}\,}}Q=a(a-4b)c$. This leaves four cases.

• If $v(a)=v(c)=0$, then reducing modulo $\mathfrak p$ we have ${{\,\mathrm{disc}\,}}Q_k = a^2c \ne 0$ so $Q_k$ is nondegenerate. By Lemma 12.3.5, $Q_k$ is isotropic. Hensel’s lemma (Lemma 12.2.18) applies to Q, showing that Q is isotropic and giving a contradiction—we omit the details.

• Suppose $v(a)=0$ and $v(c)=1$. Rescaling by a unit we may suppose $c=\pi$. The reduction $[a,a,b]_k$ modulo $\mathfrak p$ is nondegenerate, so if it is isotropic then Q is isotropic, a contradiction. Therefore by Lemma 12.3.10, we may suppose $a=1$ and $b=t$; this is the desired form.

• Next, consider the case $v(a)=1$ and $v(c)=0$; we may suppose $c=1$. If $v(b) \ge 2$, then $\pi ^{-1}[a,a,b] \sim [1,1,b/\pi ]$ so the argument in the previous case applies to show that Q is isotropic. If $v(b)=1$, then we scale z by $\pi$ and divide Q by $\pi$ to reduce to the previous case. Finally, if $v(b)=0$, we scale y by $\pi$ and $\pi ^{-1}[a,a,b\pi ^2] \sim [1,1,b(\pi ^2/a)]$ is again isotropic, a contradiction as in the case $v(b) \ge 2$.

• In the case $v(a)=v(c)=1$, if $v(b) > 0$ then dividing through by $\pi$ we reduce to the first case, so we may suppose $v(b)=0$. Then we multiply by a and replace $x \leftarrow ax$, and $y \leftarrow y/b$ and $z \leftarrow z/\pi$ to work with $[1,a/b,a/b] \boxplus \langle ac/\pi ^2 \rangle$, which after interchanging x, y reduces to the previous case $v(a)=1$, $v(c)=0$.

Having exhausted the cases and the reader, the result now follows. $\square$

Corollary 12.3.12

Let F be a nonarchimedean local field with valuation ring R and uniformizer $\pi \in R$. Let B be a quaternion algebra over F.

If ${{\,\mathrm{char}\,}}k \ne 2$, then B is a division algebra if and only if

and if ${{\,\mathrm{char}\,}}F = {{\,\mathrm{char}\,}}k = 2$, then B is a division algebra if and only if

$$ B \simeq \displaystyle {\biggl [\frac{t,\pi }{F}\biggr )}, \text { where }t \in R\text { is nontrivial in }k/\wp (k). $$

In Theorem 13.3.11, we rephrase this corollary in terms of the unramified quadratic extension of F.

Remark 12.3.13. In mixed characteristic where ${{\,\mathrm{char}\,}}F=0$ and ${{\,\mathrm{char}\,}}k=2$, in the extension $K=F[x]/(x^2+x+t)$ for t nontrivial in $k/\wp (k)$ we can complete the square to obtain $K=F(\sqrt{e})$ with $e \in F^\times \smallsetminus F^{\times 2}$.

Definition 12.3.14

Let B be a quaternion algebra over F. The Hasse invariant of B is defined to be $-1$ if B is a division algebra and $+1$ if $B \simeq {{\,\mathrm{M}\,}}_2(F)$.

4 Hilbert symbol

Let F be a local field with ${{\,\mathrm{char}\,}}F \ne 2$. We record the splitting behavior of quaternion algebras as follows.

Definition 12.4.1

We define the Hilbert symbol

$$\begin{aligned} (\phantom {a},\phantom {b})_F :F^\times \times F^\times \rightarrow \{\pm 1\} \end{aligned}$$

by the condition that $(a,b)_F=1$ if and only if the quaternion algebra is split.

The Hilbert symbol is well-defined as a map

$$\begin{aligned} F^\times /F^{\times 2} \times F^\times /F^{\times 2} \rightarrow \{\pm 1\} \end{aligned}$$

(Exercise 2.4). By Main Theorem 5.4.4(v), we have $(a,b)_F=1$ if and only if the Hilbert equation $ax^2+by^2=1$ has a solution with $x,y \in F$: this is called Hilbert’s criterion for the splitting of a quaternion algebra.

Remark 12.4.2. The similarity between the symbols and $(a,b)_F$ is intentional; but they are not the same, as the former represents an algebra and the latter takes the value $\pm 1$.

In some contexts, the Hilbert symbol $(a,b)_F$ is defined to be the isomorphism class of the quaternion algebra in the Brauer group ${{\,\mathrm{Br}\,}}(F)$, rather than $\pm 1$ according to whether or not the algebra is split. Conflating these two symbols is not uncommon and in certain contexts it can be quite convenient, but we warn that it can lead to confusion and caution against referring to a quaternion algebra or its isomorphism class as a Hilbert symbol.

Lemma 12.4.3

Let $a,b \in F^\times$. Then the following statements hold:

• (a) $(ac^2,bd^2)_F=(a,b)_F$ for all $c,d \in F^\times$.

• (b) $(b,a)_F=(a,b)_F$.

• (c) $(a,b)_F=(a,-ab)_F=(b,-ab)_F$.

• (d) $(1,a)_F=(a,-a)_F=1$.

• (e) If $a \ne 1$, then $(a,1-a)_F=1$.

• (f) If $\sigma \in {{\,\mathrm{Aut}\,}}(F)$, then $(a,b)_F=(\sigma (a),\sigma (b))_F$.

Proof. Statements (a)–(c) follow from Exercise 2.4. For (d), the Hilbert equation $x^2+ay^2=1$ has the obvious solution $(x,y)=(1,0)$. And $\langle a,-a \rangle$ is isotropic (taking $(x,y)=(1,1)$) so is a hyperbolic plane and represents 1 as in the proof of Main Theorem 5.4.4, or we argue

$$\begin{aligned} (a,-a)_F=(a,a^2)_F=(a,1)_F=(1,a)_F=1 \end{aligned}$$

by Exercise 2.4. For part (e), by Hilbert’s criterion $(a,1-a)_F=1$ since the quadratic equation $ax^2+(1-a)y^2=1$ has the solution $(x,y)=(1,1)$. Finally, part (f): the Hilbert equation $ax^2+by^2=1$ has a solution with $x,y \in F$ if and only if $\sigma (a)x^2+\sigma (b)y^2=1$ has such a solution. $\square$

Remark 12.4.4. Staring at the properties in Lemma 12.4.3 and seeking to axiomatize them, the study of symbols like the Hilbert symbol leads naturally to the definition of $K_2(F)$. In its various formulations, algebraic K -theory (K for the German “Klasse”, following Grothendieck) seeks to understand certain kinds of functors from rings to abelian groups in a universal sense, encoded in groups $K_n(R)$ for $n \in \mathbb Z _{\ge 0}$ and R a commutative ring: see e.g. Karoubi [Kar2010]. For a field F, we have $K_0(F)=\mathbb Z$ and $K_1(F)=F^\times$. By a theorem of Matsumoto [Mat69] (see also Milnor [Milno71]), the group $K_2(F)$ is the universal domain for symbols over F:

$$\begin{aligned} K_2(F) :=(F^\times \otimes _\mathbb{Z } F^\times ) / \langle a \otimes (1-a) : a \ne 0,1 \rangle . \end{aligned}$$

(The tensor product over $\mathbb Z$ views $F^\times$ as an abelian group and therefore a $\mathbb Z$-module.) The map $a \otimes b \mapsto (a,b)_F$ extends to a map $K_2(F) \rightarrow \{\pm 1\}$, a Steinberg symbol, a homomorphism from $K_2(F)$ to a multiplicative abelian group. The higher K-groups are related to deeper arithmetic of commutative rings. For an introduction, see Weibel [Weib2013] and Curtis–Reiner [CR87, Chapter 5].

We now turn to be quite explicit about the values of the Hilbert symbol. We begin with the case where F is archimedean. If $F=\mathbb C$, then the Hilbert symbol is identically 1. If $F=\mathbb R$, then

$$\begin{aligned} (a,b)_\mathbb{R } = {\left\{ \begin{array}{ll} 1, &{} \text { if }a>0\text { or }b>0; \\ -1, &{} \text { if }a<0\text { and }b<0. \end{array}\right. } \end{aligned}$$

(12.4.5)

Lemma 12.4.6

The Hilbert symbol defines a nondegenerate symmetric bimultiplicative pairing

$$\begin{aligned} (\,{},\,)_F :F^\times /F^{\times 2} \times F^\times /F^{\times 2} \rightarrow \{\pm 1\}. \end{aligned}$$

By bimultiplicativity, we mean that

$$\begin{aligned} (a,bc)_F=(a,b)_F(a,c)_F \quad \text {and} \quad (ab,c)_F=(a,c)_F(b,c)_F \end{aligned}$$

(12.4.7)

for all $a,b,c \in F^\times$ (equivalent, by symmetry).

Keeping in the vibe of this section, we give a proof under the hypothesis that ${{\,\mathrm{char}\,}}k \ne 2$; for a general proof, see Corollary 13.4.6.

Proof (${{\,\mathrm{char}\,}}k \ne 2$).  This lemma can be read off of the direct computation below (12.4.9), recording what was computed along the way in the proof of Theorem 12.3.4. $\square$

12.4.8

Since the Hilbert symbol is well-defined up to squares, the symbol $(a,b)_F$ is determined by the values with $a,b \in \{1,e,\pi ,e\pi \}$ where e is a nonsquare in $k^\times$. Let $s=(-1)^{(\#k-1)/2}$, so that $s=1,-1$ according as $-1$ is a square in k. Then:

$$\begin{aligned} \begin{array}{c|cccc} (a,b)_F &{} 1 &{} e &{} \pi &{} e\pi \\ \hline 1 &{} 1 &{} 1 &{} 1 &{} 1 \\ e &{} 1 &{} 1 &{} -1 &{} -1 \\ \pi &{} 1 &{} -1 &{} s &{} -s \\ e\pi &{} 1 &{} -1 &{} -s &{} s \end{array} \end{aligned}$$

(12.4.9)

The computation of this table is requested in Exercise 12.15. For example, if $a=e$, we showed in the proof of Theorem 12.3.4 that $(e,1)_F=(e,e)_F=1$, because $ex^2+y^2-z^2$ and $ex^2+ey^2-z^2$ are isotropic.

In general, writing $a=a_0 \pi ^{v(a)}$ and $b=b_0 \pi ^{v(b)}$ we have

(12.4.10)

where $q=\#k$ and

(12.4.11)

is the Legendre symbol: see Exercise 12.16. (Multiplicativity can also be read off of the formula 12.4.10.)

12.4.12

The following easy criteria follow from 12.4.9 (or (12.4.10)):

• (a) If $v(ab)=0$, then $(a,b)_F=1$.

• (b) If $v(a)=0$ and $v(b)=v(\pi )$, then

  

12.4.13

To compute the Hilbert symbol for a local field F with ${{\,\mathrm{char}\,}}F=0$ and ${{\,\mathrm{char}\,}}k = 2$ is significantly more involved. But we can at least compute the Hilbert symbol by hand for $F=\mathbb Q _2$.

To begin, the group $\mathbb Q _2^\times /\mathbb Q _2^{\times 2}$ is generated by $-1,-3,2$, so representatives are $\{\pm 1, \pm 3, \pm 2, \pm 6\}$. We recall Hilbert’s criterion: $(a,b)_{F}=1$ if and only if $ax^2+by^2=1$ has a solution with $x,y \in F$.

If $a,b \in \mathbb Z$ are odd, then

$$ \begin{aligned}&ax^2+by^2=z^2\text { has a nontrivial solution in }\mathbb Q _2 \\&\qquad \quad \Leftrightarrow \quad a \equiv 1 ~(\text{ mod } ~{4})\text { or }b \equiv 1 ~(\text{ mod } ~{4}); \end{aligned} $$

by homogeneity and Hensel’s lemma, it is enough to check for a solution modulo 4. This deals with all of the symbols with a, b odd: summarizing, we have in this case

$$\begin{aligned} (a,b)_2 = (-1)^{(a-1)(b-1)/4}. \end{aligned}$$

(12.4.14)

By the determination above, we see that $(-3,b)=-1$ for $b=\pm 2, \pm 6$ and $(2,2)_{2}=(-1,2)_{2}=1$ the latter by Hilbert’s criterion, as $-1+2=1$; knowing multiplicativity (Lemma 12.4.6), we have uniquely determined all Hilbert symbols, in particular, for $a \in \mathbb Z$ odd we have

$$\begin{aligned} (a,2)_2=(-1)^{(a^2-1)/8}. \end{aligned}$$

(12.4.15)

It is still useful to compute several of these symbols individually, in the same manner as (12.4.13) (working modulo 8): see Exercise 12.17. We summarize the results here:

$$\begin{aligned} \begin{array}{c|cccccccc} (a,b)_{2} &{} 1 &{} -3 &{} -1 &{} 3 &{} 2 &{} -6 &{} -2 &{} 6 \\ \hline 1 &{} 1 &{} 1 &{} 1 &{} 1 &{} 1 &{} 1 &{} 1 &{} 1 \\ -3 &{} 1 &{} 1 &{} 1 &{} 1 &{} -1 &{} -1 &{} -1 &{} -1 \\ -1 &{} 1 &{} 1 &{} -1 &{} -1 &{} 1 &{} 1 &{} -1 &{} -1 \\ 3 &{} 1 &{} 1 &{} -1 &{} -1 &{} -1 &{} -1 &{} 1 &{} 1 \\ 2 &{} 1 &{} -1 &{} 1 &{} -1 &{} 1 &{} -1 &{} 1 &{} -1 \\ -6 &{} 1 &{} -1 &{} 1 &{} -1 &{} -1 &{} 1 &{} -1 &{} 1 \\ -2 &{} 1 &{} -1 &{} -1 &{} 1 &{} 1 &{} -1 &{} -1 &{} 1 \\ 6 &{} 1 &{} -1 &{} -1 &{} 1 &{} -1 &{} 1 &{} 1 &{} -1 \end{array} \end{aligned}$$

(12.4.16)

Remark 12.4.17. Analogously, one can define a symbol $[a,b)_F$ for the splitting of quaternion algebras for F a local field with ${{\,\mathrm{char}\,}}F=2$. This symbol is no longer called the Hilbert symbol, but many properties remain: in particular, there is still an analogue of the Hilbert equation, and $\displaystyle {\biggl [\frac{a,b}{F}\biggr )}$ is split if and only if $bx^2+bxy+aby^2=1$ has a solution with $x,y \in F$.

Exercises

• 1. Let p be an odd prime. Show

  • (a) Show the equality

    $$\begin{aligned} (1-4x)^{1/2} = 1 - \sum _{n=1}^{\infty } C_n x^n \in \mathbb Z [[x]] \end{aligned}$$

    of formal series in x with coefficients in $\mathbb Z$, where

    $$\begin{aligned} C_n :=\frac{1}{2n-1}\left( {\begin{array}{c}2n\\ n\end{array}}\right) \in \mathbb Z _{>0} \end{aligned}$$

    are the Catalan numbers. [Hint: use binomial expansion.]

  • (b) Let p be an odd prime. Show that the squaring map is bijective on $1+p\mathbb Z _p$. [Hint: show that the series expansion in (a) converges in $\mathbb Z _p$.]

• 2. Recall that a topological space is $T _1$ if for every pair of distinct points, each point has an open neighborhood not containing the other.

  • (a) Show that a topological space X is $T _1$ if and only if $\{x\}$ is closed for all $x \in X$.

  • (b) Let G be a topological group. Show that G is Hausdorff if and only if G is $T _1$.

• 3. In this exercise we prove some basic facts about topological groups. Let G be a topological group.

  • (a) Let $H \le G$ be a subgroup. Show that H is open if and only if there exists $h \in H$ and an open neighborhood of h contained in H.

  • (b) Show that if $H \le G$ is an open subgroup, then H is closed.

  • (c) Show that a closed subgroup $H \le G$ of finite index is open.

  • (d) Suppose that G is compact. Show that an open subgroup $H \le G$ is of finite index, and that every open subgroup contains an open normal subgroup.

$\triangleright$ 4.:

Let G be a topological group. Let $U \ni 1$ be an open neighborhood of 1.

(a):

Show that there exists an open neighborhood $V \subseteq U$ of $1 \in V$ such that $V^2 = V\cdot V \subseteq U$. [Hint: Multiplication is continuous.]

(b):

Similarly, show that there exists an open neighborhood $V \subseteq U$ of $1 \in V$ such that $V^{-1} V \subseteq U$.

• 5. Let G be a topological group and let $H \le G$ be a closed subgroup. Equip G/H with the quotient topology. Show that G/H is Hausdorff. [Hint: Use Exercise 12.4(b).]

$\triangleright$ 6.:

Let k be a finite field and let $Q:V \rightarrow k$ be a ternary quadratic form. Show that Q is isotropic. [Hint: Reduce to the case of finding a solution to $y^2=f(x)$ where f is a polynomial of degree 2. If $\#k$ is odd, count squares and the number of distinct values taken by f(x) in k. Second approach: reduce to the case where $\#k$ is odd, and show that $x^2+y^2$ represents a nonsquare, since the squares cannot be closed under addition!] [This repeats Exercise 5.5!]

$\triangleright$ 7.:

Let k be a finite field with ${{\,\mathrm{char}\,}}k \ne 2$ and let $e \in k^\times$. Show directly that there is an isometry $\langle -1, e \rangle \simeq \langle 1, -e \rangle$.

$\triangleright$ 8.:

Let R be a DVR with field of fractions F, let $a,b,c \in F$ be nonzero and let $Q=\langle a,-b,-c \rangle$. Show that Q is similar over F to $\langle 1,-b',-c' \rangle$ with $0 = v(b') \le v(c')$. [Hint: first get $v(a),v(b),v(c) \in \{0,1\}$.]

$\triangleright$ 9.:

Let k be a finite field with even cardinality. Show that $\# k/\wp (k) = 2$, where $\wp (k)$ is the Artin-Schreier group.

• 10. By Theorem 12.2.15, a complete archimedean local field is isomorphic to $\mathbb R$ or $\mathbb C$. Extend this classification to division algebras as follows.

  The notion of absolute value (Definition 12.2.2) extends to a division algebra without modification, as does the notion of archimedean and nonarchimedean.

  • (a) Show that $\mathbb H$ has an absolute value $|\alpha \,| = \sqrt{{{\,\mathrm{nrd}\,}}(\alpha )}$ for $\alpha \in \mathbb H$.

  • (b) Let D be a division algebra equipped with an absolute value $|\phantom {x} \,|$. Show that if $|\phantom {x} \,|$ is archimedean, then ${{\,\mathrm{char}\,}}D = 0$ and if the restriction of $|\phantom {x} \,|$ to its center Z(D) is archimedean.

  • (c) Show that every division algebra complete with respect to an archimedean absolute value is isomorphic to $\mathbb R$, $\mathbb C$, or $\mathbb H$ and with the absolute value equivalent to the absolute value $|\alpha \,|=\sqrt{{{\,\mathrm{nrd}\,}}\alpha }$ in each case. [Hint: recall Theorem 3.5.1.]

$\triangleright$ 11.:

Prove Lemma 12.2.18 using Lemma 12.2.17. [Hint: let j be the index that achieves the minimal valuation among partial derivatives, and consider the restriction $f(a_1,\dots ,a_{j-1},x,a_{j+1},\dots ,a_n) \in R[x]$ to one variable.]

$\triangleright$ 12.:

In this exercise, we consider an extension of Hensel’s lemma to several polynomials (in several variables).

Let F be a nonarchimedean local field with valuation v and valuation ring R.

(a):

Let $f_1,\dots ,f_n \in R[x_1,\dots ,x_n]$ be polynomials with $n \ge 1$. Let $a \in R^n$ have $m :=\min _i v(f_i(a))$. Let

$$\begin{aligned} \mathrm {D}_f :=\left( \frac{\partial f_i}{\partial x_j}(a)\right) _{i,j} \in M_{n}(R). \end{aligned}$$

Suppose $m > 2v(\det \mathrm {D}_f(a))$. Show that there exists $\widetilde{a} \in R^n$ such that $f_i(a)=0$ for all $i=1,\dots ,n$ and $\widetilde{a} \equiv a \pmod \mathfrak{p ^m}$. [Hint: by Taylor expansion, write $f(a+\pi ^t x)=f(a) + \mathrm {D}_f(a)p^t x + p^{2t} r(x)$ in vector notation with $t \ge m$, and iteratively solve the system in a manner analogous to Newton’s method. See Conrad [con] for a complete development.][Con]

(b):

Show that (a) also holds more generally if the number of polynomials r has $r \le n$ and there exists an $r \times r$ matrix minor of $\mathrm {D}_f(a)$ whose determinant has valuation $<m/2$. [Hint: see Exercise 12.11.]

(c):

Finish the proof of Lemma 12.3.10 as follows. Let $Q,Q'$ be two binary quadratic forms over R such that their reductions $Q_k,Q_k'$ are anisotropic. Show that $Q \simeq Q'$. [Hint: reduce to the case where $Q_k=Q_k'$. If $Q_k \simeq \langle 1, -e \rangle$, rescale the basis using a plain vanilla flavor of Hensel’s lemma. If $Q_k \simeq [1,1,t]$, reduce to showing that $Q \simeq [1,1,t]$. Consider a general change of variables in ${{\,\mathrm{GL}\,}}_2(R)$ from [1, 1, t] that reduces to the identity modulo $\mathfrak p$ and apply the deluxe version of Hensel’s lemma in part (b) to the resulting system of three equations in four unknowns: there is a minor with determinant $1-4t \in R^\times$.]

• 13. Let $F {\ne } \mathbb C$ be a local field and let Q be a nondegenerate ternary quadratic form over F. Let $K {\supseteq } F$ be a quadratic field extension. Show that Q is isotropic over K.

• 14. Give another proof of Lemma 12.4.6 that the local Hilbert symbol is bimultiplicative using Example 8.2.2 and the Brauer group (section 8.3).

$\triangleright$ 15.:

Show that the table of Hilbert symbols (12.4.9) is correct.

$\triangleright$ 16.:

One can package 12.4.8 together with multiplying by squares to prove the following more general criterion. Let F be a nonarchimedean local field with uniformizer $\pi$, valuation v with $v(\pi )=1$, and residue field k. Let $q=\#k$ and suppose q is odd.

Show that for $a,b \in F^\times$, if we write $a=a_0\pi ^{v(a)}$ and $b=b_0\pi ^{v(b)}$, then

$\triangleright$ 17.:

Show that the table of Hilbert symbols (12.4.16) is correct by considering the equation $ax^2+by^2 \equiv 1 \pmod {8}$.

• 18. Prove a descent for the Hilbert symbol, as follows. Let K be a finite extension of the local field F with ${{\,\mathrm{char}\,}}F \ne 2$ and let $a,b \in F^\times$. Show that $(a,b)_K=(a,{{\,\mathrm{Nm}\,}}_{K|F}(b))_F = ({{\,\mathrm{Nm}\,}}_{K|F}(a),b)_F$.