QUATERNION ALGEBRAS WITH THE SAME SUBFIELDS

SKIP GARIBALDI AND DAVID J. SALTMAN

Abstract. G. Prasad and A. Rapinchuk asked if two quaternion division F -algebras that have the same subfields are necessarily isomorphic. The answer is known to be “no” for some very large fields. We prove that the answer is “yes” if F is an extension of a global field K so that F/K is unirational and has zero unramified Brauer group. We also prove a similar result for Pfister forms and give an application to tractable fields.

1. Introduction

Gopal Prasad and Andrei Rapinchuk asked the following question in Remark 5.4 of their paper [PR]:

(1.1)

If two quaternion division algebras over a field F have the same maximal subfields, are the algebras necessarily isomorphic?

The answer is “no” for some fields F, see §2 below. The answer is “yes” if F is a global field by the Albert-Brauer-Hasse-Minkowski Theorem [NSW, 8.1.17]. Prasad and Rapinchuk note that the answer is unknown even for fields like ℚ(x). We prove that the answer is “yes” for this field:

Theorem 1.2. Let F be a field of characteristic ≠ 2 that is transparent. If D_{1} and D_{2} are quaternion division algebras over F that have the same maximal subfields, then D_{1} and D_{2} are isomorphic.

The term “transparent” is defined in §6 below. Every retract rational extension of a local, global, real-closed, or algebraically closed field is transparent; in particular K(x_{1}, …, x_{n}) is transparent for every global field K of characteristic ≠ 2 and every n. There are many other examples.

The theorem can be viewed as a statement about symbols in the Galois cohomology group H^{2}(F, ℤ/2ℤ). We also state and prove an analogue for symbols in Galois cohomology of H^{d}(F, ℤ/2ℤ)—i.e., d-Pfister quadratic forms—in Theorem 9.1 below.

Notation. A global field is a finite extension of ℚ or of 𝔽_{p}(t) for some prime p. A local field is a completion of a global field with respect to a discrete valuation.

2000 Mathematics Subject Classification. Primary 16K20; secondary 11E04, 12G05.

2. An example

Example 2.1. Several people (in no particular order: Markus Rost, Kelly McKinnie, Adrian Wadsworth, Murray Schacher, Daniel Goldstein...) noted that some hypothesis on the field F is necessary for the conclusion of the theorem to hold. Here is an example to illustrate this.

Let F_{0} be a field of characteristic ≠ 2. We follow [Lam 05] for notation on quadratic forms, except that we define the symbol ⟨⟨a_{1}, …, a_{d}⟩⟩ to be the d-Pfister form ⊗_{i=1}^{d}⟨1, −a_{i}⟩. For a Pfister form ϕ, we put ϕ′ for the unique form such that ϕ ≅ ⟨1⟩ ⊕ ϕ′.

Suppose that F_{0} has quaternion division algebras Q_{1}, Q_{2} that are not isomorphic. Write ϕ_{i} := ⟨⟨a_{i}, b⟩⟩ for the norm form of Q_{i}. The Albert form ϕ_{1} − ϕ_{2} has anisotropic part similar to γ = ⟨⟨a_{1}a_{2}, b⟩⟩.

Fix an extension E/F_{0} and a proper quadratic extension E(\sqrt{c}) contained in Q_{i} but not Q_{i+1}, with subscripts taken modulo 2. Equivalently, −c is represented by ϕ′_{i} but not by ϕ′_{i+1}. Put q_{i,c} := ⟨c⟩ ⊕ ϕ′_{i+1}; it is anisotropic. Further, its determinant c is a nonsquare, so q_{i,c} is not similar to ϕ_{i}, ϕ_{i+1}, nor γ. All three of those forms remain anisotropic over the function field E(q_{i,c}) of q_{i,c} by [Lam 05, X.4.10(3)]. That is, Q_{1} ⊗ E(q_{i,c}) and Q_{2} ⊗ E(q_{i,c}) are division and contain E(q_{i,c})(\sqrt{c}), and the two algebras are still distinct.

We build a field F from F_{0} inductively. Suppose we have constructed the field F_{j} for some j ≥ 0. Let F_{j+1} be the composita of the extensions F_{j}(q_{i,c}) for c as in the previous paragraph with E = F_{j} and i = 1, 2. Let F be the colimit of the F_{j}.

By construction, Q_{1} ⊗F and Q_{2} ⊗F are both division, are not isomorphic to each other, and have the same quadratic subfields. This shows that some hypothesis on the field F is necessary for the conclusion of the theorem to hold.

One cannot simply omit the hypothesis “division” in the theorem, by Example 3.6.

Furthermore, Theorem 1.2 does not generalize to division algebras of degree > 2, even over number fields. One can easily construct examples using the local-global principle for division algebras over a number field, see [PR, Example 6.5] for details.

3. Discrete valuations: good residue characteristic

Fix a field F with a discrete valuation v. We write F̄ for the residue field and F̂ for the completion. (Throughout this note, for fields with a discrete valuation and elements of such a field, we use bars and hats to indicate the corresponding residue field/completion and residues/image over the completion.)

For a central simple F -algebra A, we write Â for A ⊗ F̂. It is Brauerequivalent to a central division algebra B over F̂. As v is complete on F̂, it extends uniquely to a valuation on B by setting v(b) := \frac{1}{deg B} v(Nrd_{B}(b)) for b ∈ B^{×}, see, e.g., [R, 12.6]. We say that A is unramified if v(B^{×}) = v(F^{×})— e.g., if F̂ splits A—and ramified if v(B^{×}) is strictly larger. That is, A is ramified or unramified if v is so on B. Note that these meanings for ramified and unramified are quite a bit weaker than the the usual definitions (which usually require that the residue division algebra of B be separable over F̄).

The valuation ring in F̂ is contained in a noncommutative discrete valuation ring S in B, and the residue algebra of S is a (possibly commutative) division algebra B̄ whose center contains F̄. One has the formula

dimF̄ B̄ · [v(B^{×}) : v(F^{×})] = dimF̂ B

(see [W, p. 393] for references). That is, A is unramified if and only if dimF̄ B̄ = dimF̂ B.

We use a few warm-up lemmas.

Lemma 3.1. Let A_{1}, A_{2} be central simple algebras over a field F that has a discrete valuation. If there is a separable maximal commutative subalgebra that embeds in one of the Â_{i} and not the other, then the same is true for the A_{i}.

Proof. Let L̂ be a separable maximal commutative subalgebra of Â_{1} that is not a maximal subalgebra of Â_{2}. Choose α̂ ∈ L̂ so that L̂ = F̂(α̂). Since A_{1} is dense in Â_{1}, we can choose α ∈ A_{1} as close as we want to α̂. Thus we can choose the Cayley-Hamilton polynomial of α to be as close as we need to that of α̂. In particular, by Krasner’s Lemma (applied as in [R, 33.8]), we can assume that F̂(α) ≅ F̂(α̂). In particular, F̂(α) is not a maximal subalgebra of Â_{2}, which implies that F (α) is not a maximal subalgebra of A_{2} and is by construction a maximal subfield of A_{1}. Since F̂(α) ≅ F̂(α̂), the minimal polynomial of α over F has nonzero discriminant and hence is separable. □

Lemma 3.2. Let A_{1}, A_{2} be central division algebras of prime degree over a field F with a discrete valuation. If

(1) Â_{1} is split and Â_{2} is not; or

(2) both Â_{1} and Â_{2} are division, but only one is ramified, then there is a separable maximal subfield of A_{1} or A_{2} that does not embed in the other.

Proof. The statement is obvious from Lemma 3.1. In case (1), we take take the diagonal subalgebra in Â_{1}. In case (2), we take a maximal separable subfield that is ramified. □

For the rest of this section, we fix a prime p and let F be a field with a primitive p-th root of unity ζ; we suppose that F has a discrete valuation v with residue field of characteristic different from p.

A degree p symbol algebra is an (associative) F -algebra generated by elements α, β subject to the relations α^{p} = a, β^{p} = b, and αβ = ζβα. It is central simple over F [D, p. 78] and we denote it by (a, b)_{F} or simply (a, b).

Lemma 3.3. If A is a degree p symbol algebra over F, then A is isomorphic to (a, b)_{F} for some a, b ∈ F^{×} with v(a) = 0.

Proof. Write A as (a, b) where a = xπ^{n_{a}} and b = yπ^{n_{b}} where x, y have value 0 and π is a uniformizer for F. We may assume that p divides neither n_{a} nor n_{b}, so n_{a} = sn_{b} modulo p. But A is isomorphic to (a(−b)^{−s}, b) because (−b, b) is split [D, p. 82, Cor. 5], and v(a(−b)^{−s}) is divisible by p. □

3.4. Let (a, b) be a symbol algebra as in Lemma 3.3 and write F̂ for the completion of F at v. If (a, b)⊗F̂ is not division, certainly (a, b) is unramified. Otherwise, there are two cases:

• If v(b) is divisible by p, then we can assume v(b) = 0, (a, b) is unramified, and \overline{(a, b)} is the symbol algebra (ā, b̄)F̄. (The residue algebra— which is division—contains (ā, b̄)F̄, hence they must be equal.)

• If v(b) is not divisible by p, then (a, b) is ramified and \overline{(a, b)} is the field extension F̄(\sqrt[p]{ā}). (By hypothesis, a is not a p-th power in F̂, so by Hensel’s Lemma ā is not a p-th power in F̄.)

Proposition 3.5. Let A_{1} and A_{2} be degree p symbol algebras over F that are division algebras. If

(1) A_{1} ramifies and A_{2} does not; or

(2) A_{1} and A_{2} both ramify but their ramification defines different exten-

sions of the residue field F̄, then there is a maximal subfield L of one of the A_{i} that does not split the other.

Proof. Write A_{i} as (a_{i}, b_{i}). We can assume v(a_{1}) = v(a_{2}) = 0. If both the A_{i} ramify, then the v(b_{i}) must be prime to p and the field extensions of the ramification of the A_{i} must be F̄(\sqrt[p]{ā_{i}}). If these are different fields, then the field L_{2} := F (\sqrt[p]{a_{2}}) cannot split A_{1} because it does not split the ramification of A_{1}.

Thus assume A_{2} is unramified and A_{1} ramifies. In particular, A_{1} ⊗ F̂ is division, and we are done by Lemma 3.2. □

Example 3.6. In the proposition, the hypothesis that A_{1} and A_{2} are both division is necessary. Indeed, for ℓ ≠ 2, the field ℚ_{ℓ} of ℓ-adic numbers has two quaternion algebras: one split and one division. Only one is ramified, but both are split by every quadratic extension of ℚ_{ℓ}.

4. Discrete valuations: bad residue characteristic

In this section, we consider quaternion algebras over a field with a dyadic discrete valuation. The purpose of this section is to prove:

Proposition 4.1. Let A_{1}, A_{2} be quaternion division algebras over a field F of characteristic 0 that has a dyadic discrete valuation. If A_{1}⊗A_{2} is Brauerequivalent to a ramified quaternion division algebra with residue algebra a separable quadratic extension of F̄, then there is a quadratic extension of F that embeds in A_{1} or A_{2} but not both.

What really helps nail down the structure of A_{1} ⊗ A_{2} is the hypothesis that the residue algebra is a separable quadratic extension. In particular, the division algebra underlying A_{1} ⊗ A_{2} is “inertially split” in the language of [JW].

Example 4.2. Let B be a quaternion division algebra over a field F of characteristic 0 that is complete with respect to a dyadic discrete valuation, and suppose that the residue algebra B̄ contains a separable quadratic extension L̄ of F. One can lift L̄/F̄ to a unique quadratic extension L/F and write B = (L, b) for some b ∈ F^{×} that we may assume has value 0 or 1. One finds two possibilities:

• If v(b) = 0, then B̄ is a quaternion algebra over F̄ generated by elements i, j such that

F̄(i) = L̄, i^{2} + i ∈ F̄, j^{2} = b̄, and ij = j(i + 1).

In this case B is unramified.

• If v(b) = 1, then B̄ = L̄ and B is ramified. For detailed explanation of these claims, see, e.g., [GP]. (Unlike the situation in the preceding section, there are more than just these two possibilities for general quaternion algebras over F, but in those other cases the residue algebra B̄ is a purely inseparable extension of F̄ of dimension 2 or 4 with B̄^{2} ⊆ F̄. Note that if L/F is unramified and L̄/F̄ is purely inseparable, then L/F is not uniquely determined by L̄/F̄.)

Proof of Proposition 4.1. Put B for the quaternion division F -algebra underlying A_{1} ⊗ A_{2}. By the hypothesis on its residue algebra, B remains division when we extend scalars to the completion F̂. Hence at least one of Â_{1}, Â_{2} is division; by Lemma 3.2 (using that A_{1} and A_{2} are division) we may assume that both are division and that the valuation has the same ramification index on both algebras. By Lemma 3.1, we may assume that F is complete.

The residue division algebra Ā_{1} is distinct from F̄ and we can find a quadratic subfield L of A_{1} so that the residue field L̄ is a quadratic extension of F̄. As L is a maximal subfield of A_{1}, it is also one for A_{2} and hence B. Certainly, L̄ is a subfield of the residue algebra B̄; as both have dimension 2 over F̄ they are equal. That is, L̄ is separable. We write A_{i} = (L, a_{i}) and B = (L, b) for some a_{i}, b ∈ F^{×}. As the valuation does not ramify on L we may assume that v(b) = 1. By the previous paragraph, we may write A_{i} = (L, a_{i}) for some a_{i} ∈ F^{×}. As Ā_{i} contains the separable extension L̄/F̄, we have v(a_{1}) = v(a_{2}) as in Example 4.2. Then A_{1}⊗A_{2}⊗B—which is split— is Brauer-equivalent to (L, a_{1}a_{2}b), which is ramified (because v(a_{1}a_{2}b) = 1 + 2v(a_{i}) is odd) and in particular not split. This is a contradiction, which completes the proof of the proposition. □

5. Unramified cohomology

For a field F of characteristic not 2, we write H^{d}(F) for the Galois cohomology group H^{d}(F, ℤ/2ℤ). We remind the reader that for d = 1, this group is F^{×}/F^{×2} and for d = 2 it is identified with the 2-torsion in the Brauer group of F.

Definition 5.1. Fix an integer d ≥ 2. We define H_{u}^{d}(F) to be the subgroup of H^{d}(F) consisting of classes that are unramified at every nondyadic discrete valuation of F (in the usual sense, as in [GMS, p. 19]) and are killed by H^{d}(F) → H^{d}(R) for every real closure R of F.

Example 5.2. (1) A local field F (of characteristic ≠ 2) has H_{u}^{2}(F) = 0 if and only if F is non-dyadic. Indeed, F has no orderings, so it suffices to consider the unique discrete valuation on F.

(2) A global field F (of characteristic ≠ 2) has H_{u}^{2}(F) = 0 if and only if F has at most 1 dyadic place. This is Hasse’s local-global theorem for central simple algebras [NSW, 8.1.17].

(3) A real closed field F has H_{u}^{d}(F) = 0 for all d ≥ 2; this is trivial.

A local field F (of characteristic ≠ 2) has H_{u}^{d}(F) = 0 for d ≥ 3. This is trivial because such a field has no orderings and cohomological 2-dimension 2.

Lemma 5.3. If F (\sqrt{−1}) has cohomological 2-dimension < d, then H_{u}^{d}(F) is zero.

Proof. By [A, Satz 3], an element x ∈ H^{d}(F) is zero at every real closure if and only if x · (−1)^{r} is zero for some r ≥ 0. On the other hand, there is an exact sequence for every n [KMRT, 30.12(1)]:

H^{n}(F (\sqrt{−1})) ⟶^{cor} H^{n}(F) ⟶^{·(−1)} H^{n+1}(F) ⟶^{res} H^{n+1}(F (\sqrt{−1})

For n ≥ d, the two end terms are zero, so the cup product with (−1)^{r} defines an isomorphism H^{d}(F) ⟶^{∼} H^{d+r}(F) for all r ≥ 0. □

We immediately obtain:

Corollary 5.4. A global field F of characteristic ≠ 2 has H_{u}^{d}(F) = 0 for all d ≥ 3. □

Corollary 5.5. If F has transcendence degree d over a real-closed field, then H_{u}^{d+1}(F) = 0.

Proof. F (\sqrt{−1}) has transcendence degree d over an algebraically closed field, hence has cohomological dimension ≤ d [Se, §II.4.3]. □

We define a discrete valuation of an extension F/F_{0} to be a discrete valuation on F that vanishes on F_{0}.

Proposition 5.6. For every extension F/F_{0}, the homomorphism res: H^{d}(F_{0}) → H^{d}(F) restricts to a homomorphism H_{u}^{d}(F_{0}) → H_{u}^{d}(F). If additionally F/F_{0} is unirational and

(5.7) every class in H^{d}(F) that is unramified at every discrete valuation of F/F_{0} comes from H^{d}(F_{0}),

then the homomorphism H_{u}^{d}(F_{0}) → H_{u}^{d}(F) is an isomorphism.

Proof. Fix y ∈ H_{u}^{d}(F_{0}). Every ordering of F restricts to an ordering on F_{0}; write R and R_{0} for the corresponding real closures. By hypothesis, the image of y in H^{d}(R_{0}) is zero, hence it also has zero image under the composition H^{d}(F_{0}) → H^{d}(R_{0}) → H^{d}(R). It is easy to see that res(y) is unramified at every nondyadic discrete valuation of F, and the first claim follows.

We now prove the second sentence. The natural map H^{d}(F_{0}) → H^{d}(F) is injective by [GMS, p. 28]. Fix x ∈ H_{u}^{d}(F). As it is unramified at every discrete valuation of F/F_{0}, it is the restriction of some element x_{0} ∈ H^{d}(F_{0}) by (5.7). We will show that x_{0} belongs to H_{u}^{d}(F_{0}).

Fix an extension E := F_{0}(x_{1}, x_{2}, …, x_{n}) containing F. Every ordering on F_{0} extends to an ordering v on E (and hence also on F) by a recipe as in [Lam 80, pp. 9–11]. As the map H^{d}((F_{0})_{v}) → H^{d}(F_{v}) is an injection by Lemma 5.8 below, we deduce that x_{0} is killed by every real-closure of F_{0}.

One extends a nondyadic discrete valuation v on F_{0} to E by setting v(x_{i}) = 0 and x̄_{i} to be transcendental over F̄_{0} as in [B, §VI.10.1, Prop. 2]; then the residue field of Ē is F̄_{0}(x_{1}, x_{2}, …, x_{n}) and the natural map H^{d−1}(F̄_{0}) → H^{d−1}(Ē) is injective, hence so is H^{d−1}(F̄_{0}) → H^{d−1}(F̄). It follows that the image of x_{0} in H^{d−1}(F̄_{0}) is zero, which completes the proof. □

Here is the promised lemma:

Lemma 5.8. If R_{1} ⊆ R_{2} are real-closed fields, then the natural map H^{d}(R_{1}) → H^{d}(R_{2}) is an isomorphism for every d ≥ 0.

Proof. H^{d}(R_{i}) is the usual group cohomology H^{d}(ℤ/2ℤ, ℤ/2ℤ). Obviously H^{0}(R_{i}) = H^{1}(R_{i}) = ℤ/2ℤ, and 2-periodicity for the cohomology of finite cyclic groups shows that H^{d}(R_{i}) = ℤ/2ℤ for all d ≥ 0, with nonzero element (−1)^{d}. □

Example 5.9. Recall that an extension F of F_{0} is retract rational if it is the field of fractions of an F_{0}-algebra domain S and there are F_{0}-algebra homomorphisms S → F_{0}[x_{1}, …, x_{n}](1/s) → S whose composition is the identity on S, where F_{0}[x_{1}, …, x_{n}] is a polynomial ring and s ∈ F_{0}[x_{1}, …, x_{n}] is a nonzero element. Obviously such an extension is unirational, and (5.7) holds by [Sa, 11.8] or [M, Prop. 2.15], so H_{u}^{d}(F_{0}) ≅ H_{u}^{d}(F).

6. Transparent fields

Definition 6.1. Fix a field F of characteristic ≠ 2. For each dyadic valuation v, fix a completion F̂ of F. Write K for the limit of the finite unramified extensions of F̂ with separable residue field. This extension is called the maximal unramified extension of F̂ (which highlights the fact that usually the word “unramified” is quite a bit stronger than how we are using it). There is a well-defined ramification map

H^{2}(K/F̂, μ_{2}) → H^{1}(F̄, ℤ/2ℤ) s.t. (a, b) ↦ \begin{cases}0 & if v(a) = v(b) = 0 \\ (\overline{a}) & if v(a) = 0 & v(b) = 1 \end{cases}

We say that F is transparent if for every division algebra D ≠ F of exponent 2 such that [D] belongs to H_{u}^{2}(F), there is some dyadic place v of F and completion F̂ such that

(1) D ⊗ F̂ is split by the maximal unramified extension of F̂, and

(2) the image of [D] in H^{1}(F̄, ℤ/2ℤ) is not zero. Another way to phrase these hypotheses is: D⊗F̂ is division and the residue division algebra D̄ is a separable quadratic extension of F̄.

Roughly speaking, transparent fields are those for which nonzero 2-torsion elements of the Brauer group can be detected using ramification.

Example 6.2. (0) If H_{u}^{2}(F) = 0—e.g., if F is real closed—then F is

(vacuously, trivially) transparent.

(1) If F is local, then F is transparent. If F is nondyadic, this is Example 5.2(1). Otherwise, for each nonzero x ∈ H_{u}^{2}(F), the residue division algebra is necessarily a separable quadratic extension of F̄, because F̄ is perfect with zero Brauer group.

(2) If F is global, then F is transparent. Indeed, H_{u}^{2}(F) consists of classes ramified only at dyadic places, and every such place has finite residue field.

The unramified Brauer group of an extension F/F_{0} is the subgroup of the Brauer group of F consisting of elements that are unramified at every discrete valuation of F/F_{0}.

Proposition 6.3. If F_{0} is transparent, F/F_{0} is unirational, and the unramified Brauer group of F/F_{0} is zero, then F is transparent.

Proof. Suppose there is a nonzero x ∈ H_{u}^{2}(F). By Proposition 5.6, x is the image of some nonzero x_{0} ∈ H_{u}^{2}(F_{0}) and there is a dyadic valuation v of F_{0} such that conditions (1) and (2) of Definition 6.1 holds for the division algebra D_{0} represented by x_{0}. Extend v to a valuation on F as in the proof of Proposition 5.6. The maximal unramified extension of F̂ (with respect to v) contains the maximal unramified extension of F̂_{0}, so it kills x_{0} and hence also x; this verifies condition (6.1.1). For condition (6.1.2), it suffices to note that the map H^{1}(F̄_{0}, ℤ/2ℤ) → H^{1}(F̄, ℤ/2ℤ) is injective as in the proof of Proposition 5.6. □

Example 6.4. A retract rational extension of a transparent field is transparent, cf. Example 5.9.

7. Proof of the main theorem

We now prove Theorem 1.2. We assume that D_{1} and D_{2} are not isomorphic and that there is some quadratic extension of F that is contained in both of them, hence that D_{1} ⊗ D_{2} has index 2. We will produce a quadratic extension of F that is contained in one and not the other.

Suppose first that there is a real closure R of F that does not split D_{1}⊗D_{2}. Then one of the D_{i} is split by R and the other is not; say D_{1} is split. We can write D_{1} = (a_{1}, b_{1}) where a_{1} is positive in R. The field F (\sqrt{a_{1}}) is contained in D_{1}, but not contained in D_{2} because D_{2} ⊗ R is division.

If there is a nondyadic discrete valuation of F where D_{1} ⊗ D_{2} is ramified, then Proposition 3.5 provides the desired quadratic extension.

Finally, suppose that D_{1} ⊗ D_{2} is split by every real closure and is unramified at every nondyadic place, so D_{1} ⊗ D_{2} belongs to H_{u}^{2}(F). By the original hypotheses on D_{1} and D_{2}, the class [D_{1} ⊗ D_{2}] is represented by a quaternion division algebra B. But F is transparent, so Proposition 4.1 provides a quadratic extension of F that embeds in D_{1} or D_{2} but not the other. □

8. Pfister forms and nondyadic valuations

With the theorem from the introduction proved, we now set to proving a version of it for Pfister forms. The goal of this section is the following proposition, which is an analogue of Proposition 3.5.

Proposition 8.1. Let F be a field with a nondyadic discrete valuation. Let ϕ_{1}, ϕ_{2} be d-Pfister forms over F for some d ≥ 2. If

(1) ϕ_{1} is ramified and ϕ_{2} is not; or

(2) ϕ_{1} and ϕ_{2} both ramify, but with different ramification,

then there is a (d − 1)-Pfister form γ over F such that γ divides ϕ_{1} or ϕ_{2} but not both.

The proof given below for case (1) of the proposition was suggested to us by Adrian Wadsworth, and is much simpler than our original proof for that case. It makes use of the following lemma:

Lemma 8.2. Let ψ = ⟨a_{1}, …, a_{n}⟩ be a nondegenerate quadratic form over a field F with a nondyadic discrete valuation v. If v(a_{i}) = 0 for all i and ψ is isotropic over the completion F̂ of F, then there is a 2-dimensional nondegenerate subform q of ψ such that q ⊗ F̂ is isotropic.

Proof. As ϕ⊗F̂ is isotropic, there exist x_{i} ∈ F̂ not all zero so that ∑ a_{i}x_{i}^{2} = 0. Scaling by a uniformizer, we may assume that v(x_{i}) ≥ 0 for all i and that at least one x_{i} has value 0. Let j be the smallest index such that v(x_{j}) = 0; there must be at least two such indices, so j < n.

Fix t_{i} in the valuation ring of F so that t̄_{i} = x̄_{i}, and put

r := ∑_{i=1}^{j} a_{i}t_{i}^{2} and s := ∑_{i=j+1}^{n} a_{i}t_{i}^{2}.

By construction of j, v(r) = 0. Since r + s = 0, also v(s) = 0. Now ⟨r, s⟩ is a subform of ψ (over F), and

r̄ + s̄ = \overline{r + s} = \overline{∑ a_{i}t_{i}^{2}} = \overline{∑ a_{i}x_{i}^{2}} = 0.

So ⟨r, s⟩ is isotropic over F̂ [Lam 05]. □

Proof of Proposition 8.1. Obviously, ⟨⟨xπ, yπ⟩⟩ ≅ ⟨⟨−xy, yπ⟩⟩, so we may assume that

ϕ_{i} = ⟨⟨a_{i2}, a_{i3}, …, a_{id}, b_{i}⟩⟩

where a_{ij} has value 0 and b_{i} has value 0 or 1.

In case (2), b_{1} and b_{2} have value 1 and the residue forms

r(ϕ_{i}) = ⟨⟨ā_{i2}, ā_{i3}, …, ā_{id}⟩⟩

are not isomorphic. We take γ to be ⟨⟨a_{12}, a_{13}, …, a_{1d}⟩⟩; it divides ϕ_{1}. The projective quadric X defined by γ = 0 is defined over the discrete valuation ring, and F̄(X) does not split r(ϕ_{2}) by [Lam 05, X.4.10]. Hence F (X) does not split ϕ_{2} and γ does not divide ϕ_{2}.

Now suppose we are in case (1); obviously v(b_{1}) = 1. If ϕ_{2} ⊗ F̂ is anisotropic, then as ϕ_{2} does not ramify, v(b_{2}) = 0 and the second residue form of ϕ_{2} is zero. Certainly γ := ⟨⟨a_{13}, …, a_{1d}, b_{1}⟩⟩ divides ϕ_{1}. On the other hand, every anisotropic multiple of γ ⊗ F̂ over F̂ will have a nonzero second residue form, so ϕ_{2} cannot contain γ.

Otherwise, ϕ_{2} ⊗ F̂ is hyperbolic, and there exists some 2-dimensional subform q of ϕ_{2} that becomes hyperbolic over F̂ by Lemma 8.2. It follows that there is a (d − 1)-Pfister ψ that is contained in ϕ_{2} and that becomes hyperbolic over F̂. But ϕ_{1} ⊗ F̂ is ramified, hence anisotropic, so ϕ_{1} cannot contain ψ. □

9. Theorem for quadratic forms

Theorem 9.1. Let F be a field of characteristic ≠ 2 such that H_{u}^{d}(F) = 0 for some d ≥ 2. Let ϕ_{1} and ϕ_{2} be anisotropic d-Pfister forms over F such that, for every (d − 1)-Pfister form γ, we have: γ divides ϕ_{1} if and only if γ divides ϕ_{2}. Then ϕ_{1} is isomorphic to ϕ_{2}.

We remark that for d = 2, Theorem 9.1 does not include the case where F is a retract rational extension of a global field with more than one dyadic place, a case included in Theorem 1.2.

Proof of Theorem 9.1. Suppose that ϕ_{1} and ϕ_{2} are d-Pfister forms over F for some d ≥ 2, H_{u}^{d}(F) is zero, and ϕ_{1} is not isomorphic to ϕ_{2}. Then ϕ_{1} +ϕ_{2} is in H^{d}(F) \ H_{u}^{d}(F).

If there is a nondyadic discrete valuation where ϕ_{1} + ϕ_{2} ramifies, then Proposition 8.1 gives a (d − 1)-Pfister form that divides one of the ϕ_{i} and not the other. Otherwise, there is an ordering v of F such that ϕ_{1} and ϕ_{2} are not isomorphic over the real closure F_{v}, i.e., one is locally hyperbolic and the other is not. Write ϕ_{i} = ⟨⟨a_{i1}, a_{i2}, …, a_{id}⟩⟩ and suppose that ϕ_{1} is locally hyperbolic, so some a_{1j} is positive; renumbering if necessary, we may assume that it is a_{11}. Then the form ⟨⟨a_{12}, a_{13}, …, a_{1d}⟩⟩ divides ϕ_{1}, but it is hyperbolic over F_{v} and so does not divide ϕ_{2}. □

10. Appendix: tractable fields

We now use the notion of unramified cohomology from §5 to prove some new results and give new proofs of some known results regarding tractable fields. Recall from [CTW] that a field F of characteristic ≠ 2 is called tractable if for every a_{1}, a_{2}, a_{3}, b_{1}, b_{2}, b_{3} ∈ F^{×} such that the six quaternion algebras (a_{i}, b_{j}) are split for all i ≠ j and the three quaternion algebras (a_{i}, b_{i}) are pairwise isomorphic, the algebras (a_{i}, b_{i}) are necessarily split. (This condition was motivated by studying conditions for decomposability of central simple algebras of degree 8 and exponent 2.) We prove:

Proposition 10.1. If H_{u}^{2}(F) = 0, then F is tractable.

Proof. For sake of contradiction, suppose we are given a_{i}, b_{i} as in the definition of tractable where the quaternion algebras (a_{i}, b_{i}) are not split. If the (a_{i}, b_{i}) do not split at some real closure of F, then the a_{i} and b_{i} are all negative there. Hence (a_{i}, b_{j}) for i ≠ j is also not split, a contradiction.

So the (a_{i}, b_{i}) ramify at some nondyadic discrete valuation v of F and in particular are division over a the completion of F at v. We replace F with its completion and derive a contradiction. Modifying each a_{i} or b_{i} by a square if necessary, we may assume that each one has value 0 or 1. Further, as each (a_{i}, b_{i}) is ramified, at most one of the two slots has value 0.

Suppose first that one of the a_{i} or b_{i}—say, a_{1}—has value 0. Then b_{1} has value 1 and a_{1} is not a square in F. As (a_{1}, b_{j}) is split for j ≠ 1, we deduce that v(b_{j}) = 0. Hence v(a_{2}) = v(a_{3}) = 1. As (a_{3}, b_{3}) is division, b_{3} is not a square in F. Since v(a_{2}) = 1 and v(b_{3}) = 0, it follows that (a_{2}, b_{3}) is division, a contradiction.

It remains to consider the case where the a_{i}’s and b_{i}’s all have value 1. We can write a_{i} = πα_{i} and b_{i} = πβ_{i}, where π is a prime element—say, a_{1} so α_{1} = 1—and α_{i}, β_{i} have value 0. In the Brauer group of F, we have for i ≠ j:

0 = (πα_{i}, πβ_{j}) = (πα_{i}, −α_{i}β_{j}) = (π, −α_{i}β_{j}) + (α_{i}, β_{j}).

It follows that −α_{i}β_{j} is a square in F for i ≠ j. In particular, −β_{2}, −β_{3}, α_{2}, and α_{3} are all squares. Then

(a_{2}, b_{2}) = (πα_{2}, πβ_{2}) = (π, −π) = 0,

a contradiction. □

Example 10.2. In [CTW], the authors proved that a local field is tractable if and only if it is nondyadic (ibid., Cor. 2.3) and a global field is tractable if and only if it has at most 1 dyadic place (ibid., Th. 2.10). Combining the preceding proposition and Example 5.2 gives a proof of the “if” direction for both of these statements.

Corollary 10.3 ([CTW, Th. 3.17]). Every field of transcendence degree 1 over a real-closed is tractable.

Proof. Combine the proposition and Corollary 5.5. □

Example 10.4. The converse of Proposition 10.1 is false. For every odd prime p, ℚ_{p}((x)) is tractable by [CTW, Prop. 2.5]. On the other hand, the usual discrete valuation on the formal power series ring is essentially the unique one [EP, 4.4.1], so H_{u}^{2}(ℚ_{p}((x))) = H^{2}(ℚ_{p}) = ℤ/2ℤ.

Corollary 10.5. Let F_{0} be a field of characteristic ≠ 2 that • is tractable and

• is global, local, real-closed, or has no 2-torsion in its Brauer group. Then every unirational extension F/F_{0} satisfying (5.7)—e.g., every retract rational extension F/F_{0}—is also tractable.

Proof. We note that F_{0} has H_{u}^{2}(F_{0}) = 0. In case F_{0} is global or local, then F_{0} has one dyadic place or is non-dyadic respectively by [CTW], and the claim follows, see Example 5.2. Then H_{u}^{2}(F) is also zero by Proposition 5.6, hence F is tractable. □

Of course, this corollary includes the statement that rational extensions of tractable global fields are tractable, which was a nontrivial result (Example 3.14) in [CTW]. However, a stronger result has already been proved in [Han, Cor. 2.14]: every rational extension of a tractable field is tractable.

We do get a large family of new examples of tractable fields:

Corollary 10.6. Let F be a global field with at most one dyadic place. If G is an isotropic, simply connected, absolutely almost simple linear algebraic group over F, then the function field F (G) is tractable.

Proof. The extension F (G)/F is retract rational by Theorems 5.9 and 8.1 in [Gi]. □

Acknowledgments. The first author’s research was partially supported by National Science Foundation grant no. DMS-0653502. Both authors thank Adrian Wadsworth and Kelly McKinnie for their comments on an earlier version of this paper.

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(Garibaldi) Department of Mathematics & Computer Science, Emory University, Atlanta, GA 30322, USA

   E-mail address: skip@member.ams.org
   URL: http://www.mathcs.emory.edu/~skip/

(Saltman) Center for Communications Research-Princeton, 805 Bunn Drive, Princeton, NJ 008540, USA

   E-mail address: saltman@idaccr.org