COHOMOLOGICAL INVARIANTS OF CENTRAL SIMPLE ALGEBRAS OF DEGREE 4

GRÉGORY BERHUY

Abstract. In this paper, we prove a result of Rost, which describes the cohomological invariants of central simple algebras of degree 4 with values in μ_{2} when the base field contains a square root of −1.

Introduction

In [5], Rost, Serre and Tignol defined a cohomological invariant of central simple algebras of degree 4 with values in μ_{2} when the base field contains a square root of −1. On the other hand, taking the Brauer class of the tensor square of a central simple algebra of degree 4 yields a cohomological invariant of degree 2 with values in μ_{2}. In this paper, we prove a result of Rost (unpublished), which asserts that these two invariants are essentially the only ones (see Section 4 for a more precise statement).

This paper is organized as follows. After proving some preliminary results in Section 1, we determine the cohomological invariants of cyclic and bicyclic algebras in Section 2. Section 3 is devoted to the construction of a generic central simple algebra of degree 4. Finally, in Section 4, we prove that a cohomological invariant which vanishes on cyclic algebras and biquaternion algebras is identically zero (which another result due to Rost). As a corollary, we obtain a complete description of cohomological invariants of central simple algebras of degree 4.

The proofs of all the results of this paper rely heavily on the use of valuations and residue maps. We let the reader refer to [2] for the basic definitions and results on these topics.

We are grateful to Markus Rost for permitting us to publish his results and to Alexander Merkurjev for providing us his private lecture notes on Rost’s theorem, and for pointing us a mistake in an earlier version of this paper.

1. Preliminaries

Let F be a field of characteristic different from 2. We will denote by Sets the category of sets, by Rings the category of commutative rings, by Fields_{F} the category of fields extensions of F and by Alg_{F} the category of commutative F - algebras.

For any field extension K/F, we will denote by H^{∗}(K) the cohomology ring of K with coefficients in μ_{2}. We then get a functor H^{∗} : Fields_{F} ⟶ Rings.

If K/F is a field extension, then H^{∗}(K) carries a natural structure of a H^{∗}(F)- module, given by the external law

\begin{matrix}H^{∗}(F) × H^{∗}(K) & ⟶ H^{∗}(K) \\ (a, ξ) & ⟼ a·ξ = Res_{K/F} (a) ∪ ξ. \end{matrix}

We start with some results on the cohomology of rational extensions.

Let n ≥ 1 be an integer, let t_{1}, …, t_{n} be algebraically independent indeterminates over k, and let K_{n} = k(t_{1}, …, t_{n}). Let us denote by P_{n} the set of subsets of {1, …, n}. If I = {i_{1}, …, i_{r}} ∈ P_{n}, we set

(t_{I}) = (t_{i_{1}}) ∪ ⋯ ∪ (t_{i_{r}}).

Notice that this definition does not depend on the numbering of the elements of I, since the cup-product in H^{∗}(K) is commutative.

If I = ∅, then (t_{I}) = 1 (and thus a·(t_{I}) = Res_{K/k}(a) in this case).

Lemma 1.1. The cohomology classes (t_{I}), I ∈ P_{n} are linearly independent over H^{∗}(F).

Proof. We proceed by induction on n. Assume that n = 1, and let a_{0}, a_{1} ∈ H^{∗}(F) such that a_{0}·1+a_{1}·(t_{1}) = 0. Taking residues with respect to the valuation υ_{t_{1}} shows that a_{1} = 0. Then multiplying by (−t_{1}) and taking residues shows that a_{0} = 0.

Assume now that the result is proved for n ≥ 1, and let us prove it for n + 1 indeterminates. Let a_{I} ∈ H^{∗}(F), I ∈ P_{n+1} such that ∑_{I} a_{I}·(t_{I}) = 0. Then we

have

∑_{J} a_{J}·(t_{J}) + ∑_{J} a_{J∪{n+1}}·(t_{J}) ∪ (t_{n+1}) = 0,

where J describes P_{n}. Reasoning as in the case n = 1, we get

∑_{J} a_{J}·(t_{J}) = ∑_{J} a_{J∪{n+1}}·(t_{J}) = 0,

and by induction we have a_{J} = a_{J∪{n+1}} = 0 for all J ∈ P_{n}, that is a_{I} = 0 for all I ∈ P_{n+1}. This concludes the proof. □

Proposition 1.2. Let ξ ∈ H^{∗}(K_{n}) be a cohomology class which is unramified at every discrete k-valuation υ ≠ υ_{t_{i}}, i = 1, …, n. Then there exist unique cohomology classes a_{I} ∈ H^{∗}(F) such that ξ = ∑_{I} a_{I}·(t_{I}).

Proof. We prove it by induction on n. Assume first that n = 1, and let a_{1} ∈ H^{∗}(F) be the residue of ξ with respect to the t_{1}-adic valuation. Then ξ −Res_{K/F} (a_{1})∪(t_{1}) is unramified at every discrete F -valuation υ ≠ υ_{t_{1}} by assumption on ξ and because a_{1} is constant. By choice of a_{1}, it is also unramified at υ_{t_{1}}, so ξ −Res_{K_{1}/F} (a_{1})∪(t_{1}) is a constant class, and we may finally write

ξ = Res_{K_{1}/F} (a_{0}) + Res_{K_{1}/F} (a_{1}) ∪ (t_{1}) = a_{0}·1 + a_{1}·(t_{1}).

Now assume that the result is proved for n ≥ 1, and let ξ ∈ H^{∗}(K_{n+1}) be a cohomology class which is unramified at every discrete F -valuation υ ≠ υ_{t_{i}}, i = 1, …, n + 1. In particular, it is unramified at every discrete K_{n}-valuation υ ≠ υ_{t_{n+1}}. By the case n = 1, there exist b_{0}, b_{1} ∈ H^{∗}(K_{n}) such that ξ = b_{0}·1 + b_{1}·(t_{n+1}). Let υ be a discrete F -valuation on K_{n}, υ ≠ υ_{t_{i}}, i = 1, …, n, and let w the unique F -valuation on K_{n+1} extending υ such that w(t_{n+1}) = 0. Notice that the corresponding ramification index is equal to 1. Since ξ is unramified at w, taking residues shows that Res_{κ(w)/κ(υ)}(r_{υ}(b_{1}))∪(t_{n+1}) = 0. Now κ(w) = κ(υ)(t_{n+1}). The previous equality rewrites r_{υ}(b_{1})·(t_{n+1}) = 0, and by the case n = 1 (applied to the base field κ(υ)), we get r_{υ}(b_{1}) = 0. Moreover, ξ ∪ (−t_{n+1}) = b_{0}·(−1) + b_{0}·(t_{n+1}) is also unramified at w, and reasoning as before shows that r_{υ}(b_{0}) = 0. Hence b_{0} and b_{1} are unramified at any discrete F -valuation υ on K_{n}, υ ≠ υ_{t_{i}}, i = 1, …, n. Now use the induction hypothesis to conclude that ξ = ∑_{I} a_{I}·(t_{I}) for some a_{I} ∈ H^{∗}(F).

The uniqueness of the classes a_{I} comes from the previous lemma. This completes the proof. □

Definition 1.3. Let F : Fields_{F} ⟶ Sets be a covariant functor. A cohomological invariant of F over F is a natural transformation F ⟶ H^{∗} of functors Fields_{F} from Sets.

Cohomological invariants of F form an H^{∗}(F)-module, that we will denote by Inv(F, H^{∗}).

We will need in the sequel the notion of a classifying pair for a functor F.

Definition 1.4. Let F : Alg_{F} ⟶ Sets be a covariant functor. Let R be an F -algebra, and let a ∈ F(R). We say that the pair (R, a) is classifying for F, if the following conditions hold:

(1) The ring R is a noetherian domain;

(2) For every field extension L/F, and every a′ ∈ F(L), there exists a maximal ideal 𝔪 of R and a morphism of F -algebras f : R ⟶ L such that ker(f) ⊃ 𝔪 and F(f) maps a onto a′.

The Specialization Theorem (cf. [2, Theorem 12.2]) then immediately gives:

Lemma 1.5. Let F : Alg_{F} ⟶ Sets be a subfunctor of H^{1}(_{−}, G), where G is an algebraic group over F. Assume that (R, a) is classifying for F, and let α, β ∈ Inv(F, H^{∗}). Let K be the quotient field of R. If α(a_{K}) = β(a_{K}), then α = β.

2. Invariants of cyclic and bicyclic algebras

For n ≥ 1, we will denote by CSA_{n} : Alg_{F} ⟶ Sets the functor of isomorphism classes of Azumaya algebras of rank n.

Let R be a commutative ring. Assume that n ∤ char(R) and that R contains a primitive n-th root of 1, that we will denote by ζ_{n}. For all u, v ∈ R^{×}, the R-algebra {u, v}_{n,R} generated by two elements e and f subject to the relations

e^{n} = u, f^{n} = v, fe = ζ_{n}ef

is an Azumaya R-algebra of rank n, called a symbol algebra.

For r ≥ 1, we will denote by MS_{n,r} the subfunctor of CSA_{n^{r}} of isomorphism classes of tensor products of r symbols algebras of degree n.

We will denote by 1 the constant cohomological invariant of CSA_{n} (where CSA_{n} is now viewed as a functor from Fields_{F} to Sets). If n = 2m, then for every field extension K/F and every central simple K-algebra A of exponent dividing n, the class m[A] is killed by 2, and therefore defines a cohomology class of H^{2}(K) via the usual isomorphism. This defines a cohomological invariant of CSA_{n}, as well as a cohomological invariant of MS_{n,r} for all r ≥ 1, that we will denote by f_{m} in both cases. In particular, we have

f_{m}({a_{1}, b_{1}}_{n,K} ⊗_{F} ⋯ ⊗_{F} {a_{r}, b_{r}}_{n,K}) = (a_{1}) ∪ (b_{1}) + … + (a_{r}) ∪ (b_{r})

for every field extension K/F and all a_{i}, b_{i} ∈ K^{×}.

If A is a central simple F -algebra, the trace form of A is the quadratic form

q_{A}: \begin{matrix}A & ⟶ F \\ a & ⟼ Trd_{A}(a^{2}). \end{matrix}

Lemma 2.1. Assume that n ∤ char(F) and that μ_{n} ∈ F. Assume also that n = 2m.

For all u, v ∈ F^{×}, the trace form of {u, v}_{n,F} is Witt-equivalent to

⟨n, nu, nv, (−1)^{m}nuv⟩.

Proof. One may check that Trd_{A}(e^{i}f^{j}) = 0 if (i, j) ≠ (0, 0). It easily follows that the subspaces

F ·1, F ·e^{m}, F ·f^{m}, F ·e^{m}f^{m} and F ·e^{i}f^{j} ⊕ F ·e^{n−i}f^{m−j},

where 0 ≤ i ≤ j ≤ m, (i, j) ≠ (0, 0), (0, m), (m, 0), (m, m) are mutually orthogonal. Moreover, the 2m^{2} − 2 planes above are hyperbolic since e^{i}f^{j} is isotropic. The result then follows from the fact that the reduced traces of 1, e^{m}, f^{m} and e^{m}f^{m} are respectively n, nu, nv and (−1)^{m}nuv. This concludes the proof. □

Assume that n ∤ char(F) and that μ_{n} ⊂ F. Moreover, assume that −1 ∈ F^{×2} (this condition is automatically satisfied if n ≡ 0[4]).

If A = {u_{1}, v_{1}}_{n,F} ⊗_{F} ⋯ ⊗_{F} {u_{r}, v_{r}}_{n,F}, it follows from the previous lemma that q_{A} is Witt-equivalent to ⟨n^{r}⟩⟨⟨u_{1}, v_{1}, …, u_{r}, v_{r}⟩⟩. Therefore, q_{A} ∈ I^{2r}F, and we have

e_{2r}(q_{A}) = (u_{1}) ∪ (v_{1}) ∪ ⋯ ∪ (u_{r}) ∪ (v_{r}).

This defines an element of Inv(MS_{n,r}, H^{∗}) that we denote by e_{2r}.

Proposition 2.2. Let F be a field of characteristic different from 2, and let n = 2m.

(1) Assume that n ∤ char(F), and that ζ_{n} ∈ F, where ζ_{n} is a primitive n^{th}-root

of 1. Then Inv(MS_{n,1}, H^{∗}) is a free H^{∗}(F)-module with basis 1, f_{m}.

(2) Assume moreover that −1 ∈ F^{×2}. Then Inv(MS_{n,2}, H^{∗}) is a free H^{∗}(F)-

module with basis 1, f_{m}, e_{4}.

Proof. We first prove (1). Let α ∈ Inv(MS_{n,1}, H^{∗}). Let t_{1}, t_{2} be two indeterminates over F, and let K = F (t_{1}, t_{2}). Set A = {t_{1}, t_{2}}_{n,K}, and let υ ≠ υ_{t_{i}}, i = 1, 2 be a discrete F -valuation. By assumption on υ, t_{1}, t_{2} ∈ O_{υ}^{×}, and we have A ≃ {t_{1}, t_{2}}_{n,O_{υ}} ⊗_{O_{υ}} K. It follows from [2, Theorem 11.7] that the class α_{K}({t_{1}, t_{2}}_{n,K}) is unramified at υ. By Proposition 1.2, there exist a_{0}, a_{1}, a_{2}, a_{3} ∈ H^{∗}(F) such that

α_{K}({t_{1}, t_{2}}_{n,K}) = a_{0}·1 + a_{1}·(t_{1}) + a_{2}·(t_{2}) + a_{3}·(t_{1}) ∪ (t_{2}).

Now let υ be the (t_{1} − 1)-adic valuation on F (t_{2}). Applying [2, Theorem 11.7] shows by specialization that

α_{K}({1, t_{2}}_{n,F (t_{2})}) = a_{0}·1 + a_{2}·(t_{2}),

that is

α_{K}(M_{n}(F (t_{2}))) = a_{0}·1 + a_{2}·(t_{2}).

Now M_{n}(F (t_{2})) is unramified at υ_{t_{2}}, and therefore, so is α_{K}(M_{n}(F (t_{2}))). Taking residues then yields a_{2} = 0. Similar arguments show that a_{1} = 0, and thus

α_{K}({t_{1}, t_{2}}_{n,K}) = a_{0}·1 + a_{3}·(t_{1}) ∪ (t_{2}).

Let R = F [t_{1}, t_{1}^{−1}, t_{2}, t_{2}^{−1}]. It is clear that the pair (R, {t_{1}, t_{2}}_{n,R}) is classifying for MS_{n,1}. The previous equality shows that the invariants α and a_{0}·1+a_{3}·f_{m} coincide on a classifying pair, hence they are equal by Lemma 1.5. Moreover, if a_{0}·1+a_{3}·f_{m} = 0, then applying this equality to {t_{1}, t_{2}}_{n,K} shows that a_{0}·1 + a_{3}·(t_{1}) ∪ (t_{2}) = 0. Lemma 1.1 then yields a_{0} = a_{3} = 0. This proves (1).

Assume now that −1 ∈ F^{×2}. The fact that 1, f_{m}, e_{4} are linearly independent over H^{∗}(F) easily comes from Lemma 1.1. Let α ∈ Inv(MS_{n,2}, H^{∗}), let K/F be a field extension, and let A be a symbol algebra over K. Any field extension L/K yields a field extension L/F, and the maps

β_{L}: \begin{matrix}MS_{n,r}(L) & ⟶ H^{∗}(L) \\ B & ⟼ α_{L}(B ⊗_{L} A_{L}) \end{matrix}

fit together into a cohomological invariant β ∈ Inv(MS_{n,1}, H^{∗}). By (1), there exists α_{i,K}(A) ∈ H^{∗}(K) such that

β = α_{0,K}(A)·1 + α_{1,K}(A)·f_{m}.

By linear independence of 1, f_{m} (over the base field K), these classes are unique. It easily follows that α_{i,K}(A) only depends on the isomorphism class of A. Moreover, it follows from the uniqueness of α_{i,K}(A) and the fact that α is a cohomological invariant that the maps α_{i,K} fit together into a cohomological invariant α_{i} ∈ Inv(MS_{n,1}, H^{∗}). By (1), we may write α_{i} = a_{i}·1 + b_{i}·f_{m}, for some a_{i}, b_{i} ∈ H^{∗}(F).

Taking K = F (u_{1}, v_{1}, u_{2}, v_{2}), where u_{i}, v_{i} are indeterminates, B = {u_{1}, v_{1}}_{n,K} and A = {u_{2}, v_{2}}_{n,K}, we get

\begin{aligned}α_{K}(\{u_{1}, v_{1}\}_{n,K} ⊗_{K} \{u_{2}, v_{2}\}_{n,K}) & = & a_{0}·1 + b_{0}·(u_{2}) ∪ (v_{2}) \\ & & +(a_{1}·1 + b_{1}·(u_{2}) ∪ (v_{2})) ∪ (u_{1}) ∪ (v_{1}). \end{aligned}

The F -automorphism of K which exchanges u_{1} and u_{2}, and exchanges v_{1} and v_{2}, induces maps MS_{n,2}(K) ⟶ MS_{n,2}(K) and H^{∗}(K) ⟶ H^{∗}(K). Notice that the first one maps the isomorphism class of {u_{1}, v_{1}}_{n,K} ⊗_{K} {u_{2}, v_{2}}_{n,K} onto itself. Since α commutes with induced maps, we then get

\begin{aligned}α_{K}(\{u_{1}, v_{1}\}_{n,K} ⊗_{K} \{u_{2}, v_{2}\}_{n,K}) & = & a_{0}·1 + b_{0}·(u_{1}) ∪ (v_{1}) \\ & & +(a_{1}·1 + b_{1}·(u_{1}) ∪ (v_{1})) ∪ (u_{2}) ∪ (v_{2}). \end{aligned}

Comparing with the previous equality and using Lemma 1.1, we get that b_{0} = a_{1}. Therefore, {u_{1}, v_{1}}_{n,K} ⊗_{K} {u_{2}, v_{2}}_{n,K} has same image by α and by a_{0}·1 + a_{1}·f_{m} + b_{1}·e_{4}. Now let R = F [u_{i}, u_{i}^{−1}, v_{i}, v_{i}^{−1}, i = 1, 2]. Since the pair (R, {u_{1}, v_{1}}_{n,R} ⊗_{R} {u_{2}, v_{2}}_{n,R}) is classifying for MS_{n,2}, it follows from the previous considerations that the invariants α and a_{0}·1 + a_{1}·f_{m} + b_{2}·e_{4} are equal. This concludes the proof.

□

3. Central simple algebras of degree 4

We start this section with a parametrization of central simple algebras of degree 4.

Let R be a commutative ring such that 2 ∈ R^{×}. If a ∈ R^{×}, we will denote by R[\sqrt{a}] the étale quadratic R-algebra R[X]/(X^{2} − a).

Let L be a biquadratic étale R-algebra, that is an R-algebra L generated by two elements α, β subject to the relations

α^{2} = d, β^{2} = d′, αβ = βα,

for some d, d′ ∈ R^{×}. Such an algebra will be denoted by F [\sqrt{d}, \sqrt{d′}].

The group G of automorphisms of the R-algebra L = F [\sqrt{d}, \sqrt{d′}] is isomorphic to ℤ/2ℤ × ℤ/2ℤ, generated by the automorphisms σ and τ which are uniquely determined by the following relations:

σ(α) = α, σ(β) = −β, τ(α) = −α, τ(β) = β.

Let a ∈ R[\sqrt{d}]^{×}, b ∈ R[\sqrt{d′}]^{×}, c ∈ R[\sqrt{dd′}]^{×}, and set

n_{a} = N_{R[\sqrt{d}]/R}(a), n_{b} = N_{R[\sqrt{d′}]/R}(b), n_{c} = N_{R[\sqrt{dd′}]/R}(c).

Notice that n_{a}, n_{b}, n_{c} ∈ R^{×} and that we have

n_{a} = aτ(a), n_{b} = bσ(b), n_{c} = cσ(c) = cτ(c).

Assume that n_{c} = n_{a}n_{b}, and let 1, e_{σ}, e_{τ} , e_{στ} be the canonical basis of the L-vector space Map(G, L). We define a product law on Map(G, L) by imposing the relations

e_{σ}^{2} = a, e_{τ}^{2} = b, e_{στ}^{2} = c, e_{σ}e_{τ} = e_{στ} ,
e_{σ}λ = σ(λ)e_{σ}, e_{τ} λ = τ(λ)e_{τ} , for all λ ∈ L,

and extending by distributivity.

By definition, the elements e_{ρ}, ρ ∈ G are invertible with respect to this product law. Notice that the missing products e_{ρ}e_{ρ′}, ρ, ρ′ ∈ G may be obtained using the relations above, so that the product law on Map(G, L) is completely determined. For example, we have

e_{τ} e_{στ} = e_{σ}^{−1}e_{σ}e_{τ} e_{στ} = e_{σ}^{−1}(e_{στ})^{2} = a^{−1}e_{σ}c = a^{−1}σ(c)e_{σ}.

One can check that we obtain an Azumaya R-algebra of rank 4, that we will denote by (a, b, c, L/R). Notice that if we set e = e_{σ} and f = e_{τ}, then we have (a, b, c, L/R) = L ⊕ Le ⊕ Lf ⊕ Lef, and

e^{2} = a, f^{2} = b, (ef)^{2} = c, eλ = σ(λ)e, fλ = τ(λ)f, for all λ ∈ L.

Notice also for later use that we have fe = a^{−1}σ(cb^{−1})ef, since

fe = e^{−1}(ef)^{2}f^{−1} = a^{−1}ecb^{−1}f = a^{−1}σ(cb^{−1})ef.

Lemma 3.1. Let F be a field of characteristic different from 2. Then every central simple F -algebra of degree 4 is isomorphic to some (a, b, c, L/F). Moreover, one may assume that a, b, c ∉ F^{×}.

Proof. Assume first that A is a division F -algebra. By a theorem of Albert (see [1, Theorem 11.9] for example), A contains a biquadratic étale F -algebra L. By Skolem-Noether’s theorem, σ and τ extend to inner automorphisms Int(e) and Int(f) of A, for some e, f ∈ A^{×}. This rewrites

eλ = σ(λ)e, fλ = τ(λ)f, for all λ ∈ L.

Since Int(e)_{|_{L}} = σ, we have

Int(e^{2})_{|_{L}} = (Int(e)_{|_{L}})^{2} = σ^{2} = Id_{L},

so a = e^{2} ∈ C_{A}(L) = L (since L is a maximal subfield of A) . Since e ∈ A^{×}, we have a ∈ L^{×}. Similarly, b = f^{2} ∈ L^{×} and c = (ef)^{2} ∈ L^{×}.

Associativity of the product law of A implies that

σ(a) = a, τ(b) = b, στ(c) = c, n_{c} = n_{a}n_{b}.

Indeed, we have

e^{3} = ee^{2} = ea = σ(a)e = e^{2}e = ae,

and since e is invertible, we get σ(a) = a. Similar arguments show that τ(b) = b and στ(c) = c. We also have

be = f^{2}e
= f(fe)
= fa^{−1}σ(cb^{−1})ef
= τ(a^{−1}σ(cb^{−1}))f(ef)
= cτ(a)^{−1}σ(b)^{−1}(fe)f
= cτ(a)^{−1}σ(b)^{−1}a^{−1}σ(cb^{−1})(ef)f
= cτ(a)^{−1}σ(b)^{−1}a^{−1}σ(c)σ(b)^{−1}eb
= cτ(a)^{−1}σ(b)^{−1}σ(c)a^{−1}σ(b)^{−1}σ(b)e
= n_{c}n_{a}^{−1}σ(b)^{−1}e.

Since e is invertible, we get b = n_{c}n_{a}^{−1}σ(b)^{−1}, that is n_{c} = n_{a}n_{b}. The proof of the linear independence of 1, e, f, ef over L is straightforward and left to the reader. Hence A contains a subalgebra isomorphic to (a, b, c, L/F), and therefore A ≃ (a, b, c, L/F) since they have same dimension over F.

Assume now that A is not a division algebra. Then A ≃ M_{2}(Q), where Q = (d, d′) is a (not necessarily division) quaternion F -algebra.

Let L = F [\sqrt{d}, \sqrt{d′}] be a biquadratic extension with generators α and β, and let us consider the F -algebra (d, 1, d, L/F). We then have

e^{2} = d, f^{2} = 1, (ef)^{2} = d, eα = αe, eβ = −βe, fα = −αf, fβ = βf.

Notice that we also have fe = dσ(d^{−1})ef = ef.

Set i = e, j = βf, i′ = f, j′ = αe. It is straightforward to check that the F -algebra B generated by i and j is isomorphic to Q, and that the F -algebra B′ generated by i′ and j′ is isomorphic to (1, d^{2}) ≃ M_{2}(F). Moreover one may check that B′ is the centralizer of B. Since B′ is a central simple F -algebra, we get

(d, 1, d, L/F) ≃ B ⊗_{F} B′ ≃ Q ⊗_{F} M_{2}(F) ≃ M_{2}(Q) ≃ A.

We now prove the last part. Write A ≃ (a, b, c, L/F), for some biquadratic étale F -algebra L = F [\sqrt{d}, \sqrt{d′}]. Replacing f by (1 + α + β)f, one may assume that b ∉ F^{×}. Indeed, if b ∈ F^{×}, we then have

[(1 + α + β)f]^{2} = (1 + α + β)(1 − α + β)f^{2} = b(1 + d′ − d + 2β) ∉ F^{×}.

Write a = a_{0} + a_{1}α. Notice that, for all s ∈ F^{×}, we have

\begin{aligned}[(s + α + β)e]^{2} & = & (s^{2} + d − d′ + 2sα)(a_{0} + a_{1}α) \\ & = & (a_{0}(s^{2} + d′ − d) + 2sa_{1}d) + (a_{1}(s^{2} + d − d′) + 2a_{0}s)α, \end{aligned}

and

[(s + α + β)ef]^{2} = (s + α + β)(s − α − β)c = c(s^{2} − d − d′ − 2αβ).

Thus, after replacing e by (s + α + β)e for a suitable s ∈ F^{×}, one may also assume that a, c ∉ F^{×}. This concludes the proof. □

Remark 3.2. For any biquadratic F -algebra L, we have M_{4}(F) ≃ (1, 1, 1, L/F).

Indeed, the subalgebra B generated by e and β is isomorphic to (1, d′) ≃ M_{2}(F), and its centralizer C is the subalgebra generated by f and αe, which is isomorphic to (1, d^{2}) ≃ M_{2}(F). Since B is a central simple F -algebra, we have

(1, 1, 1, L/F) ≃ B ⊗_{F} C ≃ M_{2}(F) ⊗_{F} M_{2}(F) ≃ M_{4}(F).

Proposition 3.3. Let F be a field of characteristic different from 2, and let A = (a, b, c, L/F). Then

2[A] = [(d, n_{c})] ∈ Br(F).

Proof. Recall from [4] that for every central simple F -algebra of degree n, we have

det(q_{A}) = det(q_{M_{n}(F)}), w_{2}(q_{A}) = w_{2}(q_{M_{n}(F)}) + \frac{n(n − 1)}{2}[A].

Since q_{A} and q_{M_{n}(F)} have same dimension and determinant, the quadratic form q_{A} ⊥ −q_{M_{n}(F)} lies in I^{2}(F), and the second equality may be rewritten as

c(q_{A} ⊥ −q_{M_{n}(F)}) = \frac{n(n − 1)}{2}[A],

where c denotes the Clifford invariant. In particular, if n = 4, we have

c(q_{A} ⊥ −q_{M_{n}(F)}) = 6[A] = 2[A].

Let us introduce some notation. Let E be an F -algebra. For u ∈ E^{×}, we denote by q_{E,u} the quadratic form

q_{E,u}: \begin{matrix}E & ⟶ F \\ x & ⟼ Tr_{E/F} (ux^{2}). \end{matrix}

If u = 1, we denote it by q_{E}.

Let ρ be an F -automorphism of L satisfying ρ^{2} = Id_{E}. If u ∈ E^{×}, ρ(u) = u, we denote by q_{E,ρ,u} the quadratic form

q_{E,ρ,u}: \begin{matrix}E & ⟶ F \\ x & ⟼ Tr_{E/F} (uxρ(x)) \end{matrix}

If u = 1, we denote it by q_{E,ρ}.

Now assume that A = (a, b, u, L/F). It is easy to check that we have

Trd_{A}(x + ye + zf + tef) = Tr_{L/F} (x), for all x, y, z, t, ∈ L.

This implies that the subspaces L, Le, Lf and Lef are mutually orthogonal. It follows from the previous observation that we have

q_{A} ≃ q_{L} ⊥ q_{L,σ,a} ⊥ q_{L,τ,b} ⊥ q_{L,στ,c}.

Let ρ be an F -automorphism of L, and assume that ρ ≠ Id_{L}, so we may write L^{⟨ρ⟩} = F [\sqrt{Δ}], and L = L^{⟨ρ⟩}[\sqrt{Δ′}]. Notice that uxρ(x) ∈ L^{⟨ρ⟩}, so that we have Tr_{L/F} (uxρ(x)) = 2Tr_{L^{⟨ρ⟩}/F} (uxρ(x)).

Now if x = x_{0} + x_{1}α′ where α^{′2} = Δ′ and x_{i} ∈ L^{⟨ρ⟩}, we have xρ(x) = x_{0}^{2} − Δ′x_{1}^{2}, and therefore

q_{L,ρ,u} ≃ ⟨2, −2Δ′⟩ ⊗ q_{L^{⟨ρ⟩},u}.

Write u = u_{0} + u_{1}\sqrt{Δ}, u_{i} ∈ F, and set n_{u} = u_{0}^{2} − Δu_{1}^{2}. The representative matrix of the previous quadratic form in the basis 1, \sqrt{Δ} is

( \begin{matrix}2u_{0} & 2u_{1}Δ \\ 2u_{1}Δ & 2u_{0}Δ \end{matrix} ) .

If u_{0} = 0, this 2-dimensional quadratic form is isotropic, hence hyperbolic. If u_{0} ≠ 0, this form then represents 2u_{0} and has determinant 4n_{u}Δ, so it is isomorphic to ⟨2u_{0}⟩ ⊗ ⟨1, n_{u}Δ⟩. We then get

q_{L,ρ,u} ≃ ⟨u_{0}⟩ ⊗ ⟨⟨−n_{u}Δ, Δ′⟩⟩,

where this form has to be understood as the hyperbolic form of dimension 4 if u_{0} = 0. Taking u = 1, we get

q_{L,ρ} ≃ ⟨⟨−Δ, Δ′⟩⟩,

and thus

q_{L,ρ,u} ⊥ −q_{L,ρ} ≃ ⟨1, −Δ′⟩ ⊗ ⟨u_{0}, u_{0}n_{u}Δ, 1, Δ⟩.

Since c(φ ⊗ φ′) = [(disc(φ), disc(φ′))] for every even-dimensional forms φ, φ′, we get

c(q_{L,ρ,u} ⊥ −q_{L,ρ}) = [(Δ′, n_{u})].

Remark 3.2 implies that q_{A} ⊥ −q_{M_{4}(F)} is Witt-equivalent to

(q_{L,σ,a} ⊥ −q_{L,σ}) ⊥ (q_{L,τ,b} ⊥ −q_{L,τ}) ⊥ (q_{L,στ,c} ⊥ −q_{L,στ}).

We then finally get

c(q_{A} ⊥ −q_{M_{4}(F)}) = [(d′, n_{a})] + [(d, n_{b})] + [(d, n_{c})].

Since the quaternion algebras (d, n_{a}), (d′, n_{b}) and (dd′, n_{c}) are split, we get

c(q_{A} ⊥ −q_{M_{4}(F)}) = [(dd′, n_{a})] + [(dd′, n_{b})] + [(d, n_{c})]
= [(dd′, n_{a}n_{b})] + [(d, n_{c})]
= [(dd′, n_{c})] + [(d, n_{c})]
= [(d, n_{c})].

This concludes the proof. □

We now define a classifying pair for CSA_{4}. Let us consider the affine variety V ⊂ 𝔸^{6} defined by the equation

x^{2} − uy^{2} − vz^{2} + uvt^{2} + uv = 0.

This is a rational variety, with coordinate ring

F [V] = F [X, Y, Z, T, U, V]/(X^{2} − UY^{2} − V Z^{2} + UV T^{2} + UV).

Let us denote by x, y, z, t, u, v the images of X, Y, Z, T, U, V in F [V] respectively. Notice that u and v are algebraically independent over F. Indeed, we have a surjective F -algebra morphism

\begin{matrix}F [V] & ⟶ F [V, Z] \\ P (x, y, z, t, u, v) & ⟼ P (0, i, Z, i, V Z^{2}, V) \end{matrix}

which maps u and v onto V Z^{2} and V respectively. Since V Z^{2} and V are algebraically independent over F, so are u and v.

Now let us consider the open subset U of V defined by the equations

u ≠ 0, v ≠ 0, t ≠ 0, y^{2} − vt^{2} ≠ 0, z^{2} − u ≠ 0.

Then U is also a rational variety, whose coordinate ring F [U] is the localization of F [V] at u, v, t, y^{2} − vt^{2} and z^{2} − u. Set R_{0} = F [U]. Then R_{0} is a noetherian ring, whose quotient field F_{0} is a rational extension of F. Moreover, u, v ∈ R_{0} are algebraically independent over F. In particular, there exists a transcendence basis of F_{0}/F containing u and v.

Set w = u(y^{2} − vt^{2}) ∈ R_{0}^{×} and let L_{0} = R_{0}[\sqrt{u}, \sqrt{uw}]. We now define three elements a_{0}, b_{0}, c_{0} ∈ L_{0}^{×} by

a_{0} = z + \sqrt{u} ∈ R_{0}[\sqrt{u}]^{×}, b_{0} = \frac{y}{t} + \frac{\sqrt{uw}}{tu} ∈ R_{0}[\sqrt{uw}]^{×}, c_{0} = x + \sqrt{w} ∈ R_{0}[\sqrt{w}]^{×}.

Let us check that n_{a_{0}}n_{b_{0}} = n_{c_{0}}. We have

n_{a_{0}}n_{b_{0}} = \frac{(z^{2} − u)(y^{2}u^{2} − uw)}{t^{2}u^{2}} = \frac{(z^{2} − u)(y^{2}u − w)}{t^{2}u}.

Since y^{2}u − w = uvt^{2}, we get n_{a_{0}}n_{b_{0}} = (z^{2} − u)v. Now z^{2}v − uv = x^{2} − uy^{2} + uvt^{2} = x^{2} − w = n_{c_{0}}, and we are done. Hence we may consider the Azumaya algebra

A_{0} = (a_{0}, b_{0}, c_{0}, L_{0}/R_{0}).

Lemma 3.4. The pair (R_{0}, A_{0}) is classifying for CSA_{4}.

Proof. Let E/F be a field extension, and let A be a central simple E-algebra of degree 4. By Lemma 3.1, we have A ≃ (a, b, c, L/E), with a, b, c ∉ E^{×}. Write L = E[\sqrt{d}, \sqrt{d′}], with generators α, β. Write

a = λ_{0} + λ_{1}α, b = μ_{0} + μ_{1}β, c = γ_{0} + γ_{1}αβ.

We set

\overline{x} = γ_{0}, \overline{y} = \frac{μ_{0}γ_{1}}{λ_{1}μ_{1}} , \overline{z} = λ_{0}, \overline{t} = \frac{γ_{1}}{λ_{1}μ_{1}} , \overline{u} = λ_{1}^{2}d, \overline{v} = μ_{0}^{2} − μ_{1}^{2}d′.

Using the equality

(λ_{0}^{2} − λ_{1}^{2}d)(μ_{0}^{2} − μ_{1}^{2}d) = (γ_{0}^{2} − γ_{1}^{2}dd′),

and the fact that λ_{1}, μ_{1}, γ_{1} ∈ F^{×}, one may check that p = (\overline{x}, \overline{y}, \overline{z}, \overline{t}, \overline{u}, \overline{v}) ∈ U(E). Then evaluation at p yields an F -algebra morphism f : R_{0} ⟶ E such that

A_{0} ⊗_{R_{0}} E ≃ A.

Now it suffices to notice that ker(f) = (x − \overline{x}, y − \overline{y}, z − \overline{z}, t − \overline{t}, u − \overline{u}, v − \overline{v}) is a maximal ideal to conclude. □

4. Cohomological invariants of CSA_{4}

We now prove the following theorem, due to Rost (unpublished):

Theorem 4.1 (Rost). Let F be a field of characteristic different from 2. Assume that −1 ∈ F^{×2}. Then the map

Inv(CSA_{4}, H^{∗}) ⟶ Inv(MS_{2,2}, H^{∗}) × Inv(MS_{4,1}, H^{∗})

is injective. In other words, a cohomological invariant which is zero on biquaternion algebras and cyclic algebras is identically zero.

Proof. Let α ∈ Inv(CSA_{4}, H^{∗}). Let (R_{0}, A_{0}) be the classifying pair of CSA_{4} defined in the previous section, and let F_{0} be the quotient field of R_{0}. Finally, set

A_{0} = A_{0} ⊗_{R_{0}} F_{0} = (z + \sqrt{u}, \frac{y}{t} + \frac{\sqrt{uw}}{tu} , x + \sqrt{w}, F_{0}[\sqrt{u}, \sqrt{uw}]/F_{0}).

Notice that we have 2[A_{0}] = [(u, v)] in Br(F_{0}). Indeed, by Proposition 3.3, we have

2[A_{0}] = [(u, x^{2} − w)]
= [(u, x^{2} − uy^{2} + uvt^{2})]
= [(u, vz^{2})]
= [(u, v)].

Let υ be a discrete F -valuation on F_{0}. If υ ≠ υ_{π}, with π = u, v, t, y^{2} − vt^{2}, z^{2} − u,

then u, v, w, t ∈ O_{υ}^{×} and z + \sqrt{u}, \frac{y}{t} + \frac{\sqrt{uw}}{tu} , x + \sqrt{w} ∈ O_{υ}[\sqrt{u}, \sqrt{uw}]^{×}. We then get that

A_{0} ≃ (z + \sqrt{u}, \frac{y}{t} + \frac{\sqrt{uw}}{tu} , x + \sqrt{w}, O_{υ}[\sqrt{u}, \sqrt{uw}]/O_{υ}) ⊗_{O_{υ}} F_{0},

and therefore α_{F_{0}}(A_{0}) is unramified at υ by [2, Theorem 11.7] .

Assume now that υ = υ_{π}, with π = t, z^{2} − u or y^{2} − tv^{2}. Then u, v ∈ O_{υ}^{×}, and thus we have

(u) ∪ (v) = j_{υ}((\overline{u}) ∪ (\overline{v})),

where j_{υ} : H^{2}(κ(υ)) ↪ H^{2}((F_{0})_{υ}) is the canonical injection. Now in all three cases, the quadratic form ⟨1, −\overline{u}, −\overline{v}, \overline{uv}⟩ is isotropic over κ(υ). It is clear if π = z^{2} − u or y^{2} − vt^{2}, and it follows from the equalities

0 = x^{2} − uy^{2} − vz^{2} + uvt^{2} + uv = x^{2} − uy^{2} − vz^{2} + uv mod t,

if π = t. Thus (\overline{u}) ∪ (\overline{v}) = 0, and therefore (u) ∪ (v) = 0 as well. Hence the quaternion algebra (u, v) splits over (F_{0})_{υ}, and thus 2[A_{0}] splits over (F_{0})_{υ}. By a theorem of Albert, A_{0} ⊗_{F_{0}} (F_{0})_{υ} is a biquaternion algebra. By assumption, we get

Res_{(F_{0})_{υ}/F_{0}}(α_{F_{0}}(A_{0})) = α_{(F_{0})_{υ}}(A_{0} ⊗_{F_{0}} (F_{0})_{υ}) = 0.

Consequently, r_{υ}(α_{F_{0}}(A_{0})) = 0 and α_{F_{0}}(A_{0}) is once again unramified at υ.

Finally, α_{F_{0}}(A_{0}) is unramified at every discrete F -valuation υ ≠ υ_{u}, υ_{v}. In particular, it is unramified at every F_{1}-valuation, where F_{1} = F (u, v). Since there exists a transcendence basis of F_{0}/F containing u and v, the field extension F_{0}/F_{1} is rational, and we deduce that α_{F_{0}}(A_{0}) = Res_{F_{0}/F_{1}}(β), for some β ∈ H^{∗}(F_{1}). Let υ_{1} be a discrete F -valuation on F_{0}, which different from the u-adic and the v-adic valuations. Let us show that β is unramified at υ_{1}.

Let u, v, t_{1}, t_{2}, t_{3} be a transcendence basis of F_{0}/F, and let us extend υ_{1} to a discrete F -valuation υ on F_{0} by setting

υ_{F_{1}} = υ_{1}, υ(t_{i}) = 0, i = 1, 2, 3.

Then the corresponding ramification index is 1 and κ(υ_{1}) = κ(υ)(t_{1}, t_{2}, t_{3}), so that κ(υ)/κ(υ_{1}) is a rational extension. Since α_{F_{0}}(A_{0}) is unramified at υ, we get

0 = r_{υ}(α_{F_{0}}(A_{0})) = Res_{κ(υ)/κ(υ_{1})}(r_{υ_{1}}(β)).

Since κ(υ)/κ(υ_{1}) is rational, the map Res_{κ(υ)/κ(υ_{1})} is injective, and we get that β is unramified at υ_{1}. By Proposition 1.2, we have

β = a_{0}·1 + a_{1}·(u) + a_{2}·(v) + a_{3}·(u) ∪ (v), a_{i} ∈ H^{∗}(F).

Hence we get

α_{F_{0}}(A_{0}) = a_{0}·1 + a_{1}·Res_{F_{0}/F_{1}}((u)) + a_{2}·Res_{F_{0}/F_{1}}((v)) + a_{3}·Res_{F_{0}/F_{1}}((u) ∪ (v)).

Let us consider the (v − 1)-adic valuation on F_{0}, and let F_{2} its residue field. Since A_{0} is unramified at this valuation, one may specialize A_{0} to a central simple F_{2}- algebra B. Then 2[A_{0}] specializes to 2[B]. But 2[A_{0}] = [(u, v)] specializes to 0, and therefore B is a biquaternion algebra. Then α_{F_{2}}(B) = 0 by assumption, and therefore

a_{0}·1 + a_{1}·Res_{F_{2}/F_{1}}((u)) = 0.

In other words, Res_{F_{2}/F_{1}}(a_{0}·1 + a_{1}·(u)) = 0. Since F_{1} ⊂ F_{2} ⊂ F_{0}, and F_{0}/F_{1} is rational, the map Res_{F_{2}/F_{1}} is injective and thus

a_{0}·1 + a_{1}·(u) = 0 in H^{∗}(F_{1}).

By Lemma 1.1, we get a_{0} = a_{1} = 0. Considering the (u − 1)-adic valuation yields that a_{2} = 0 in a similar way, so that

α_{F_{0}}(A_{0}) = a_{3}·Res_{F_{0}/F_{1}}((u) ∪ (v)).

Since Res_{F_{0}/F_{1}}((u) ∪ (v)) corresponds to [(u, v)] = 2[A_{0}] in Br_{2}(F_{0}) via the usual isomorphism, the previous equality rewrites

α_{F_{0}}(A_{0}) = a_{3}·f_{2,F_{0}}(A_{0}).

By Lemma 1.5, we get α = a_{3}·f_{2}. By assumption on α, we then have

0 = α_{F_{1}}({u, v}_{4,F_{1}}) = a_{3}·(u) ∪ (v) ∈ H^{∗}(F_{1}).

By Lemma 1.1, we get a_{3} = 0, and thus α = 0. This concludes the proof. □

We now describe the cohomological invariants of CSA_{4}. Let F be a field of characteristic different from 2. Assume that −1 ∈ F^{×2}, and let K/F be a field extension. By [5], for every central simple K-algebra A of degree 4, there exists a unique 2-fold Pfister form q_{2} and a unique 4-fold Pfister form q_{4} such that

q_{A} ∼ q_{2} + q_{4} ∈ W (K).

Taking the cohomology class e_{4}(q_{4}) ∈ H^{4}(K) yields a cohomological invariant of Inv(CSA_{4}, H^{∗}). This invariant restricts to zero on cyclic algebras, and its restriction to MS_{2,2} obviously coincide with the invariant e_{4} defined in Section 2. Therefore, we still denote this invariant by e_{4}.

Corollary 4.2. Let F be a field of characteristic different from 2. Assume that −1 ∈ F^{×2}. Then Inv(CSA_{4}, H^{∗}) is a free H^{∗}(F)-module with basis 1, f_{2}, e_{4}.

Proof. We first prove that 1, f_{2}, e_{4} are linearly independent. Assume that we have

a_{0}·1 + a_{1}·f_{2} + a_{2}·e_{4} = 0 for some a_{i} ∈ H^{∗}(F).

Since the restriction of e_{4} to MS_{4,1} is zero, it follows that a_{0} + a_{1}·f_{2} = 0 ∈ Inv(MS_{4,1}, H^{∗}). By Proposition 2.2 (1), we get a_{0} = a_{1} = 0. Thus, we get a_{2}·e_{4} = 0 ∈ Inv(CSA_{4}, H^{∗}). In particular, a_{2}·e_{4} = 0 ∈ Inv(MS_{2,2}, H^{∗}). By Proposition 2.2 (2), we get a_{2} = 0.

We now prove that 1, f_{2}, e_{4} span Inv(CSA_{4}, H^{∗}) as an H^{∗}(F)-module. Let α ∈ Inv(CSA_{4}, H^{∗}). By Proposition 2.2, we have

α_{|_{MS_{2,2}}} = a_{0}·1 + a_{1}·f_{2} + a_{2}·e_{4},

for some a_{i} ∈ H^{∗}(F), and also

α_{|_{MS_{4,1}}} = b_{0}·1 + b_{1}·f_{2},

for some b_{i} ∈ H^{∗}(F). Since M_{4}(F) ≃ {1, 1}_{4,F} ≃ (1, 1) ⊗_{F} (1, 1), we may apply M_{4}(F) to both equalities to get a_{0} = b_{0}.

Let u, v be two independent indeterminates over F. We have

M_{2}((u, v)) ≃ (1, 1) ⊗_{F (u,v)} (u, v) ≃ {u^{2}, v^{2}}_{4,F (u,v)},

and applying M_{2}((u, v)) to both equalities yields

a_{1}·(u) ∪ (v) = b_{1}·(u) ∪ (v) ∈ H^{∗}(F (u, v)).

It follows from Lemma 1.1 that a_{1} = b_{1}. Since e_{4} is zero on MS_{4,1}, we conclude that α and a_{0}·1 + a_{1}·f_{2} + a_{2}·e_{4} coincide on MS_{4,1} and MS_{2,2}. By the previous theorem, we get α = a_{0}·1 + a_{1}·f_{2} + a_{2}·e_{4}, and this concludes the proof. □

References

[1] A.A.Albert, Structure of Algebras. A.M.S. Coll.Pub. 24, New York, A.M.S., 1939.

[2] R. S. Garibaldi, A. S. Merkurjev, J.-P. Serre, Cohomological invariants in Galois cohomology, University Lecture Series 28, A.M.S., Providence, RI, 2003.

[3] M.-A. Knus, A.S. Merkurjev, M. Rost, J.-P. Tignol, The book of involutions, A.M.S. Coll. Pub. 44, 1998.

[4] D. W. Lewis, J. Morales, The Hasse invariant of the trace form of a central simple algebra. Pub. Math. Besancon, Théorie des nombres, Année 1993-94, 1–6.

[5] M. Rost, J.-P. Serre, J.-P. Tignol, La forme trace d’une algèbre simple centrale de degré 4. C.R. Acad. Sci. Paris, Sér. I 342 (2006), 83–87.

Email: berhuy@ujf-grenoble.fr

Address: Université Joseph Fourier, UFR de Mathématiques, Institut Fourier 100 rue des maths, BP 74, F-38402 St Martin d’Hères Cedex, France