ON THE CLASSIFICATION OF ALBERT ALGEBRAS

MARKUS ROST

                         preliminary version

1. Introduction

This is a very preliminary text. Our goal is to prove Corollary 4 at the very end of this text.

I am indebted to H. Petersson for comments on earlier versions of this text.

2. Jordan symbol algebras of degree 3

Let R be a commutative ring in which 6 is invertible. Let ζ ∈ R with ζ^{2}+ζ+1 = 0 and let a, b, c ∈ R. The aim of this section is to define Jordan algebras J_{ζ}(R; a, b, c) generalizing the first Tits construction of Albert algebras (=exceptional Jordan algebras) over fields to possibly degenerate Jordan algebras over rings. The condition 1/2 ∈ R is probably not necessary for this, but I haven’t thought about details.

2.1. First Notations. J(R; a) = R[X]/(X^{3} − a) is the cubic Kummer extension associated to a. The algebra J(R; a) is a free R-module with basis X^{i}, 0 ≤ i ≤ 2.

A_{ζ}(R; a, b) is the associative R-algebra with generators X, Y and relations X^{3} = a, Y^{3} = b and Y X = ζXY. The algebra A_{ζ}(R; a, b) is a free R-module with basis X^{i}Y^{j}, 0 ≤ i, j ≤ 2.

J(R; a, b) = A_{ζ}(R; a, b)^{+} is the Jordan R-algebra associated to A_{ζ}(R; a, b). The algebra J(R; a, b) is a free R-module with basis X^{i} · Y^{j}, 0 ≤ i, j ≤ 2, where x · y = (xy + yx)/2 is the Jordan product. Since A_{ζ}(R; a, b)^{op} = A_{ζ^{2}}(R; a, b), one has J(R; a, b) = A_{ζ^{2}}(R; a, b)^{+}.

2.2. The first Tits construction. It is possible to define the Jordan algebras J_{ζ}(R; a, b, c) in terms of generators and relations, see proposition 1. However to carry out the details is somewhat tedious. Therefore we use an indirect way by referring to the first Tits construction over fields.

Let F be a field of characteristic different from 2, 3. Moreover let A be a central simple F -algebra of degree 3 and let c ∈ F^{∗}, let Nrd, Trd: A → F be the reduced norm and reduced trace of A and let ×: A × A → A be the symmetric bilinear product defined by (x × x)x = Nrd(x). For x ∈ A, set \overline{x} = \frac{1}{2}(Trd(x) − x). Tits [2, 5] defined an Albert algebra

J(A, c) = A_{0} ⊕ A_{1} ⊕ A_{2}

Date: September 3, 2002, last revision: October 5, 2002.

as the sum of three copies of A with the product

x_{0}  y_{1}  z_{2}
x′_{0} \frac{1}{2}(xx′ + x′x)_{0} (\overline{x′}y)_{1}  (z\overline{x′})_{2}
y′_{1}  (\overline{x}y′)_{1}  c(y × y′)_{2} (\overline{y′}z)_{0}
z′_{2}  (z′\overline{x})_{2}  (\overline{yz′})_{0}  \frac{1}{c}(z × z′)_{1}

Now assume that F contains a primitive cube root ζ of 1 and let a, b, c ∈ F^{∗}. We put

J_{ζ}(F; a, b, c) = J(A_{ζ}(a, b), c)

Note that the algebra J(F; a, b) is a subalgebra of J_{ζ}(F; a, b, c) via x ↦ (x, 0, 0). Let

X_{1} = (X, 0, 0), X_{2} = (Y, 0, 0), X_{3} = (0, 1, 0).

Let R_{0} = Z[1/6, ζ, a, b, c]/(ζ^{2} + ζ + 1) and let F_{0} = Q(μ_{3})(a, b, c) be the fraction field of R_{0}. Let

J_{ζ}(R_{0}; a, b, c) ⊂ J_{ζ}(F_{0}; a, b, c)

be the R_{0}-subalgebra generated by X_{1}, X_{2}, X_{3}.

Now let R be a commutative ring in which 6 is invertible, let ζ ∈ R with ζ^{2} +ζ + 1 = 0 and let a_{1}, a_{2}, a_{3} ∈ R. We define J_{ζ}(R, a_{1}, a_{2}, a_{3}) as the R-algebra obtained via specialization from the universal example: Let R_{0} → R be the homomorphism sending ζ, a, b, c to the elements ζ, a_{1}, a_{2}, resp. a_{3} of R and let

J_{ζ}(R, a_{1}, a_{2}, a_{3}) = J_{ζ}(R_{0}, a, b, c) ⊗_{R_{0}} R

Let Γ = (Z/3)^{3} and let ω: Alt(3, Γ) → Z/3 be the isomorphism with

ω((1, 0, 0) ∧ (0, 1, 0) ∧ (0, 0, 1)) = 1

For r ∈ Γ we denote by r_{1}, r_{2}, r_{3} ∈ Z the integers with 0 ≤ r_{i} ≤ 2 and r = (r_{1}, r_{2}, r_{3}) mod 3.

For r ∈ Γ let L_{r} be the R-submodule of J_{ζ}(R; a_{1}, a_{2}, a_{3}) generated by the element (X_{1}^{r_{1}}X_{2}^{r_{2}})X_{3}^{r_{3}}.

Proposition 1. The R-modules L_{r} are free of rank 1. One has

J_{ζ}(R; a_{1}, a_{2}, a_{3}) = ⨁_{r∈Γ} L_{r}

and L_{r}L_{s} ⊂ L_{r+s}.

The R-algebra J_{ζ}(R; a_{1}, a_{2}, a_{3}) is generated by X_{1}, X_{2}, X_{3}.

For r ∈ Γ one has

((X_{1}^{r_{1}}X_{2}^{r_{2}})X_{3}^{r_{3}})^{3} = \frac{1}{8^{n}} a_{1}^{r_{1}}a_{2}^{r_{2}}a_{3}^{r_{3}}

where n = 2 if the r_{i} are all nonzero, n = 1 if exactly one of the r_{i} is equal to 0 and n = 0 otherwise.

Let r, s, t ∈ Γ be linearly independent elements (i. e., r ∧ s ∧ t ≠ 0) and let U ∈ L_{r}, V ∈ L_{s}, W ∈ L_{t}. Then

U^{2}(UV) = U(U^{2}V) = \frac{1}{4}U^{3}V
U(UV) = −\frac{1}{2}U^{2}V
(UV)(UW) = −\frac{1}{2}U^{2}(V W)
(UV)(U^{2}W) = \frac{1}{4}U^{3}(V W)
(UV)W = ζ^{ω(r∧s∧t)}(UW)V

Proof. This all follows by inspection of the product in J_{ζ}(F_{0}; a, b, c). □

The significance of proposition 1 is that it provides a complete description of J_{ζ}(R; a_{1}, a_{2}, a_{3}) as commutative (but not necessarily associative) Γ-graded Ralgebra with generators X_{1}, X_{2}, X_{3} and certain relations. We mention one particular consequence:

Corollary 1.

J_{ζ}(R; a_{3}, a_{1}, a_{2}) ≃ J_{ζ}(R; a_{1}, a_{2}, a_{3})
J_{ζ}(R; a_{2}, a_{1}, a_{3}) ≃ J_{ζ^{2}}(R; a_{1}, a_{2}, a_{3})

3. The H^{3}(Z/3)-invariant

We summarize results from [6, 10]. See also [5].

Proposition 2. For Albert algebras J over fields F (of characteristic different from 2, 3) there exist a unique cohomological invariant g_{3}(J) ∈ H^{3}(F, Z/3) such that: For a field F containing a primitive cube root ζ of 1 and for a_{1}, a_{2}, a_{3} ∈ F^{∗} one has

g_{3}(J_{ζ}(F; a_{1}, a_{2}, a_{3})) ⊗ ζ = (a_{1}) ∪ (a_{2}) ∪ (a_{3})

in H^{3}(F, Z/3) ⊗ μ_{3} = H^{3}(F, μ_{3}^{⊗3}) (note that μ_{3} ⊗ μ_{3} ≡ Z/3).

4. F_{4}-torsors

We assume that R contains a subfield k with ζ ∈ k. The algebraic group F_{4} = Aut(J_{ζ}(k; 1, 1, 1)) (= group of k-algebra automorphisms) is a split group over k of type F_{4}.

Lemma 1. Let a_{1}, a_{2}, a_{3} ∈ R be invertible.

Then Isom(J_{ζ}(R; 1, 1, 1), J_{ζ}(R; a_{1}, a_{2}, a_{3})) is an F_{4}-torsor over R (in the etale topology).

Proof. It suffices to note that after adjoining cube roots of the a_{i} the Jordan Ralgebras J_{ζ}(R; a_{1}, a_{2}, a_{3}) and J_{ζ}(R; 1, 1, 1) become isomorphic. □

We will use the following corollary of the theorem of Raghunathan-Ramanathan [9] (together with a theorem of Steinberg) applied to F_{4}-torsors over the affine line:

Proposition 3. Let k be perfect and let J be a algebra over k[t] which is locally (in the etale topology) isomorphic to the split Albert algebra. Then J ≃ J_{0} ⊗ k[t] for some Albert algebra J_{0} over k.

By Lemma 1, this condition is satisfied if for every irreducible polynomial P ∈ k[t] there exists a_{1}, a_{2}, a_{3} ∈ (k[t]_{(P)})^{∗} with J ⊗ k[t]_{(P)} ≃ J_{ζ}(k[t]_{(P)}; a_{1}, a_{2}, a_{3}).

  Remark added on Sept. 10:

Actually, we will need only the following:

Proposition 4. Let char k ≠ 3 with μ_{3} ⊂ k, let J be an Albert divsion algebra of first Tits construction over k, and let J′ ⊂ J ⊗ k(t) be a k[t]-subalgebra which is locally (in the etale topology) isomorphic to the split Albert algebra. Then J′ ≃ J ⊗ k[t].

Proof. (Sketch) One uses, I guess, Harder’s method used for quadratic forms. The argument simplifies for the anisotropic case.

Extend the order J′ to an order J̄ on the projective line by taking the order J ⊗k[1/t] at infinity. Let x ∈ H^{0}(P^{1}, Hom(O_{P^{1}}(1), J̄)) be a global section. Since J′ is a vector bundle (in fact, a direct sum of line bundles O_{P^{1}}(r), [1, p. 516]), one has H^{0}(P^{1}, Hom(O_{P^{1}}(n), J̄)) = 0 for n ≫ 0. Thus x is nilpotent, and since J(t) is a field, it follows x = 0. Hence H^{0}(P^{1}, Hom(O_{P^{1}}(1), J̄)) = 0. This shows that J̄ is trivial as a vector bundle, since the vector bundle J̄ is self dual (via the trace form). Thus J_{0} = H^{0}(P^{1}, J̄) is an Albert algebra over k with J̄ = J_{0} ⊗ O_{P^{1}}. It suffices to note J_{0} = J̄(∞) = J. □

5. Orders

Let R = k[t] and F = k(t). Let further f_{1}, f_{2}, f_{3} ∈ R be nonzero polynomials. Then

J_{ζ}(R; f_{1}, f_{2}, f_{3}) ⊂ J_{ζ}(F; f_{1}, f_{2}, f_{3})

is an R-order (see [3, 4, 8]). Our goal in this section is to show that if J_{ζ}(F; f_{1}, f_{2}, f_{3}) is unramified on the affine line, then there exists a regular R-order J′ with

J_{ζ}(R; f_{1}, f_{2}, f_{3}) ⊂ J′ ⊂ J_{ζ}(F; f_{1}, f_{2}, f_{3})

[By a regular order we understand here an order which is locally (in the etale topology) isomorphic to a split Albert algebra. For a regular order the quadratic form is nondegenerate. The converse should be also true, but we won’t need this.]

By proposition 3 it follows then that if J_{ζ}(F; f_{1}, f_{2}, f_{3}) ≃ J_{0}(t) for some Albert algebra J_{0} over k, then J_{ζ}(R; f_{1}, f_{2}, f_{3}) admits an embedding into J_{0} ⊗k[t] (at least if k is perfect). This gives rise to some sort of Cassels-Pfister theorem for Albert algebras.

One is tempted to take for J′ a maximal R-order, as in the case of associative central simple algebras or of quadratic forms. However Knebusch [4] has given an example of a maximal order of an Albert algebra which is not regular. It is not clear to me whether such phenomena can appear in our specific applications. Anyway, we will construct J′ locally by means of a sequence of elementary operations on the generators of Jordan symbol algebras.

Let v: F^{∗} → Z be a discrete valuation, let R be the valuation ring, let κ be the residue field and let π be a prime element of v. (F can be any field with char F ≠ 2, 3 and μ_{3} ⊂ F; we assume also char κ ≠ 3). Let

∂_{v}: H^{3}(F, μ_{3}) → H^{2}(κ, Z/3)

be the residue map.

Proposition 5. Let f_{1}, f_{2}, f_{3} ∈ R be nonzero elements with

∂_{v}((f_{1}) ∪ (f_{2}) ∪ (f_{3})) = 0

Then there exists an R-order

J_{ζ}(R; f_{1}, f_{2}, f_{3}) ⊂ J′ ⊂ J_{ζ}(F; f_{1}, f_{2}, f_{3})

such that J′ ≃ J_{ζ}(R; g_{1}, g_{2}, g_{3}) with g_{i} ∈ R^{∗}.

Remark 1. Actually, what we will need in the proof is that J_{ζ}(F; f_{1}, f_{2}, f_{3}) is unramified, that is, it is extended from some regular Albert algebra over R. In the application this will be obvious, since then J_{ζ}(F; f_{1}, f_{2}, f_{3}) is extended from an Albert algebra over the ground field.

We will also use the following fact: If J_{ζ}(F; f_{1}, f_{2}, f_{3}) is unramified, and if f_{1} is a prime element for R and f_{2}, f_{3} are R-units, then the reduction J(f̄_{2}, f̄_{3}) of J(f_{2}, f_{3}) is the split algebra.

Moreover, in this case, the norm of J(f_{2}, f_{3}) represents a prime element.

Namely, if J(f̄_{2}, f̄_{3}) is split, then f̄_{2} = N_{L̄/κ}(λ̄) where L = R[t]/(t^{3} − f_{3}) and λ ∈ L^{∗}. After replacing f_{2} by f_{2}N_{L/κ}(λ)^{−1} we may assume that f̄_{2} = 1. Then, for e ∈ R and a prime element π one has with S = R[s]/(s^{3} − f_{2}) ⊂ J(f_{2}, f_{3})

N_{S/R}(1 + eπ − s) = (1 + eπ)^{3} − f_{2} ≡ π(3e + \frac{1 − f_{2}}{π} ) mod π^{2}

It is clear that for e = 0 or for e = 1 the right hand side is nonzero. Then N(1 + eπ − s) is a prime element.

Proof of Proposition 5. Let X_{1}, X_{2}, X_{3} be the generators of J_{ζ}(R; f_{1}, f_{2}, f_{3}).

We first reduce to the case v(f_{2}) = v(f_{3}) = 0. Suppose that v(f_{1}) ≥ v(f_{2}) > 0. Then X_{1}X_{2}^{−1}, X_{2}, X_{3} generate an order J′ isomorphic to J_{ζ}(R; f_{1}f_{2}^{−1}, f_{2}, f_{3}). Moreover J_{ζ}(R; f_{1}, f_{2}, f_{3}) is contained in J′ since X_{1} = 4(X_{1}X_{2}^{−1})X_{2}, see proposition 1. Iterating this argument, possibly using the symmetry relations of corollary 1, one may indeed arrange v(f_{2}) = v(f_{3}) = 0.

Now write v(f_{1}) = r + 3m with 0 ≤ r ≤ 2 and m ≥ 0. Then X_{1}π^{−m}, X_{2}, X_{3} generate an order J′ isomorphic to J_{ζ}(R; f_{1}π^{−3m}, f_{2}, f_{3}) and containing J_{ζ}(R; f_{1}, f_{2}, f_{3}). We may therefore assume m = 0.

Suppose v(f_{1}) = 0. Then we may (and must) take J′ = J_{ζ}(R; f_{1}, f_{2}, f_{3}).

Suppose v(f_{1}) = 2. Then X_{1}^{2}π^{−1}, X_{2}, X_{3} generate an order J′ isomorphic to J_{ζ^{2}}(R; f_{1}^{2}π^{−3}, f_{2}, f_{3}). Moreover J_{ζ}(R; f_{1}, f_{2}, f_{3}) is contained in J′ since X_{1} = (X_{1}^{2}π^{−1})^{2}u with u = π^{2}/f_{1} ∈ R^{∗}, see proposition 1. Since v(f_{1}^{2}π^{−3}) = 1, we are reduced to the case v(f_{1}) = 1.

Finally suppose v(f_{1}) = 1. Then

(f̄_{2}) ∪ (f̄_{3}) = ∂_{v}((f_{1}) ∪ (f_{2}) ∪ (f_{3})) = 0

Hence A_{ζ}(R; f_{2}, f_{3}) is an Azumaya algebra over R which is split over the residue class field. It follows that the reduced norm of A_{ζ}(R; f_{2}, f_{3}) represents a prime element. Hence there exist an element α in the subalgebra of J_{ζ}(R; f_{1}, f_{2}, f_{3}) generated by X_{2}, X_{3} with v(N(α)) = 1. Let J′ be the order generated by X′_{1} = X_{1}(T (α) − 2α)^{−1} and X_{2}, X_{3}. Note that N(X′_{1}) = N(X_{1})N(α)^{−1} = f_{1}N(α)^{−1} and therefore J′ ≃ J_{ζ}(R; f_{1}N(α)^{−1}, f_{2}, f_{3}). Clearly f_{1}N(α)^{−1} ∈ R^{∗}. Moreover one has X_{1} = X′_{1}(T (α) − 2α) and therefore J′ contains J_{ζ}(R; f_{1}, f_{2}, f_{3}). □

Now let again R = k[t] and F = k(t).

Proposition 6. Let f_{1}, f_{2}, f_{3} ∈ R be nonzero elements with

(f_{1}) ∪ (f_{2}) ∪ (f_{3}) ∈ H^{3}(k, μ_{3}) ⊂ H^{3}(F, μ_{3})

Then there exists a regular R-order

J_{ζ}(R; f_{1}, f_{2}, f_{3}) ⊂ J′ ⊂ J_{ζ}(F; f_{1}, f_{2}, f_{3})

Remark 2. Actually, what we will need in the proof is that J_{ζ}(F; f_{1}, f_{2}, f_{3}) is everywhere unramified. In the application this will be obvious, since then J_{ζ}(F; f_{1}, f_{2}, f_{3}) is extended from an Albert algebra over the ground field.

Proof. Take the regular order J_{ζ}(R[(f_{1}f_{2}f_{3})^{−1}]; f_{1}, f_{2}, f_{3}) and extend it to the affine line using proposition 5. □

Corollary 2. Let a_{1}, a_{2}, a_{3} ∈ k^{∗} and let f_{1}, f_{2}, f_{3} ∈ k[t] be nonzero elements with

J_{ζ}(k(t); f_{1}, f_{2}, f_{3}) ≃ J_{ζ}(k(t); a_{1}, a_{2}, a_{3})

Then there exists a homomorphism

J_{ζ}(k[t]; f_{1}, f_{2}, f_{3}) → J_{ζ}(k[t]; a_{1}, a_{2}, a_{3})

of k[t]-algebras. □

6. A Cassels-Pfister type theorem for Albert algebras

Let J be a Albert algebra over k of first Tits construction. We write J(t) = J ⊗_{k} k(t) and J[t] = J ⊗_{k} k[t]. We assume μ_{3} ⊂ k.

We recall the following fact [2], [7, Corollary 3]:

Lemma 2. Let a ∈ k^{∗} and suppose that J ⊗ k(\sqrt[3]{a}) has zero divisors. Then there exist an element x ∈ J with T (x) = T (x^{2}) = 0 and N(x) = a.

Theorem 1. Let f ∈ k[t] and suppose that there exists α ∈ J(t) with T (α) = T (α^{2}) = 0 and N(α) = f. Then there exists α ∈ J[t] with T (α) = T (α^{2}) = 0 and N(α) = f.

Proof. There exists g, h ∈ k(t)^{∗} such that

J(t) ≃ J_{ζ}(k(t); f, g, h)

This follows from the fact that the invariant f_{3}(J) ∈ H^{3}(k, Z/2) is trivial. See the proof of [5, Proposition (40.5)].

After multiplication with cubes, we may assume that g, h ∈ k[t]. The claim follows from corollary 2. □

Corollary 3. Let a, b ∈ k^{∗} and suppose that J has zero divisors over the function field of the curve

r^{3} + at^{3} + b = 0

Then there exist x, y ∈ J such that

T (x) = T (x^{2}) = 0
N(x) = a
T (y) = T (y^{2}) = 0
N(y) = b
T (xy) = T (yx^{2}) = T (xy^{2}) = 0

Remark 3. Note that if we have the additional relation T (x^{2}y^{2}) = 0, then x, y would be the standard generators of a subalgebra J(a, b) of J.

Proof. We may assume that J is a division algebra. By Lemma 2 and Theorem 1 there exists α ∈ J[t] with T (α) = T (α^{2}) = 0 and N(α) = at^{3} + b. Since J is a division algebra, one must have deg(α) = 1. Write α = xt + y. The claimed properties are now easy to verify, by noting that N(α) = α^{3}, whence

(xt + y)^{3} = at^{3} + b

and therefore

x^{2}y + xyx + yx^{2} = y^{2}x + yxy + xy^{2} = 0
□

7. The main result

Let a, b ∈ k^{∗} and let K = k(s, t)[u] with u^{3} = at^{3} + b(1 − as^{3}). Then

b = \frac{u^{3} − at^{3}}{1 − as^{3}} = N_{K⊗k_{a}/K} (\frac{u − αt}{1 − αs})

with k_{a} = k[α], α^{3} = a. In particular, the algebras A_{ζ}(a, b) and J(a, b) and the symbol (a) ∪ (b) become trivial over K.

Remark 4 (We don’t need this in the following). The field K is the function field of the Severi-Brauer Y variety of A_{ζ}(a, b). Indeed, K is the function field of the cubic surface

u^{3} − at^{3} + bv^{3} − abw^{3} = 0

which turns out to be the blow up of Y in specific subschemes Spec k_{a} → Y, Spec k_{b} → Y (recall that the blow up of P^{2} in 6 points is a cubic surface).

Theorem 2. Let J be a Albert algebra over k (char k ≠ 2, 3) and assume that J has zero divisors over K. Then there exists a finite field extension ℓ of k of degree prime to 3 and an embedding of J(a, b) into J over ℓ.

In particular, if ℓ contains a primitive root ζ of 1, then

J_{ℓ} ≃ J_{ζ}(a, b, c)

for some c ∈ ℓ^{∗} (in fact, one may choose c ∈ k^{∗}).

Proof. We may assume that k has no proper extensions of degree prime to 3. This means that every irreducible polynomial over k has degree divisible by 3 or is linear.

We may further assume that J is a Jordan field.

By Corollary 3, there exist x, y ∈ J(s) such that

T (x) = T (x^{2}) = 0
N(x) = a
T (y) = T (y^{2}) = 0
N(y) = b(1 − as^{3})
T (xy) = T (yx^{2}) = T (xy^{2}) = 0

It suffices (see Remark 3) to find x_{0}, y_{0} ∈ J such that

T (x_{0}) = T (x_{0}^{2}) = 0
N(x_{0}) = a
T (y_{0}) = T (y_{0}^{2}) = 0
N(y_{0}) = bN_{k_{a}/k}(λ) for some λ ∈ k_{a}^{∗}
T (x_{0}y_{0}) = T (y_{0}x_{0}^{2}) = T (x_{0}y_{0}^{2}) = 0
T (x_{0}^{2}y_{0}^{2}) = 0

Write

x = \frac{X}{P} , y = \frac{Y}{Q}

where X, Y ∈ J[s] and P, Q ∈ k[s] with X, P resp. Y, Q relatively prime.

Then N(X) = aP^{3}. Let P′ be an irreducible factor of P. Then X mod P′ is nonzero and a zero divisor of J over k[s]/(P′). Since J is a field, P′ cannot be linear. By the assumption made at the beginning of this proof, it follows that P′ has degree divisible by 3. Hence deg P = 3n. Since J is a field, it follows also that deg X = deg P.

Similarly one sees from N(Y) = b(1 − as^{3})Q^{3} that each factor of Q has degree divisible by 3 and that deg Y = 1 + deg Q = 1 + 3m.

Consider the polynomial f(s) = T (X^{2}Y^{2}). One has deg f ≤ d with d = 2(1 + 3m + 3n).

If deg f = d, then there exists a zero s_{0} of f, by our assumption on k. In this case we take x_{0} = x(s_{0}) and y_{0} = y(s_{0}).

If deg f < d, we take for x_{0} and y_{0} the leading coefficients of X, Y (after arranging that P, Q have leading coefficient 1). □

By Springer’s theorem (and other things) we get

Corollary 4. Let k be a field with char k ≠ 2, 3 and let J, J′ be Albert algebras over k with isomorphic trace forms and with g_{3}(J) = g_{3}(J′). Then there exist a finite field extension F of k of degree prime to 3 and a finite field extension L of k of degree dividing 3 such that J_{F} ≃ J′_{F} and J_{L} ≃ J′_{L}.

Of course one would like to know J ≃ J′ over k.

References

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[3] M. Knebusch, Der Begriff der Ordnung einer Jordanalgebra, Abh. Math. Sem. Univ. Hamburg 28 (1965), 168–184.

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[5] M.-A. Knus, A. Merkurjev, M. Rost, and J.-P. Tignol, The book of involutions, American Mathematical Society, Providence, RI, 1998, With a preface in French by J. Tits.

[6] H. P. Petersson and M. L. Racine, An elementary approach to the Serre-Rost invariant of Albert algebras, Indag. Math. (N.S.) 7 (1996), no. 3, 343–365.

[7] , Cubic subfields of exceptional simple Jordan algebras, Proc. Amer. Math. Soc. 91 (1984), no. 1, 31–36.

[8] M. L. Racine, The arithmetics of quadratic Jordan algebras, American Mathematical Society, Providence, R.I., 1973, Memoirs of the American Mathematical Society, No. 136.

[9] M. S. Raghunathan and A. Ramanathan, Principal bundles on the affine line, Proc. Indian Acad. Sci. Math. Sci. 93 (1984), no. 2-3, 137–145.

[10] M. Rost, A (mod 3) invariant for exceptional Jordan algebras, C. R. Acad. Sci. Paris Sér. I Math. 313 (1991), no. 12, 823–827.

Department of Mathematics, The Ohio State University, 231 W 18th Avenue, Columbus, OH 43210, USA

E-mail address: rost@math.ohio-state.edu
URL: http://www.math.ohio-state.edu/~rost