On The Tensor Product of Two Composition Algebras

Patrick J. Morandi  S. Pumplün

1 Introduction

Let C_{1} ⊗_{F} C_{2} be the tensor product of two composition algebras over a field F with char(F) ≠ 2. R. Brauer [7] and A. A. Albert [1], [2], [3] seemed to be the first mathematicians who investigated the tensor product of two quaternion algebras. Later their results were generalized to this more general situation by B. N. Allison [4], [5], [6] and to biquaternion algebras over rings by Knus [12].

In the second section we give some new results on the Albert form of these algebras. We also investigate the F -quadric defined by this Albert form, generalizing a result of Knus ([13]).

Since Allison regarded the involution σ = γ_{1} ⊗ γ_{2} as an essential part of the algebra C = C_{1} ⊗ C_{2}, he only studied automorphisms of C which are compatible with σ. In the last section we show that any automorphism of C that preserves a certain biquaternion subalgebra also is compatible with σ. As a consequence, if C is the tensor product of two octonion algebras, we show that C does not satisfy the Skolem-Noether Theorem.

Let F be a field and C a unital, nonassociative F -algebra. Then C is a composition algebra if there exists a nondegenerate quadratic form n : C → F such that n(x · y) = n(x)n(y) for all x, y ∈ C. The form n is uniquely determined by these conditions and is called the norm of C. We will write n = n_{C}. Composition algebras only exist in ranks 1, 2, 4 or 8 (see [10]). Those of rank 4 are called quaternion algebras, and those of rank 8 octonion algebras. A composition algebra C has a canonical involution γ given by γ(x) = t(x)1_{C} − x, where the trace map t : C → F is given by t(x) = n(1, x).

An example of an 8-dimensional composition algebra is Zorn’s algebra of vector matrices Zor(F) (see [14, p. 507] for the definition). The norm form of Zor(F) is given by the determinant and is a hyperbolic form.

Composition algebras are quadratic; i.e., they satisfy the identities

x^{2} − t(x)x + n(x)1_{C} = 0 for all x ∈ C,
n(1_{C}) = 1,

and are alternative algebras; i.e., xy^{2} = (xy)y and x^{2}y = x(xy) for all x, y ∈ C. In particular, n(x) = γ(x)x = xγ(x) and t(x)1_{C} = γ(x) + x.

For any composition algebra D over F with dim_{F} (D) ≤ 4, and any μ ∈ F^{×}, the F -vector space D ⊕ D becomes a composition algebra via the multiplication

(u, v)(u′, v′) = (uu′ + μγ(v′)v, v′u + vγ(u′))

for all u, v, u′, v′ ∈ D, with norm

n((u, v)) = n_{D}(u) − μn_{D}(v).

This algebra is denoted by Cay(D, μ). Note that the embedding of D into the first summand of Cay(D, μ) is an algebra monomorphism. The norm form of Cay(D, μ) is obviously isometric to ⟨1, −μ⟩ ⊗ n_{D}. Since two composition algebras are isomorphic if and only if their norm forms are isometric, we see that if C is a composition algebra whose norm form satisfies n_{C} ≅ ⟨1, −μ⟩ ⊗ n_{D} for some D, then C ≅ Cay(D, μ). In particular, Zor(F) ≅ Cay(D, 1) for any quaternion algebra D since ⟨1, −1⟩ ⊗ n_{D} is hyperbolic. A composition algebra is split if it contains an isomorphic copy of F ⊕ F as a composition subalgebra, which is the case if and only if it contains zero divisors.

2 Albert Forms

From now on we consider only fields F with char(F) ≠ 2. It is well known that any norm of a composition algebra is a Pfister form, and conversely, any Pfister form is the norm of some composition algebra.

Let C be a composition algebra. Define C′ = ⟨F 1⟩^{⊥} = {x ∈ C : t_{1}(x) = n_{1}(x, 1) = 0}. Then n′ = n|_{C′} is the pure norm of C. Note that

C′ = {x ∈ C : x = 0 or x ∉ F 1_{C} and x^{2} ∈ F 1_{C}}
= {x ∈ C : γ(x) = −x}.

Moreover, C is split if and only if its norm n is hyperbolic, two composition algebras are isomorphic if and only if their norms are isometric, and C is a division algebra if and only if n is anisotropic.

We now investigate tensor products of two composition algebras. Following Albert, we associate to the tensor product C = C_{1} ⊗ C_{2} of two composition algebras with dim(C_{i}) = r_{i} and n_{C_{i}} = n_{i} the (r_{1} + r_{2} − 2)-dimensional form n′_{1} ⊥ ⟨−1⟩n′_{2} of determinant −1. This definition, for C_{1} or C_{2} an octonion algebra, was first given by Allison in [5]. In the Witt ring W (F), obviously this form is equivalent to n_{1} −n_{2}. Like the norm form of a composition algebra, this Albert form contains crucial information about the tensor product algebra C. For biquaternion algebras, this is well-known ([1, Thm. 3], [11, Thm. 3.12]). We introduce some notation and terminology. If q is a quadratic form and if ℍ = ⟨1, −1⟩ is the hyperbolic plane, then q = q_{0} ⊥ iℍ for some anisotropic form q_{0} and integer i. The integer i is called the Witt index of q and is denoted i_{W} (q). In the proof of the following proposition, we use the notion of linkage of Pfister forms (see [8, Sec. 4]). Recall that two n-fold Pfister forms q_{1} and q_{2} are r-linked if there is an r-fold Pfister form h with q_{i} = h ⊗ q′_{i} for some Pfister forms q′_{i}. Finally, we call a two-dimensional commutative F -algebra that is separable over F a quadratic étale algebra. Note that any quadratic étale algebra is either a quadratic field extension of F or is isomorphic to F ⊕ F. Part of the following result has been proved in [9, Thm. 5.1].

Proposition 2.1. Let C_{1} and C_{2} be octonion algebras over F with norms n_{1} and n_{2}, and let i = i_{W} (N) be the Witt index of the Albert-form N = n′_{1} ⊥ ⟨−1⟩n′_{2}.

(i) i = 0 ⟺ C_{1} and C_{2} do not contain isomorphic quadratic étale subalgebras.

(ii) i = 1 ⟺ C_{1} and C_{2} contain isomorphic quadratic étale subalgebras, but no

isomorphic quaternion subalgebras.

(iii) i = 3 ⟺ C_{1} and C_{2} contain isomorphic quaternion subalgebras, but C_{1} and C_{2} are not isomorphic.

(iv) i = 7 ⟺ C_{1} ≅ C_{2}.

Proof. By [8, Props. 4.4, 4.5], the Witt index of n_{1} ⊥ ⟨−1⟩ n_{2} is either 0 or 2^{r}, where r is the linkage number of n_{1} ⊥ ⟨−1⟩n_{2}. Note that the Witt index of N is one less than the Witt index of n_{1} ⊥ ⟨−1⟩n_{2} since n_{1} ⊥ ⟨−1⟩n_{2} = ℍ ⊥ N. If C_{1} ≅ C_{2}, then n_{1} ≅ n_{2}, so i = 7. Conversely, if i = 7, then n_{1} ⊥ ⟨−1⟩n_{2} is hyperbolic, so n_{1} ≅ n_{2}, which forces C_{1} ≅ C_{2}. If C_{1} and C_{2} are not isomorphic but contain a common quaternion algebra Q, then C_{i} = Cay(Q, μ_{i}) for some i. Therefore, n_{1} = n_{Q} ⊗ ⟨1, −μ_{1}⟩ and n_{2} = n_{Q} ⊗ ⟨1, −μ_{2}⟩. These descriptions show that n_{1} and n_{2} are 2-linked, so i = 3. Conversely, if i = 3, then n_{1} and n_{2} are 2-linked but not isometric. If ⟨⟨a, b⟩⟩ is a factor of both n_{1} and n_{2}, then n_{1} = ⟨⟨a, b, c⟩⟩ and n_{2} = ⟨⟨a, b, d⟩⟩ for some c, d ∈ F^{×}. If Q = (−a, −b), we get C_{1} = Cay(Q, −c) and Cay(Q, −d), so C_{1} and C_{2} contain a common quaternion algebra. If C_{1} and C_{2} contain a common quadratic étale algebra F [t]/(t^{2} − a) but no common quaternion algebra, then ⟨1, −a⟩ is a factor of n_{1} and n_{2}, which means they are 1-linked. If n_{1} and n_{2} are 2-linked, then the previous step shows that C_{1} and C_{2} have a common quaternion subalgebra, which is false. Conversely, if n_{1} and n_{2} are 1-linked but not 2-linked, then C_{1} and C_{2} do not have a common quaternion subalgebra, and if ⟨⟨a⟩⟩ is a common factor to n_{1} and n_{2}, then C_{1} and C_{2} both contain the étale algebra F [t]/(t^{2} − a).

Proposition 2.2. Let C_{1} be an octonion algebra over F and C_{2} be a quaternion algebra over F, with norms n_{1} and n_{2}. Again consider the Witt index i of the Albert form N = n′_{1} ⊥ ⟨−1⟩n′_{2}.

(i) i = 0 ⟺ C_{1} and C_{2} do not contain isomorphic quadratic étale subalgebras.

(ii) i = 1 ⟺ C_{1} and C_{2} contain isomorphic quadratic étale subalgebras, but C_{2} is

not a quaternion subalgebra of C_{1}.

(iii) i = 3 ⟺ C_{1} ≅ Cay(C_{2}, μ) for a suitable μ ∈ F^{×}, and C_{2} is a division algebra.

(iv) i = 5 ⟺ C_{1} ≅ Zor(F) and C_{2} ≅ M_{2}(F).

Proof. In the case that both algebras C_{1} and C_{2} are division algebras, this is an immediate consequence of [9, Lemma 3.2]. If both C_{1} and C_{2} is split, then clearly N has Witt index 5. If C_{2} is a division algebra and C_{1} = Cay(C_{2}, μ) for some μ, then n_{2} is anisotropic and N ⊥ ℍ = n_{2} ⊗ ⟨1, −μ⟩ ⊥ ⟨−1⟩ n_{2} = 4ℍ ⊥ ⟨−μ⟩ n_{2}, so N has Witt index 3. Note that the converse is easy, since if i = 3, then n_{2} is isomorphic to a subform of n_{1}, which forces n_{2} to be a factor of n_{1}. If n_{1} = ⟨1, a⟩ ⊗ n_{2}, then C_{1} ≅ Cay(C_{2}, −a), so C_{2} is a subalgebra of C_{1}. If C_{1} and C_{2} contain a common quadratic étale algebra F [t]/(t^{2} −a) but C_{2} is not a quaternion subalgebra of C_{1}, then n_{1} and n_{2} have ⟨⟨a⟩⟩ as a common factor, so i = 1. Finally, if N is isotropic, there are x_{i} ∈ C_{i}, both skew, with n_{1}(x_{1}) = n_{2}(x_{2}). Then, as t_{1}(x_{1}) = 0 = t_{2}(x_{2}), the algebras F [x_{1}] and F [x_{2}] are isomorphic, so C_{1} and C_{2} share a common quadratic étale subalgebra. This finishes the proof.

If C is a biquaternion algebra; i.e., C ≅ C_{1} ⊗ C_{2} for two quaternion algebras C_{1} and C_{2}, then the Albert form n′_{1} ⊥ ⟨−1⟩n′_{2} is determined up to similarity by the isomorphism class of the algebra C ([11, Thm. 3.12]). Allison generalizes this result ([5, Thm. 5.4]) to tensor products of arbitrary composition algebras. However, he always considers the involution σ = γ_{1}⊗γ_{2} as a crucial part of the algebra C = C_{1}⊗C_{2}. Allison proves that (C_{1}⊗C_{2}, γ_{1}⊗γ_{2}) and (C_{3} ⊗ C_{4}, γ_{3} ⊗ γ_{4}) are isotopic algebras if and only if they have similar Albert forms, for the cases that C_{1}, C_{3} are octonion and C_{2}, C_{4} quaternion or octonion algebras.

The fact that any F -algebra isomorphism φ : (C_{1} ⊗ C_{2}, γ_{1} ⊗ γ_{2}) → (C_{3} ⊗ C_{4}, γ_{3} ⊗ γ_{4}) between arbitrary products of composition algebras yields an isometry n′_{1} ⊥ ⟨−1⟩n′_{2} ≅ μ(n′_{3} ⊥ ⟨−1⟩n′_{4}) for a suitable μ ∈ F^{×} is easy to see. Also, since for C = C_{1} ⊗ C_{2} the map ⟨ , ⟩ : C ×C → F given by ⟨x_{1} ⊗x_{2}, y_{1} ⊗y_{2}⟩ = n_{1}(x_{1}, y_{1})⊗n_{2}(x_{2}, y_{2}) is a nondegenerate symmetric bilinear form on C such that ⟨σ(x), σ(y)⟩ = ⟨x, y⟩, the equation ⟨zx, y⟩ = ⟨x, σ(z)y⟩ holds (that is, an invariant form cf. [6, p. 144]), and τ : C × C → k, τ(x, y) = ⟨x, σ(y)⟩ an associative nondegenerate symmetric bilinear form which is proper, it follows easily that n_{1} ⊗ n_{2} ≅ n_{3} ⊗ n_{4}.

Suppose that we have two algebras that each are a tensor product of an octonion algebra and a quaternion algebra. We obtain a necessary and sufficient condition for when their Albert forms are similar. We use the notation D(q) to denote the elements of F^{×} represented by a quadratic form q.

Proposition 2.3. Let C_{1}, C_{2} be octonion algebras and Q_{1}, Q_{2} quaternion algebras over F. Let N_{1} and N_{2} be the Albert forms of C_{1}⊗Q_{1} and C_{2}⊗Q_{2}, respectively. If N_{1} ≅ μN_{2} for some μ ∈ F^{×}, then Q_{1} ≅ Q_{2}. Moreover, there is a quaternion algebra Q and elements c, d ∈ F^{×} such that C_{1} ≅ Cay(Q, c), C_{2} ≅ Cay(Q, d), Cay(Q_{1}, μ) ≅ Cay(Q, cd), and −μc ∈ D(n_{C_{2}}). Conversely, if there is a quaternion algebra Q and elements c, d, μ ∈ F^{×} such that C_{1}, Q_{1} = Q_{2}, and C_{2} satisfy the conditions of the previous sentence, then N_{1} ≅ μN_{2}.

Proof. Suppose that N_{1} ≅ μN_{2} for some μ ∈ F^{×}. If c : W (F) → Br(F) is the Clifford invariant, then c(N_{1}) = c(μN_{2}) = c(N_{2}). Since c is trivial on I^{3}(F), we have c(N_{1}) = c(−n_{Q_{1}}) and c(N_{2}) = c(−n_{Q_{2}}) (see [15, Ch. 5.3]). Therefore, c(n_{Q_{1}}) = c(n_{Q_{2}}). However, the Clifford invariant of the norm form of a quaternion algebra is the class of the quaternion algebra, by [15, Cor. V.3.3]. This implies that c(N_{1}) = [Q_{1}] and c(μN_{2}) = [Q_{2}]. Since [Q_{1}] = [Q_{2}], we get Q_{1} ≅ Q_{2}. As a consequence of this, n_{Q_{1}} ≅ n_{Q_{2}}. Thus,

n_{C_{1}} ⊥ − ⟨1, −μ⟩ ⊗ n_{Q_{1}} ≅ n_{C_{1}} ⊥ (−n_{Q_{1}} ⊥ μn_{Q_{1}})
≅ μ (n_{C_{2}} ⊥ −n_{Q_{2}}) ⊥ μn_{Q_{1}} ≅ μn_{C_{1}} ⊥ (−μn_{Q_{1}} ⊥ μn_{Q_{2}})
≅ μn_{C_{2}} ⊥ 4ℍ.

The forms n_{C_{1}} and ⟨1, −μ⟩ ⊗ n_{Q_{1}} are Pfister forms. The line above shows that these Pfister forms are 2-linked, in the terminology of [8]. Therefore, there is a 2-fold Pfister form ⟨⟨−a, −b⟩⟩ with n_{C_{1}} ≅ ⟨⟨−a, −b, −c⟩⟩ and ⟨1, −μ⟩⊗n_{Q_{1}} ≅ ⟨⟨−a, −b, −e⟩⟩ for some c, e ∈ F^{×}. An elementary calculation shows that

⟨⟨−a, −b, −c⟩⟩ ⊥ − ⟨⟨−a, −b, −e⟩⟩ ≅ 4ℍ ⊥ ⟨−c, e⟩ ⊗ ⟨⟨−a, −b⟩⟩ .

Therefore, μn_{C_{2}} ≅ ⟨−c, e⟩ ⊗ ⟨⟨−a, −b⟩⟩. Thus, n_{C_{2}} ≅ −μc ⟨⟨−a, −b, −ce⟩⟩. Since n_{C_{2}} and ⟨⟨−a, −b, −ce⟩⟩ are Pfister forms, we get n_{C_{2}} ≅ ⟨⟨−a, −b, −ce⟩⟩. If we set d = ce and let Q be the quaternion algebra (a, b)_{F}, then the isomorphisms n_{C_{1}} ≅ ⟨⟨−a, −b, −c⟩⟩ and n_{C_{2}} ≅ ⟨⟨−a, −b, −d⟩⟩ give C_{1} ≅ Cay(Q, c) and C_{2} ≅ Cay(Q, d). Moreover, n_{C_{2}} ≅ −μcn_{C_{2}}, so−μc ∈ D(n_{C_{2}}). Finally, the isomorphism ⟨1, −μ⟩ ⊗ n_{Q_{2}} ≅ ⟨⟨−a, −b, −e⟩⟩ gives Cay(Q_{1}, μ) ≅ Cay(Q, e) ≅ Cay(Q, cd).

It is a short calculation to show that if C_{1} = Cay(Q, c), C_{2} = Cay(Q, d), and Q_{1} = Q_{2} is a quaternion algebra with Cay(Q_{1}, μ) ≅ Cay(Q, −dc), then n′_{C_{1}} ⊥ ⟨−1⟩ n′_{Q_{1}} ≅ μ(n′_{C_{2}} ⊥ ⟨−1⟩ n′_{Q_{2}}).

The argument of the previous proposition does not work for a tensor product of two octonion algebras since the Albert form is an element of I^{3}(F), whose Clifford invariant is trivial.

Corollary 2.4. With the notation in the previous proposition, suppose that N_{1} ≅ μN_{2} for some μ ∈ F^{×}. If one of C_{1} and C_{2} is split, then the other algebra is isomorphic to Cay(Q_{1}, μ).

Proof. We saw in the proof of the previous proposition that

n_{C_{1}} ⊥ − ⟨1, −μ⟩ ⊗ n_{Q_{1}} ≅ μn_{C_{2}} ⊥ 4ℍ.

Suppose that C_{2} is split. Then n_{C_{1}} ⊥ − ⟨1, −μ⟩ ⊗ n_{Q_{1}} is hyperbolic, so n_{C_{1}} ≅ ⟨1, −μ⟩ ⊗ n_{Q_{2}}. Therefore, C_{1} ≅ Cay(Q_{1}, μ). On the other hand, if C_{1} is split, then n_{C_{1}} ≅ 4ℍ, so by Witt cancellation, −μn_{C_{2}} ≅ ⟨1, −μ⟩ ⊗ n_{Q_{1}}. Since n_{C_{2}} and ⟨1, −μ⟩ ⊗ n_{Q_{1}} are both Pfister forms, this implies that n_{C_{2}} ≅ ⟨1, −μ⟩ ⊗ n_{Q_{1}}, and so C_{2} ≅ Cay(Q_{1}, μ).

In Proposition 2.3 above, it is possible for N_{1} ≅ μN_{2} without C_{1} ≅ C_{2}. Moreover, the quaternion algebra Q of the proposition need not be isomorphic to Q_{1}. We verify both these claims in the following example.

Example 2.5. In this example we produce nonisomorphic octonion algebras C_{1} and C_{2} and a quaternion algebra Q_{1} that is not isomorphic to a subalgebra of either C_{1} or C_{2}, and such that the Albert forms of C_{1} ⊗_{F} Q_{1} and C_{2} ⊗_{F} Q_{1} are similar. To do this we produce nonisometric Pfister forms ⟨⟨x, y, z⟩⟩ and ⟨⟨x, y, w⟩⟩ and elements u, v, μ with ⟨⟨x, y, zw⟩⟩ ≅ ⟨⟨u, v, μ⟩⟩ such that the Witt index of ⟨⟨x, y, z⟩⟩ ⊥ − ⟨⟨u, v, μ⟩⟩ and ⟨⟨x, y, w⟩⟩ ⊥ − ⟨⟨u, v, μ⟩⟩ are both 2, and μz ∈ D(⟨⟨x, y, w⟩⟩). We then set Q = (−x, −y), C_{1} = Cay(Q, −z), C_{2} = Cay(Q, −w), and Q_{1} = (−u, −v). From Proposition 2.3, we have N_{1} ≅ μN_{2}. However, Proposition 2.2 shows that Q_{1} is not isomorphic to a subalgebra of either C_{1} or C_{2}. Moreover, C_{1} and C_{2} are not isomorphic since their norm forms are not isometric. Note that Q and C_{2} are not isomorphic since Q_{1} is not a subalgebra of C_{1}.

Let k be a field of characteristic not 2, and let F = k(x, y, z, w) be the rational function field in 4 variables over k. Set μ = xyzw, n_{1} = ⟨⟨x, y, z⟩⟩, and n_{2} = ⟨⟨x, y, w⟩⟩. By embedding F in the Laurent series field k(x, y, z)((w)), we see that n_{1} and n_{2} = ⟨⟨x, y⟩⟩ ⊥ w ⟨⟨x, y⟩⟩ are not isomorphic over this field by Springer’s theorem [15, Prop. VI.1.9], so n_{1} and n_{2} are not isomorphic over F. Also, μz = z^{2}(xyw), which is clearly represented by n_{3}. Set Q_{1} = (−zw, −xzw). A short calculation shows that ⟨⟨x, y, zw⟩⟩ = ⟨⟨zw, xzw, μ⟩⟩. Finally, for the Witt indices, we have

n_{1} ⊥ −n_{Q_{1}} = ⟨1, x, y, xy, z, xz, yz, xyz⟩ ⊥ − ⟨1, zw, xzw, x⟩
= 2ℍ ⊥ ⟨y, xy, z, xz, yz, xyz, −zw, −xzw⟩
= 2ℍ ⊥ ⟨y, xy, z, xz, yz, xyz⟩ ⊥ w ⟨−z, −xz⟩ .

The Springer theorem shows that this form has Witt index 2. Similarly,

n_{3} ⊥ −n_{Q_{1}} = ⟨1, x, y, xy, w, xw, yw, xyw⟩ ⊥ − ⟨1, zw, xzw, x⟩
= 2ℍ ⊥ ⟨y, xy, w, xw, yw, xyw, −zw, −xzw⟩
= 2ℍ ⊥ ⟨y, xy⟩ ⊥ w ⟨1, x, y, xy, −z, −xz⟩

has Witt index 2.

For the remainder of this section we will also consider the case that char(F) = 2. Let C_{1} and C_{2} be composition algebras over F of dim_{F} (C_{i}) = r_{i} ≥ 2, and let n_{i} be the norm form of C_{i}. Using the notation of [13], the subspace Q(C_{1}, C_{2}) = {u = x_{1}⊗1−1⊗x_{2} : t_{1}(x_{1}) = t_{2}(x_{2})} has dimension r_{1} + r_{2} − 2, and Q(C_{1}, C_{2}) = {z − (γ_{1} ⊗ γ_{2})(z) : z ∈ C_{1} ⊗ C_{2}} is the set of alternating elements of C_{1} ⊗ C_{2} with respect to γ_{1} ⊗ γ_{2}. The nondegenerate quadratic form N : Q(C_{1}, C_{2}) → F given by N(x_{1} ⊗ 1 − 1 ⊗ x_{2}) = n_{1}(x_{1}) − n_{2}(x_{2}) is isometric to the Albert form n′_{1} ⊥ ⟨−1⟩n′_{2} of C_{1} ⊗ C_{2}.

Let V_{N} ⊂ ℙ^{r_{1}+r_{2}−3} be the F -quadric defined via N. In the case that char(F) ≠ 2, V_{N} coincides with the open subvariety U_{N} of closed points x_{1} ⊗ 1 − 1 ⊗ x_{2} with x_{1} ∉ F 1 and x_{2} ∉ F 1. We now generalize [13, Prop.] in the following two propositions. We will make use of the following fact that comes from Galois theory: Let F [z_{i}] be the commutative F - subalgebra of dimension two of C_{i} generated by z_{i} ∈ C_{i} for i = 1, 2. Then there exists an isomorphism α : F [z_{1}] →^{∼} F [z_{2}] such that α(z_{1}) = z_{2} if and only if n_{1}(z_{1}) = n_{2}(z_{2}) and t_{1}(z_{1}) = t_{2}(z_{2}).

Proposition 2.6. There exists a bijection Φ between the set of F -rational points of U_{N} and the set of triples (K_{1}, K_{2}, α), where K_{i} is a two-dimensional commutative subalgebra of C_{i}, and where α : K_{1} →^{∼} K_{2} is an F -algebra isomorphism:

Φ : {P ∈ U_{N} | P an F -rational point} →^{∼} {(K_{1}, K_{2}, α) | K_{1}, K_{2}, α as above}

P = z_{1} ⊗ 1 − 1 ⊗ z_{2} ⟼ ( \begin{matrix}F [z_{1}], F [z_{2}], α : F [z_{1}] →^{∼} F [z_{2}] \\ z_{1} ↦ z_{2} \end{matrix} )

Proof. Any F -rational point P ∈ U_{N} corresponds with an element x_{1}⊗1−1⊗x_{2} ∈ Q(C_{1}, C_{2}) with t_{1}(x_{1}) = t_{2}(x_{2}) and n_{1}(x_{1}) = n_{2}(x_{2}). Then there exists an F -algebra isomorphism α : F [x_{1}] →^{∼} F [x_{2}] with x_{1} ↦ x_{2}. For x_{1} ⊗ 1 − 1 ⊗ x_{2} = z_{1} ⊗ 1 − 1 ⊗ z_{2} it can be easily verified that

( \begin{matrix}F [x_{1}], F [x_{2}], α : F [x_{1}] →^{∼} F [x_{2}] \\ x_{1} ↦ x_{2} \end{matrix} ) = ( \begin{matrix}F [z_{1}], F [z_{2}], β : F [z_{1}] →^{∼} F [z_{2}] \\ z_{1} ↦ z_{2} \end{matrix} )

Therefore, the mapping Φ is well defined.

Given a triple (K_{1}, K_{2}, α), there are elements z_{i} ∈ C′_{i} such that K_{i} = F [z_{i}] and α : F [z_{1}] →^{∼} F [z_{2}] with z_{1} ↦ z_{2}. By the remark before the proposition, we have n_{1}(z_{1}) = n_{2}(z_{2}) and t_{1}(z_{1}) = t_{2}(z_{2}); thus N(z_{1} ⊗ 1 − 1 ⊗ z_{2}) = 0, and the triple defines the F -rational point P ∈ U_{N} corresponding to z_{1} ⊗ 1 − 1 ⊗ z_{2}. So Φ is surjective.

To prove injectivity, suppose that Φ(x_{1} ⊗ 1 − 1 ⊗ x_{2}) = Φ(z_{1} ⊗ 1 − 1 ⊗ z_{2}). Then F [x_{1}] = F [z_{1}], F [x_{2}] = F [z_{2}] and the maps α : F [x_{1}] →^{∼} F [x_{2}], x_{1} ↦ x_{2} and β : F [z_{1}] →^{∼} F [z_{2}], z_{1} ↦ z_{2} are equal. Since F [x_{i}] = F [z_{i}], write x_{1} = a + bz_{1} and x_{2} = c + dz_{2} with a, b, c, d ∈ F. We have a = c since t_{1}(x_{1}) = t_{2}(x_{2}). Therefore, we may replace x_{1} with bz_{1} and x_{2} by dz_{2} without changing x_{1} ⊗ 1 − 1 ⊗ x_{2}. Thus, x_{1} ⊗ 1 − 1 ⊗ x_{2} = bz_{1} ⊗ 1 − 1 ⊗ dz_{2}, and n_{1}(x_{1}) = n_{2}(x_{2}), n_{1}(z_{1}) = n_{2}(z_{2}) imply that n_{1}(x_{1}) = b^{2}n_{1}(z_{1}) and n_{2}(x_{2}) = d^{2}n_{2}(z_{2}). Therefore, b^{2} = d^{2}, so b = ±d. Now x_{2} = α(x_{1}) = α(bz_{1}) = bz_{2} yields b = d and we get x_{1} ⊗ 1 − 1 ⊗ x_{2} = b(z_{1} ⊗ 1 − 1 ⊗ z_{2}) which shows that Φ is injective.

In the case that char(F) = 2, the set U_{N} = {x_{1} ⊗ 1 − 1 ⊗ x_{2} : x_{1} ∉ F 1, x_{2} ∉ F 1} is a proper open subvariety of V_{N}. The proof of the previous proposition shows that Φ again is a bijection between the F -rational points of U_{N} and the triples (K_{1}, K_{2}, α), where K_{i} is a two-dimensional commutative F -subalgebra of C_{i} and α : K →^{∼} L an F -algebra isomorphism. We can say more in this situation. Proposition 2.7. Let char(F) = 2. There exists an F -rational point in V_{N} if and only if there exists a triple (K_{1}, K_{2}, α) such that K_{i} is a quadratic étale subalgebra of C_{i} and α : K_{1} ⟶^{∼} K_{2} an F -algebra isomorphism. In addition, there exists an F -rational point in V_{N} ∩ {t_{1}(x_{1}) = 0} if and only if there exists a triple (K_{1}, K_{2}, α) such that K_{1} and K_{2} are purely inseparable quadratic extensions and α : K_{1} ⟶^{∼} K_{2} an F -algebra isomorphism. Proof. As pointed out before the proposition, there is a bijection between F -rational points in U_{N} and triples (K_{1}, K_{2}, α) with K_{i} ⊆ C_{i} commutative subalgebras of dimension 2 over F. To prove the first statement, only one half needs further argument. Suppose V_{N} has an F -rational point. Since V_{N} is a quadric hypersurface, V_{N} is then birationally equivalent to ℙ^{r_{1}+r_{2}−3}. The F -rational points of projective space are dense, so there is an F -rational point in U_{N}. Therefore, we get a triple (K_{1}, K_{2}, α) with K_{i} a quadratic étale subalgebra of C_{i}.

For the second statement, an F -rational point in V_{N} ∩ {t_{1}(x_{1}) = 0} corresponds with an element x_{1} ⊗ 1 − 1 ⊗ x_{2} such that n_{1}(x_{1}) = n_{2}(x_{2}) and t_{1}(x_{1}) = t_{2}(x_{2}) = 0, so F [x_{i}] is a purely inseparable extension. There exists an isomorphism α : F [x_{1}] →^{∼} F [x_{2}] with α(x_{1}) = x_{2} and thus a triple (F [x_{1}], F [x_{2}], α). Conversely, if there is a triple (K_{1}, K_{2}, α) with K_{i} purely inseparable, there are x_{i} ∈ C_{i} with t_{i}(x_{i}) = 0 such that K = F [x_{1}], L = F [x_{2}] and α : F [x_{1}] →^{∼} F [x_{2}], x_{1} ↦ x_{2}, so n_{1}(x_{1}) = n_{2}(x_{2}), and x_{1} ⊗ 1 − 1 ⊗ x_{2} defines an F -rational point in V_{N} ∩ {t_{1}(x_{1}) = 0}.

3 Automorphisms

Let (Q_{i}, γ_{i}) be a quaternion algebra with its standard involution, and let C_{i} = Cay(Q_{i}, μ_{i}) be an octonion algebra. We also write γ_{i} for the standard involution on C_{i}. Let C = C_{1} ⊗_{F} C_{2} and σ = γ_{1} ⊗ γ_{2}. Let A = Q_{1} ⊗_{F} Q_{2}, an F -central associative subalgebra of C. In this section we will prove that any automorphism of C that preserves A is compatible with the involution σ. As a consequence, we show that the Skolem-Noether theorem does not hold for A.

Let

e = (0, 1) ⊗ (1, 0),
f = (1, 0) ⊗ (0, 1),
g = (0, 1) ⊗ (0, 1).

We can decompose C as an F -vector space as

C = A ⊕ Ae ⊕ Af ⊕ Ag.

Note that an automorphism of C is determined by its action on A and on {e, f, g}. Elementary calculations show that Ae = eA = A(Ae) = (Ae)A, and similar relations hold for f and g.

We point out some properties of this decomposition. We have, for a ∈ A,

ea = (γ_{1} ⊗ id)(a)e
fa = (id ⊗ γ_{2})(a)f
ga = σ(a)g.

For example, if a = (x, 0) ⊗ (y, 0), then

ga = (0, 1) ⊗ (0, 1) · (x, 0) ⊗ (y, 0) = (0, γ_{1}(x)) ⊗ (0, γ_{2}(y))
= (γ_{1}(x), 0) ⊗ (γ_{2}(y), 0)g = σ(a)g.

The other properties follow similarly.

Lemma 3.1. For any d, a ∈ A, we have (dg)a = σ(a)(dg). Also, a(dg) = (da)g.

Proof. Let d = ∑_{i}(u_{i}, 0) ⊗ (v_{i}, 0). For a = (x, 0) ⊗ (y, 0), we have

(dg)a = ∑_{i}(0, u_{i}) ⊗ (0, v_{i}) · (x, 0) ⊗ (y, 0) = ∑_{i}(0, u_{i}γ_{1}(x)) ⊗ (0, v_{i}γ_{2}(y))

and

σ(a)(dg) = (γ_{1}(x), 0) ⊗ (γ_{2}(x), 0) ∑_{i}(0, u_{i}) ⊗ (0, v_{i}) = ∑_{i}(0, u_{i}γ_{1}(x)) ⊗ (0, v_{i}γ_{2}(y)).

This proves the first assertion. For the second, we see that

a(dg) = (x, 0) ⊗ (y, 0) · ∑_{i}(0, u_{i}) ⊗ (0, v_{i}) = ∑_{i}(0, u_{i}x) ⊗ (0, v_{i}y)

and

(da)g = (∑_{i}(u_{i}, 0) ⊗ (v_{i}, 0) · (x, 0) ⊗ (y, 0)) g

= (∑_{i}(u_{i}x, 0) ⊗ (v_{i}y, 0)) g = ∑_{i}(0, u_{i}x) ⊗ (0, v_{i}y).

Lemma 3.2. The right multiplication maps R_{e}, R_{f}, and R_{g} are injective. If φ ∈ Aut_{F} (C), then R_{φ(e)}, R_{φ(f)}, and R_{φ(g)} are injective.

Proof. The second statement is obvious from the first, so we prove the first. The arguments are similar, so we only prove that R_{e} is injective. Let c = ∑_{i}(u_{i}, v_{i}) ⊗ (w_{i}, x_{i}) ∈ C. Then

ce = ∑_{i}(u_{i}, v_{i}) ⊗ (w_{i}, x_{i}) · (0, 1) ⊗ (1, 0)
= ∑_{i}(μ_{1}v_{i}, u_{i}) ⊗ (w_{i}, x_{i}),

and

(ce)e = ∑_{i}(μ_{1}v_{i}, u_{i}) ⊗ (w_{i}, x_{i}) · (0, 1) ⊗ (1, 0)
= ∑_{i}(μ_{1}u_{i}, μ_{1}v_{i}) ⊗ (w_{i}, x_{i}) = μ_{1}c.

Since R_{e} ∘ R_{e} is multiplication by the scalar μ_{1}, we see that R_{e} is injective. By similar calculations, we see that R_{f}^{2} is multiplication by μ_{2} and R_{g}^{2} is multiplication by μ_{1}μ_{2}, so R_{f} and R_{g} are also injective.

One fact we will use often is that if Q is a quaternion algebra with standard involution γ, and if a ∈ Q satisfies ab = γ(b)a for all b ∈ Q, then a = 0. This can be verified by an elementary calculation. The following lemma is a slight generalization of this fact.

Lemma 3.3. If x ∈ A satisfies xa = (γ_{1} ⊗ id)(a)x for all a ∈ A, then x = 0. Similarly, if xa = (id ⊗ γ_{2})(a)x for all a, then x = 0.

Proof. Suppose x satisfies the first condition above. If a ∈ Q_{2}, then xa = ax, so x ∈ C_{A}(Q_{2}) = Q_{1}. We then let a ∈ Q_{1}, so xa = γ_{1}(a)x. Since this is true for all a ∈ Q_{1}, the remark before the lemma shows that x = 0. A similar argument works for the second statement.

In the proof of the following theorem, we make use of the following well known result: if A = Q_{1} ⊗_{F} Q_{2} is a biquaternion algebra with involution σ = γ_{1} ⊗γ_{2}, where γ_{i} is the standard involution on Q_{i}, then any automorphism of A that is compatible with σ either preserves the Q_{i} or interchanges them. To aid the reader we recall one proof of this fact. If φ is such an automorphism, then φ restricts to a Lie algebra automorphism of Skew(A, σ), the space of skew-symmetric elements of A with respect to σ. This Lie algebra is the direct sum of the two simple Lie subalgebras Skew(Q_{1}, γ_{1}) and Skew(Q_{2}, γ_{2}). Since these are the unique simple Lie ideals of Skew(A, σ), the map φ must preserve them or interchange them. As Q_{i} = F ⊕ Skew(Q_{i}, γ_{i}), this clearly shows that φ preserves the Q_{i} or interchanges them.

We now give the main result of this section.

Theorem 3.4. If φ ∈ Aut_{F} (C) satisfies φ(A) = A, then φ ∘ σ = σ ∘ φ.

Proof. We first prove that either

φ(e) ∈ Q_{1}e,
φ(f) ∈ Q_{2}f,
φ(g) ∈ Ag,

or

φ(e) ∈ Q_{2}f,
φ(f) ∈ Q_{1}e,
φ(g) ∈ Ag.

For simplicity, we write τ_{1} = (γ_{1} ⊗ id) |_{A} and τ_{2} = (id ⊗ γ_{2}) |_{A}. We begin by writing

φ(e) = a_{1} + b_{1}e + c_{1}f + d_{1}g,
φ(f) = a_{2} + b_{2}e + c_{2}f + d_{2}g,
φ(g) = a_{3} + b_{3}e + c_{3}f + d_{3}g

with the coefficients in A. Let b ∈ A. Applying φ to the equations eb = τ_{1}(b)e, fb = τ_{2}(b)f, and gb = σ(b)g and setting a = φ(b) gives

φ(e)a = τ′_{1}(a)φ(e),
φ(f)a = τ′_{2}(a)φ(f),
φ(g)a = σ′(a)φ(g),

where σ′ = (φσφ^{−1}) |_{A} and τ′_{i} = (φτ_{i}φ^{−1}) |_{A}. In particular, looking at the g-coefficients, we get

(d_{1}g)a = τ′_{1}(a)(d_{1}g),
(d_{2}g)a = τ′_{2}(a)(d_{2}g),
(d_{3}g)a = σ′(a)(d_{3}g).

Now, we have (dg)a = σ(a)(dg) = (dσ(a))g for all d ∈ A by Lemma 3.1. So, we get

d_{1}σ(a) = d_{1}τ′_{1}(a),  (1)
d_{2}σ(a) = d_{2}τ′_{2}(a),
d_{3}σ(a) = d_{3}σ′(a).

It is clear that σ ≠ τ′_{i} as τ′_{i} is not an involution. If A is a division algebra, then this forces d_{1} = d_{2} = 0. We prove that d_{1} = d_{2} = 0 in general in Corollary 3.9 below. From this we see that imφ ⊆ A ⊕ Ae ⊕ Af ⊕ (d_{3}A) g. Since φ is surjective, we get d_{3}A = A (using Lemma 9), so d_{3} is a unit in A. From Equation 1, this forces σ′ = σ. Thus, (φσφ^{−1}) |_{A} = σ|_{A}, so φ|_{A} is compatible with σ|_{A}. Note that this forces φ either to fix the Q_{i} or to interchange them, as pointed out before the theorem. So, (φτ_{1}φ^{−1}) |_{A} is either τ_{1} or τ_{2}, and similarly for (φτ_{2}φ^{−1}) |_{A}.

We next consider the coefficients a_{3}, b_{3} and c_{3}. Since φ ∘ σ|_{A} = σ ∘ φ|_{A}, we have

a_{3}a = σ(a)a_{3},
(b_{3}e)a = σ(a)(b_{3}e),
(c_{3}f)a = σ(a)(c_{3}f).

for all a ∈ A. Clearly a_{3} = 0. Writing b_{3} = ∑_{i}(u_{i}, 0) ⊗ (v_{i}, 0) with {u_{i}} an F -independent set, for a = (x, 0) ⊗ (y, 0) we have

(b_{3}e)a = ∑_{i}(0, u_{i}) ⊗ (v_{i}, 0) · (x, 0) ⊗ (y, 0)
= ∑_{i}(0, u_{i}γ_{1}(x)) ⊗ (v_{i}y, 0),
σ(a)(b_{3}e) = (γ_{1}(x), 0) ⊗ (γ_{2}(y), 0) · ∑_{i}(0, u_{i}) ⊗ (v_{i}, 0)
= ∑_{i}(0, u_{i}γ_{1}(x)) ⊗ (γ_{2}(y)v_{i}, 0).

Setting these two equal with x = 1, and using the independence of {u_{i}} yields v_{i}y = γ_{2}(y)v_{i}, so v_{i} = 0 for all i. Thus, b_{3} = 0. Similarly, c_{3} = 0. We have thus proven that φ(g) = d_{3}g.

We next show that φ(e) ∈ Q_{1}e or φ(e) ∈ Q_{2}f. From the condition eb = τ_{1}(b)e, applying φ, we see that

a_{1}a = τ′_{1}(a)a_{1},
(b_{1}e)a = τ′_{1}(a)(b_{1}e),
(c_{1}f)a = τ′_{1}(a)(c_{1}f),

and τ′_{1} is either τ_{1} or τ_{2}, as we saw above. We see that a_{1} = 0 by Lemma 3.3. Suppose that τ′_{1} = τ_{1}. For c_{1}, write c_{1} = ∑_{i}(u_{i}, 0) ⊗ (v_{i}, 0). We have (c_{1}f)a = τ_{1}(a)(c_{1}f). For a = (x, 0) ⊗ (y, 0) we have

∑_{i}(u_{i}, 0) ⊗ (0, v_{i}) · (x, 0) ⊗ (y, 0) = ∑_{i}(u_{i}x, 0) ⊗ (0, v_{i}γ_{2}(y)),
(γ_{1}(x), 0) ⊗ (y, 0) ∑_{i}(u_{i}, 0) ⊗ (0, v_{i}) = ∑_{i}(γ_{1}(x)u_{i}, 0) ⊗ (0, v_{i}y).

By assuming that the {v_{i}} are F -independent, and setting y = 1, we get u_{i}x = γ_{1}(x)u_{i}, which forces each u_{i} = 0, so c_{1} = 0. Since a_{1} = c_{1} = d_{1} = 0, we have φ(e) ∈ Ae. On the other hand, if τ′_{1} = τ_{2}, a similar calculation shows that φ(e) = c_{1}f ∈ Af. We further need to show that in the first case that b_{1} ∈ Q_{1} and that c_{1} ∈ Q_{2} in the second case. For the first case write b_{1} = ∑_{i}(u_{i}, 0) ⊗ (v_{i}, 0) with {u_{i}} an F -independent set. Since we have (b_{1}e)a = τ_{1}(a)(b_{1}e), for a = (x, 0) ⊗ (y, 0),

(b_{1}e)a = ∑_{i}(0, u_{i}) ⊗ (v_{i}, 0) · (x, 0) ⊗ (y, 0) = ∑_{i}(0, u_{i}γ_{1}(x)) ⊗ (v_{i}y, 0)

and

τ_{1}(a)(b_{1}e) = (γ_{1}(x), 0) ⊗ (y, 0) ∑_{i}(0, u_{i}) ⊗ (v_{i}, 0) = ∑_{i}(0, u_{i}γ_{1}(x)) ⊗ (yv_{i}, 0).

By setting x = 1 and using the independence of the {u_{i}}, we see that v_{i}y = yv_{i} for all y ∈ Q_{2}. Thus, each v_{i} ∈ F. Therefore, b_{1} = ∑_{i}(u_{i}, 0) ⊗ (v_{i}, 0) = (∑_{i} u_{i}v_{i}, 0) ⊗ (1, 0) ∈ Q_{1}. The argument for c_{1} is similar. Finally, similar calculations show that φ(f) ∈ Q_{1}e or φ(f) ∈ Q_{2}f.

We now have proved that either

φ(e) = be for some b ∈ Q_{1},
φ(f) = cf for some c ∈ Q_{2},
φ(g) = dg for some d ∈ A.

or

φ(e) = cf for some c ∈ Q_{2},
φ(f) = be for some b ∈ Q_{1},
φ(g) = dg for some d ∈ A.

We claim that either of these cases implies that φ∘σ = σ∘φ. Note that σ|_{A}∘τ_{1} = τ_{1}∘σ|_{A} = τ_{2} and σ|_{A} ∘ τ_{2} = τ_{2} ∘ σ|_{A} = τ_{1}. We consider the first case. Note that σ(e) = −e, σ(f) = −f, and σ(g) = g. We have

φ(σ(x + ye + zf + wg)) = φ(σ(x) − eσ(y) − fσ(z) + gσ(w))
= φ(σ(x) − τ_{2}(y)e − τ_{1}(z)f + wg)
= φσ(x) − φτ_{2}(y)(be) − φτ_{1}(z)(cf) + φ(w)(dg)

and

σ(φ(x + ye + zf + wg)) = σ(φ(x) + φ(y)(be) + φ(z)(cf) + φ(w)(dg))
= σφ(x) − (eσ(b))σφ(y) − (fσ(c))σφ(z) + (gσ(d)σφ(w))
= σφ(x) − (τ_{2}(b)e)σφ(y) − (τ_{1}(c)f)σφ(z) + (dg)σφ(w)
= σφ(x) − (be)σφ(y) − (cf)σφ(z) + (dg)σφ(w)
= σφ(x) − τ_{2}φ(y)(be) − τ_{1}φ(z)(cf) + φ(w)(dg).

We used b ∈ Q_{1} and c ∈ Q_{2} to simplify the fourth line of the display above. Also, because b ∈ Q_{1}, we can see by a short calculation that (be)a = τ_{1}(a)(be) for all a ∈ A. Similarly, since c ∈ Q_{2}, we have (cf)a = τ_{2}(a)(cf). This shows how we obtained the fifth line of the display.

We will have φ∘σ = σ∘φ once we know that (φ ∘ σ) |_{A} = (σ ∘ φ) |_{A} and φ|_{A}∘τ_{i} = τ_{i}∘φ|_{A}. We saw earlier that (φσφ^{−1}) |_{A} = σ|_{A}, so (φ ∘ σ) |_{A} = (σ ∘ φ) |_{A}. Moreover, we obtained the first case above by having (φτ_{i}φ^{−1}) |_{A} = τ_{i}. So, we have φ ∘ σ = σ ∘ φ. The second case is similar.

We now need to prove that Equation 1 above forces d_{1} = d_{2} = 0. To do this it suffices to go to a splitting field of A. Thus, let A = M_{4}(F). Recall that the first line of Equation 1 says that d_{1}σ(a) = d_{1}τ′_{1}(a) for all a ∈ A, and where τ′_{1} = φτ_{1}φ^{−1}. We see that the left multiplication map L_{d_{1}} annihilates X = {σ(a) − τ′_{1}(a) : a ∈ A}, and so it annihilates the right ideal I generated by X. Let

S = Sym(A, σ),
K = Skew(A, σ),
P = {a ∈ A : τ′_{1}(a) = −a} ,
T = {a ∈ A : τ′_{1}(a) = a} .

For simplicity, we write τ = τ′_{1}. It is a short argument to see that T = φ(Q_{2}), so T is a quaternion algebra. Our approach will be to show that the assumption I ≠ A forces T to be a commutative algebra, a contradiction to T = φ(Q_{2}). Thus, A = I ⊆ ker(L_{d_{1}}), which forces d_{1} = 0. Similarly we get d_{2} = 0.

We now get information about I. We have P = Skew(φ(Q_{1}), φγ_{1}φ^{−1})⊗Q_{2}, so dim_{F} (P) = 12. By dimension count, dim_{F} (S ∩ P) ≥ 6. If a ∈ S ∩ P, then σ(a) − τ(a) = 2a ∈ X. So S ∩ P ⊆ X ⊆ I. The ideal I thus contains a 6-dimensional subspace of Sym(A, σ). We claim, and prove in the lemma below, that this forces dim_{F} (I) ≥ 12 and, if dim_{F} (I) = 12, that I ∩ S = S ∩ P has dimension 6. Furthermore, the kernel of the map P → S given by x ↦ σ(x) + x is P ∩ K. Since the image is contained in I ∩ S = S ∩ P, we see that the image has dimension 6, so the kernel has dimension 12 − 6 = 6. Thus, K ⊆ P. Moreover, by the descriptions of P and T above, we see that T P ⊆ P. Therefore, T K ⊆ P. Since dim_{F} (S ∩ P) = 6 = dim(K ∩ P) = \frac{1}{2} dim_{F} (P), we have P = (S ∩ P) ⊕ K. From this, we see that if t ∈ T and k ∈ K, then tk ∈ P and tk + kσ(t) ∈ K ⊆ P, so tk − kσ(t) ∈ S ∩ P. We consider this situation after the following lemma.

Lemma 3.5. Let I be a right ideal of M_{4}(F). If dim_{F} (I) = 12, then dim_{F} (I ∩ S) = 6. If dim_{F} (I) < 12, then dim_{F} (I ∩ S) < 6.

Proof. We first show that we may assume σ is the transpose T. Since σ is orthogonal, σ = Int(s) ∘ T for some s with s^{T} = s. Then S = sSym(A, T). So,

dim_{F} (I ∩ S) = dim_{F} (I ∩ sSym(A, T)) = dim_{F} (s^{−1}I ∩ Sym(A, T)).

Thus, by replacing I by s^{−1}I, we may assume σ = T.

First suppose that dim_{F} (I) = 12. Since I is generated by a single element, we see that there are α, β, γ, δ ∈ F, not all 0, with

I = {a ∈ A : ( α β γ δ ) a = 0} .

Suppose that s ∈ I ∩ S. Write

s = ( \begin{matrix}a & b & c & d \\ b & e & f & g \\ c & f & h & i \\ d & g & i & j \end{matrix} )

for appropriate entries in F. The condition ( α β γ δ ) s = ( 0 0 0 0 ) corresponds to a homogeneous system of 4 equations in the 10 variables a, …, j. This system has coefficient matrix

( \begin{matrix}α & β & γ & δ & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & α & 0 & 0 & β & γ & δ & 0 & 0 & 0 \\ 0 & 0 & α & 0 & 0 & β & 0 & γ & δ & 0 \\ 0 & 0 & 0 & α & 0 & 0 & β & 0 & γ & δ \end{matrix} ) .

Since at least one of α, β, γ, δ is nonzero, this matrix clearly has rank 4. Therefore, its kernel is dimension 6. However, its kernel is I ∩ S, so dim_{F} (I ∩ S) = 6, as desired.

If dim_{F} (I) < 12, then dim_{F} (I) ∈ {0, 4, 8}. If dim_{F} (I) = 8, then we can write I = {a ∈ A : xa = 0} for some 2 × 4 matrix x of rank 2. By writing xs = 0 for a symmetric matrix s, we then obtain a coefficient matrix of size 8 × 10, and by writing it out we see that it has rank 8. Thus, its kernel has dimension 2, so dim_{F} (I ∩ S) = 2 < 6. Clearly if dim_{F} (I) ≤ 4, then dim_{F} (I ∩ S) < 6. This completes the proof of the lemma.

We are now consider the following situation. Let I be a right ideal of dimension 12 in A, let σ be an orthogonal involution on A, and define

R_{I,σ} = {a ∈ A : ak − kσ(a) ∈ I for all k ∈ K} .

Lemma 3.6. If σ and τ are orthogonal involutions on A, then R_{I,σ} = R_{I,τ} for any I.

Proof. Since τ and σ are both orthogonal, there is an s ∈ A^{∗} with τ = Int(s) ∘ σ. Then Skew(A, τ)s = Skew(A, σ). Let a ∈ R_{I,τ}. Then ay − yτ(a) ∈ I for all y ∈ Skew(A, τ). So,ay − ysσ(a)s^{−1} ∈ I, so a(ys) − (ys)σ(a) ∈ Is = I. Thus, a ∈ R_{I,σ}. The reverse inclusion is similar.

Lemma 3.7. If J = αI for some α ∈ A^{∗}, then R_{J,σ} = αR_{I,σ}α^{−1}.

Proof. We point out any right ideal of A of dimension 12 can be written in the form αI for some α ∈ A^{∗}. Let a ∈ R_{I,σ} and x ∈ Skew(A, σ). Then α^{−1}xσ(α)^{−1} ∈ Skew(A, σ). Therefore,

(αaα^{−1})x − x(αaα^{−1}) = αaα^{−1}x − xσ(α)^{−1}σ(a)σ(α)
= α (aα^{−1}xσ(α)^{−1} − α^{−1}xσ(α)^{−1}σ(a)) σ(α)
∈ αIσ(α) = αI = J.

Thus, αR_{I,σ}α^{−1} ⊆ R_{J,σ}. The reverse inclusion is similar.

Proposition 3.8. With the notation above, R_{I,σ} is a 4-dimensional commutative F -subalgebra of M_{4}(F) and R has a two-sided ideal J of F -dimension 3 with J^{2} = 0.

Proof. By Lemma 3.6, we may assume that σ = T is the transpose involution. By Lemma 3.7, the set R_{I,σ} is, up to conjugation, independent of the ideal I, so we may choose I to be the right ideal consisting of all matrices whose first row is zero. We determine R_{T,I} for this right ideal. Let a = ∑_{i,j} α_{ij}e_{ij} ∈ R_{I,σ}. We note that {e_{ij} − e_{ji} : i > j} is a basis for Skew(A, T). Let k = e_{uv} − e_{vu} with u > v. Then

ak − ka^{T} = ∑_{i,j} α_{ij}e_{ij}(e_{uv} − e_{vu}) − ∑_{i,j} α_{ij}(e_{uv} − e_{vu})e_{ji}
= ∑_{i} α_{iu}e_{iv} − ∑_{i} α_{iv}e_{iu} − ∑_{i} α_{iv}e_{ui} + ∑_{i} α_{iu}e_{vi.}

The first row of this matrix is α_{1u}e_{1v} − α_{1v}e_{1u} if v > 1, and α_{1u}e_{1v} − α_{1v}e_{1u} + ∑_{i} α_{iu}e_{1i} if v = 1. In any case, the first row is zero. Thus, α_{1u} = α_{1v} = 0 for all u > v > 1, and if v = 1, we get (α_{1u} + α_{1u})e_{11} + (α_{uu} − α_{11})e_{1u} + ∑_{i≠u,1} α_{iu}e_{1i} = 0, so α_{iu} = 0 if i ≠ u, 1, and α_{uu} = α_{11}. Thus, the diagonal entries of a are all the same. Also, the only nonzero non-diagonal entries are in the first column. There is no restriction on α_{i1} for i > 1, so we have shown that

R = \{( \begin{matrix}a & 0 & 0 & 0 \\ b & a & 0 & 0 \\ c & 0 & a & 0 \\ d & 0 & 0 & a \end{matrix} ) : a, b, c, d ∈ F \} .

From this description, it is clear that dim_{F} (R) = 4 and that R is a commutative F -subalgebra of A. Moreover, if J = I ∩ R = {a ∈ R : trace(a) = 0}, then we see that J is an ideal of R of dimension 3, and that J^{2} = 0. This completes the proof.

We can now finish the proof of Theorem 3.4.

Corollary 3.9. In Equation 1 above, we have d_{1} = d_{2} = 0.

Proof. The first line of Equation 1 is that d_{1}(σ(a) − τ′_{1}(a)) = 0 for all a ∈ A. We have shown above that the right ideal I generated by {σ(a) − τ′_{1}(a) : a ∈ A} has dimension at least 12, and that if it is dimension 12, then we have φ(Q_{2}) ⊆ R_{I,σ}. However, Proposition 3.8 shows that this set is a 4-dimensional commutative F -algebra. This is a contradiction since φ(Q_{2}) is a quaternion algebra. Thus, I = A. Since 0 = d_{1}I = d_{1}A, we get d_{1} = 0. Similarly, d_{2} = 0.

From this result we can show that the Skolem-Noether theorem does not hold for the tensor product of two octonion algebras.

Corollary 3.10. There exists simple F -subalgebras B and B′ of C and an F -algebra isomorphism f : B → B′ such that there is no F -algebra automorphism φ of C with φ|_{B} = f.

Proof. Let f be an F -algebra automorphism of A that is not compatible with σ|_{A}; such maps exist since we can take f to be the inner automorphism of an element t ∈ A with σ(t)t ∉ F. The condition σ(t)t ∈ F is precisely the condition needed to ensure that f is compatible with σ|_{A}. For example, we can take t = 1 + i_{1}i_{2} ∈ A = Q_{1} ⊗_{F} Q_{2} (where the standard generators of Q_{r} are i_{r} and j_{r}). If f extends to an automorphism φ of C, then φ(A) = A, so φ is compatible with σ. This forces φ|_{A} = f to be compatible with σ|_{A}, and f is chosen so that this does not happen.

Remark 3.11. Consider the slightly more general situation that φ ∈ Aut(C) satisfies φ(A) = Ã with Ã = Q̃_{1} ⊗ Q̃_{2} a biquaternion algebra, where Q̃_{i} is a quaternion subalgebra of C_{i}, i = 1, 2, or of C_{j}, j = 1, 2, i ≠ j. Then a similar argument shows that also φ ∘ σ = σ ∘ φ, i.e., φ ∈ Aut(C, σ) = (G_{2} × G_{2}) ⋊ ℤ_{2}. In particular, this means that for any φ ∈ Aut(C) which is not compatible with σ, the image of any biquaternion subalgebra A of the type investigated above (i.e., each factor a subalgebra of one of the octonion algebras) cannot be of this type again.

References

[1] A. A. Albert, On the Wedderburn norm condition for cyclic algebras, Bull. Amer. Math. Soc. 37 (1931), 301-312.

[2] A. A. Albert, A construction of non-cyclic normal division algebras, Bull. Amer. Math. Soc. 38 (1932), 449-456.

[3] A. A. Albert, Tensor products of quaternion algebras, Proc. Amer. Math. Soc. 35 (1972), 65-66.

[4] B. N. Allison, Structurable division algebras and relative rank one simple Lie algebras, Canadian Math. Soc. Conference Proc. 5 (1986), 149-156.

[5] B. N. Allison, Tensor products of composition algebras, Albert forms and some exceptional simple Lie algebras. Trans. Amer. Math. Soc. 306 (2) (1988), 667-695.

[6] B. N. Allison, A class of nonassociative algebras with involution containing the class of Jordan algebras. Math. Ann. 237 (1978), 133-156.

[7] R. Brauer, Untersuchungen über die arithmetischen Eigenschaften von Gruppen bilinearer Substitutionen, II, Math. Zeit. 31 (1929), 733-747.

[8] R. Elman, T.-Y. Lam, Pfister forms and K-theory of fields. J. Algebra 23, (1972) 181-213.

[9] D. Hoffmann, Twisted Pfister forms. Doc. Math. 1 (03) (1996), 67-102.

[10] N. Jacobson, Composition algebras and their automorphisms, Rend. Circ. Mat. Palermo (2) 7 (1958), 55-80.

[11] N. Jacobson, Some applications of Jordan norms to involutorial simple associative algebras. Adv. in Math. 48 (1983), 149-165.

[12] M.-A. Knus, Quadratic and Hermitian forms over rings, Grundlehren der Mathematischen Wissenschaften, vol. 294, Springer-Verlag, Berlin, 1991.

[13] M.-A. Knus, Sur la forme d’Albert et le produit tensoriel de deux algébres de quaternions. Bull. Soc. Math. Belg. 45 (3) (1993), 332-337.

[14] M.-A. Knus, A. Merkurjev, M. Rost, J.-P. Tignol, The book of involutions, Colloquium Publications, vol. 44, American Mathematical Society, Providence, RI, 1998.

[15] T.-Y. Lam, The Algebraic Theory of Quadratic Forms, Mathematics Lecture Notes Series, Benjamin, Reading, MA, 1973.

Department of Mathematical Sciences, New Mexico State University, Las Cruces NM 88003 USA

E-mail address: pmorandi@nmsu.edu

Facultat für Mathematik, Universität Regensburg, Universitatstrasse 31, D-93040 Regensburg, Germany

E-mail address: susanne.pumpluen@mathematik.uni-regensburg.de