’

isomorphism of algebras. Then u = ±(1) ∈ F · 1 is such that u3 = 1 and β(x) =

∼ ∼

±(x)u2 is an algebra isomorphism C1 ’ C2 . Conversely, if β : C1 ’ C2 is an

’ ’

isomorphism of algebras, then, for any u ∈ C2 such that u3 = 1, the map ± de¬ned

∼

by ±(x) = β(x)u is an isomorphism (C1 , ) ’ (C2 , ) of para-quadratic algebras.

’

∼

In particular an isomorphism C1 ’ C2 of para-Hurwitz algebras which is not an

’

464 VIII. COMPOSITION AND TRIALITY

F [X]/(X 2 + X + 1), i.e., C1 is

isomorphism of algebras can only occur if C1

isomorphic to F (u) with u3 = 1.

Proof : The proof of (??) shows that u = ±(1) ∈ F . We show that u3 = 1. We have

u u = u2 = u. Thus multiplying by u and conjugating gives u3 = uu = n2 (u) =

n2 ±(1) = n1 (1) = 1 by Proposition (??). It then follows that C2 F (u). The

condition ±(x) ±(y) = ±(x y) with y = 1 gives ±(x)u = ±(x) and, replacing x

by xy,

±(xy)u = ±(xy) = ±(x)±(y).

By conjugating and multiplying both sides with u4 = u we obtain

[±(x)u2 ][±(y)u2 ] = ±(xy)u2 ,

so that the map β : C1 ’ C2 de¬ned by β(x) = ±(x)u2 is an isomorphism of

∼

C2 = F (u) with u3 = 1 and if β : C1 ’ C2 is

algebras. Conversely, if C1 ’

an isomorphism, then ± : C1 ’ C2 de¬ned by ±(x) = β(x)u is an isomorphism

∼

(C1 , ) ’ (C2 , ).

’

Observe that ru : x ’ x u is an automorphism of F (u), of order 3. In

fact we have AutF F (u), = S3 , generated by the conjugation and ru . This is in

contrast with the quadratic algebra F (u), · for which AutF F (u) = Z/2Z.

(34.6) Corollary. The map P : (C, ) ’ (C, ) is an equivalence Hurw m ≡ Hurw m

of groupoids if m = 4, 8, and P is bijective on isomorphism classes if m = 2.

In view of Corollary (??) we call a n-dimensional para-Hurwitz composition

algebra of type A1 if n = 4 and of type G2 if n = 8.

(34.7) Remark. It follows from Corollary (??) that

AutF (C, ) = AutF (C, )

for any Hurwitz algebra C of dimension ≥ 4. Thus the classi¬cation of twisted

forms of para-Hurwitz algebras is equivalent to the classi¬cation of Hurwitz alge-

bras in dimensions ≥ 4. In particular any twisted form of a para-Hurwitz algebra of

dimension ≥ 4 is again a para-Hurwitz algebra. The situation is di¬erent in dimen-

sion 2: There exist forms of para-quadratic algebras which are not para-quadratic

algebras (see Theorem (??)).

The identity 1 of a Hurwitz algebra C plays a special role also for the associated

para-Hurwitz algebra: it is an idempotent and satis¬es 1 x = x 1 = ’x for all

x ∈ C such that bnC (x, 1) = 0. Let (S, , n) be a symmetric composition algebra.

An idempotent e of S (i.e., an element such that e e = e) is called a para-unit if

e x = x e = ’x for x ∈ S, bn (e, x) = 0.

(34.8) Lemma. A symmetric composition algebra is para-Hurwitz if and only if

it admits a para-unit.

Proof : If (S, ) is para-Hurwitz, then 1 ∈ S is a para-unit. Conversely, for any

para-unit e in a symmetric composition algebra (S, , n), we have n(e) = 1 and

x y = (e x) (y e)

de¬nes a multiplication with identity element e on S. We have x y = x y where

x = bn (e, x)e ’ x.

§34. SYMMETRIC COMPOSITIONS 465

34.B. Petersson algebras. Let (C, , n) be a Hurwitz algebra and let • be

an F -automorphism of C such that •3 = 1. Following Petersson [?] we de¬ne a

new multiplication on C by

x y = •(x) •2 (y).

This algebra, denoted C• , is a composition algebra for the same norm n and we

call it a Petersson algebra. It is straightforward to check that

(x y) x = n(x)y = x (y x)

so that Petersson algebras are symmetric composition algebras. Observe that • is

automatically an automorphism of (C, ). For • = 1, (C, ) is para-Hurwitz.

Conversely, symmetric composition algebras with nontrivial idempotents are

Petersson algebras (Petersson [?, Satz 2.1], or Elduque-P´rez [?, Theorem 2.5]):

e

(34.9) Proposition. Let (S, , n) be a symmetric composition algebra and let e ∈

S be a nontrivial idempotent.

(1) The product x y = (e x) (y e) gives S the structure of a Hurwitz algebra

with identity e, norm n, and conjugation x ’ x = bn (x, e)e ’ x.

(2) The map

•(x) = e (e x) = bn (e, x)e ’ x e = x e

is an automorphism of (S, ) (and (S, )) of order ¤ 3 and (S, ) = S• is a Petersson

algebra with respect to •.

Proof : (??) is easy and left as an exercise.

(??) Replacing x by e x and z by e in the identity (??):

bn (x, z)y = x (y z) + z (y x)

gives

x y = bn (e, x)y ’ e y (e x)

hence

(x y) e = y bn (x, e)e ’ e x = e (y e) (x e) e = (y e) (x e).

Thus • is an automorphism of (S, ), •3 (x) = x = x, x •2 (y) and

y = •(x)

(S, ) = S• as claimed.

In general a symmetric composition may not contain an idempotent. However:

(34.10) Lemma. Let (S, , n) be a symmetric composition algebra.

(1) If the cubic form bn (x x, x) is isotropic on S, then (S, ) contains an idempo-

tent. In particular there always exists a ¬eld extension L/F of degree 3 such that

(S, )L contains an idempotent e.

(2) For any nontrivial idempotent e ∈ S we have n(e) = 1.

Proof : (??) It su¬ces to ¬nd f = 0 with f f = »f , » ∈ F — so that e = f »’1

then is an idempotent. Let x = 0 be such that bn (x x, x) = 0. We have

(x x) (x x) = ’n(x)(x x)

by (??), so we take f = x x if n(x) = 0. If n(x) = 0, we may also assume that

x x = 0: if x x = 0 we replace x by x x and use again (??). Since x is isotropic

466 VIII. COMPOSITION AND TRIALITY

and n is nonsingular, there exists some y ∈ S such that n(y) = 0 and bn (x, y) = ’1.

A straightforward computation using (??) shows that

(x y + y x) (x y + y x) = (x y + y x) + 3bn (y, y x)x,

and

e = x y + y x + bn (y, y x)x

is an idempotent and is nonzero since

e x = (y x) x = bn (x, y)x = ’x.

(??) Since e = (e e) e = n(e)e, we have n(e) = 1.

(34.11) Remark. Lemma (??.??) is in fact a special case of Theorem (??) and

its proof is copied from the proof of implication (??) ’ (??) of (??).

Assume that char F = 3 and that F contains a primitive cube root of unity ω.

The existence of an automorphism of order 3 on a Hurwitz algebra C is equivalent

with the existence of a Z/3Z-grading:

(34.12) Lemma. Suppose that F contains a primitive cube root of unity ω.

(1) If • is an automorphism of C of order 3, then C (or C• ) admits a decomposition

C = C • = S0 • S1 • S2 ,

with

Si = { x ∈ C | •(x) = ω i x }

and such that

(a) Si Sj ‚ Si+j (resp. Si Sj ‚ Si+j ), with subscripts taken modulo 3,

(b) bn (Si , Sj ) = 0 unless i + j ≡ 0 mod 3.

In particular (S0 , , n) ‚ C• is a para-Hurwitz algebra of even dimension and S1

(resp. S2 ) is a maximal isotropic subspace of S1 • S2 .

(2) Conversely, any Z/3Z-grading of C de¬nes an automorphism • of order 3 of C,

hence a Petersson algebra C• .

Proof : Claim (??) follows easily from the fact that ω i , i = 0, 1, 2, are the eigen-

values of the automorphism •. For (??) we take the identity on degree 0 elements,

multiplication by ω on degree 1 elements and multiplication by ω 2 on degree 2

elements.

If • = 1, S0 in Lemma (??) must have dimension 2 or 4 (being a para-Hurwitz

algebra). We show next that C• is para-Hurwitz if dim S0 = 2. The case dim S0 = 4

and dim C• = 8 corresponds to a di¬erent type of symmetric composition, discussed

in the next subsection.

(34.13) Proposition (Elduque-P´rez). Let F be a ¬eld of characteristic not 3, let

e

C be a Hurwitz algebra over F , let • be an F -automorphism of C of order 3. Then

S0 = { x ∈ C | •(x) = x }

is a para-Hurwitz algebra of dimension 2 or 4. The Petersson algebra C • is iso-

morphic to a para-Hurwitz algebra if and only if dim S0 = 2.

§34. SYMMETRIC COMPOSITIONS 467

Proof : The ¬rst claim is clear. For the second claim we use an argument in Elduque-

P´rez [?, proof of Proposition 3.4]. If dim C = 2 there is nothing to prove. Thus

e

by Remark (??) we may assume that F contains a primitive cube root of unity. To

simplify notations we denote the multiplication in C by (x, y) ’ xy and we put

n = NC for the norm of C. Let xi ∈ Si , i = 1, 2; we have x2 = n(xi ) = 0 by

i

Lemma (??), so that

bn (x1 x2 , x2 x1 ) = bn (x1 , x2 )2

by (??). Furthermore (x1 x2 )(x2 x1 ) = x1 (x2 x2 )x1 = 0 (by Artin™s theorem, see (??))

and

(x1 x2 )2 ’ bn (x1 x2 , 1)x1 x2 + n(x1 x2 ) · 1 = 0

implies that (x1 x2 )2 = ’bn (x1 , x2 )x1 x2 . Choosing x1 , x2 such that bn (x1 , x2 ) =

’1, we see that e1 = x1 x2 is an idempotent of C and it is easily seen that e2 =

1 ’ e1 = x2 x1 . We claim that if dim S0 = 2, then e1 = y1 y2 for any pair (y1 , y2 ) ∈

S1 — S2 such that bn (y1 , y2 ) = ’1. We have S0 = F · e1 • F · e2 if dim S0 = 2, so

that the claim will follow if we can show that bn (e1 , y1 y2 ) = 0. Let y1 = »x1 + x1

with bn (x1 , x2 ) = 0. By using (??) and the fact that n(Si ) = 0 for i = 1, 2, we

deduce

bn (e1 , y1 y2 ) = bn x1 x2 , (»x1 + x1 )y2

= n(x1 )bn (x2 , »y2 ) + bn (x1 x2 , x1 y2 )

= ’bn (x1 y2 , x1 x2 ).

However x1 x2 satis¬es

(x1 x2 )2 ’ bn (x1 x2 , 1)x1 x2 + n(x1 x2 ) = 0

hence (x1 x2 )2 = 0, since bn (x1 x2 , 1) = ’bn (x1 , x2 ) = 0. Since the algebra S0 is

´tale, we must have x1 x2 = 0 and, as claimed, bn (e1 , y1 y2 ) = 0. Similarly we have

e

e2 = y2 y1 for (y1 , y2 ) ∈ S1 — S2 such that bn (y1 , y2 ) = ’1. It follows that

e1 y1 = (1 ’ e2 )y1 = (1 ’ y2 y1 )y1 = y1 ,

y1 e1 = y1 (y1 y2 ) = 0 = e2 y1 ,

y1 e2 = y 1 , e2 y2 = y 2 = y 2 e1 and e1 y2 = 0 = y2 e2 ,

so that

S1 = { x ∈ C | e1 x = x = xe2 }

and

S2 = { x ∈ C | e2 x = x = xe1 }.

The element

e = ω 2 e1 + ωe2

is a para-unit of C• , since

e x = (ω 2 e1 + ωe2 )(ω 2i x) = ’(ωe1 + ω 2 e2 )ω 2i x = ’x

for x ∈ Si and since

e (ωe1 ’ ω 2 e2 ) = (ω 2 e1 + ωe2 )(’ω 2 e1 + ω 2 e2 ) = (’ωe1 + ω 2 e2 )

for ωe1 ’ ω 2 e2 ∈ e⊥ ‚ S0 . The claim then follows from Lemma (??).

468 VIII. COMPOSITION AND TRIALITY

34.C. Cubic separable alternative algebras. Following Faulkner [?] we

now give another approach to symmetric composition algebras over ¬elds of char-

acteristic not 3. We ¬rst recall some useful identities holding in cubic alternative

algebras. Let A be a ¬nite dimensional unital separable alternative F -algebra of

degree 3 and let

PA,a (X) = X 3 ’ TA (a)X 2 + SA (a)X ’ NA (a)1

be its generic minimal polynomial. The trace TA is linear, the form SA is quadratic

and the norm NA is cubic. As was observed in the introduction to this chapter we

have

NA (X ’ a · 1) = PA,a (X), NA (xy) = NA (x)NA (y), and TA (xy) = TA (yx).

Let

bSA (x, y) = SA (x + y) ’ SA (x) ’ SA (y),

x# = x2 ’ TA (x)x + SA (x) · 1

and

x — y = (x + y)# ’ x# ’ y # .

Note that

NA (x) = xx# = x# x.

Observe that the #-operation and the —-product are de¬ned for any cubic

algebra. They will be systematically used in Chapter IX for cubic Jordan algebras.

(34.14) Lemma. (1) NA (1) = 1, SA (1) = TA (1) = 3, 1# = 1, 1 — 1 = 2,

(2) (xy)# = y # x# ,

(3) SA (x) = TA (x# ), SA (x# ) = TA (x)NA (x), NA (x# ) = NA (x)2 ,

(4) bSA (x, y) = TA (x — y).

(5) NA (x + »y) = »3 NA (y) + »2 TA (x · y # ) + »TA (x# · y) + NA (x) for x, y ∈ A,

and » ∈ F .

(6) The coe¬cient of ±βγ in NA (±x + βy + γz) is TA x(y — z) and TA x(y — z)

is symmetric in x, y, and z.

(7) bSA (x, 1) = 2TA (x),

(8) x — 1 = TA (x) · 1 ’ x,

(9) TA (xy) = TA (x)TA (y) ’ bSA (x, y),

(10) TA (xz)y = TA x(zy) .

Proof : We may assume that F is in¬nite and identify polynomials through their

coe¬cients. NA (1) = 1 follows from the multiplicativity of NA , so that 1# = 1

and 1 — 1 = 2. Putting a = 1 in NA (X ’ a · 1) = PA,a (X) gives PA,1 (X) =

(X ’ 1)3 NA (1) = (X ’ 1)3 , hence SA (1) = TA (1) = 3.

By density it su¬ces to prove (??) for x, y such that NA (x) = 0 = NA (y).

Then (xy)# = (xy)’1 NA (xy) = y ’1 NA (y)x’1 NA (x) = y # x# .

Again by density, it su¬ces to prove (??) for x such that NA (x) = 0. We then

have NA (x ’ ») = NA (1 ’ »x’1 )NA (x). Comparing the coe¬cients of » gives (??),

and (??) follows by linearizing (??).

(??) follows by computing NA (x + »y) = NA (xy ’1 + »)NA (y).

The ¬rst claim of (??) follows by computing the coe¬cient of ±βγ in NA ±x +

(βy + γz) (and using (??)). The last claim of (??) then is clear by symmetry.

§34. SYMMETRIC COMPOSITIONS 469

(??) follows from (??), since

bSA (x, 1) = TA (x — 1) = TA x(1 — 1) = 2TA (x)

and (??) implies (??).

For (??) we have

bSA (x, y) = TA (x — y)1 = TA (x — 1)y

= TA TA (x) · 1 ’ x y = TA (x)TA (y) ’ TA (xy).

Finally, by linearizing

TA x(yx) = TA (xy)x = TA (yx2 ) = TA y x# + xTA (x) ’ SA (x)1 ,

we obtain

TA x(yz) + TA z(yx) = TA (xy)z + TA (zy)x

= TA y(x — z) + TA (yx)TA (z) + TA (yz)TA (x)

’ TA (y)bSA (x, z)

= TA y(x — z) + TA (yx)TA (z) + TA (yz)TA (x)

+ TA (xz)TA (y) ’ TA (x)TA (y)TA (z),

so that by (??) TA x(yz) + TA z(yx) = TA (xy)z + TA (zy)x is symmetric in

x, y and z. It follows that

TA x(yz) + TA z(yx) = TA y(xz) + TA z(xy)

and TA (xz)y = TA x(zy) , as claimed.

(34.15) Proposition. A cubic alternative algebra is separable if and only if the

bilinear trace form T (x, y) = TA (xy) is nonsingular.

Proof : By (??) T is associative, hence the claim follows from Dieudonn´™s Theo-

e

rem (??).

We recall:

∼

(34.16) Proposition. For any isomorphism ± : A ’ A of cubic unital alterna-

’

tive algebras we have

TA ±(x) = TA (x), SA ±(x) = SA (x), NA ±(x) = NA (x).

Proof : The polynomial pA ,±(a) (X) is a minimal generic polynomial for A, hence

the claim by uniqueness.

(34.17) Theorem. Let A be a cubic separable unital alternative algebra over F of

dimension > 1. Then either :

(1) A L, for some unique (up to isomorphism) cubic ´tale algebra L over F ,

e

(2) A F —Q where Q is a unique (up to isomorphism) quaternion algebra over F ,

(3) A F — C where C is a unique (up to isomorphism) Cayley algebra over F ,

(4) A is isomorphic to a unique (up to isomorphism) central simple associative

algebra of degree 3.

In particular such an algebra has dimension 3, 5 or 9. In case (??) the generic

minimal polynomial is the characteristic polynomial, in case (??) and (??) the

product of the generic minimal polynomial pF,a (X) = X ’ a of F with the generic

minimal polynomial pC,c (X) = X 2 ’ TC (c)X + NC (c) · 1 of C = Q or C = C and

in case (??) the reduced characteristic polynomial.

470 VIII. COMPOSITION AND TRIALITY

Proof : The claim is a special case of Theorem (??).

Let 1An denote the category of central simple algebras of degree n + 1 over

F . Let I : Sepalt m’1 (2) ’ Sepalt m (3) be the functor C ’ F — C. Theorem

(??) gives equivalences of groupoids Sepalt 3 (3) ≡ S3 , Sepalt 5 (3) ≡ I(1A1 ), and

Sepalt 9 (3) I(G2 ) 1A2 .

We assume from now on (and till the end of the section) that F is a ¬eld of

characteristic di¬erent from 3. Let A be cubic alternative separable over F and let

A0 = { x ∈ A | TA (x) = 0 }.

1 1 1

Since x = 3 TA (x) · 1 + x ’ 3 TA (x) · 1 and TA x ’ 3 TA (x) = 0 we have A =

F · 1 • A0 and the bilinear trace form T : (x, y) ’ TA (xy) is nonsingular on A0 . By

Lemma (??) the polar of the quadratic form SA on A0 is ’T . Thus the restriction

of SA to A0 is a nonsingular quadratic form.

We further assume that F contains a primitive cube root of unity ω and set

µ = 1’ω . We de¬ne a multiplication on A0 by

3

1

(34.18) x y = µxy + (1 ’ µ)yx ’ 3 TA (yx)1.

This type of multiplication was ¬rst considered by Okubo [?] for matrix algebras

and by Faulkner [?] for cubic alternative algebras.

(34.19) Proposition. The algebra (A0 , ) is a symmetric composition algebra with

1

norm n(x) = ’ 3 SA (x).

Proof : The form n is nonsingular, since SA is nonsingular. We check that

1

(x y) x = x (y x) = ’ 3 SA (x)y = n(x)y.

Lemma (??) will then imply that (A0 , ) is a symmetric composition algebra. We

have 3µ(1 ’ µ) = 1. It follows that

(34.20)

(x y) x = x (y x) = µ2 xyx + (1 ’ µ)2 xyx + µ(1 ’ µ)(yx2 + x2 y)

’ 3 TA (xy)x ’ 3 µTA (xyx)1 ’ 1 (1 ’ µ)TA (xyx)1

1 1

3

= [1 ’ 2µ(1 ’ µ)]xyx + µ(1 ’ µ)(yx2 + x2 y)

’ 1 TA (xy)x ’ 3 TA (xyx)1

1

3

= 1 (xyx + yx2 + x2 y) ’ 3 TA (xy)x ’ 3 TA (xyx)1.

1 1

3

By evaluating TA on the generic polynomial, we obtain 3NA (x) = TA (x3 ) for ele-

ments in A0 . Thus

x3 + SA (x)x ’ 1 TA (x3 )1 = 0

(34.21) 3

holds for all x ∈ A0 . Since it su¬ces to prove (??) over a ¬eld extension, we may

assume that F is in¬nite. Replacing x by x + »y in (??), the coe¬cient of » must

then be zero. Hence we are lead to the identity

xyx + yx2 + x2 y ’ TA (xy)x + SA (x)y ’ TA (xyx)1 = 0

for all x, y ∈ A0 , taking into account that bSA (x, y) = ’TA (xy) on A0 . Combining

this with equation (??) shows that

(x y) x = x (y x) = ’ 1 SA (x)y = n(x)y

3

as claimed.

§34. SYMMETRIC COMPOSITIONS 471

Hence we have a functor 1 C : Sepalt m+1 (3) ’ Scomp m for m = 2, 4 and 8

given by A ’ (A0 , n). We now construct a functor 1A in opposite direction; a

straightforward computation shows that (??) is equivalent to

(34.22) xy = (1 + ω)x y ’ ωy x + bn (x, y) · 1

for the multiplication in A0 ‚ A. Thus, given a symmetric composition (S, ), it is

natural to de¬ne a multiplication (x, y) ’ x · y = xy on A = F • S by (??) for x,

y ∈ S, and by 1 · x = x = x · 1. Let 1A be the functor (S, ) ’ (F • S, ·).

(34.23) Theorem (Elduque-Myung). The functors 1 C and 1A de¬ne an equiva-

lence of groupoids

Sepalt m+1 (3) ≡ Scomp m

for m = 2, 4 and 8.

Proof : We ¬rst show that A = 1A(S) = F • S is a separable alternative algebra of

degree 3: Let x = ±1 + a, ± ∈ F and a ∈ S. We have

x2 = ±2 + bn (a, a) 1 + 2±a + a a

and

xx2 = x2 x = [±3 + 3±bn (a, a) + bn (a a, a)]1 + [3±2 + 3n(a)]a + 3±(a a).

It follows that

x3 ’ 3±x2 + 3±2 ’ 3n(a) x = [±3 ’ 3n(a)± + bn (a a, a)]1

so that elements of A satisfy a polynomial condition of degree 3

pA,x (X) = X 3 ’ TA (x)X 2 + SA (x)X ’ NA (x)1 = 0

with

SA (x) = 3±2 ’ 3n(a)

TA (x) = 3±,

and

NA (x) = ±3 ’ 3±n(a) + bn (a a, a)

for x = ±1 + a. To show that A is of degree 3 we may assume that the ground

¬eld F is in¬nite and we need an element x ∈ A such that the set {1, x, x2 } is

linearly independent. Because x2 = x x + (x, x)1 for x ∈ S, it su¬ces to have

x ∈ S such that {1, x, x x} is linearly independent. Since TA (1) = 3, while

TA (x) = 0 = TA (x x), the only possible linear dependence is between x and x x.

If {x, x x} is linearly dependent for all x ∈ S, there is a map f : S ’ F such

that x x = f (x)x for x ∈ S. By the following Lemma (??) f is linear. Since

x x x = n(x)x we get n(x) = f (x)2 . This is only possible if dimF S = 1. We

next check that A is alternative. It su¬ces to verify that

a2 b = a(ab) and ba2 = (ba)a for a, b ∈ S.

We have

a2 b = [a a + (a, a)]b = (1 + ω)(a a) b ’ ωb (a a) + bn (a a, b)

472 VIII. COMPOSITION AND TRIALITY

and

a(ab) = a[(1 + ω)a b ’ ωb a + bn (a, b)]

= (1 + ω)[a (a b) ’ ω(a b) a + bn (a, a b)]

’ ω[(1 + ω)a (b a) ’ ω(b a) a + bn (a, b a)] + bn (a, b)a.

By (??) we have

(a a) b + (b a) a = bn (a, b)a = b (a a) + a (a b).

This, together with the identities

bn (a, a b) = bn (a a, b) = bn (b, a a) = bn (b a, a) = bn (a, b a)

which follow from the associativity of n, implies that a2 b = a(ab). The proof of

ba2 = (ba)a is similar. Thus A is alternative of degree 3. We next check that A is

separable. We have for x = ± + a, y = β + b,

xy = ±β + βa + ±b + (1 + ω)a b ’ ωb a + bn (a, b)

so that

T (x, y) = TA (xy) = 3±β + 3bn (a, b)

is a nonsingular bilinear form. Since the trace form of a cubic alternative algebra is

associative (Lemma (??)), A is separable by Dieudonn´™s Theorem (??). We ¬nally

e