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semilinear with respect to the ¬rst variable and ρ2 -semilinear with respect to the
second variable and such that β1 (x, x) = β(x) for all x ∈ V : the di¬erence β of
two such maps would be such that β(x, x) = 0, hence β(x, y) = ’β(y, x). This
is incompatible with the semilinearity properties of β. Over a separable closure
Fsep of F such a map exists, since L — Fsep is split and, by Proposition (??),
VFsep is of Cayley type. Thus, by descent, such a β1 resp. β2 exists. Furthermore
x —i y = βi (x — y) satis¬es the identities of a cyclic composition, since it does
over Fsep . The last claim follows from Example (??).

It follows from Proposition (??) that the twisted composition (S — L, L, Q, β)
is independent of the choice of a generator of Gal(L/F ).
(36.13) Proposition. Cyclic compositions over F are classi¬ed by the pointed set
H 1 (F, Spin8 Z/3Z).
Proof : In view of Propositions (??) and (??) any cyclic composition over a sepa-
rable closure Fsep of F is isomorphic to a Cayley composition with multiplication —
as described in Example (??) and ρ either given by (x0 , x1 , x2 ) ’ (x1 , x2 , x0 ) or
(x0 , x1 , x2 ) ’ (x2 , x0 , x1 ). It then follows as in the proof of Proposition (??) that
the group of automorphisms of the cyclic composition
C — (F — F — F ), n ⊥ n ⊥ n, F — F — F, ρ, — R

is isomorphic to Spin(ns )(R) Z/3Z. The claim then follows by constructing a
representation G ’ GL(W ) such that Spin(ns ) Z/3Z = AutG (w) as in the proof
of (??), (??). We let this construction as an exercise.

Let (V, L, Q, ρ, —) be a cyclic composition. We have a homomorphism
p : Aut(V, L, ρ, —) ’ Aut(L) = A3
induced by restriction. Assume that p is surjective and has a section s. Then s
de¬nes an action of A3 on (V, L, Q, ρ, —).
496 VIII. COMPOSITION AND TRIALITY


(36.14) Proposition. Let F be a ¬eld of characteristic not equal to 3. For any
faithful ρ-semilinear action of A3 on (V, L, Q, ρ, —), V0 = V A3 carries the structure
of a symmetric composition algebra and V = V0 — L. In particular, if dimL V = 8,
then p has only two possible sections (up to isomorphism) over F sep .
Proof : The ¬rst claim follows by Galois descent and the second from the fact
that over Fsep we have only two types of symmetric composition algebras (The-
orem (??)).
(36.15) Corollary. Datas (V, L, Q, ρ, —), p, s where (V, L, Q, ρ, —) is a cyclic com-
position, are classi¬ed either by H 1 (F, G — A3 ) where G is of type G2 , if the sec-
tion s de¬nes a para-Hurwitz composition, or by H 1 F, PGU3 (K) — A3 where
K = F [X]/(X 2 + X + 1).
Proof : This corresponds to the two possible structures of symmetric composition
algebras over Fsep .
Let (V, L, ρ, —) be a cyclic composition of dimension 8. We write ρ V for the
L-space V with the action of L twisted through ρ and put ρ Q(x) = ρ Q(x) . Let
x (y) = x — y and rx (y) = y — x.

(36.16) Proposition. The map
0 2
x
∈ EndL (ρ V • ρ V ),
x’ x∈V
rx 0
extends to an isomorphism of algebras with involution
2

±V : C(V, Q), „ ’ EndL (ρ V • ρ V ), σQ

2
where σQ is the involution associated with the quadratic form Q = ρ Q ⊥ ρ Q. In
particular ±V restricts to an isomorphism
2

C0 (V, Q), „ ’ EndL (ρ V ), σρ Q — EndL (ρ V ), σρ2Q .

Proof : The existence of ±V follows from the universal property of the Cli¬ord
2
algebra, taking the identities (x — y) — x = ρ Q(x)y and x — (y — x) = ρ Q(x)y into
account. It is an isomorphism because C(V, Q) is central simple over F . The
fact that ± is compatible with involutions is a consequence of the formula (??)
above.
Proposition (??) is a twisted version of Proposition (??) and can be used to
deduce analogues of Propositions (??) and (??). The proofs of the following two
results will only be sketched.
(36.17) Proposition. Let t be a proper similitude of (V, Q), with multiplier µ(t).
There exist proper similitudes u, v of (V, Q) such that
µ(v)’1 v(x — y) = u(x) — t(y),
µ(u)’1 u(x — y) = t(x) — v(y),
µ(t)’1 t(x — y) = v(x) — u(y)
for all x, y ∈ V . If t ∈ RL/F O+ (V, Q) (F ) is such that Sn(t) = 1, then u, v can
be chosen in RL/F (O+ (V, Q))(F ).
§36. TWISTED COMPOSITIONS 497

2
Proof : We identify C(V, Q) with EndL (ρ V • ρ V ), σQ through ±V . The map
0 2
tx
∈ EndL (ρ V • ρ V )
x’ ’1
µ(t) rtx 0
extends to an automorphism t of C(V, Q) whose restriction to C0 (V, Q) is C0 (t).
Thus we may write t = Int u v where u , v are similitudes of Q and
0
0
’1
u 0 0 u 0 0
x tx
= ’1
0 v rx 0 0 v µ(t) rtx 0
or
and v (y — x) = u (y) — t(x)µ(t)’1 .
u (x — y) = t(x) — v (y)
Putting u = u µ(u )’1 and v = v and using that
2
µ(u)’1 = µ(u ) = ρµ(t) ρ µ(v)
gives the ¬rst two formulas. The third formula follows by replacing x by y — x in
the ¬rst one (compare with the proof of Proposition (??)).
(36.18) Proposition. Assume that char F = 2. For R ∈ Alg F we have

RL/F Spin(V, Q) (R)
3
{ (v, u, t) ∈ RL/F O+ (V ) (R) | v(x — y) = u(x) — t(y) },
the group A3 operates on Spin(V, Q), and Spin(V, Q)A3 is a group scheme over F
such that RL/F [Spin(V, Q)A3 ]L = Spin(V, Q).
Proof : Proposition (??) follows from Proposition (??) as Proposition (??) follows
from Proposition (??).
The computation given above of the Cli¬ord algebra of a cyclic composition can
be used to compute the even Cli¬ord algebra of an arbitrary twisted composition:
(36.19) Proposition. Let “ = (V, L, Q, β) be a twisted composition. There exists
an isomorphism
±V : C0 (V, Q) = C EndL (V ), σQ ’ ρ EndL (V ) — ∆
where ∆(L) is the discriminant algebra of L and ρ is a generator of the group
Gal L — ∆(L)/L A3 .
Proof : By Proposition (??) there exists exactly one cyclic composition (with re-
spect to ρ) on (V, L, Q, β)—∆ and by Proposition (??) we then have an isomorphism
ρ2

±V —∆ : C0 (V, Q) — ∆ ’ EndL—∆ ρ (V — ∆) — EndL—∆
’ (V — ∆)
of L — ∆-algebras with involution. By composing with the canonical map
C0 (V, Q) ’ C0 (V, Q) — ∆
2
and the projection ρ EndL (V ) — ∆ — ρ EndL (V ) — ∆ ’ ρ
EndL (V ) — ∆ we
obtain a homomorphism of central simple algebras
±V : C0 (V, Q) = C EndL (V ), σQ ’ ρ EndL (V ) — ∆
which is an isomorphism by dimension count.
498 VIII. COMPOSITION AND TRIALITY


36.C. Twisted Hurwitz compositions. In this section we ¬rst extend the
construction of a twisted composition C — L given in Example (??) for L cyclic to
an arbitrary cubic ´tale algebra L and a para-Hurwitz algebra (C, , n). We refer
e
to ?? for details on cubic ´tale algebras.
e
Let ∆ be the discriminant (as a quadratic ´tale algebra) of L, let ι be the
e
generator of Gal(∆/F ) and let C be a Hurwitz algebra over F . Since L — ∆ is
cyclic over ∆, by Proposition (??) there exists a cyclic composition (x, y) ’ x — y
on C — L — ∆ for each choice of a generator ρ of Gal(L — ∆/∆). The automorphism
˜ = π — 1 — ι of C — L — ∆ is ι-semilinear and satis¬es ˜2 = 1 and ˜(x — y) = ˜(y) —˜(x)
ι ι ι ι ι
for x, y ∈ C — L — ∆. Let
V = { x ∈ C — L — ∆ | ˜(x) = x }
ι
be the corresponding descent (from L—∆ to L). Since ˜(x—y) = ˜(y)—˜(x), the map
ι ι ι
β(x) = x—x restricts to a quadratic map β on V . The restriction Q of n—1—1 to V
takes values in L and is nonsingular. Thus (V, L, Q, β) is a twisted composition.
We write it “(C, L). A twisted composition similar to a composition “(C, L) is
called a twisted Hurwitz composition over L. If C is a Cayley algebra we also say
that “(C, L) is a twisted composition of type G2 . The underlying quadratic space
(V, Q) of “(C, L) is extended from the quadratic space (V0 , Q0 ) over F with
V0 = { x ∈ C — ∆ | π — ι(x) = x }
and Q0 is the restriction of Q to V0 . The space (V0 , Q0 ) is called the ∆-associate
of (C, n) by Petersson-Racine [?] (see also Loos [?]) and is denoted by (C, N )∆ . In
[?] a K-associate (U, q)K is attached to any pointed quadratic space (U, q, e) and
any ´tale quadratic algebra K:
e
(U, q)K = { x ∈ U — K | π — ιK (x) = x }
where π is the re¬‚ection with respect to the point e. For any pair of quadratic ´tale
e
algebras K1 , K2 with norms n1 , n2 , (K1 , n1 )K2 = (K2 , n2 )K1 is the ´tale quadratic
e
algebra
K1 — K2 = { x ∈ K1 — K2 | (ι1 — ι2 )(x) = x }.
Recall that K — K F — F.
(36.20) Lemma. Let (U, p) be a pointed quadratic space. There are canonical
isomorphisms
(U, p)K1 (U, p)K1 —K2 (U, p)F —F (U, p).
and
K2

Reference: See Loos [?].
In what follows we use the notation [a] for the 1-dimensional regular quadratic
form q(x) = ax2 (and, as usual, a for the bilinear form b(x, y) = axy, a ∈ F — ).
(36.21) Lemma. Let (C, n) be a Hurwitz algebra (with unit 1) and let (C 0 , n0 ) =
1⊥ be the subspace of (C, n) of elements of trace zero. We have, with the above
notations, (V0 , Q0 ) = (C, n)∆ and :
(1) If char F = 2, V0 = F · 1 • C 0 , Q0 = [1] ⊥ δ — (C 0 , n0 ).
(2) If char F = 2, let u ∈ C, ξ ∈ ∆ with TC (u) = 1 = T∆ (ξ) and let w =
u — 1 + 1 — ξ. Then V0 = F · w • C 0 , Q0 (w) = n(u) + ξ 2 + ξ, bQ0 (w, x) = bn (x, u)
and Q0 (x) = n(x) for x ∈ C 0 .
§36. TWISTED COMPOSITIONS 499


Proof : Claims (??) and (??) can be checked directly. The ¬rst claim then follows
easily.
(36.22) Proposition. Two twisted Hurwitz compositions “(C1 , L1 ) and “(C2 , L2 )
are isomorphic if and only if L1 L2 and C1 C2 .
Proof : The if direction is clear. We show the converse. If the compositions
“(C1 , L1 ) and “(C2 , L2 ) are isomorphic, we have by de¬nition an isomorphism

L1 ’ L2 and replacing the L2 -action by the L1 -action through this isomorphism,

we may assume that L1 = L2 = L. In particular the quadratic forms Q1 and Q2
are then isometric as quadratic spaces over L. By Springer™s theorem there exists
an isometry (C1 , n1 )∆ (C2 , n2 )∆ . Since ∆ — ∆ F — F , (C1 , n1 ) (C2 , n2 )
follows from Lemma (??), hence C1 C2 by Theorem (??).
Let G be the automorphism group of the split Cayley algebra. Since G =
SpinS3 (see ??), we have a homomorphism
8
G — S3 ’ Spin8 S3 .
(36.23) Proposition. Twisted compositions of type G2 are classi¬ed by the image
of H 1 (F, G — S3 ) in H 1 (F, Spin8 S3 ).
Proof : A pair (φ, ψ) where φ is an automorphism of C and ψ is an automorphism
of L de¬nes an automorphism “(φ, ψ) of “(C, L). The map
H 1 (F, G — S3 ) = H 1 (F, G) — H 1 (F, S3 ) ’ H 1 (F, Spin8 S3 )
corresponds to [C] — [L] ’ [“(C, L)].
The following result is due to Springer [?] for L cyclic and V of dimension 8.
(36.24) Theorem. Let (V, L, Q, β) be a twisted composition and let N be the cubic
form N (x) = Q x, β(x) , x ∈ V . The following conditions are equivalent:
(1) The twisted composition (V, L, Q, β) is similar to a twisted Hurwitz composition.
(2) The cubic form N (x) is isotropic.
(3) There exists e ∈ V with β(e) = »e, » = 0, and Q(e) = »# .
Proof of (??) ’ (??): We may assume that (V, L, Q, β) is a Hurwitz composition.
The element x = c — 1 — ξ ∈ C — L — ∆ lies in V if tC (c) = 0 = t∆ (ξ). For such
an element we have β(x) = c2 — 1 — ξ 2 ∈ L. Since tC (c) = 0, c2 = nC (c) and so
N (x) = 0 if char F = 2. If char F = 2, x = 1 — 1 — 1 lies in V , β(x) = x, and
N (x) = bQ (x, x) = 0.
For the proof of (??) ’ (??) we need the following lemma:
(36.25) Lemma. Any element x of a twisted composition (V, L, Q, β) satis¬es the
identity
β 2 (x) + Q(x)β(x) = N (x)x.
Proof : Since over the separable closure Fsep any composition is cyclic, it su¬ces to
prove the formula for a cyclic composition and for β(x) = x—x. In view of Formulas
(??) and (??) (p. ??), we have
(x — x) — (x — x) + (x — x) — x — x = ρ bQ (x — x, x) x
hence the assertion, since (x—x)—x = ρ2 Q(x) x (and the product — is ρ-semilinear
with respect to the ¬rst variable).
500 VIII. COMPOSITION AND TRIALITY


Proof of (??) ’ (??): First let x = 0 be such that β(x) = 0 and N (x) = 0. If
Q(x) = 0 we take e = β(x) and apply Lemma (??). If Q(x) = 0, we replace x by
β(x) and apply (??) to see that our new x satis¬es Q(x) = 0 and β(x) = 0. Thus
we may assume that Q(x) = 0 and β(x) = 0. Let y ∈ V be such that Q(y) = 0 and
bQ (x, y) = ’1. Let
β(u, v) = β(u + v) ’ β(u) ’ β(v).
We claim that
(36.26) β β(x, y) = β(x, y) + f · x
for some f ∈ F . We extend scalars to L — ∆ and so assume that β(x) = x — x is
cyclic. It su¬ces to check that f ∈ ∆ © L. Since β(x, y) = x — y + y — x, we have
β β(x, y) = (x — y + y — x) — (x — y + y — x).
Applying formulas (??) and (??) (p. ??), we ¬rst obtain
(y — x) — x + (x — x) — y = (y — x) — x = ’x
x — (x — y) + y — (x — x) = x — (x — y) = ’x
since β(x) = x — x = 0 and bQ (x, y) = ’1. Applying again formulas (??) and (??)
then give
(x — y) — (x — y) = ρ bQ (x — y, y) x + y — x
(x — y) — (y — x) = 0
(y — x) — (x — y) = ρ bQ (y — x, y) x
(y — x) — (y — x) = ρ2 bQ (y, y — x) x + x — y


and
β β(x, y) = β(x, y) + f · x
with
f = ρ bQ (x — y, y) + ρ bQ (y — x, y) + ρ2 bQ (y, y — x)
= ρ2 bQ β(y), x + bQ β(y), x + ρ bQ β(y), x = TL—∆/∆ bQ β(y), x
hence f ∈ ∆ © L. Let µ ∈ L be such that TL (µ) = f ; we claim that e = µx + β(x, y)
satis¬es β(e) = e:
β(e) = β(µx) + β β(x, y) + β µx, β(x, y)
= β(x, y) + f · x + β µx, β(x, y) .
To compute β µx, β(x, y) we assume that β(x) = x — x is cyclic. We have:
β µx, β(x, y) = (µx) — (x — y + y — x) + (x — y + y — x) — (µx)
= ρ(µ)[x — (x — y + y — x)] + ρ2 (µ)[(x — y + y — x) — x]
= ’ρ(µ)x ’ ρ2 (µ)x = (µ ’ f )x.
This implies that β(e) = β(x, y) + µx = e. The relation »# = Q(e) follows from
β(e) = »e and Lemma (??), since replacing β(e) by »e we see that »# »e+Q(e)»e =
2Q(e)»e.
§36. TWISTED COMPOSITIONS 501


Proof of (??) ’ (??): We ¬rst construct the Hurwitz algebra C. Replacing β by
the similar composition »’1 β, we may assume that β(e) = e. This implies that
Q(e) = 1. Let ∆ be the discriminant algebra of L. We also call β the extension
β — 1 of β to L — ∆. Let ρ1 , ρ2 be the generators of Gal(L — ∆/∆) and let ι be
the generator of Gal(∆/F ), so that (putting ι = 1 — ι on L — ∆ and V — ∆), we
have ρ2 = ρi+1 and ιρi = ρi+1 ι. Let βi (x, y) = x —i y be the two unique extensions
i
of β to L — ∆ as cyclic compositions with respect to ρi . By uniqueness we have
(see Examples (??) and (??))
β1 (x, y) = β2 (y, x) and ιβ1 (x, y) = β2 (ιx, ιy).
Let π : V ’ V be the hyperplane re¬‚ection
π(x) = x = ’x + bQ (e, x)e

as well as its extension to V — ∆ and let •i : V — ∆ ’ V — ∆ be the ρi -semilinear

map given by •i (x) = βi (x, e).
(36.27) Lemma. The following identities hold for the maps •i :
(1) •i π = π•i
•2 = •i+1
(2) i
(3) ι•i ι = •i+1

Proof : (??) We have β1 (x, e) = ’β1 (x, e) + ρ1 bQ (e, x) e so that
β1 (x, e) = β1 (x, e) ’ bQ e, β1 (x, e) e
’ ρ1 bQ (e, x) e + bQ e, ρ1 bQ (e, x) e e = β1 (x, e)
(??) By (??)
β1 β1 (x, e), e = β1 β1 (x, e), e = ρ2 bQ (e, x) e ’ β1 (e, x) = β1 (e, x)
using that β1 β1 (x, y), z + β1 β1 (z, y), x = ρ2 bQ (x, z) y, a formula which follows
from (??) (p. ??).
(??) follows from ιβ1 (x, y) = β2 (ιx, ιy).
We next de¬ne a multiplication γ1 on V — ∆ by
γ1 (x, y) = β1 •2 (x), •1 (y) .
(36.28) Lemma. The multiplication γ1 satis¬es the following properties:
(1) Q γ1 (x, y) = Q(x)Q(y)
(2) γ1 (x, e) = γ1 (e, x) = x
(3) γ1 (x, y) = γ1 (y, x)
(4) ιγ1 (x, y) = γ1 (ιy, ιx)
(5) γ1 •1 (x), •1 (y) = •1 γ1 (x, y)

Proof : (??) We have
Q γ1 (x, y) = Q β1 •2 (x), •1 (y) = ρ1 Q •2 (x) ρ2 Q •1 (y)
= ρ1 ρ2 Q(x) ρ2 ρ1 Q(y) = Q(x)Q(y)
(??) We have γ1 (x, e) = β1 β2 (x, e), e = x and similarly γ1 (e, x) = x.
502 VIII. COMPOSITION AND TRIALITY


(??) In view of (??) and (??) γ1 is a Hurwitz multiplication with conjugation
x ’ x and (??) holds for any multiplication of a Hurwitz algebra, see (??).
(??) We have
ιγ1 (x, y) = ιβ1 •2 (x), •1 (y) = β2 ι•2 (x), ι•1 (y)
= β2 •1 ι(x), •2 ι(y) = β1 •2 ι(y), •1 ι(x) = γ1 (ιy, ιx)
(??) Using (??) we have
•1 γ1 (x, y) = β1 γ1 (y, x), e = β1 β1 •2 (y), •1 (x) , e
= ρ2 bQ •2 (y), e •1 (x) ’ β1 β1 e, •1 (x) , •2 (y)
= β1 β1 e, •1 (x) , bQ •2 (y), e e ’ β1 β1 e, •1 (x) , •2 (y)
= β1 β1 e, •1 (x) , •2 (y) = β1 •2 •1 (x), •2 (y)
= γ1 •1 (x), •1 (y)


We go back to the proof of Theorem (??). It follows from Lemma (??) that
γ1 ιπ(x), ιπ(y) = ιπ γ1 (x, y)
Since ιπ•1 = •2 ιπ, the automorphisms {•1 , ιπ} of V — ∆ de¬ne a Galois action
1
of the group Gal(L — ∆/F ) = S3 on the Hurwitz algebra V — ∆. Let C be the
descended Hurwitz algebra over F , so that
V —∆ =C —L—∆
Observe that V on the left is the subspace of elements of V — ∆ ¬xed by Id V — ι
and that C — L on the right is the subspace of elements of C — L — ∆ ¬xed by ιπ.
We only consider the case where char F = 2 and leave the case char F = 2 as an
exercise. Let V = Le ⊥ V 0 , V 0 = e⊥ , and let d ∈ L — ∆ be a generator of the
discriminant algebra ∆ such that (1 — ι)(d) = ’d. The map
φ: V — ∆ ’ C — L — ∆
given by φ ( e + v ) — s = 1 — — s + v — ds is such that φ —¦ ι = (π — ι) —¦ φ. Thus
the image of V in C — L — ∆ can be identi¬ed with L — 1 ⊥ C 0 — L — 1 and β is
the restriction of the twisted Hurwitz composition on C — L — ∆. This shows that
(??) implies (??).
(36.29) Corollary. If L is not a ¬eld, any twisted composition over L is similar
to a twisted Hurwitz composition “(C, L).
Proof : The claim for compositions over F — F — F follows from Proposition (??) (a
result used in the proof of Theorem (??)). Let (V, L, Q, β) be a twisted composition
over L = F — ∆, ∆ a quadratic ¬eld extension of F . We have to check that
(V, L, Q, β) is similar to a composition with an element e such that β(e) = e.
By decomposing V = (V0 , V1 ), Q = (Q0 , Q1 ) and β = (β0 , β1 ) according to the
decomposition F —∆, we see (compare with the proof of (??)) that β0 is a quadratic
map V1 ’ V0 such that Q0 β0 (v1 ) = nK/F Q1 (v1 ) and β1 is a ∆-linear map
ι
V1 — V0 ’ V1 such that Q1 β1 (v1 — v0 ) = ι Q1 (v1 ) Q0 (v0 ). Let bQ1 (x, y) be
the ∆-bilinear polar of Q1 and let bQ0 (x, y) be F -bilinear polar of Q0 as well as its
extension to V0 —∆ as a ∆-bilinear form. By an argument similar to the argument in
the proof of proposition (??), there exists a unique extension β0 : V1 —∆ι V1 ’ V0 —∆
§36. TWISTED COMPOSITIONS 503


of β0 as a ∆-hermitian map. Property (??) of the de¬nition of a twisted composition
implies that

(36.30) bQ0 v0 , β0 (v1 ) = bQ1 v1 , β1 (v1 — v0 ) for v0 ∈ V0 , v1 ∈ V1 .

For v = (v0 , 0), v0 = 0 in V0 we have bQ v, β(v) = 0, so that the claim follows
from Theorem (??).

(36.31) Remark. In the next section we give examples of twisted compositions
of dimension 8 over a ¬eld L which are not induced by Cayley algebras.

(36.32) Remark. Let “ = (V, L, Q, β) be a twisted composition. Since L — L
L—L—∆ (where the ¬rst projection is multiplication), “—L is similar to a Hurwitz
twisted composition “(C, L — L) for some unique Hurwitz algebra C(“) over L.
The algebra C(“) over L is in fact extended from a Hurwitz algebra C over F .
In dimension 8, the algebra C is determined by a cohomological invariant f3 (See
Proposition (??)).

36.D. Twisted compositions of type A2 . We ¬nish this chapter by showing
how to associate twisted compositions to symmetric compositions arising from cen-
tral simple algebras of degree 3 or cubic ´tale algebras. We assume that char F = 3
e
and ¬rst suppose that F contains a primitive cube root of unity ω. Let » ∈ F — and
√ √
let L = F [X]/(X 3 ’») = F ( 3 ») where 3 » is the class of X modulo (X 3 ’»). Since

char F = 3 and µ3 ‚ F — , F ( 3 ») is a cyclic cubic F -algebra and u ’ ωu de¬nes a
generator of Gal(L/F ). Let A be a central simple F -algebra of degree three or a
cubic ´tale algebra and let (A0 , n, ) be the corresponding composition algebra as
e
de¬ned in Proposition (??). As in Example (??) we de¬ne a cyclic composition on

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