f00

L = 0 k 0 f ∈ F , k ∈ K .

00k

Proposition (??) shows that ∆(L) K.

For the rest of the proof, we may thus assume B is a division algebra. Let „ be

a distinguished involution on B. By (??), there exists a subspace U ‚ Sym(B, „ )

of dimension 2 such that u3 = Nrd(u) for all u ∈ U . Pick a nonzero element u ∈ U .

Since dim U = 2, the linear map U ’ F which carries x ∈ U to Trd(u’1 x) has a

nonzero kernel; we may therefore ¬nd a nonzero v ∈ U such that Trd(u’1 v) = 0.

Consider the reduced characteristic polynomial of u’1 v:

Nrd(X ’ u’1 v) = X 3 + Srd(u’1 v)X ’ Nrd(u’1 v).

By substituting 1 and ’1 for X and multiplying by Nrd(u), we obtain

Nrd(u ’ v) = Nrd(u) + Nrd(u) Srd(u’1 v) ’ Nrd(v)

and

Nrd(u + v) = Nrd(u) + Nrd(u) Srd(u’1 v) + Nrd(v),

hence

Nrd(u + v) ’ Nrd(u ’ v) = 2 Nrd(v).

On the other hand, since u, v, u + v, u ’ v ∈ U and x3 = Nrd(x) for all x ∈ U , we

obtain

Nrd(u + v) = (u + v)3

= Nrd(u) + (u2 v + uvu + vu2 ) + (uv 2 + vuv + v 2 u) + Nrd(v)

and

Nrd(u ’ v) = (u ’ v)3

= Nrd(u) ’ (u2 v + uvu + vu2 ) + (uv 2 + vuv + v 2 u) ’ Nrd(v),

30 See Exercise ?? for the case where char F = 3.

§19. CENTRAL SIMPLE ALGEBRAS OF DEGREE THREE 309

hence

Nrd(u + v) ’ Nrd(u ’ v) = 2(u2 v + uvu + vu2 ) + 2 Nrd(v).

By comparing the expressions above for Nrd(u + v) ’ Nrd(u ’ v), it follows that

u2 v + uvu + vu2 = 0.

De¬ne

t1 = u’1 v = Nrd(u)’1 u2 v,

t2 = u’1 t1 u = Nrd(u)’1 uvu,

t3 = u’1 t2 u = Nrd(u)’1 vu2 ,

so that

t1 + t 2 + t 3 = 0

and conjugation by u permutes t1 , t2 , and t3 cyclically. Moreover, since u and v

are „ -symmetric, we have „ (t2 ) = t2 and „ (t1 ) = t3 .

Let w = t’1 t3 . Suppose ¬rst that w ∈ K. Since Nrd(t2 ) = Nrd(t3 ), we have

2

3

Nrd(w) = w = 1. If w = 1, then t2 = t3 , hence also t3 = t1 , a contradiction to

t1 + t2 + t3 = 0. Therefore, w is a primitive cube root of unity. Conjugating each

side of the relation t3 = wt2 by u, we ¬nd t2 = wt1 ; hence K(t1 ) = K(t2 ) = K(t3 )

and conjugation by u is an automorphism of order 3 of this sub¬eld. Cubing the

equations t2 = wt1 and t3 = wt2 , we obtain t3 = t3 = t3 , hence this element is

1 2 3

invariant under conjugation by u and therefore t3 = t3 = t3 ∈ K — . Since „ (t2 ) = t2 ,

1 2 3

we have in fact t3 ∈ F — , hence Proposition (??) shows that F (t2 ) is a sub¬eld of B

2

F [X]/(X 2 + X + 1). On the other hand, by applying

with discriminant ∆ F (t2 )

„ to each side of the equation t2 = wt1 , we ¬nd t2 = t3 „ (w). Since t3 = wt2 , it

follows that „ (w) = w ’1 , so that K = F (w) and therefore K F [X]/(X 2 +X +1).

The theorem is thus proved if w ∈ K, since then ∆ F (t2 ) K.

Suppose next that w ∈ K, hence K(w) is a cubic extension of K. Since

t1 + t2 + t3 = 0, we have

Int(u’1 )(w) = t’1 t1 = ’t’1 (t2 + t3 ) = ’1 ’ w’1 ∈ K(w),

3 3

hence Int(u’1 ) restricts to a K-automorphism θ of K(w). If θ = Id, then w =

’1 ’ w’1 , hence w is a root of an equation of degree 2 with coe¬cients in K, a

contradiction. Therefore, θ is nontrivial; it is of order 3 since u3 ∈ F — .

Now consider the action of „ :

„ (w) = t1 t’1 = t2 (t’1 t1 )t’1 = ’t2 (1 + w)t’1 .

2 2 2 2

This shows that the involution „ = Int(t’1 ) —¦ „ satis¬es

2

„ (w) = ’1 ’ w ∈ K(w).

Therefore, „ de¬nes an automorphism of order 2 of K(w). We claim that θ and „

generate a group of automorphisms of K(w) isomorphic to the symmetric group S3 .

Indeed,

’w

„ —¦ θ(w) = ’1 + (1 + w)’1 =

1+w

and

θ2 —¦ „ (w) = ’(1 + w ’1 )’1 = „ —¦ θ(w).

310 V. ALGEBRAS OF DEGREE THREE

Therefore, K(w)/F is a Galois extension with Galois group S3 . Let L = K(w)„ ,

the sub¬eld of elements ¬xed by „ . This sub¬eld is a cubic ´tale extension of F .

e

Since L/F is not cyclic, we have ∆(L) F — F , by Corollary (??). However,

LK = L — K K(w) is a cyclic extension of K, hence ∆(LK ) K — K. Therefore,

∆(L) K.

Suppose L ‚ B is a cubic ´tale F -algebra with discriminant ∆(L) isomorphic

e

to K. Let LK = L—K L—∆(L). By (??), we have L—∆(L) Σ(L), hence L—K

can be given a Galois S3 -algebra structure over F . Under any of these S3 -algebra

structures, the automorphism Id — ι gives the action of some transposition, and K

is the algebra of invariant elements under the action of the alternating group A3 . It

follows that LK can be given a Galois C3 -algebra structure, since A3 C3 = Z/3Z.

We ¬x such a structure and set ρ = 1 + 3Z ∈ C3 , as in §??. Since conjugation by

a transposition yields the nontrivial automorphism of A3 , we have

(Id — ι) ρ(x) = ρ2 Id — ι(x) for x ∈ LK .

By (??), there exist involutions „ of the second kind on B ¬xing the elements of

L. We proceed to describe these involutions in terms of the cyclic algebra structure

of B. It will be shown below (see (??)) that these involutions are all distinguished.

(This property also follows from (??)).

(19.15) Proposition. Suppose „ is an involution of the second kind on B such

that L ‚ Sym(B, „ ). The algebra B is a cyclic algebra:

B = L K • LK z • L K z 2

where z is subject to the relations: „ (z) = z, zx = ρ(x)z for all x ∈ L K and

z3 ∈ F —.

Proof : We ¬rst consider the case where B is split. We may then assume B =

EndK (LK ) and identify x ∈ LK with the endomorphism of multiplication by x. The

involution „ is the adjoint involution with respect to some nonsingular hermitian

form

h : LK — LK ’ K.

Since HomK (LK , K) is a free module of rank 1 over LK , the linear form x ’ h(1, x)

is of the form x ’ TLK /K ( x) for some ∈ LK . For x, y ∈ LK , we then have

h(x, y) = h 1, „ (x)y = TLK /K „ (x)y .

If were not invertible, then we could ¬nd x = 0 in LK such that x = 0. It follows

that h „ (x), y = 0 for all y ∈ LK , a contradiction. So, ∈ L— . Moreover, since

K

h(y, x) = ι h(x, y) for all x, y ∈ LK , we have

TLK /K „ (y)x = „ TLK /K „ (x)y for x, y ∈ LK ,

hence „ ( ) = since the bilinear trace form on LK is nonsingular. Therefore,

∈ L— .

Note that the restriction of „ to LK is Id — ι, hence

„ —¦ ρ(x) = ρ2 —¦ „ (x) for x ∈ LK .

’1

Consider β = ρ ∈ EndK (LK ). For x, y ∈ LK we have

h β(x), y = TLK /K „ —¦ ρ(x)y = TLK /K ρ2 —¦ „ (x)y

§19. CENTRAL SIMPLE ALGEBRAS OF DEGREE THREE 311

and

h x, β(y) = TLK /K „ (x)ρ(y) .

Since TLK /K ρ2 (u) = TLK /K (u) for all u ∈ LK , we also have

h x, β(y) = TLK /K ρ2 —¦ „ (x)y ,

hence h(β(x), y) = h x, β(y) for all x, y ∈ LK and therefore „ (β) = β. Clearly,

β —¦ x = ρ(x) —¦ β for x ∈ L, and β 3 = ’1 ρ( ’1 )ρ2 ( ’1 ) ∈ F — . Therefore, we may

choose z = β. This proves the proposition in the case where B is split.

If B is not split, consider the F -vector space:

S = { z ∈ Sym(B, „ ) | zx = ρ(x)z for x ∈ LK }.

The invertible elements in S form a Zariski-open set. Extension of scalars to a

splitting ¬eld of B shows that this open set is not empty. Since B is not split, the

¬eld F is in¬nite, hence the rational points in S are dense. We may therefore ¬nd

an invertible element in S.

If z ∈ S, then z 3 centralizes LK , hence z 3 ∈ LK . Since z 3 commutes with z

and z ∈ Sym(B, „ ), we have z 3 ∈ F . Therefore, every invertible element z ∈ S

satis¬es the required conditions.

´

Etale subalgebras and the invariant π(„ ). We now ¬x an involution of

the second kind „ on the central simple K-algebra B of degree 3 and a cubic ´tale e

F -algebra L ‚ Sym(B, „ ). We assume throughout that char F = 2. We will give a

special expression for the quadratic form Q„ , hence also for the P¬ster form π(„ ),

taking into account the algebra L (see Theorem (??)). As an application, we prove

the following statements: if an involution is the identity on a cubic ´tale F -algebra

e

of discriminant isomorphic to K, then it is distinguished; moreover, every cubic

´tale F -subalgebra in B is stabilized by some distinguished involution.

e

The idea to obtain the special form of Q„ is to consider the orthogonal decom-

position Sym(B, „ ) = L ⊥ M where M = L⊥ is the orthogonal complement of L

for the quadratic form Q„ :

M = { x ∈ Sym(B, „ ) | TrdB (x ) = 0 for ∈ L }.

We show that the restriction of Q„ to M is the transfer of some hermitian form

HM on M . This hermitian form is actually de¬ned on the whole of B, with values

in LK —K LK , where LK = L — K ‚ B.

We ¬rst make B a right LK —LK -module as follows: for b ∈ B and 1 , 2 ∈ LK ,

we set:

b—( — 2) = 1 b 2.

1

(19.16) Lemma. The separability idempotent e ∈ LK — LK satis¬es the following

properties relative to —:

(1) — e = for all ∈ L, and Trd(x) = Trd(x — e) for all x ∈ B;

(2) (x — e) = (x ) — e = ( x) — e = (x — e) for all x ∈ B, ∈ LK ;

(3) B — e = LK ;

(4) x — e = 0 for all x ∈ MK = M — K.

From (??) and (??) it follows that multiplication by e is the orthogonal projection

B ’ LK for the trace bilinear form.

312 V. ALGEBRAS OF DEGREE THREE

3

Proof : (??) Let e = i=1 ui — vi . Proposition (??) shows that ui vi = 1; since

i

LK is commutative, it follows that

3 3

—e= ui v i = ( ui v i ) = for ∈ LK .

i=1 i=1

Moreover, for x ∈ B we have

3 3

Trd(x — e) = Trd( ui xvi ) = Trd ( vi ui )x = Trd(x).

i=1 i=1

(??) For x ∈ B and ∈ LK ,

(x — e) = (x — e) — (1 — ) = x — e(1 — ) .

By (??), we have e(1 — ) = (1 — )e = ( — 1)e, hence, by substituting this in the

preceding equality:

(x — e) = (x ) — e = ( x) — e.

Similarly, (x — e) = (x — e) — ( — 1) = x — e( — 1) and e( — 1) = ( — 1)e, hence

we also have

(x — e) = ( x) — e.

(??) Property (??) shows that B — e centralizes LK , hence B — e = LK .

(??) For x ∈ MK and ∈ LK we have by (??) and (??):

Trd (x — e) = Trd (x ) — e = Trd(x ) = 0.

From (??), we obtain (x — e) ∈ MK © LK = {0}.

(19.17) Example. The split case. Suppose B = EndK (V ) for some 3-dimensional

vector space V and „ = „h is the adjoint involution with respect to some hermitian

form h on V . Suppose also that L F — F — F and let e1 , e2 , e3 ∈ L be the

primitive idempotents. There is a corresponding direct sum decomposition of V

into K-subspaces of dimension 1:

V = V1 • V2 • V3

such that ei is the projection onto Vi with kernel Vj • Vk for {i, j, k} = {1, 2, 3}.

Since e1 , e2 , e3 are „ -symmetric, we have for x, y ∈ V and i, j = 1, 2, 3, i = j:

h ei (x), ej (y) = h x, ei —¦ ej (y) = 0.

Therefore, the subspaces V1 , V2 , V3 are pairwise orthogonal with respect to h. For

i = 1, 2, 3, pick a nonzero vector vi ∈ Vi and let h(vi , vi ) = δi ∈ F — . We may use

the basis (v1 , v2 , v3 ) to identify B with M3 (K); the involution „ is then given by

« «

’1 ’1

ι(x11 ) δ1 ι(x21 )δ2 δ1 ι(x31 )δ3

x11 x12 x13

„ x21 x22 x23 = δ2 ι(x12 )δ1 δ2 ι(x32 )δ3 ,

’1 ’1

ι(x22 )

’1 ’1

x31 x32 x33 δ3 ι(x13 )δ1 δ3 ι(x23 )δ2 ι(x33 )

so that

±«

x δ2 a δ3 b

δ1 ι(a) δ 3 c

y

Sym(B, „ ) = x, y, z ∈ F , a, b, c ∈ K .

δ1 ι(b) δ2 ι(c) z

Under this identi¬cation,

« « «

100 000 000

e1 = 0 0 0 , e2 = 0 1 0 , e3 = 0 0 0 ,

000 000 001

§19. CENTRAL SIMPLE ALGEBRAS OF DEGREE THREE 313

and L is the F -algebra of diagonal matrices in Sym(B, „ ). The separability idem-

potent is e = e1 — e1 + e2 — e2 + e3 — e3 . A computation shows that

±«

0 δ2 a δ3 b

M = δ1 ι(a) δ3 c a, b, c ∈ K .

0

δ1 ι(b) δ2 ι(c) 0

The K-algebra LK is the algebra of diagonal matrices in B and MK is the space of

matrices whose diagonal entries are all 0. Let

«

111

m = 1 1 1 ∈ B.

111

For x = (xij )1¤i,j¤3 ∈ B, we have

3

m—( xij ei — ej ) = x,

i,j=1

hence B is a free LK — LK -module of rank 1.

We now return to the general case, and let θ be the K-automorphism of LK —LK

which switches the factors:

θ( — 2) = — for 1, ∈ LK .

1 2 1 2

As in (??), we call e ∈ LK — LK the separability idempotent of LK . The charac-

terization of e in (??) shows that e is invariant under θ.

(19.18) Proposition. Consider B as a right LK —LK -module through the —-multi-

plication. The module B is free of rank 1. Moreover, there is a unique hermitian

form

H : B — B ’ L K — LK

with respect to θ such that for all x ∈ B,

(19.19)

2 2 2

1

H(x, x) = e(x — x ) + (1 ’ e) (x — e) — 1 + 1 — (x — e) ’ Trd(x ) ,

2

2

where x = x — e and x = x — (1 ’ e). (Note that x and x — e lie in LK , by (??).)

Proof : Since LK — LK is an ´tale F -algebra, it decomposes into a direct product

e

of ¬elds by (??). Let LK — LK L1 — · · · — Ln for some ¬elds L1 , . . . , Ln . Then

B B1 — · · · — Bn where Bi is a vector space over Li for i = 1, . . . , n. To see that

B is a free LK — LK -module of rank 1, it su¬ces to prove that dimLi Bi = 1 for

i = 1, . . . , n. Since dimF B = dimF (LK — LK ), it actually su¬ces to show that

dimLi Bi = 0 for i = 1, . . . , n, which means that B is a faithful LK — LK -module.

This property may be checked over a scalar extension of F . Since it holds in the

split case, as was observed in (??), it also holds in the general case. (For a slightly

di¬erent proof, see Jacobson [?, p. 44].)

Now, let b ∈ B be a basis of B (as a free LK — LK -module). We de¬ne a

hermitian form H on B by

H(b — »1 , b — »2 ) = θ(»1 )H(b, b)»2 for »1 , »2 ∈ LK — LK ,

where H(b, b) is given by formula (??). This is obviously the unique hermitian

form on B for which H(b, b) takes the required value. In order to show that the

hermitian form thus de¬ned satis¬es formula (??) for all x ∈ B, we may extend

314 V. ALGEBRAS OF DEGREE THREE

scalars to a splitting ¬eld of B and L, and assume we are in the split case discussed

in (??). With the same notation as in (??), de¬ne:

3

(19.20) H (x, y) = xji yij ei — ej ∈ LK — LK

i,j=1

for x = (xij )1¤i,j¤3 , y = (yij )1¤i,j¤3 ∈ B. Straightforward computations show that

H is hermitian and satis¬es formula (??) for all x ∈ B. In particular, H (b, b) =

H(b, b), hence H = H. This proves the existence and uniqueness of the hermitian

form H.

The hermitian form H restricts to hermitian forms on Sym(B, „ ) and on MK

which we discuss next.

Let ω be the K-semilinear automorphism of LK — LK de¬ned by

ω( — k1 ) — ( — k2 ) = — ι(k2 ) — — ι(k1 )

1 2 2 1

for 1, ∈ L and k1 , k2 ∈ K. The following property is clear from the de¬nition:

2

„ (x — ») = „ (x) — ω(») for x ∈ B and » ∈ LK — LK .

This shows that Sym(B, „ ) is a right module over the algebra (LK — LK )ω of ω-

invariant elements in LK — LK . Moreover, by extending scalars to a splitting ¬eld

of B and using the explicit description of H = H in (??), one can check that

H „ (x), „ (y) = ω H(x, y) for x, y ∈ B.

Therefore, the hermitian form H restricts to a hermitian form

HS : Sym(B, „ ) — Sym(B, „ ) ’ (LK — LK )ω .

Now, consider the restriction of H to MK . By (??), we have x — e = 0 for all

x ∈ MK , hence x — (1 ’ e) = x and the LK — LK -module action on B restricts to

an action of (1 ’ e) · (LK — LK ) on MK . Moreover, for x, y ∈ MK ,

H(x, y) = H x — (1 ’ e), y = (1 ’ e)H(x, y) ∈ (1 ’ e) · (LK — LK ),

hence H restricts to a hermitian form

HMK : MK — MK ’ (1 ’ e) · (LK — LK ).

Recall from (??) the embedding 3 : LK ’ (1 ’ e) · (LK — LK ). By (??) and (??),

3 induces a canonical isomorphism

∼

3: LK — ∆(LK ) ’ (1 ’ e) · (LK — LK )

’

which we use to identify (1’e)·(LK —LK ) with LK —∆(LK ). Since the image of 3

is the subalgebra of (1 ’ e) · (LK — LK ) of elements ¬xed by θ, the automorphism

of LK — ∆(LK ) corresponding to θ via 3 is the identity on LK and the unique

nontrivial K-automorphism on ∆(LK ). We call this automorphism also θ. Thus,

we may consider the restriction of H to MK as a hermitian form with respect to θ:

HMK : MK — MK ’ LK — ∆(LK ).

In particular, HMK (x, x) ∈ LK for all x ∈ MK .

(19.21) Lemma. For all x ∈ MK ,

2TLK /K HMK (x, x) = TrdB (x2 )

and

2

H x2 — (1 ’ e), x2 — (1 ’ e) .

NLK /K HMK (x, x) =N 3 (LK )/K

§19. CENTRAL SIMPLE ALGEBRAS OF DEGREE THREE 315

For all x ∈ MK and ∈ LK ,

#

HMK ( x, x) = HMK (x, x),

#

: LK ’ LK is the quadratic map de¬ned in (??).

where

Proof : It su¬ces to verify these formulas when B and L are split. We may thus

assume B and L are as in (??). For

«

0 x12 x13

x = x21 0 x23 ∈ MK ,

x31 x32 0

we have, using (??) and (??),

H(x, x) = x12 x21 (e1 — e2 + e2 — e1 ) + x13 x31 (e1 — e3 + e3 — e1 )

+ x23 x32 (e2 — e3 + e3 — e2 )

= 3 (x23 x32 e1 + x13 x31 e2 + x12 x21 e3 ),

hence HMK (x, x) = x23 x32 e1 + x13 x31 e2 + x12 x21 e3 ∈ LK . It follows that

Trd(x2 )

1

TLK /K HMK (x, x) = x23 x32 + x13 x31 + x12 x21 = 2

and

NLK /K H(x, x) = x23 x32 x13 x31 x12 x21 .

On the other hand,

«

0 x13 x32 x12 x23

x — (1 ’ e) = x23 x31 x21 x13 ,

2

0

x32 x21 x31 x12 0

hence

H x2 — (1 ’ e), x2 — (1 ’ e) = (x23 x32 x13 x31 x12 x21 )2 .

N 3 (LK )/K

For = 1 e1 + 2 e2 + 3 e3 ∈ LK , we have

«

0 1 x12 1 x13

2 x21 2 x23

0

x=

3 x31 3 x32 0

and

H( x, x) = 3 ( 2 3 e1 + 1 3 e2 + 1 2 e3 )H(x, x),

proving the last formula of the lemma, since # = 2 3 e1 + 1 3 e2 + 1 2 e3 . Alter-

nately, the last formula follows from the fact that x = x—( —1), hence H( x, x) =

( — ) · H(x, x), together with the observation that ( — )(1 ’ e) = 3 ( # ).

The ¬rst formula also follows from the de¬nition of H(x, x) and of 3 , since

(??) shows that TL/F (x2 — e) = Trd(x2 ).

Finally, we combine the restrictions HS and HMK of H to Sym(B, „ ) and to

MK to describe the restriction of H to Sym(B, „ ) © MK = M . The automorphism