symplectic basis of L for the bilinear form bS , so that the matrix of bS with respect

to this basis is

«

J 0

01

¬ ·

..

A= where J = 1 0 .

.

0 J

Assume moreover that e1 = 1. For i, j = 1, . . . , n, de¬ne

1¤k< ¤n ξk (ei )ξ (ej ) if i > j,

bij =

0 if i ¤ j.

Let B = (bij )1¤i,j¤n ∈ Mn (Fsep ) and let

m

u = tr(A’1 B) = i=1 b2i,2i’1 .

Under the transposition permutation of ξ1 , . . . , ξn which exchanges ξr and ξr+1

and ¬xes ξi for i = r, r + 1, the element bij is replaced by bij + ξr (ei )ξr+1 (ej ) +

ξr (ej )ξr+1 (ei ), hence u becomes u + µ where

m

µ= ξr (e2i )ξr+1 (e2i’1 ) + ξr (e2i’1 )ξr+1 (e2i ) .

i=1

Claim. µ = 1.

Since (ei )1¤i¤n is a symplectic basis of L for the bilinear form bS , it follows

from (??) that TL/F (e2 ) = 1 and TL/F (ei ) = 0 for i = 2. The dual basis for the

bilinear form T is then (fi )1¤i¤n where

f1 = e1 + e2 , f2 = e1 , and f2i’1 = e2i , f2i = e2i’1 for i = 2, . . . , m.

Proposition (??) shows that n ei — fi ∈ L — L is the separability idempotent

i=1

of L. From the de¬nition of this idempotent, it follows that

n n

(ξr , ξr+1 ) ei — f i = i=1 ξr (ei )ξr+1 (fi ) = 0.

i=1

(See (??) for the notation.) On the other hand, the formulas above for f1 , . . . , fn

show:

n m

i=1 ei — fi = e1 — e1 + i=1 (e2i — e2i’1 + e2i’1 — e2i ).

Since e1 = 1, the claim follows.

Since the symmetric group is generated by the transpositions (r, r + 1), the

claim shows that u is transformed into u + 1 by any odd permutation of ξ1 , . . . , ξn

and is ¬xed by any even permutation. Therefore,

u if γ ∈ “0 ,

γ(u) =

u + 1 if γ ∈ “ “0 .

We proceed to show that u2 + u represents det SL/F ∈ F/„˜(F ).

Let C = 1¤k< ¤n ξk (ei )ξ (ej ) 1¤i,j¤n ∈ Mn (Fsep ) and let

D = C + B + Bt.

We have D = (dij )1¤i,j¤n where

±

0 if i > j,

dij = ξ (e )ξ (ei ) = SL/F (ei ) if i = j,

1¤k< ¤n k i

1¤k< ¤n ξk (ei )ξ (ej ) + ξk (ej )ξ (ei ) = bS (ei , ej ) if i < j,

´

§18. ETALE AND GALOIS ALGEBRAS 295

hence D ∈ Mn (F ) is a matrix of the quadratic form SL/F with respect to the basis

(ei )1¤i¤n and D + Dt = A. Therefore, s2 (A’1 D) ∈ F represents the determinant

of SL/F . Since D = C + B + B t , Lemma (??) yields

s2 (A’1 D) = s2 (A’1 C) + „˜ tr(A’1 B) = s2 (A’1 C) + u2 + u.

To complete the proof, it su¬ces to show s2 (A’1 C) = 0.

Let M = ξi (ej ) 1¤i,j¤n ∈ Mn (Fsep ) and let V = (vij )1¤i,j¤n where

0 if i ≥ j,

vij =

1 if i < j.

The matrix M is invertible, since M t · M = T (ei , ej ) 1¤i,j¤n and T is nonsingular.

Moreover, C = M t · V · M , hence A = C + C t = M t (V + V t )M and therefore

s2 (A’1 C) = s2 (V + V t )’1 V .

Observe that (V + V t + 1)2 = 0, hence (V + V t )’1 = V + V t . It follows that

s2 (V + V t )’1 V = s2 (V + V t )V = s2 (V + V t + 1)V + V .

Using the relations between coe¬cients of characteristic polynomials (see (??)), we

may expand the right-hand expression to obtain:

s2 (V + V t )’1 V = s2 (V + V t + 1)V + s2 (V ) +

tr (V + V t + 1)V tr(V ) + tr (V + V t + 1)V 2 .

Since V + V t + 1 has rank 1, we have s2 (V + V t + 1)V = 0. Since V is nilpotent,

we have tr(V ) = s2 (V ) = 0. A computation shows that tr (V + V t + 1)V 2 = 0,

hence s2 (V + V t )’1 V = 0 and the proof is complete in the case where char F = 2

and n is even.

Assume ¬nally that char F = 2 and n is odd: n = 2m + 1. Let (ei )1¤i¤n’1

be a symplectic basis of L0 for the bilinear form bS 0 . We again denote by A

the matrix of the bilinear form with respect to this basis and de¬ne a matrix

B = (bij )1¤i,j¤n’1 ∈ Mn’1 (Fsep ) by

1¤k< ¤n ξk (ei )ξ (ej ) if i > j,

bij =

0 if i ¤ j.

Let

m

u = tr(A’1 B) = i=1 b2i,2i’1 .

In order to extend (ei )1¤i¤n’1 to a basis (ei )1¤i¤n of L, de¬ne en = 1. Since

bS 0 (x, y) = T (x, y) for all x, y ∈ L0 , by (??), the dual basis (fi )1¤i¤n for the

bilinear form T is given by

f2i’1 = e2i , f2i = e2i’1 for i = 1, . . . , m, fn = en = 1.

Proposition (??) shows that the separability idempotent of L is

m

e= i=1 (e2i — e2i’1 + e2i’1 — e2i ) + en — en

hence for r = 1, . . . , n ’ 2,

m

ξr (e2i )ξr+1 (e2i’1 ) + ξr (e2i’1 )ξr+1 (e2i ) = ξr (en )ξr+1 (en ) = 1.

i=1

296 V. ALGEBRAS OF DEGREE THREE

The same argument as in the preceding case then shows

u if γ ∈ “0 ,

γ(u) =

u + 1 if γ ∈ “ “0 .

Mimicking the preceding case, we let

C= 1¤k< ¤n ξk (ei )ξ (ej ) ∈ Mn’1 (Fsep )

1¤i,j¤n’1

and

D = C + B + B t = (dij )1¤i,j¤n’1 ,

where

±

0 if i > j,

dij = S 0 (ei ) if i = j,

bS 0 (ei , ej ) if i < j.

The element s2 (A’1 D) represents the determinant det S 0 ∈ F/„˜(F ) and

s2 (A’1 D) = s2 (A’1 C) + „˜ tr(A’1 B) = s2 (A’1 C) + u2 + u,

hence it su¬ces to show s2 (A’1 C) = 1 (n ’ 1) to complete the proof.

2

For j = 1, . . . , n’1 we have ej ∈ L0 , hence ξn (ej ) = n’1 ξ (ej ), and therefore

0

=1

n’1 n’1

cij = 1¤k< ¤n’1 ξk (ei )ξ (ej ) + k=1 ξk (ei ) ξ (ej )

=1

= n’1≥k≥ ≥1 ξk (ei )ξ (ej ).

The matrix M = ξi (ej ) ∈ Mn’1 (Fsep ) is invertible since

1¤i,j¤n’1

M t · M = T (ei , ej ) = bS 0 (ei , ej )

1¤i,j¤n’1 1¤i,j¤n’1

and bS 0 is nonsingular. Moreover, C = M t · W · M where W = (wij )1¤i,j¤n’1 is

de¬ned by

1 if i ≥ j,

wij =

0 if i < j,

hence s2 (A’1 C) = s2 (W + W t )’1 W . Computations similar to those of the

preceding case show that s2 (W + W t )’1 W = 1 (n ’ 1).

2

18.C. Cubic ´tale algebras. Cubic ´tale algebras, i.e., ´tale algebras of di-

e e e

mension 3, have special features with respect to the Galois S3 -closure and discrim-

inant: if L is such an algebra, we establish below canonical isomorphisms

L —F L L — Σ(L) and Σ(L) L — ∆(L).

Moreover, we show that if F is in¬nite of characteristic di¬erent from 3, these

algebras have the form F [X]/(X 3 ’ 3X + t) for some t ∈ F , and we set up an exact

sequence relating the square class group of L to the square class group of F .

´

§18. ETALE AND GALOIS ALGEBRAS 297

The Galois closure and the discriminant. Let L be a cubic ´tale F - e

algebra. Let Σ(L) be the Galois S3 -closure of L and let e ∈ L—L be the separability

idempotent of L.

(18.25) Proposition. There are canonical F -algebra isomorphisms

∼ ∼

˜ : (1 ’ e) · (L — L) ’ Σ(L)

’ ˜ : L — L ’ L — Σ(L)

’

and

related by ˜(x — y) = xy, ˜ (1 ’ e) · (x — y) for x, y ∈ L.

Proof : Consider the disjoint union decomposition of “-sets

X(L — L) = X(L) — X(L) = D(L) E(L)

where D(L) is the diagonal of X(L) — X(L) and E(L) is its complement. By

de¬nition, e corresponds to the characteristic function on D(L) under the canonical

“

isomorphism L — L Map X(L — L), Fsep , hence

E(L) = { ξ ∈ X(L — L) | ξ(1 ’ e) = 1 }.

Therefore, we may identify

X (1 ’ e) · (L — L) = E(L),

“

hence also (1 ’ e) · (L — L) = Map E(L), Fsep . On the other hand, there is a

canonical bijection

∼

Σ X(L) ’ E(L)

’

which maps (ξi , ξj , ξk ) ∈ Σ X(L) to (ξi , ξj ) ∈ E(L). Under the anti-equivalence

Et F ≡ Sets “ , this bijection induces an isomorphism

∼

˜ : (1 ’ e) · (L — L) ’ Σ(L).

’

By decomposing L — L = e · (L — L) • (1 ’ e) · (L — L) , and combining ˜ with

∼

the isomorphism µ’1 : e · (L — L) ’ L of (??) which maps e · (x — y) = e · (xy — 1)

’

to xy, we obtain the isomorphism

∼

˜ : L — L ’ L — Σ(L).

’

∼

Note that there are actually three canonical bijections Σ X(L) ’ E(L), since

’

(ξi , ξj , ξk ) ∈ Σ X(L) may alternately be mapped to (ξi , ξk ) or (ξj , ξk ) instead of

∼

(ξi , ξj ); therefore, there are three canonical isomorphisms (1 ’ e) · (L — L) ’ Σ(L)

’

∼

and L — L ’ L — Σ(L).

’

In view of the proposition above, there is an S3 -algebra structure on (1 ’ e) ·

(L — L) and there are three embeddings 1 , 2 , 3 : L ’ (1 ’ e) · (L — L), which we

now describe: for ∈ L, we set

1( ) = (1 ’ e) · ( — 1), 2( ) = (1 ’ e) · (1 — )

and

(18.26) 3( ) = (1 ’ e) · TL/F ( )1 — 1 ’ — 1 ’ 1 — .

These isomorphisms correspond to the maps E(L) ’ X(L) which carry an

element (ξi , ξj ) ∈ E(L) respectively to ξi , ξj and to the element ξk such that

X(L) = {ξi , ξj , ξk }. Indeed, for ∈ L,

(ξi , ξj )[(1 ’ e) · ( — 1)] = ξi ( ), (ξi , ξj )[(1 ’ e) · (1 — )] = ξj ( )

298 V. ALGEBRAS OF DEGREE THREE

and

(ξi , ξj ) (1 ’ e) · TL/F ( )1 — 1 ’ — 1 ’ 1 — = TL/F ( ) ’ ξi ( ) ’ ξj ( ) = ξk ( ).

An action of S3 on (1 ’ e) · (L — L) is de¬ned by permuting the three copies of L

in (1 ’ e) · (L — L), so that σ —¦ i = σ(i) for σ ∈ S3 .

We now turn to the discriminant algebra ∆(L):

(18.27) Proposition. Let L be a cubic ´tale algebra. The canonical embeddings

e

µ1 , µ2 , µ3 : L ’ Σ(L) de¬ne isomorphisms:

∼

µ1 , µ2 , µ3 : ∆(L) —F L ’ Σ(L).

’

Proof : Consider the transpositions in S3 :

„1 = (2, 3) „2 = (1, 3) „3 = (1, 2)

and the subgroups of order 2:

Hi = {Id, „i } ‚ S3 for i = 1, 2, 3.

For i = 1, 2, 3, the canonical map

Σ X(L) ’ Σ X(L) /A3 — Σ X(L) /Hi

which carries ζ ∈ Σ X(L) to the pair (ζ A3 , ζ Hi ) consisting of its orbits under A3

and under Hi is a “-equivariant bijection, since ζ A3 © ζ Hi = {ζ}. Moreover, pro-

jection on the i-th component πi : Σ X(L) ’ X(L) factors through Σ X(L) /Hi ;

we thus get three canonical “-equivariant bijections:

∼

πi : Σ X(L) ’ ∆ X(L) — X(L).

’

Under the anti-equivalence Et F ≡ Sets “ , these bijections yield the required isomor-

phisms µi for i = 1, 2, 3.

Combining (??) and (??), we get:

(18.28) Corollary. For every cubic ´tale F -algebra L, there are canonical F -

e

algebra isomorphisms:

(Id — µi ’1 ) —¦ ˜: L —F L ’ L — ∆(L) —F L

∼

’ for i = 1, 2, 3.

The isomorphism (Id — µ2 ’1 ) —¦ ˜ is L-linear for the action of L on L —F L and on

∆(L) —F L by multiplication on the right factor.

Proof : The ¬rst assertion follows from (??) and (??). A computation shows that

˜(1 — ) maps (ξ1 , ξ2 , ξ3 ) ∈ Σ X(L) to ξ2 ( ), for all ∈ L. Similarly, µ2 (1 — )

maps (ξ1 , ξ2 , ξ3 ) to ξ2 ( ), hence µ2 ’1 —¦ ˜ is L-linear.

We conclude with two cases where the discriminant algebra can be explicitly

calculated:

(18.29) Proposition. For every quadratic ´tale F -algebra K,

e

∆(F — K) K.

Proof : The projection on the ¬rst component ξ : F — K ’ F is an element of

X(F — K) which is invariant under the action of “. If X(K) = {·, ζ}, then X(F —

K) = {ξ, ·, ζ}, and the map which carries · to (ξ, ·, ζ)A3 and ζ to (ξ, ζ, ·)A3 de¬nes

∼

an isomorphism of “-sets X(K) ’ ∆ X(F —K) . The isomorphism ∆(F —K) K

’

follows from the anti-equivalence Et F ≡ Sets “ .

´

§18. ETALE AND GALOIS ALGEBRAS 299

(18.30) Proposition. A cubic ´tale F -algebra L can be given an action of the

e

alternating group A3 which turns it into a Galois A3 -algebra over F if and only if

∆(L) F — F .

Proof : Suppose ∆(L) F — F ; then Proposition (??) yields an isomorphism

Σ(L) L — L. The action of A3 on Σ(L) preserves each term and therefore induces

an action on each of them. The induced actions are not the same, but each of them

de¬nes a Galois A3 -algebra structure on L since Σ(L) is a Galois S3 -algebra.

Conversely, suppose L has a Galois A3 -algebra structure; we have to show that

“ acts by even permutations on X(L). By way of contradiction, suppose γ ∈ “

induces an odd permutation on X(L): we may assume X(L) = {ξ1 , ξ2 , ξ3 } and

γ

ξ1 = ξ1 , γ ξ2 = ξ3 , γ ξ3 = ξ2 . Since L is a Galois A3 -algebra, X(L) is an A3 -torsor;

we may therefore ¬nd σ ∈ A3 such that

σ σ σ

ξ1 = ξ 2 , ξ2 = ξ 3 , ξ3 = ξ 1 ;

then γ (ξ1 ) = ξ3 whereas (γ ξ1 )σ = ξ2 , a contradiction. Therefore, “ acts on X(L)

σ

by even permutations, hence ∆(L) F — F .

Reduced equations. Let L be a cubic ´tale F -algebra. As a ¬rst step in

e

¬nding a reduced form for L, we relate the quadratic form S 0 on the subspace

L0 = { x ∈ L | TL/F (x) = 0 } and the bilinear form T on L to the discriminant

algebra ∆(L). Proposition (??) shows that S 0 is nonsingular if char F = 3; the

bilinear form T is nonsingular in every characteristic, since L is ´tale.

e

(18.31) Lemma. (1) The quadratic form S 0 is isometric to the quadratic form Q

on ∆(L) de¬ned by

Q(x) = N∆(L)/F (x) ’ T∆(L)/F (x)2 for x ∈ ∆(L).

(2) Suppose char F = 2 and let δ ∈ F — be such that ∆(L) F [t]/(t2 ’ δ). The

bilinear form T on L has a diagonalization

T 1, 2, 2δ .

Proof : (??) By a theorem of Springer (see Scharlau [?, Corollary 2.5.4], or Baeza

[?, p. 119] if char F = 2), it su¬ces to check that S 0 and Q are isometric over an

odd-degree scalar extension of F . If L is a ¬eld, we may therefore extend scalars to

L; then L is replaced by L — L, which is isomorphic to L — ∆(L) — L , by (??). In

all cases, we may thus assume L F — K, where K is a quadratic ´tale F -algebra.

e

Proposition (??) shows that we may identify K = ∆(L).

The generic polynomial of (±, x) ∈ F — K = L is

(X ’ ±) X 2 ’ TK/F (x)X + NK/F (x) =

X 3 ’ TL/F (±, x)X 2 + SL/K (±, x)X ’ NL/F (±, x),

hence

TL/F (±, x) = ± + TK/F (x) and SL/F (±, x) = ±TK/F (x) + NK/F (x).

Therefore, the map which carries x ∈ K = ∆(L) to ’TK/F (x), x ∈ L0 is an

∼

isometry ∆(L), Q ’ (L0 , S 0 ).

’

(??) As in (??), we may reduce to the case where L = F — K, with K

∆(L) F [t]/(t2 ’ δ). Let t be the image of t in K. A computation shows that

T (x1 , x2 + x3 t), (y1 , y2 + y3 t) = x1 y1 + 2(x2 y2 + δx3 y3 ),

300 V. ALGEBRAS OF DEGREE THREE

hence T has a diagonalization 1, 2, 2δ with respect to the basis (1, 0), (0, 1), (0, t) .

(18.32) Proposition. Every cubic ´tale F -algebra L is isomorphic to an algebra

e

of the form F [X]/(f ) for some polynomial f , unless F = F2 and L F — F — F .

If char F = 3 and F is in¬nite, the polynomial f may be chosen of the form

f = X 3 ’ 3X + a for some a ∈ F , a = ±2;

then ∆(L) F [t]/ t2 +3(a2 ’4) if char F = 2 and ∆(L) F [t]/(t2 +t+1+a’2) if

char F = 2. (Note that X 3 ’ 3X ± 2 = (X 1)2 (X ± 2), hence F [X]/(X 3 ’ 3X ± 2)

is not ´tale.)

e

If char F = 3 (and card F is arbitrary), the polynomial f may be chosen of the

form f = X 3 ’ b for some b ∈ F — if and only if ∆(L) F [t]/(t2 + t + 1).

If char F = 3, let δ ∈ F — be such that ∆(L) F [t]/(t2 ’ δ); then f may be

chosen of the form

f = X 3 ’ δX + a for some a ∈ F .

Proof : If L is a ¬eld, then it contains a primitive element x (see for instance

Bourbaki [?, p. V.39]); we then have L F [X]/(f ) where f is the minimal poly-

nomial of x. Similarly, if L F — K for some quadratic ¬eld extension K/F ,

then L F [X]/(Xg) where g is the minimal polynomial of any primitive element

of K. If L F — F — F and F contains at least three distinct elements a, b, c,

then L F [X]/(f ) where f = (X ’ a)(X ’ b)(X ’ c). This completes the proof

of the ¬rst assertion. We now show that, under suitable hypotheses, the primitive

element x may be chosen in such a way that its minimal polynomial takes a special

form.

Suppose ¬rst that char F = 3. The lemma shows that S 0 represents ’3, since

Q(1) = N∆(L)/F (1) ’ T∆(L)/F (1)2 = ’3.

Let x ∈ L0 be such that S 0 (x) = ’3. Since F ‚ L0 , the element x is a primitive

element if L is a ¬eld, and its minimal polynomial coincides with its generic poly-

nomial, which has the form X 3 ’ 3X + a for some a ∈ F . If L F — F — F , the

nonprimitive elements in L0 have the form (x1 , x2 , x3 ) where x1 + x2 + x3 = 0 and

two of the xi are equal. The conic S 0 (X) = ’3 is nondegenerate by (??). Therefore

it has only a ¬nite number of intersection points with the lines x1 = x2 , x1 = x3

and x2 = x3 . If F is in¬nite we may therefore ¬nd a primitive element x such that

S 0 (x) = ’3. Similarly, if L F — K where K is a ¬eld, the nonprimitive elements

in L0 have the form (x1 , x2 ) where x1 , x2 ∈ F and x1 + 2x2 = 0. Again, the conic

S 0 (X) = ’3 has only a ¬nite number of intersection points with this line, hence

we may ¬nd a primitive element x such that S 0 (x) = ’3 if F is in¬nite.

Example (??) shows how to compute ∆(L) for L = F [X]/(f ). If x1 , x2 , x3 are

the roots of f in an algebraic closure of F , we have

F [t]/(t2 ’ d) with d = (x1 ’ x2 )2 (x1 ’ x3 )2 (x2 ’ x3 )2

∆(L) if char F = 2,

and

x1 x2 x1 x3 x2 x3

F [t]/(t2 + t + d) with d =

∆(L) 2 + x2 + x 2 + x2 + x 2 if char F = 2.

x2+ x2

1 1 3 2 3

If f = X 3 +pX +q, then x1 +x2 +x3 = 0, x1 x2 +x1 x3 +x2 x3 = p and x1 x2 x3 = ’q,

and a computation shows that

(x1 ’ x2 )2 (x1 ’ x3 )2 (x2 ’ x3 )2 = ’4p3 ’ 27q 2 .

´