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Suppose next that char F = 2 and n is even: n = 2m. Let (ei )1¤i¤n be a
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 .
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