elements.

We start with Galois descent, followed by a general discussion of ´tale and

e

Galois algebras. These are tools which will also be used in subsequent chapters.

´

§18. Etale and Galois Algebras

Throughout this section, let F be an arbitrary base ¬eld, let Fsep be a separable

closure of F and let “ be the absolute Galois group of F :

“ = Gal(Fsep /F ).

The central theme of this section is a correspondence between ´tale F -algebras

e

and “-sets, which is set up in the ¬rst subsection. This correspondence is then

restricted to Galois algebras and torsors. The ¬nal subsection demonstrates the

special features of ´tale algebras of dimension 3.

e

The key tool for the correspondence between ´tale F -algebras and “-sets is the

e

following Galois descent principle. Let V0 be a vector space over F . The left action

of “ on V = V0 — Fsep de¬ned by γ — (u — x) = u — γ(x) for u ∈ V0 and x ∈ Fsep is

semilinear with respect to “, i.e.,

γ — (vx) = (γ — v)γ(x)

for v ∈ V and x ∈ Fsep ; the action is also continuous since for every vector v ∈ V

there is a ¬nite extension M of F in Fsep such that Gal(Fsep /M ) acts trivially on v.

The space V0 can be recovered as the set of ¬xed elements of V under “. More

generally:

(18.1) Lemma (Galois descent). Let V be a vector space over Fsep . If “ acts

continuously on V by semilinear automorphisms, then

V “ = { v ∈ V | γ — v = v for all γ ∈ “ }

is an F -vector space and the map V “ — Fsep ’ V , v — x ’ vx, is an isomorphism

of Fsep -vector spaces.

279

280 V. ALGEBRAS OF DEGREE THREE

Proof : It is clear that V “ is an F -vector space. To prove surjectivity of the canon-

ical map V “ — Fsep ’ V , consider an arbitrary vector v ∈ V . Since the action of

“ on V is continuous, we may ¬nd a ¬nite Galois extension M of F in Fsep such

that Gal(Fsep /M ) acts trivially on v. Let (mi )1¤i¤n be a basis of M over F and

let (γi )1¤i¤n be a set of representatives of the left cosets of Gal(Fsep /M ) in “, so

that the orbit of v in V consists of γ1 — v, . . . , γn — v, with γ1 — v = v, say. Let

n

vj = (γi — v)γi (mj ) for j = 1, . . . , n.

i=1

Since for every γ ∈ “ and i ∈ {1, . . . , n} there exists ∈ {1, . . . , n} and γ ∈

Gal(Fsep /M ) such that γγi = γ γ , the action of γ on the right-hand side of the

expression above merely permutes the terms of the sum, hence vj ∈ V “ for j =

1, . . . , n. On the other hand, the (n — n) matrix γi (mj ) 1¤i,j¤n with entries in

M is invertible, since γ1 , . . . , γn are linearly independent over M in EndF (M )

(Dedekind™s lemma). If (mij )1¤i,j¤n is the inverse matrix, we have

n

v = γ1 — v = vi mi1 ,

i=1

hence v lies in the image of the canonical map V “ — Fsep ’ V .

To prove injectivity of the canonical map, it su¬ces to show that F -linearly

independent vectors in V “ are mapped to Fsep -linearly independent vectors in V .

Suppose the contrary; let v1 , . . . , vr ∈ V “ be F -linearly independent vectors for

which there exist nonzero elements m1 , . . . , mr ∈ Fsep such that r vi mi =

i=1

0. We may assume r is minimal, r > 1, and m1 = 1. The mi are not all in

F , hence we may assume m2 ∈ F . Choose γ ∈ “ satisfying γ(m2 ) = m2 . By

applying γ to both sides of the linear dependence relation and subtracting, we

r

obtain i=2 vi γ(mi ) ’ mi = 0, a nontrivial relation with fewer terms. This

contradiction proves that the canonical map V “ — Fsep ’ V is injective.

(18.2) Remark. Assume that V in Lemma (??) admits an Fsep -bilinear multipli-

cation m : V —V ’ V and that “ acts on V by (semilinear) algebra automorphisms;

then the restriction of m to V “ is a multiplication on V “ , hence V “ is an F -algebra.

Similarly, if V = V1 ⊃ V2 ⊃ · · · ⊃ Vr is a ¬nite ¬‚ag in V , i.e., Vi is a subspace

of Vj for i > j, and the action of “ on V preserves V2 , . . . , Vr , then the ¬‚ag in V

descends to a ¬‚ag V “ = V1“ ⊃ V2“ ⊃ · · · ⊃ Vr“ in V “ .

´

18.A. Etale algebras. Let Alg F be the category of unital commutative asso-

ciative F -algebras with F -algebra homomorphisms as morphisms. For every ¬nite

dimensional commutative F -algebra L, let X(L) be the set of F -algebra homomor-

phisms from L to Fsep :

X(L) = HomAlg F (L, Fsep ).

For any ¬eld extension K/F , let LK be the K-algebra L —F K. If K ‚ Fsep ,

then Fsep also is a separable closure of K, and every F -algebra homomorphism

L ’ Fsep extends in a unique way to a K-algebra homomorphism LK ’ Fsep ; we

may therefore identify:

X(LK ) = X(L).

The following proposition characterizes ´tale F -algebras:

e

´

§18. ETALE AND GALOIS ALGEBRAS 281

(18.3) Proposition. For a ¬nite dimensional commutative F -algebra L, the fol-

lowing conditions are equivalent:

(1) for every ¬eld extension K/F , the K-algebra LK is reduced, i.e., LK does not

contain any nonzero nilpotent elements;

(2) L K1 — · · · — Kr for some ¬nite separable ¬eld extensions K1 , . . . , Kr of F ;

(3) LFsep Fsep — · · · — Fsep ;

(4) the bilinear form T : L — L ’ F induced by the trace:

T (x, y) = TL/F (xy) for x, y ∈ L

is nonsingular;

(5) card X(L) = dimF L;

(6) card X(L) ≥ dimF L.

If the ¬eld F is in¬nite, the conditions above are also equivalent to:

(7) L F [X]/(f ) for some polynomial f ∈ F [X] which has no multiple root in an

algebraic closure of F .

References: The equivalences (??) ” (??) ” (??) are proven in Bourbaki [?, Th´o- e

r`me 4, p. V.34] and (??) ” (??) ” (??) in [?, Corollaire, p. V.29]. The equivalence

e

of (??) with the other conditions is shown in [?, Proposition 1, p. V.47]. (See also

Waterhouse [?, §6.2] for the equivalence of (??), (??), (??), and (??)). Finally,

to see that (??) characterizes ´tale algebras over an in¬nite ¬eld, see Bourbaki [?,

e

Proposition 3, p. V.36 and Proposition 1, p. V.47].

If L F [X]/(f ), every F -algebra homomorphism L ’ Fsep is uniquely deter-

mined by the image of X, which is a root of f in Fsep . Therefore, the maps in X(L)

are in one-to-one correspondence with the roots of f in Fsep .

A ¬nite dimensional commutative F -algebra satisfying the equivalent condi-

tions above is called ´tale. From characterizations (??) or (??), it follows that ´tale

e

e

algebras remain ´tale under scalar extension.

e

Another characterization of ´tale algebras is given in Exercise ??.

e

´

Etale F -algebras and “-sets. If L is an ´tale F -algebra of dimension n,

e

Proposition (??) shows that X(L) consists of exactly n elements. The absolute

Galois group “ = Gal(Fsep /F ) acts on this set as follows:

γ

ξ =γ—¦ξ for γ ∈ “, ξ ∈ X(L).

This action is continuous since it factors through a ¬nite quotient Gal(M/F ) of “:

we may take for M any ¬nite extension of F in Fsep which contains ξ(L) for all

ξ ∈ X(L).

The construction of X(L) is functorial, since every F -algebra homomorphism

of ´tale algebras f : L1 ’ L2 induces a “-equivariant map X(f ) : X(L2 ) ’ X(L1 )

e

de¬ned by

ξ X(f ) = ξ —¦ f for ξ ∈ X(L2 ).

Therefore, writing Et F for the category of ´tale F -algebras and Sets “ for the cate-

e

gory of ¬nite sets endowed with a continuous left action of “, there is a contravariant

functor

X : Et F ’ Sets “

which associates to L ∈ Et F the “-set X(L).

282 V. ALGEBRAS OF DEGREE THREE

We now de¬ne a functor in the opposite direction. For X ∈ Sets “ , consider

the Fsep -algebra Map(X, Fsep ) of all functions X ’ Fsep . For f ∈ Map(X, Fsep )

and ξ ∈ X, it is convenient to write f, ξ for the image of ξ under f . We de¬ne a

semilinear action of “ on Map(X, Fsep ): for γ ∈ “ and f ∈ Map(X, Fsep ), the map

γ

f is de¬ned by

’1

γ

f, ξ = γ f, γ ξ for ξ ∈ X.

If γ acts trivially on X and ¬xes f, ξ for all ξ ∈ X, then γ f = f . Therefore, the

action of “ on Map(X, Fsep ) is continuous. Let Map(X, Fsep )“ be the F -algebra of

“-invariant maps. This algebra is ´tale, since by Lemma (??)

e

Map(X, Fsep )“ —F Fsep Map(X, Fsep ) Fsep — · · · — Fsep .

Every equivariant map g : X1 ’ X2 of “-sets induces an F -algebra homomor-

phism

M (g) : Map(X2 , Fsep )“ ’ Map(X1 , Fsep )“

de¬ned by

M (g)(f ), ξ = f, ξ g for f ∈ Map(X2 , Fsep )“ , ξ ∈ X1 ,

hence there is a contravariant functor

M: Sets “ ’ Et F

which maps X ∈ Sets “ to Map(X, Fsep )“ .

(18.4) Theorem. The functors X and M de¬ne an anti-equivalence of categories

Et F ≡ Sets “ .

Under this anti-equivalence, the dimension for ´tale F -algebras corresponds to the

e

cardinality for “-sets: if L ∈ Et F corresponds to X ∈ Sets “ , i.e., X X(L) and

L M(X), then

dimF L = card X.

Moreover, the direct product (resp. tensor product) of ´tale F -algebras corresponds

e

to the disjoint union (resp. direct product) of “-sets: for L1 , . . . , Lr ´tale F -

e

algebras,

X(L1 — · · · — Lr ) = X(L1 ) ··· X(Lr )

(where is the disjoint union) and

X(L1 — · · · — Lr ) = X(L1 ) — · · · — X(Lr ),

where “ acts diagonally on the right side of the last equality.

Proof : For L ∈ Et F , the canonical F -algebra homomorphism

“

¦ : L ’ Map X(L), Fsep

carries ∈ L to the map e de¬ned by

e , ξ = ξ( ) for ξ ∈ X(L).

Since card X(L) = dimF L, we have dimFsep Map X(L), Fsep = dimF L, hence

“

dimF Map X(L), Fsep = dimF L

´

§18. ETALE AND GALOIS ALGEBRAS 283

by Lemma (??). In order to prove that ¦ is an isomorphism, it therefore su¬ces

to show that ¦ is injective. Suppose ∈ L is in the kernel of ¦. By the de¬nition

of e , this means that ξ( ) = 0 for every F -algebra homomorphism L ’ Fsep . It

follows that the isomorphism LFsep Fsep — · · · — Fsep of (??) maps — 1 to 0,

hence = 0.

For X ∈ Sets “ , there is a canonical “-equivariant map

Ψ : X ’ X Map(X, Fsep )“ ,

which associates to ξ ∈ X the homomorphism eξ de¬ned by

for f ∈ Map(X, Fsep )“ .

eξ (f ) = f, ξ

Since X Map(X, Fsep )“ = X Map(X, Fsep )“sep = X Map(X, Fsep ) , the map Ψ

F

is easily checked to be bijective. The other equations are clear.

Since direct product decompositions of an ´tale F -algebra L correspond to

e

disjoint union decompositions of X(L), it follows that L is a ¬eld if and only if

X(L) is indecomposable, which means that “ acts transitively on X(L). At the

other extreme, L F — · · · — F if and only if “ acts trivially on X(L).

Traces and norms. Let L be an ´tale F -algebra of dimension n. Besides the

e

trace TL/F and the norm NL/F , we also consider the quadratic map

SL/F : L ’ F

which yields the coe¬cient of X n’2 in the generic polynomial (see (??)).

(18.5) Proposition. Let X(L) = {ξ1 , . . . , ξn }. For all ∈ L,

TL/F ( ) = ξi ( ), SL/F ( ) = ξi ( )ξj ( ), NL/F ( ) = ξ1 ( ) · · · ξn ( ).

1¤i¤n 1¤i<j¤n

Proof : It su¬ces to check these formulas after scalar extension to Fsep . We may

thus assume L = F —· · ·—F and ξi (x1 , . . . , xn ) = xi . With respect to the canonical

basis of L over F , multiplication by (x1 , . . . , xn ) is given by the diagonal matrix

with entries x1 , . . . , xn , hence the formulas are clear.

When the ´tale algebra L is ¬xed, we set T and bS for the symmetric bilinear

e

forms on L de¬ned by

(18.6) T (x, y) = TL/F (xy) and bS (x, y) = SL/F (x + y) ’ SL/F (x) ’ SL/F (y)

for all x, y ∈ L. From (??), it follows that

T (x, y) = ξi (x)ξi (y) and bS (x, y) = ξi (x)ξj (y).

1¤i¤n 1¤i=j¤n

Therefore,

(18.7) T (x, y) + bS (x, y) = TL/F (x)TL/F (y) for x, y ∈ L.

(This formula also follows readily from the general relations among the coe¬cients

of the characteristic polynomial: see (??).) By putting y = x in this equation, we

obtain:

TL/F (x2 ) + 2SL/F (x) = TL/F (x)2 for x ∈ L,

hence the quadratic form TL/F (x2 ) is singular if char F = 2 and n ≥ 2. Proposi-

tion (??) shows however that the bilinear form T is always nonsingular.

284 V. ALGEBRAS OF DEGREE THREE

Let L0 be the kernel of the trace map:

L0 = { x ∈ L | TL/F (x) = 0 }

and let S 0 : L0 ’ F be the restriction of SL/F to L0 . We write bS 0 for the polar

form of S 0 .

(18.8) Proposition. Suppose L is an ´tale F -algebra of dimension n.

e

(1) The bilinear form bS is nonsingular if and only if char F does not divide n ’ 1.

If char F divides n ’ 1, then the radical of bS is F .

(2) The bilinear form bS 0 is nonsingular if and only if char F does not divide n. If

char F divides n, then the radical of bS 0 is F .

(3) If char F = 2, the quadratic form L/F is nonsingular if and only if n ≡ 1

mod 4; the quadratic form S 0 is nonsingular if and only if n ≡ 0 mod 4.

Proof : (??) It su¬ces to prove the statements after scalar extension to Fsep . We

may thus assume that L = F — · · · — F , hence

bS (x1 , . . . , xn ), (y1 , . . . , yn ) = xi y j

1¤i=j¤n

for x1 , . . . , xn , y1 , . . . , yn ∈ F . The matrix M of bS with respect to the canonical

basis of L satis¬es:

M + 1 = (1)1¤i,j¤n .

Therefore, (M + 1)2 = n(M + 1). If char F divides n, it follows that M + 1 is

nilpotent. If char F does not divide n, the matrix n’1 (M + 1) is an idempotent of

rank 1. In either case, the characteristic polynomial of M + 1 is X n’1 (X ’ n), so

that of M is (X + 1)n’1 (X + 1) ’ n ; hence,

det M = (’1)n’1 (n ’ 1).

It follows that bS is nonsingular if and only if char F does not divide n ’ 1.

If char F divides n ’ 1, then the rank of M is n ’ 1, hence the radical of bS has

dimension 1. This radical contains F , since (??) shows that for ± ∈ F and x ∈ L,

bS (±, x) = TL/F (±)TL/F (x) ’ TL/F (±x) = (n ’ 1)±TL/F (x) = 0.

Therefore, the radical of bS is F .

(??) Equation (??) shows that bS 0 (x, y) = ’T (x, y) for all x, y ∈ L0 and that

bS (±, x) = 0 = T (±, x) for ± ∈ F and x ∈ L0 .

If char F does not divide n, then L = F • L0 ; the elements in the radical of bS 0

then lie also in the radical of T . Since T is nonsingular, it follows that bS 0 must

also be nonsingular.

If char F divides n, then F is in the radical of bS 0 . On the other hand, the ¬rst

part of the proposition shows that bS is nonsingular, hence the radical of bS 0 must

have dimension 1; this radical is therefore F .

(??) Assume char F = 2. From (??), it follows that the quadratic form SL/F is

singular if and only if n is odd and SL/F (1) = 0. Similarly, it follows from (??) that

S 0 is singular if and only if n is even and SL/F (1) = 0. Since SL/F (1) = 1 n(n ’ 1),

2

the equality SL/F (1) = 0 holds for n odd if and only if n ≡ 1 mod 4; it holds for n

even if and only if n ≡ 0 mod 4.

´

§18. ETALE AND GALOIS ALGEBRAS 285

The separability idempotent. Let L be an ´tale F -algebra. Recall from

e

(??) that we may identify X(L —F L) = X(L) — X(L): for ξ, · ∈ X(L), the

F -algebra homomorphism (ξ, ·) : L —F L ’ Fsep is de¬ned by

(18.9) (ξ, ·)(x — y) = ξ(x)·(y) for x, y ∈ L.

Theorem (??) yields a canonical isomorphism:

“

∼

L —F L ’ Map X(L) — X(L), Fsep

’ .

The characteristic function on the diagonal of X(L)—X(L) is invariant under “; the

corresponding element e ∈ L —F L is called the separability idempotent of L. This

element is indeed an idempotent since every characteristic function is idempotent.

By de¬nition, e is determined by the following condition: for all ξ, · ∈ X(L),

0 if ξ = ·,

(ξ, ·)(e) =

1 if ξ = ·.

(18.10) Proposition. Let µ : L —F L ’ L be the multiplication map. The sepa-

rability idempotent e ∈ L —F L is uniquely determined by the following conditions:

µ(e) = 1 and e(x — 1) = e(1 — x) for all x ∈ L. The map µ : L ’ e(L —F L) which

carries x ∈ L to e(x — 1) is an F -algebra isomorphism.

Proof : In view of the canonical isomorphisms

“ “

∼ ∼

L ’ Map X(L), Fsep

’ and L —F L ’ Map X(L) — X(L), Fsep

’ ,

the conditions µ(e) = 1 and e(x — 1) = e(1 — x) for all x ∈ L are equivalent to

ξ µ(e) = 1 and (ξ, ·) e(x — 1) = (ξ, ·) e(1 — x)

for all ξ, · ∈ X(L) and x ∈ L. We have

(ξ, ·) e(x — 1) = (ξ, ·)(e)ξ(x) and (ξ, ·) e(1 — x) = (ξ, ·)(e)·(x).

Therefore, the second condition holds if and only if (ξ, ·)(e) = 0 for ξ = ·.

On the other hand, ξ µ(e) = (ξ, ξ)(e), hence the ¬rst condition is equivalent

to: (ξ, ξ)(e) = 1 for all ξ ∈ X(L). This proves that the separability idempotent is

uniquely determined by the conditions of the proposition.

The map µ is injective since µ—¦µ = IdL . It is also surjective since the properties

of e imply:

e(x — y) = e(xy — 1) = µ(xy)

for all x, y ∈ L.

(18.11) Example. Let L = F [X]/(f ) for some polynomial

f = X n + an’1 X n’1 + · · · + a1 X + a0

with no repeated roots in an algebraic closure of F . Let x = X + (f ) be the image

of X in L and let

m’1

xi — xm’1’i ∈ L —F L for m = 1, . . . , n.

tm =

i=0

(In particular, t1 = 1.) The hypothesis on f implies that its derivative f is rela-

tively prime to f , hence f (x) ∈ L is invertible.

We claim that the separability idempotent of L is

e = (tn + an’1 tn’1 + · · · + a1 t1 ) f (x)’1 — 1 .

286 V. ALGEBRAS OF DEGREE THREE

Indeed, we have µ(e) = 1 since

µ(tn + an’1 tn’1 + · · · + a1 t1 ) = nxn’1 + (n ’ 1)an’1 xn’1 + · · · + a1 = f (x).

Also, tm (x — 1 ’ 1 — x) = xm — 1 ’ 1 — xm , hence

(tn + an’1 tn’1 + · · · + a1 t1 )(x — 1 ’ 1 — x) =

f (x) ’ a0 — 1 ’ 1 — f (x) ’ a0 = 0.

Therefore, e(x — 1 ’ 1 — x) = 0 and, for m = 1, . . . , n ’ 1,

e(xm — 1 ’ 1 — xm ) = e(x — 1 ’ 1 — x)tm = 0.

Since (xi )0¤i¤n’1 is a basis of L over F , it follows that e( — 1 ’ 1 — ) = 0 for all

∈ L, proving the claim.

An alternate construction of the separability idempotent is given in the follow-

ing proposition:

(18.12) Proposition. Let L be an ´tale F -algebra of dimension n = dimF L and

e

let (ui )1¤i¤n be a basis of L. Suppose (vi )1¤i¤n is the dual basis for the bilinear

form T of (??), in the sense that

T (ui , vj ) = δij (Kronecker delta) for i, j = 1, . . . , n.

n

The element e = ui — vi ∈ L — L is the separability idempotent of L.

i=1

Proof : Since (ui )1¤i¤n and (vi )1¤i¤n are dual bases, we have for x ∈ L

n n

(18.13) x= ui T (vi , x) = vi T (ui , x).

i=1 i=1

In particular, vj = ui T (vi , vj ) and uj = vi T (ui , uj ) for all j = 1, . . . , n,

i i

hence

e= ui — uj T (vi , vj ) = vi — vj T (ui , uj ).

i,j i,j

Using this last expression for e, we get for all x ∈ L:

e(x — 1) = vi x — vj T (ui , uj ).

i,j

By (??), we have vi x = uk T (vi x, vk ), hence

k

e(x — 1) = uk — vj T (ui , uj )T (vi x, vk ).

i,j,k

Since T (vi x, vk ) = TL/F (vi xvk ) = T (vi , vk x), we have

T (ui , uj )T (vi x, vk ) = T ui T (vi , vk x), uj = T (vk x, uj ),

i i

hence

e(x — 1) = uk — vj T (vk x, uj ).

j,k

Similarly, by using the expression e = ui — uj T (vi , vj ), we get for all x ∈ L:

i,j

e(1 — x) = ui — vk T (uk x, vi ).