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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:
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§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
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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.
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Another characterization of ´tale algebras is given in Exercise ??.
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´
Etale F -algebras and “-sets. If L is an ´tale F -algebra of dimension n,
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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 )
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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 (??)
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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
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§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
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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.
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(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.
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§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 ).

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