HOMOMORPHISMS OF ABELIAN VARIETIES by Yuri G. Zarhin

28 Pages·2005·0.31 MB·English

S´eminaires & Congr`es
11,2005,p.189–215
HOMOMORPHISMS OF ABELIAN VARIETIES
by
Yuri G. Zarhin
Abstract. — We study Galois properties of points of prime order on an abelian va-
riety that imply the simplicity of its endomorphism algebra. Applications of these
propertiestohyperellipticjacobiansarediscussed.
Résumé (Homomorphismesdesvariétésabéliennes). — Nous´etudions les propri´et´es
galoisiennes des points d’ordre fini des vari´et´es ab´eliennes qui impliquent la sim-
plicit´e de leur alg`ebre d’endomorphismes. Nous discutons ceux-ci par rapport aux
jacobiennes hyperelliptiques.
It is well-known that an abelian variety is (absolutely) simple or is isogenous to a
self-product of an (absolutely) simple abelian variety if and only if the center of its
endomorphism algebra is a field. In this paper we prove that the center is a field if
the field of definition of points of prime order ` is“big enough”.
The paper is organized as follows. In 1 we discuss Galois properties of points of
§
order ` on an abelian variety X that imply that its endomorphism algebra End0(X)
is a central simple algebra over the field of rational numbers. In 2 we prove that
§
similar Galois properties for two abelian varieties X and Y combined with the linear
disjointness of the corresponding fields of definitions of points of order ` imply that
X andY arenon-isogenous(andevenHom(X,Y)=0). In 3wegiveapplicationsto
§
endomorphism algebras of hyperelliptic jacobians. In 4 we prove that if X admits
§
multiplications by a number field E and the dimension of the centralizer of E in
End0(X) is“as large as possible”then X is an abelian variety of CM-type isogenous
to a self-product of an absolutely simple abelian variety.
Throughout the paper we will freely use the following observation [21, p.174]: if
an abelian variety X is isogenous to a self-product Zd of an abelian variety Z then
a choice of an isogeny between X and Zd defines an isomorphism between End0(X)
and the algebra M (End0(Z)) of d d matrices over End0(Z). Since the center of
d
×
2000MathematicsSubjectClassification. — 14H40,14K05.
Keywordsandphrases. — Hyperellipticjacobians,homomorphismsofabelianvarieties.
(cid:13)c S´eminairesetCongr`es11,SMF2005
190 YU.G. ZARHIN
End0(Z) coincides with the center of M (End0(Z)), we get an isomorphism between
d
thecenterofEnd0(X)andthecenterofEnd0(Z)(thatdoesnotdependonthechoice
of an isogeny). Also dim(X) = d dim(Z); in particular, both d and dim(Z) divide
·
dim(X).
1. Endomorphism algebras of abelian varieties
Throughout this paper K is a field. We write K for its algebraic closure and
a
Gal(K) for the absolute Galois group Gal(K /K). We write ` for a prime different
a
fromchar(K). IfX is anabelianvarietyofpositivedimensionoverK thenwe write
a
End(X)forthe ringofallits K -endomorphismsandEnd0(X)forthe corresponding
a
Q-algebra End(X) Q. If Y is (may be, another) abelian variety over K then we
a
⊗
write Hom(X,Y) for the group of all K -homomorphisms from X to Y. It is well-
a
known that Hom(X,Y)=0 if and only if Hom(Y,X)=0.
If n is a positive integer that is not divisible by char(K) then we write X for the
n
kernelofmultiplicationbyn inX(K ). Itis well-known[21] thatX is a free Z/nZ-
a n
module of rank 2dim(X). In particular, if n = ` is a prime then X is an F -vector
` `
space of dimension 2dim(X).
If X is defined over K then X is a Galois submodule in X(K ). It is known
n a
that all points of X are defined over a finite separable extension of K. We write
n
ρ : Gal(K) Aut (X ) for the corresponding homomorphism defining the
n,X,K → Z/nZ n
structure of the Galois module on X ,
n
G Aut (X )
n,X,K Z/nZ n
⊂
foritsimageρ (Gal(K))anedK(X )forthefieldofdefinitionofallpointsofX .
n,X,K n n
Clearly,K(X )isafinite GaloisextensionofK withGaloisgroupGal(K(X )/K)=
n n
G . If n=` then we get a natural faithful linear representation
n,X,K
e G Aut (X )
`,X,K ⊂ F` `
of G in the F -vector spaceeX .
`,X,K ` `
e
Remark1.1. — If n=`2 then there is the natural surjective homomorphism
τ :G G
`,X `2,X,K `,X,K
−−→→
corresponding to the field inclusioneK(X ) K(eX ); clearly, its kernel is a finite `-
` `2
⊂
group. Clearly,everyprimedividing #(G )either divides #(G )or is equal
`2,X,K `,X,K
to `. If A is a subgroup in G of index N then its image τ (A) in G is
`2,X,K e `,eX `,X,K
isomorphic to A/A ker(τ ). It follows easily that the index of τ (A) in G
`,Xe `,X e `,X,K
equals N/`j whereT`j is the index of A ker(τ ) in ker(τ ). In particular, j is a
`,X `,X e
nonnegative integer. T
SE´MINAIRES & CONGRE`S11
HOMOMORPHISMS OF ABELIAN VARIETIES 191
We write End (X) for the ring of all K-endomorphisms of X. We have
K
Z=Z 1 End (X) End(X)
X K
· ⊂ ⊂
where 1 is the identity automorphism of X. Since X is defined over K, one may
X
associate with every u End(X) and σ Gal(K) an endomorphism σu End(X)
∈ ∈ ∈
such that σu(x)=σu(σ−1x) for x X(K ) and we get the group homomorphism
a
∈
κ :Gal(K) Aut(End(X)); κ (σ)(u)=σu σ Gal(K),u End(X).
X X
−→ ∀ ∈ ∈
It is well-known that End (X) coincides with the subring of Gal(K)-invariants in
K
End(X), i.e., End (X) = u End(X) σu = u σ Gal(K) . It is also
K
{ ∈ | ∀ ∈ }
well-known that End(X) (viewed as a group with respect to addition) is a free com-
mutative group of finite rank and End (X) is its pure subgroup, i.e., the quotient
K
End(X)/End (X) is also a free commutative group of finite rank. All endomor-
K
phismsofX aredefinedoverafinite separableextensionofK. Moreprecisely[31],if
n>3 is a positive integer not divisible by char(K) then all the endomorphisms of X
are defined over K(X ); in particular,
n
Gal(K(X )) ker(κ ) Gal(K).
n X
⊂ ⊂
Thisimplies thatifΓ :=κ (Gal(K)) Aut(End(X))thenthereexistsasurjective
K X
⊂
homomorphism κ :G (cid:16)Γ such that the composition
X,n n,X K
κ
e X,n
Gal(K) Gal(K(X )/K)=G Γ
n n,X K
−−→→ −−→→
coincides with κ and e
X
End (X)=End(X)ΓK.
K
Clearly, End(X) leaves invariant the subgroup X X(K ). It is well-known that
` a
⊂
u End(X) kills X (i.e. u(X ) = 0) if and only if u ` End(X). This gives us a
` `
∈ ∈ ·
natural embedding
End (X) Z/`Z End(X) Z/`Z , End (X );
K ⊗ ⊂ ⊗ −→ F` `
the image of End (X) Z/`Z lies in the centralizer of the Galois group, i.e., we get
K
⊗
an embedding
EndK(X)⊗Z/`Z,−→EndGal(K)(X`)=EndGe`,X,K(X`).
The next easy assertion seems to be well-known (compare with Prop.3 and its proof
on pp.107–108 in [19]) but quite useful.
Lemma1.2. — If EndGe`,X,K(X`)=F` then EndK(X)=Z.
Proof. — It follows that the F -dimension of End (X) Z/`Z does not exceed 1.
` K
⊗
This meansthatthe rankofthe free commutativegroupEnd (X)does notexceed1
K
and therefore is 1. Since Z 1 End (X), it follows easily that End (X) =
X K K
· ⊂
Z 1 =Z.
X
·
SOCIE´TE´MATHE´MATIQUEDEFRANCE2005
192 YU.G. ZARHIN
Lemma1.3. — If EndGe`,X,K(X`) is a field then EndK(X) has no zero divisors, i.e.,
End (X) Q is a division algebra over Q.
K
⊗
Proof. — It follows that End (X) Z/`Z is also a field and therefore has no zero
K
⊗
divisors. Suppose that u,v are non-zeroelements of End (X) with uv =0. Dividing
K
(if possible) u and v by suitable powers of ` in End (X), we may assume that both
K
u and v do not lie in `End (X) and induce non-zero elements in End (X) Z/`Z
K K
⊗
with zero product. Contradiction.
Let us put End0(X) := End(X) Q. Then End0(X) is a semisimple finite-
⊗
dimensional Q-algebra [21, 21]. Clearly, the natural map Aut(End(X))
§ →
Aut(End0(X)) is an embedding. This allows us to view κ as a homomorphism
X
κ :Gal(K) Aut(End(X)) Aut(End0(X)),
X
−→ ⊂
whose image coincides with Γ Aut(End(X)) Aut(End0(X)); the subalgebra
K
⊂ ⊂
End0(X)ΓK of ΓK-invariants coincides with EndK(X) Q.
⊗
Remark1.4
(i) Let us split the semisimple Q-algebra End0(X) into a finite direct product
End0(X) = D of simple Q-algebras D . (Here is identified with the set of
s∈I s s I
minimal twoQ-sided ideals in End0(X).) Let e be the identity element of D . One
s s
may view e as an idempotent in End0(X). Clearly,
s
1 = e End0(X), e e =0 s=t.
X s s t
∈ ∀ 6
Xs∈I
There exists a positive integer N such that all N e lie in End(X). We write X for
s s
·
the image X := (Ne )(X); it is an abelian subvariety in X of positive dimension.
s s
Clearly, the sum map
π : X X, (x ) x
X s s s
−→ 7−→
Ys Xs
is an isogeny. It is also clear that the intersection D End(X) leaves X X
s s
⊂
invariant. This gives us a natural identification D = ETnd0(X ). One may easily
s ∼ s
checkthateachX isisogenoustoaself-productof(absolutely)simpleabelianvariety.
s
Clearly, if s=t then Hom(X ,X )=0.
s t
6
(ii) We write C for the center of D . Then C coincides with the center of
s s s
End0(X )andisthereforeeitheratotallyrealnumberfieldofdegreedividingdim(X )
s s
or a CM-field of degree dividing 2dim(X ) [21, p.202]; the center C of End0(X) co-
s
incides with s∈ICs =⊕s∈SCs.
(iii) All thQe sets
es s s∈IQ es s∈ICs =C
{ | ∈I}⊂⊕ · ⊂⊕
SE´MINAIRES & CONGRE`S11
HOMOMORPHISMS OF ABELIAN VARIETIES 193
are stable under the Galois action Gal(K) κX Aut(End0(X)). In particular, there
−→
is a continuous homomorphism from Gal(K) to the group Perm( ) of permutations
I
of such that its kernel contains ker(κ ) and
X
I
e =κ (σ)(e )=σe , σ(C ) =C , σ(D )=D σ Gal(K),s .
σ(s) X s s s σ(s) s σ(s)
∀ ∈ ∈I
It follows that X = Ne (X) = σ(Ne (X)) = σ(X ); in particular, abelian
σ(s) σ(s) s s
subvarieties X and X have the same dimension and u σu gives rise to an
s σ(s)
7→
isomorphism of Q-algebras End0(X )=End0(X ).
σ(s) ∼ s
(iv) If J is a non-empty Galois-invariant subset in then the sum Ne
J s∈J s
is Galois-invariant and therefore lies in End (X). If J0 is another GaloisP-invariant
K
subset of that does not meet J then Ne also lies in End (X) and
I s∈J s K
Ne Ne = 0. Assume that EnPd (X) has no zero divisors. It follows
s∈J s s∈J0 s K
tPhat musPtconsistofoneGaloisorbit;inparticular,allX havethe samedimension
s
I
equal to dim(X)/#( ). In addition, if t , Gal(K) is the stabilizer of t in Gal(K)
t
I ∈I
and F is the subfield of Gal(K) -invariants in the separable closure of K then it
t t
follows easily that Gal(K) is an open subgroup of index #( ) in Gal(K), the field
t
I
extension F /K is separable of degree #( ) and X is isomorphic over K
t I s∈S s a
to the Weil restriction Res (X ). This impliesQthat X is isogenous over K to
Ft/K t a
Res (X ).
Ft/K t
Theorem1.5. — Suppose that ` is a prime, K is a field of characteristic =`. Suppose
6
that X is an abelian variety of positive dimension g defined over K. Assume that
G`,X,K contains a subgroup such EndG(X`) is a field.
G
Then one of the following conditions holds:
e
(a) The center of End0(X) is a field. In other words, End0(X) is a simple Q-
algebra.
(b)
(i) The prime ` is odd;
(ii) there exist a positive integer r>1 dividing g, a field F with
K K(X )G =:L F K(X ), [F :L]=r
` `
⊂ ⊂ ⊂
and a g/r-dimensional abelian variety Y over F such that End0(Y) is a simple
Q-algebra, the Q-algebra End0(X) is isomorphic to the direct sum of r copies
of End0(Y) and the Weil restriction Res (Y) is isogenous over K to X.
F/L a
In particular, X is isogenous over K to a product of g/r-dimensional abelian
a
varieties. In addition, contains a subgroup of index r;
G
(c)
(i) The prime `=2;
(ii) there exist a positive integer r>1 dividing g, fields L and F with
K K(X )G L F K(X ), [F :L]=r
4 4
⊂ ⊂ ⊂ ⊂
SOCIE´TE´MATHE´MATIQUEDEFRANCE2005
194 YU.G. ZARHIN
and a g/r-dimensional abelian variety Y over F such that End0(Y) is a simple
Q-algebra, the Q-algebra End0(X) is isomorphic to the direct sum of r copies
of End0(Y) and the Weil restriction Res (Y) is isogenous over K to X.
F/L a
In particular, X is isogenous over K to a product of g/r-dimensional abelian
a
varieties.In addition, there exists a nonnegative integer j such that 2j divides r
and contains a subgroup of index r/2j >1.
G
Proof. — WewillusenotationsofRemark1.4. Letusputn=`if`isoddandn=4
if `=2. Replacing K by K(X )G, we may and will assume that
`
G = .
`,X,K
G
e
If ` is odd then let us put L=K and H :=Gal(K(X )/K)= =Gal(L(X )/L).
` `
G
If ` = 2 then we choose a subgroup G of smallest possible order such
4,X,K
H ⊂
that τ ( ) = G = and put L := K(X )H K(X ). It follows easily that
2,X H 2,X,K G e 4 ⊂ 4
L(X )=K(X ) and Gal(L(X )/L)=Gal(K(X )/K), i.e.,
4 4 e 2 2
=G , G = .
4,X,L 2,X,L
H G
e e
The minimality propertyof combinedwith Remark 1.1implies thatif H G
4,X,L
H ⊂
is a subgroup of index r > 1 then τ (H) has index r/2j > 1 in G for some
2,X 2,X,L e
nonnegative index j.
e
In light of Lemma 1.3, End (X) has no zero divisors. It follows from Remark
L
1.4(iv) that Gal(L) acts on transitively. Let us put r = #( ). If r = 1 then
I I I
is a singleton and = s ,X = X ,End0(X) = D ,C = C . This means that
s s s
I { }
assertion (a) of Theorem 1.5 holds true.
Further we assume that r >1. Let us choose t and put Y :=X . If F :=F is
t t
∈I
the subfield of Gal(L) -invariants in the separable closure of K then it follows from
t
Remark1.4(iv)thatF /Lisaseparabledegreerextension,Y isdefinedoverF andX
t
is isogenous over L =K to Res (Y).
a a F/L
Recall(Remark1.4(iii))thatker(κ )actstriviallyon . ItfollowsthatGal(L(X ))
X n
I
actstriviallyon . Thisimplies thatGal(L(X ))liesinGal(L) . RecallthatGal(L)
n t t
I
is anopensubgroupofindexr inGal(L)andGal(L(X )) is a normalopensubgroup
n
in Gal(L). It follows that H :=Gal(L) /Gal(L(X )) is a subgroup of index r in
t n
Gal(L)/Gal(L(X ))=Gal(L(X )/L)=G .
n n n,X,L
e
If ` is odd then n= ` and G =G = contains a subgroup of index r >1.
n,X,L `,X,L
G
It follows from Remark 1.4 that assertion (b) of Theorem 1.5 holds true.
e e
If ` = 2 then n = 4 and G = G contains a subgroup H of index r > 1.
n,X,L 4,X,L
But in this case we know (see the very beginning of this proof) that G = and
e e 2,X,L G
τ (H) has index r/2j >1 in G for some nonnegativeintegerj. It followsfrom
2,X 2,X,L e
Remark 1.4 that assertion(c) of Theorem 1.5 holds true.
e
SE´MINAIRES & CONGRE`S11
HOMOMORPHISMS OF ABELIAN VARIETIES 195
Before stating our next result, recall that a perfect finite group with center
G Z
is called quasi-simple if the quotient / is a simple nonabelian group. Let H be a
G Z
non-centralnormal subgroup in quasi-simple . Then the image of H in simple /
G G Z
is a non-trivial normal subgroup and therefore coincides with / . This means that
G Z
= H. Since is perfect, = [ , ] = [H,H] H. It follows that = H. In
G Z G G G G ⊂ G
other words, every proper normal subgroup in a quasi-simple group is central.
Theorem1.6. — Suppose that ` is a prime, K is a field of characteristic different
from `. Suppose that X is an abelian variety of positive dimension g defined over K.
Let us assume that G contains a subgroup that enjoys the following properties:
`,X,K
G
(i) EndG(X`)=Fe`;
(ii) The group does not contain a subgroup of index 2.
G
(iii) The only normal subgroup in of index dividing g is itself.
G G
Then one of the following two conditions (a) and (b) holds:
(a) There exists a positive integer r >2 such that:
(a0) r divides g and X is isogenous over K to a product of g/r-dimensional
a
abelian varieties;
(a1) If ` is odd then contains a subgroup of index r;
G
(a2) If `=2 then there exists a nonnegative integer j such that contains a
G
subgroup of index r/2j >1.
(b)
(b1) The center of End0(X) coincides with Q. In other words, End0(X) is a
matrix algebra either over Q or over a quaternion Q-algebra.
(b2) If is perfect and End0(X) is a matrix algebra over a quaternion Q-
G
algebra H then H is unramified at every prime not dividing #( ).
G
(b3) Let be the center of . Suppose that is quasi-simple, i.e. it is per-
Z G G
fect and the quotient / is a simple group. If End0(X) = Q then there exist
G Z 6
a perfect finite (multiplicative) subgroup Π End0(X)∗ and a surjective homo-
⊂
morphism Π(cid:16) / such that every prime dividing #(Π) also divides #( ).
G Z G
Proof. — Let us assume that the center C of End0(X) is not a field. Applying
Theorem 1.5, we conclude that the condition (a) holds.
Assume now that the center C of End0(X) is a field. We need to prove (b). Let
us define n and L as in the beginning of the proof of Theorem 1.5. We have
G =G`,X,L, EndGe`,X,L(X`)=F`.
In addition, if ` = 2 and H eG is a subgroup of index r > 1 then τ (H) has
4,X,L 2,X
⊂
index r/2j > 1 in G = for some nonnegative integer j. This implies that the
2,X,L G e
only normal subgroup in G = G of index dividing g is G itself. It is
e n,X,L 4,X,L n,X,L
alsoclearthatG doesnotcontainasubgroupofindex2. ItfollowsfromRemark
n,X,L e e e
1.1thatif isperfectthenG isalsoperfectandeveryprimedividing#(G )
G e 4,X,L 4,X,L
e e
SOCIE´TE´MATHE´MATIQUEDEFRANCE2005
196 YU.G. ZARHIN
must divide #( ), because (thanks to a celebrated theoremof Feit-Thompson) #( )
G G
must be even. (If ` is odd then n=` and G = .)
n,X,L
G
It follows from Lemma 1.2 that End (X) = Z and therefore End (X) Q = Q.
L e L ⊗
Recall that End (X) Q = End0(X)Gal(L) and κ : Gal(L) Aut(End0(X)) kills
L X
⊗ →
Gal(L(X )). This gives rise to the homomorphism
n
κ :G =Gal(L(X )/L)=Gal(L)/Gal(L(X )) Aut(End0(X))
X,n n,X,L n n
−→
withκX,n(Gne,X,L)=κX(Gal(L)) Aut(End0(X))andEnd0(X)Gen,X,L =Q. Clearly,
⊂
theactionofG onEnd0(X)leavesinvariantthecenterC andthereforedefinesa
e n,X,L
e
homomorphisemGn,X,L →Aut(C) withCGn,X,L =Q. Itfollowsthat C/Qis a Galois
extension and the corresponding map
e
G Aut(C)=Gal(C/Q)
n,X,L
−→
is surjective. Recall that Ceis either a totally real number field of degree dividing g
or a purely imaginary quadratic extension of a totally real number field C+ where
[C+ :Q]divides g . In the caseof totally realC let us put C+ :=C. Clearly,in both
cases C+ is the largest totally real subfield of C and therefore the action of G
n,X,L
leavesC+ stable,i.e. C+/Qis alsoa Galoisextension. Letus put r :=[C+ :Q]. Itis
e
known[21, p.202]thatr dividesg. Clearly,theGaloisgroupGal(C+/Q)hasorderr
and we have a surjective homomorphism (composition)
G Gal(C/Q) Gal(C+/Q)
n,X,L
−−→→ −−→→
of G onto order regroup Gal(C+/Q). Clearly, its kernel is a normal subgroup
n,X,L
of index r in G . This contradicts our assumption if r > 1. Hence r = 1,
e n,X,L
i.e. C+ = Q. It follows that either C = Q or C is an imaginary quadratic field and
e
Gal(C/Q)isagroupoforder2. Inthelattercasewegetthesurjectivehomomorphism
from G onto Gal(C/Q), whose kernel is a subgroup of order 2 in G , which
n,X,L n,X,L
does not exist. This proves that C = Q. It follows from Albert’s classification [21,
e e
p.202] that End0(X) is either a matrix algebra Q or a matrix algebra M (H) where
d
H is a quaternion Q-algebra. This proves assertion(b1) of Theorem 1.6.
Assume, in addition, that is perfect. Then, as we have already seen, G
n,X,L
G
is also perfect. This implies that Γ := κ (G ) is a finite perfect subgroup of
X,n n,X,L e
Aut(End0(X)) and every prime dividing #(Γ) must divide #(G ) and therefore
e n,X,L
divides #( ). Clearly,
G e
(1) Q=End0(X)Γ.
Assume that End0(X) = Q. Then Γ = 1 . Since End0(X) is a central simple
6 6 { }
Q-algebra, all its automorphisms are inner, i.e., Aut(End0(X)) = End0(X)∗/Q∗.
Let ∆ (cid:16) Γ be the universal central extension of Γ. It is well-known that ∆ is a
finite perfect group and the set of prime divisors of #(∆) coincides with the set of
prime divisors of #(Γ). The universality property implies that the inclusion map
Γ End0(X)∗/Q∗ lifts (uniquely) to a homomorphism π : ∆ End0(X)∗. The
⊂ →
SE´MINAIRES & CONGRE`S11
HOMOMORPHISMS OF ABELIAN VARIETIES 197
equality (1) means that the centralizer of π(∆) in End0(X) coincides with Q and
thereforeker(π) doesnotcoincidewith ∆. Itfollowsthatthe image Γ ofker(π) inΓ
0
doesnotcoincidewiththewholeΓ. ItalsofollowsthatifQ[∆]isthegroupQ-algebra
of ∆ then π induces the Q-algebra homomorphism π : Q[∆] End0(X) such that
→
the centralizer of the image π(Q[∆]) in End0(X) coincides with Q.
I claim that π(Q[∆])=End0(X) and therefore End0(X) is isomorphic to a direct
summandofQ[∆]. This claimfollowseasilyfromthe nextlemma thatwillbe proven
later in this section.
Lemma1.7. — Let E be a field of characteristic zero, T a semisimple finite-
dimensional E-algebra, S a finite-dimensional central simple E-algebra, β : T S
→
an E-algebra homomorphism that sends 1 to 1. Suppose that the centralizer of the
image β(T) in S coincides with the center E. Then β is surjective, i.e. β(T)=S.
In order to prove (b2), let us assume that End0(X) = M (H) where H is a
d
quaternion Q-algebra. Then M (H) is isomorphic to a direct summand of Q[∆].
d
On the other hand, it is well-known that if q is a prime not dividing #(∆) then
Q [∆]=Q[∆] Q is a direct sum of matrix algebrasover (commutative) fields. It
q Q q
⊗
follows that M (H) Q also splits. This proves the assertion (b2).
d Q q
⊗
In order to prove (b3), let us assume that is a quasi-simple finite group with
G
center . Let us put Π:=π(∆) End0(X)∗. We are going to construct a surjective
Z ⊂
homomorphism Π (cid:16) / . In order to do that, it suffices to construct a surjective
G Z
homomorphism Γ(cid:16) / . Recall that there are surjective homomorphisms
G Z
τ :G G = , κ :G Γ.
n,X,L `,X,L X,n n,X,L
−−→→ G −−→→
(If ` is odd then τ isethe identityemap; if `=2 then τe=τ .) Let H be the kernel
2,X 0
of κ :G (cid:16)Γ. Clearly,
X,n n,X,L
(2) e G /H =Γ.
n,X,L 0 ∼
Since Γ = 1 , we have H =G e . It follows that τ(H )= . The surjectivity of
0 n,X,L 0
6 { } 6 6 G
τ :G (cid:16) implies that τ(H ) is normal in and therefore lies in the center .
n,X,L G e 0 G Z
This gives us the surjective homomorphisms
e
G /H τ(G )/τ(H )= /τ(H ) / ,
n,X,L 0 n,X,L 0 0
−−→→ G −−→→G Z
whose compositeion is a surjectiveehomomorphism G /H (cid:16) / . Using (2), we
n,X,L 0
G Z
get the desired surjective homomorphism Γ(cid:16) / .
G Ze
Proof of Lemma 1.7. — Replacing E by its algebraic closure E and tensoring T
a
andS byE ,wemayandwillassumethatE isalgebraicallyclosed. ThenS =M (E)
a n
for some positive integer n. Clearly, β(T) is a direct sum of say, b matrix algebras
over E and the center of β(T) is isomorphic to a direct sum of b copies of E. In
particular,ifb>1thenthecentralizerofβ(T)inS containstheb-dimensionalcenter
of β(T) which gives us the contradiction. So, b = 1 and β(T) = M (E) for some
∼ k
SOCIE´TE´MATHE´MATIQUEDEFRANCE2005
198 YU.G. ZARHIN
positive integer k. Clearly, k 6 n; if the equality holds then we are done. Assume
that k<n: we need to get a contradiction. So, we have
1 E β(T)=M (E), M (E)=S.
∈ ⊂ ∼ k −→ n
This provides En with a structure of faithful β(T)-module in such a way that En
does not contain a non-zero submodule with trivial (zero) action of β(T). Since
β(T) = M (E), the β(T)-module En splits into a direct sum of say, e copies of a
∼ k
simple faithful β(T)-module W with dim (W)=k. Clearly, e=n/k >1. It follows
E
easily that the centralizer of β(T) in S =M (E) coincides with
n
End (We)=M (End (W))=M (E)
β(T) e β(T) e
and has E-dimension e2 >1. Contradiction.
Corollary1.8. — Suppose that ` is a prime, K is a field of characteristic different
from `. Suppose that X is an abelian variety of positive dimension g defined over K.
Let us assume that G contains a perfect subgroup that enjoys the following
`,X,K
G
properties:
e
(a) EndG(X`)=F`;
(b) The only subgroup of index dividing g in is itself.
G G
If g is odd then either End0(X) is a matrix algebra over Q or p = char(K) > 0 and
End0(X) is a matrix algebra M (H ) over a quaternion Q-algebra H that is ramified
d p p
exactly at p and and d > 1. In particular, if char(K) does not divide #( ) then
∞ G
End0(X) is a matrix algebra over Q.
Proof of Corollary 1.8. — LetusassumethatEnd0(X)isnot isomorphictoamatrix
algebra over Q. Then End0(X) is (isomorphic to) a matrix algebra M (H) over a
d
quaternion Q-algebra H. This means that there exists an absolutely simple abelian
varietyY overK suchthatX isisogenoustoYd andEnd0(Y)=H. Clearly,dim(Y)
a
is odd. It follows from Albert’s classification [21, p.202] that p := char(K ) =
a
char(K) > 0. By Lemma 4.3 of [23], if there exists a prime q = p such that H is
6
unramified at q then 4 = dim H divides 2dim(Y). Since dim(Y) is odd, 2dim(Y)
Q
is not divisible by 4 and therefore H is unramified at all primes different from p. It
follows from the theorem of Hasse-Brauer-Noetherthat H=H .
∼ p
Now, assume that d = 1, i.e. End0(X) = H . We know that End0(X)∗ = H∗
p p
contains a nontrivial finite perfect group Π. But this contradicts to the following
elementary statement, whose proof will be given later in this section.
Lemma1.9. — Every finite subgroup in H∗ is solvable.
p
Hence End0(X)=H , i.e. d>1.
p
6
Assume now that p does not divide #( ). It follows from Theorem 1.6 that H
G
is unramified at p. This implies that H can be ramified only at which could not
∞
be the case. The obtained contradiction proves that End0(X) is a matrix algebra
over Q.
SE´MINAIRES & CONGRE`S11