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