HODGE GROUPS OF CERTAIN SUPERELLIPTIC JACOBIANS JIANGWEIXUEANDYURIG.ZARHIN 0 1 0 2 Throughoutthis paper K is a field ofcharacteristiczero, K¯ its algebraicclosure n and Gal(K) = Aut(K¯/K) the absolute Galois group of K. If X is an abelian a variety over K¯ then we write End(X) for the ring of all its K¯-endomorphisms and J 0 End0(X)for the correspondingQ-algebraEnd(X)⊗Q;the notation1X standsfor the identity automorphism of X. 2 Let f(x) ∈ K[x] be a polynomial of degree n ≥ 3 with coefficients in K and ] without multiple roots, Rf ⊂ K¯ the (n-element) set of roots of f and K(Rf) ⊂ G K¯ the splitting field of f. We write Gal(f) = Gal(f/K) for the Galois group A Gal(K(R )/K) of f; it permutes roots of f and may be viewed as a certain per- f . mutation group of R , i.e., as as a subgroup of the group Perm(R ) ∼= S of h f f n permutation of R . (Gal(f) is transitive if and only if f is irreducible.) t f a Suppose that p is a prime that does not divide n and a positive integer q = pr m is a power of p. We write C for the superelliptic K-curve yq =f(x) and J(C ) f,q f,q [ for its jacobian. Clearly, J(C ) is an abelian variety that is defined over K and f,q 3 (n−1)(q−1) v dim(J(C ))= . f,q 6 2 7 Assume that K contains a primitive qth root of unity ζ . In a series of papers 6 q 2 [12,14,15,16], one ofthe authors(Y.Z.)discussedthe structureofEnd0(J(C )), f,q . assuming that n ≥ 5 and the Galois group Gal(f) of f(x) over K is, at least, 0 1 doublytransitive. Inparticular,heprovedthatifn≥5andGal(f)coincideseither 9 with full symmetric group S or with alternating group A then End0(J(C )) is n n f,q 0 (canonically) isomorphic to a product r Q(ζ ) of cyclotomic fields. (If q = p : i=1 pi v then we proved that End(J(Cf,p)) = QZ[ζp].) More precisely, if q 6= p then the i map (x,y) → (x,yp) defines the map of curves C → C , which induces X f,q f,q/p (by Albanese functoriality) the surjective homomorphism J(C ) → J(C ) of r f,q f,q/p a abelian varieties over K; we write J(f,q) for the identity component of its kernel. (If q = p then we put J(f,q) = J(C ).) One may check [16] that J(C ) is K- f,p f,q isogenoustotheproduct r J(f,pi)andtheautomorphismδ :(x,y)7→(x,ζ y)of i=1 q q Cf,q gives rise to an embQedding Z[ζq]֒→End(J(f,q)), ζq 7→δq. What was actually proved in [16, 18] is that Z[ζ ]∼=Z[δ ]=End(J(f,q)) q q if Gal(f)=S , p≥3 and n≥4 or Gal(f)=A and n≥5. n n LetusassumethatK ⊂CandletthefieldK¯ bethealgebraicclosureofK inC. ThisallowsustoconsiderJ(C )andJ(f,q) ascomplexabelianvarieties. Ourgoal f,q is to study the (reductive Q-algebraic connected) Hodge group Hdg=Hdg(J(f,q)) of J(f,q). Notice that when q = 2 (i.e., in the hyperelliptic case) this group was completely determined in [13] (when f(x) has “large” Galois group). When q > 2 1 2 JIANGWEIXUEANDYURIG.ZARHIN we determined in our previous paper [9] the center of Hdg(J(f,q)), also assuming that the Galois group of f(x) is “large”. Let us assume that q >2. In order to describe our results let us recall that the jacobianJ(C )carriesthe canonicalprincipalpolarizationthatisinvariantunder f,q all automorphisms (induced by automorphisms) of C . This implies that the in- f,q ducedpolarizationontheabeliansubvarietyJ(f,q) isδ -invariant. Thispolarization q gives rise to the δ -invariant nondegenerate alternating Q-bilinear form q ψ :H (J(f,q),Q)×H (J(f,q),Q)→Q q 1 1 on the first rational homology group of the complex abelian variety J(f,q). On the other hand, H (J(f,q),Q) carries the natural structure of Q[δ ]∼=Q(ζ )-vector 1 q q space. The δ -invariance of ψ implies that q q ψ (ex,y)=ψ (x,e¯y) ∀e∈Q(ζ ); x,y ∈H (J(f,q),Q). q q q 1 Here e7→e¯stands for the complex conjugation map. Let Q(ζ )+ ={e∈Q(ζ )|e¯=e} q q be the maximal totally real subfield of the cyclotomic CM field Q(ζ ) and let q Q(ζq)− ={e∈Q(ζq)|e¯=−e}. Pick a non-zero element α ∈ Q(ζq)−. Now the standard construction (see, for instance, [5, p. 531]) allows us to define the non-degenerate Q(ζ )-sesquilinear q Hermitian form φ :H (J(f,q),Q)×H (J(f,q),Q)→Q(ζ ) q 1 1 q such that ψ (x,y)=Tr (αφ (x,y)) ∀x,y ∈H (J(f,q),Q). q Q(ζq)/Q q 1 WewriteU(H (J(f,q),Q),φ )fortheunitarygroupofφ oftheQ(ζ )-vectorspace 1 q q q H (J(f,q),Q), viewed as an algebraic Q-subgroup of GL(H (J(f,q),Q)) (via Weil’s 1 1 restrictionof scalars from Q(ζ )+ to Q (ibid). Since the Hodge group respects the q polarization and commutes with endomorphisms of J(f,q), Hdg(J(f,q))⊂U(H (J(f,q),Q),φ ). 1 q Our main result is the following statement. Theorem 0.1. Suppose that n ≥ 4 and p is a prime that does not divide n. Let f(x) ∈ C[x] be a degree n polynomial without multiple roots. Let r be a positive integer and q =pr. Suppose that there exists a subfield K of C that contains all the coefficients of f(x). Let us assume that f(x) is irreducible over K and the Galois group Gal(f) of f(x) over K is either S or A . Assume additionally that either n n n≥5 or n=4 and Gal(f)=S . 4 Suppose that n>q and one of the following three conditions holds: (A) q <n<2q; (B) p is odd and n6≡1 mod q; (C) p=2, n6≡1 mod q and n6≡q−1 mod 2q. Then Hdg(J(f,q))=U(H (J(f,q),Q),φ ). 1 q Remark 0.2. The case of q =p=3 was earlier treated in [14]. HODGE GROUPS OF CERTAIN SUPERELLIPTIC JACOBIANS 3 Lefschetz’s theorem about algebraicity of 2-dimensionalHodge classes and clas- sical invariant theory for the unitary groups [5, Theorem 0 on p. 524; see also pp. 531–532] imply the following corollary to Theorem 0.1. Corollary0.3. Let(n,p,q,f(x))satisfytheconditionsofTheorem0.1. Thenevery Hodge class on each self-product (J(f,q))m of J(f,q) can be presented as a linear combination with rational coefficients of products of divisor classes. In particular, the Hodge conjecture holds true for (J(f,q))m. The paperis organizedas follows. InSection 1we discuss Lie algebrasofHodge groups of complex abelian varieties. Its main result, Theorem 1.1 (that may be of independent interest) asserts that under certain conditions the semisimple part of theHodgegroup(anditsLiealgebra)is“aslargeaspossible”. WededuceTheorem 0.1fromTheorem1.1,using auxiliaryresults fromSection2 concerningdivisibility properties of certain arithmetic functions. In Section 3 we discuss linear reductive Lie algebras. The last Section contains the proof of Theorem 1.1. 1. Complex abelian varieties Let Z be a complex abelian variety of positive dimension and let Ω1(Z) be the dim(Z)-dimensional complex vector space of regular differential 1-forms on Z. We write C for the center of the semisimple finite-dimensional Q-algebra End0(Z). Z We write H (Z,Q) for its first rational homology group. It is well known that 1 H (Z,Q) is a 2dim(Z)-dimensional Q-vector space. 1 The Q-algebra End0(Z) acts faithfully on H (Z,Q). In particular, if E is a 1 subfieldofEnd0(Z)thatcontainstheidentitymapthenH (Z,Q)carriesthenatural 1 structure of E-vector space of dimension 2dim(Z) d(Z,E)= . [E :Q] Let Σ be the set of all field embeddings σ :E ֒→C. It is well-known that E C :=E⊗ C=C, E =E⊗ C= E⊗ C= C . σ E,σ C Q E,σ σ σY∈ΣE σY∈ΣE If σ ∈ Σ then we write σ¯ for the complex-conjugate of σ. We write X for the E E Q-vector space of functions φ:Σ →Q with E φ(σ¯)+φ(σ)=0 ∀σ ∈Σ . E If E/Q is Galois then X carries the natural structure of Gal(E/Q)-module. E LetLie(Z)bethetangentspacetotheoriginofZ;itisadim(Z)-dimensionalC- vectorspace. Byfunctoriality,End0(Z)andthereforeE actonLie(Z)andtherefore provide Lie(Z) with a natural structure of E⊗ C-module. Clearly, Q Lie(Z)= CσLie(Z)=⊕σ∈ΣELie(Z)σ σM∈ΣE where Lie(Z) := C Lie(Z) = {x ∈ Lie(Z) | ex = σ(e)x ∀e ∈ E}. Let us σ σ put n = n (Z,E) = dim Lie(Z) = dim Lie(Z) . It is well-known that the σ σ Cσ σ C σ natural map Ω1(Z) → Hom (Lie(Z),C) is an isomorphism. This allows us to C define via duality the natural homomorphism E → End (Hom (Lie(Z),C)) = C C 4 JIANGWEIXUEANDYURIG.ZARHIN End (Ω1(Z)). This providesΩ1(Z) with a naturalstructure ofE⊗ C-module in C Q such a way that Ω1(Z)σ :=CσΩ1(Z)∼=HomC(Lie(Z)σ,C). In particular, n =dim (Lie(Z) )=dim (Ω1(Z) ) (1). σ C σ C σ Let us consider the first complex homology group of Z H (Z,C)=H (Z,Q)⊗ C, 1 1 Q which is a 2dim(Z)-dimensional complex vector space. If E is as above then H (Z,C) carries the natural structure of a free E := E ⊗ C-module of rank 1 C Q d(Z,E). We have H1(Z,C)= CσH1(Z,C)=⊕σ∈ΣEH1(Z,Q)σ σM∈ΣE where H (Z,Q) :=C H (Z,C)={x∈H (Z,C)|ex=σ(e)x ∀e∈E}=H (Z,Q)⊗ C. 1 σ σ 1 1 1 E,σ Clearly, every H (Z,Q) is a d(Z,E)-dimensional C-vector subspace that is also a 1 σ E -submodule of H (Z,C). C 1 There is a canonical Hodge decomposition ([3, chapter 1], [1, pp. 52–53]) H (Z,C)=H−1,0⊕H0,−1 1 whereH−1,0 =H−1,0(Z)andH0,−1 =H0,−1(Z)aremutually“complexconjugate” dim(Z)-dimensional complex vector spaces. This splitting is End0(Z)-invariant and the End0(Z)-module H−1,0 is canonically isomorphic to the commutative Lie algebra Lie(Z) of Z. This implies that H−1,0 and Lie(Z) are isomorphic as E- modules and even as E -modules. C Let f0 =f0 :H (Z,C)→H (Z,C) H H,Z 1 1 be the C-linear operator in H (Z,C) defined as follows. 1 1 1 f (x)=− x ∀ x∈H−1,0; f0 (x)= x ∀ x∈H0,−1. H 2 H 2 Clearly,f0 commuteswithEnd0(Z). Inparticular,everyH (Z,Q) isf0 -invariant. H 1 σ H More precisely, the linear operator f0 :H (Z,Q) →H (Z,Q) is semisimple and H 1 σ 1 σ its spectrum lies in the two-element set {1/2,−1/2}. Taking into account that the E -modules H−1,0 and Lie(Z) are isomorphic, we conclude that the multiplicity C of eigenvalue −1/2 is n = n (Z,E) while the multiplicity of eigenvalue 1/2 is σ σ d(Z,E)−n (Z,E). Let σ¯ : E ֒→ C be the composition of σ : E ֒→ C and the σ complex conjugation C→C. It is known ([1], [2]) that n +n =d(Z,E). σ σ¯ This implies that the multiplicity of eigenvalue 1/2 is n . σ¯ We refer to [5], [10, Sect. 6.6.1 and 6.6.2] for the definition and basic proper- ties of the Hodge group (aka special Mumford–Tate group) Hdg = Hdg of (the Z rationalHodgestructureH (Z,Q)andof)Z. RecallthatHdgisaconnectedreduc- 1 tive algebraic Q-subgroup of GL(H (Z,Q)), whose centralizer in End (H (Z,Q)) 1 Q 1 coincides with End0(Z). Let hdg=hdg ⊂End (H (Z,Q)) Z Q 1 HODGE GROUPS OF CERTAIN SUPERELLIPTIC JACOBIANS 5 betheQ-LiealgebraofHdg;itisareductiveQ-LiesubalgebraofEnd (H (Z,Q)), Q 1 its natural representation in H (Z,Q) is completely reducible and its centralizer 1 there coincides with End0(Z). Notice also that its complexification hdg =hdg⊗ C⊂End (H (Z,Q))⊗ C=End (H (Z,C)) C Q Q 1 Q C 1 contains f0 [9, Sect. 3.4]. H Suppose that E = End0(Z) is a CM field. Choose a polarization on Z. The correspondingRosatiinvolutiononEnd0(Z)coincideswiththecomplexconjugation e7→e¯onE. ThepolarizationgivesrisetothenondegeneratealternatingQ-bilinear form ψ :H (Z,Q)×H (Z,Q)→Q 1 1 such that ψ(ex,y)=ψ(x,e¯y) ∀x,y ∈H (Z,Q); e∈E. 1 Let E+ ={e∈E |e¯=e} be the maximal totally real subfield of the CM field E and let E− ={e∈E |e¯=−e}. Pick a non-zero element α ∈ Q(ζq)−. Now the standard construction (see, for in- stance,[5,p. 531])allowsustodefinethenon-degenerateE-sesquilinearHermitian form φ:H (Z,Q)×H (Z,Q)→E 1 1 such that ψ(x,y)=Tr (αφ(x,y)) ∀x,y ∈H (Z,Q). E/Q 1 WewriteU(H (Z,Q),φ)fortheunitarygroupofφoftheE-vectorspaceH (Z,Q), 1 1 viewedasanalgebraicQ-subgroupofGL(H (Z,Q))(viaWeil’srestrictionofscalars 1 from E+ to Q (ibid). It is well-known that U(H (Z,Q),φ) is reductive and its Q- 1 dimension is 1 [E+ :Q]d(Z,E)2 = [E :Q]d(Z,E)2. 2 Let u(H (Z,Q),φ) be the Q-Lie algebra of U(H (Z,Q),φ): it is a reductive 1 1 Q-Lie subalgebra of End (H (Z,Q)). Explicitly, Q 1 u(H (Z,Q),φ)={u∈End (H (Z,Q))|φ(ux,y)+φ(x,uy)=0∀x,y ∈H (Z,Q)}. 1 E 1 1 The reductive Q-Lie algebra u(H (Z,Q),φ) splits into a direct sum 1 u(H1(Z,Q),φ)=E−⊕su(H1(Z,Q),φ) of its center E− and the semisimple Q-Lie algebra su(H (Z,Q),φ)={u∈u(H (Z,Q),φ)|Tr (u)=0.} 1 1 E Here Tr :End (H (Z,Q))→E E E 1 is the trace map. One may easily check that 1 dim (su(H (Z,Q),φ))= [E :Q]{d(Z,E)2−1}. Q 1 2 Since the Hodge group respects the polarization and commutes with endomor- phisms of Z, Hdg(Z)⊂U(H (Z,Q),φ) 1 6 JIANGWEIXUEANDYURIG.ZARHIN (ibid). This implies that hdg⊂u(H (Z,Q),φ)⊂End (H (Z,Q)). 1 Q 1 Thisimpliesthatthesemisimpleparthdgss =[hdg,hdg]ofhdgliesinsu(H (Z,Q),φ). 1 In particular, 1 dim (hdgss)≤ [E :Q]{d(Z,E)2−1}; Q 2 the equality holds if and only if hdgss =su(H (Z,Q),φ). 1 The following statement may be viewed as a partial generalization of Theorem 3 in [5, p. 526]. Theorem1.1. SupposethatE =End0(Z)isaCMfield. Assumethatalln (Z,E) σ aredistinctpositiveintegers. Assumeadditionallythatthereexistsafieldembedding σ :E ֒→C such that n (Z,E) and d(Z,E) are relatively prime. σ Then hdgss =su(H (Z,Q),φ). 1 Remark1.2. Clearly,inthecourseoftheproofofTheorem1.1,itsufficestocheck that 1 dim (hdgss)≥ [E :Q]{d(Z,E)2−1}. Q 2 We prove Theorem 1.1 in Section 4. Proof of Theorem 0.1. Let us put Z =J(f,q) and E =Q(ζ ). Clearly, q d(Z,E)=n−1. We know that End0(Z) = E = Q(ζ ) [16, 18] and if σ = σ : Q(ζ ) ֒→ C is a q i q field embedding that sends ζ to ζ−i with 1 ≤ i < q, (i,p) = 1 then n = [ni/q] q q σ [16,17]. (Clearly,everyfieldembeddingQ(ζ )֒→Cisoftheformσ .) Sincen>q, q i all integers [ni/q] are positive and distinct. Propositions 2.1 and 2.2 below imply that under our assumptions on (p,q,n) there exists a positive integer i < q such that (i,p) = 1 and [ni/q] and n−1 are relatively prime. This allows us to apply Theorem 1.1 and conclude that hdgss =su(H (Z,Q),φ). On the other hand, by a 1 resultfrom[9],thecenterofhdgcoincideswithE−. Thisimpliesthatthereductive Q-Lie algebra hdg coincides with the direct sum E−⊕su(H1(Z,Q),φ)=u(H1(Z,Q),φ). Now the connectedness of the Hodge group and the unitary group implies that Hdg(Z)=U(H (Z,Q),φ), i.e., 1 Hdg(J(f,q))=U(H (J(f,q),Q),φ ). 1 q (cid:3) 2. Divisibility properties of integral parts Proposition 2.1. Suppose that p is a prime, r a positive integer, q =pr and n is a positive integer that is not divisible by p. Suppose that one of the following conditions holds: (i) q <n<2q; (ii) The prime p is odd. In addition, either p∤(n−1) or n<2q. HODGE GROUPS OF CERTAIN SUPERELLIPTIC JACOBIANS 7 Then there exists an integer i such that 1≤i≤q−1, (i,p)=1 and integers [ni/p] and n−1 are relatively prime. Proof. If q <n<2q then [n·1/q]=[n/q]=1 and we may take i=1. So, further we assume that p is odd and either n<q or n−1 is not divisible by p. If q >n>q/2 then [2n/q]=1 and we may take i=2. If 0<n<q/2 then there exists a positive integer µ such that q ≤µn<(µ+1)n<2q. Since q is a power of p and n is not divisible by p, q <µn<(µ+1)n<2q. Clearly, 1=[nµ/q]=[n(µ+1)/q]. So, we take as i either µ or µ+1, depending on which one is not divisible by p. Now let us assume that n−1 is not divisible by q. Let us put k=[n/q], c=n−kq, d=c−1. We have c=d+1, n=qk+c, n−1=qk+d; 2≤c≤q−1,1≤d≤q−2, (c,p)=1, (d,p)=1. Let i be an integer such that 1≤i≤q−1 and (i,p)=1. Put j =[ci/q]. Clearly, j is a nonnegative integer such that qj <ic<(q+1)j. (The first inequality holds, because neither c nor i are divisible by p.) In other words, 0<ic−qj <q. In addition, [ni/q]=[(kq+c)i/q]=ik+[ci/q]=ik+j. Supposethatn−1and[ni/q]arenotrelativelyprime. Thenthereexistsaprime ℓ that divides both n−1 and[ni/q]. This implies that q·k+d·1=0 in Z/ℓZ and i·k+j·1=0inZ/ℓZ. So,wegetthe homogeneoussystemoftwolinearequations overthe field Z/ℓZthatadmits the nontrivialsolution(k,1)6=(0,0). By Cramer’s rule,the determinant id−qj is zero in Z/ℓZ, i.e., the integer id−qj is divisible by ℓ. In particular, id−qj 6=±1. So, we prove the Proposition if we find such i that id−qj is either 1 or −1. In order to do that, notice that id−qj =i(c−1)−qj =(ic−qj)−i. This implies that −i<id−qj <q−i. Since 1≤i≤q−1, 1−q <id−qj <q−1. 8 JIANGWEIXUEANDYURIG.ZARHIN Nowifwechooseiinsuchawaythat1≤i≤q−1andidiscongruentto1modulo q (such choice is possible, because d is not divisible by p) then id−qj is congruent to 1 modulo q and therefore id−qj = 1. In addition, the latter equality implies that i is not divisible by p. (cid:3) Proposition 2.1 admits the following (partial) generalization. Proposition 2.2. Suppose that p is a prime, r a positive integer, q =pr and n is a positive integer that is not divisible by p. Suppose that n−1 is not divisible by q. If p=2, assume additionally that q =2r >2 and n6≡q−1 mod 2q. Then there exists an integer i such that 1≤i≤q−1, (i,p)=1 and integers [ni/p] and n−1 are relatively prime. Remark 2.3. If q =p=2 then every odd n is congruent to 1 modulo 2. If p=2,q =4 then (n,q) satisfy the conditions of Proposition 2.2 if and only if n−7 is divisible by 8. Proof of Proposition 2.2. As in the proof of Proposition 2.1, let us put k =[n/q], c=n−kq, d=c−1. We have (1) (c,p)=1, 2≤c≤q−1, 1≤d≤q−2. We are given that q does not divide d. However, in the light of Proposition 2.1, we may and will assume that p divides d; in particular, q >p and d≥p≥2>1. Let ′ ′ t=(d,q), d =d/t, q =q/t. Then ′ ′ ′ q >1, t>1, (d,q )=1 and both t and q′ are powers of p. This implies that ′ ′ (d,p)=1, t≥p≥2, q ≥p≥2>1. Since q′ >1, d′ ≥ 1 and (d′,q′) =1, there exists a unique pair of integers (i,j) such that ′ ′ ′ di−q j =1, 0<i≤q −1, j ≥0. Clearly, (i,p) = 1 and therefore (i+q′,p) = 1, because q′ is a power of p. Since t≥2, we have i+q′ <2q′ ≤tq′ =q. We will treat the case p = 2, n ≡−1 mod q and n 6≡q−1 mod 2q separately atthe end. So wefurther assumethat ifp=2,then n+1is not divisible by q. We will show that either i or i+q′ is the desired integer, i.e., either [ni/q] and n−1 are relatively prime or [n(i+q′)/q] and n−1 are relatively prime. It is convenient to consider both integers [ni/q] and [n(i+q′)/q] as [n(i+ǫq′)/q] with ǫ=0,1. For every nonnegative integer ǫ we have ′ ′ ′ ′ (2) d(i+ǫq )−q (j+ǫd)=1. Multiplying both side of (2) by t, we get ′ ′ (3) d(i+ǫq )−q(j+ǫd)=t. HODGE GROUPS OF CERTAIN SUPERELLIPTIC JACOBIANS 9 Since c=d+1, it follows that c(i+ǫq′) (d+1)(i+ǫq′) t+i+ǫq′ ′ = =j+ǫd + q q q and therefore c(i+ǫq′) t+i+ǫq′ ′ (4) −(j+ǫd)= . (cid:20) q (cid:21) (cid:20) q (cid:21) The following Lemma will be proven at the end of this Section. Lemma 2.4. We keep the assumptions of Proposition 2.2. If p = 2, we assume additionally that q ∤n+1. Then [(t+i+q′)/q]=0 and therefore [(t+i)/q]=0. Let us assume that either p is odd or p = 2 and n+1 is not divisible by q. Combining Lemma 2.4 with (4), we conclude that c(i+ǫq′) ′ (5) =(j+ǫd) (cid:20) q (cid:21) if ǫ=0 or 1. It follows that if ǫ=0 or 1 then ′ ′ ′ ′ ′ ′ [n(i+ǫq )/q]=[(kq+c)(i+ǫq )/q]=k(i+ǫq )+[c(i+ǫq )/q]=k(i+ǫq )+(j+ǫd). Now suppose that n−1 and [n(i+ǫq′)]/q] are not relatively prime for some ǫ=0 or 1. Then there exists a prime ℓ that divides both n−1 and [n(i+ǫq′)/q]. This implies that q·k+d·1 = 0 in Z/ℓZ and (i+ǫq′)·k+(j +ǫd′)·1 = 0 in Z/ℓZ. So we get the homogeneous system of linear equations over the field Z/ℓZ with determinant ′ ′ d(i+ǫq )−q(j+ǫd)=t (by (3)), which admits a non-trivial solution (k,1) 6= (0,0). By Cramer’s rule the determinant t is zero in Z/ℓZ, i.e., ℓ | t. Since t is a power of p, we conclude that ℓ=p. Since(i+ǫq′)·k+(j+ǫd′)·1=0inZ/ℓZandℓ=p,theinteger(i+ǫq′)·k+(j+ǫd′) is divisible by p. Nowsupposethatn−1and[n(i+ǫq′)/q]arenot relativelyprimeforbothǫ=0 and1. This implies thatbothintegersi·k+j and(i+q′)·k+(j+d′) aredivisible byp. Thereforetheir difference q′k+d′ is alsodivisible by p, whichis notthe case, because q′ is a power of p while (d′,p) = 1. The obtained contradiction proves Proposition when either p is odd or p=2 and n+1 is not divisible by q. Atlast,letustreattheremainingcasewhenp=2,q=2r withr ≥2,theinteger n+1 is divisible by q but n 6≡ q−1 mod 2q. Then q divides n+1 and the ratio k := (n+1)/q is an even integer. We have n = 2rk−1. Let us put i = 2r−1−1. Since r ≥2, the integer i is odd. We have ni (2rk−1)(2r−1−1) = =(2r−1−1)k−1. (cid:20) q (cid:21) (cid:20) 2r (cid:21) It follows that [ni/q]≡−1 mod k. In particular, [ni/q] is odd, since k is even. Notice that (n−1)/2=[ni/q]+k. 10 JIANGWEIXUEANDYURIG.ZARHIN Combining all those assertions, we get (n−1,[ni/q])=((n−1)/2,[ni/q])=([ni/q]+k,[ni/q]) =(k,[ni/q])=(k,−1)=1. (cid:3) Proof of Lemma 2.4. Recall that q′ and t are powers of p. This implies that t ≥ p≥3 if p is odd; if p=2 then either t=2 or t≥4. First, let us assume that p is odd and therefore t ≥ p ≥ 3 and q′ ≥ p ≥ 3. It follows that ′ q (t−2)≥3(t−2)=3t−6≥t. Since i≤q′−1, ′ ′ ′ q =tq ≥t+2q >t+i+q . This implies that [(t+i+q′)/q]=0. Second, assume that p=2 and t≥4. Then q′ ≥2 and therefore ′ q (t−2)≥2(t−2)=2t−4≥t. As above, i≤q′−1, ′ ′ ′ ′ ′ q =tq =(t−2)q +2q ≥t+2q >t+i+q and therefore [(t+i+q′)/q]=0. Third, assume that p=2 and t=2. Then q t=2, d=2d′, q′ = =2r−1. 2 Recall that (i,p)=1, i. e., i is odd. Since i≤q′−1, the sum t+i+q′ =2+i+2r−1 is greater or equal than q = 2r only if i = q′ −1 = 2r−1−1. By (2), d′ ·i ≡ 1 mod q′. If i=q′−1, then d′ =q′−1, since 1≤d′ <q′. This implies that ′ ′ c=d+1=2d +1=2q −1=q−1, which contradicts the assumption that n+1=(kq+c)+1=kq+(c+1) is not divisible by q. So, this case does not occur. (cid:3) 3. Linear reductive Lie algebras Throughoutthis Section, Q is a field of characteristiczero, C is an algebraically closed field containing Q. If W is a Q-vector space (resp. Q-algebra or Q-Lie algebra) then we write W for W ⊗ C provided with the natural structure of a C Q C-vector space (resp. C-algebra or C-Lie algebra). Let W be a nonzero finite-dimensional Q-vector space. Let E ⊂End (W) be a Q subfieldthatcontainsthe scalarsQ·Id . ThenE/Qisa finite algebraicextension W and W carries the natural structure of E-vector space; in addition, dim (W)=[E :Q]·dim (W). Q E We write Σ for the set of all Q-linearfield embedding σ :E ֒→C. If σ ∈Σ then we write W for the C-vector space W ⊗ C; clearly, σ E,σ dim (W )=dim (W); C σ E