1 A Proof of the Hodge Conjecture Kazuhisa MAEHARA∗ Nov.14, 2006 1 Introduction In this paper we show that the Hodge conjecture and a part of the Tate conjec- ture hold. Since it is difficult to find algebraic cycles in general, the strategy is to proceed by induction argument to vanish a certain subspace of a cohomology of an open affine subvariety of an affine variety which is obtained excluding a general hyperplane from a given variety. 2 In Case of Non Sigular Varieties Let k beafieldwithanalgebraicclosurek and X a smooth geometrically irreducible variety over k. There exists the canonical cycle map for =char k r r 2r cl :CH(X) −→ Het (Xk, Q(r)) This image is included in the fixed part 2r Gk Het (Xk, Q(r)) where Gk = Gal(k/k).Tatesconjecturesaysthatifk is finitely generated r 2r Gk as a field, the image of cl generates Het (Xk, Q(r)) . Fix an isomorphism ι : Q → C. Let k = C. One has the canonical cycle map clr :CHr(X) −→ H2r(X(C), Q(2π i)r) This image is included into H2r(X(C), Q(2π i)r) ∩ Hr,r(X(C)) ∗Associate Professor, General Education and Research Center, Tokyo Polytechnic University Received Sept. 28, 2006 2 ACADEMIC REPORTS Fac. Eng. Tokyo Polytech. Univ. Vol. 29 No.1 (2006) Hodge conjecture says that the image of clr generates H2r(X(C), Q(2π i)r) ∩ Hr,r(X(C)). Let U be a smooth quasiprojective variety over k. The images of the canonical cycle maps are 2r Gk Het (Uk, Q(r)) for finitely generated k r 2r 2r F H (U, C) ∩ W2rH (U, Q(r)) for k = C Let U be a smooth quasi-projective variety over k and X a smooth projective compactification of U. One denotes by cl∗ the following cycle maps clDR, cl, clH ; 2r 2r 0 2r (a) ΓDR(HDR(U)(r)) = W0(HDR(U)(r)) ∩ F (HDR(U)(r)) 2r 2r Gk 2r (b) Γ(H (U)(r)) = H (U)(r) ∩ W0(H (U)(r))) 2r 2r 0 2r (c) ΓH (Hσ (U)(r)) = W0(Hσ (U)(r) ∩ F (Hσ (U)(r)) ⊗ C) Lemma 1 It suffices to prove it for a smooth affine variety over k. Proof. It is well known that it is enough to treat it in the case of dim X =2d. Choose a smooth irreducible hyperplane Y on X.ThusX −Y is a smooth affine variety. One obtains the following commutative diagram: d−1 2d CH (Y) −→ Γ∗H∗,Y(X)(d) ↓↓ d 2d CH (X) −→ Γ∗H∗ (X)(d) ↓↓ d 2d CH (X − Y) −→ Γ∗H∗ (X − Y)(d) ↓↓ 00 2d ∼ Here the vertical sequences are exact. By duality, one has H∗,Y(X)(d) = 2d−2 H∗ (Y)(d − 1). Assume conjectures hold for X − Y . Induction hypothesis for d 2d−2 CH (Y) −→ Γ∗H∗ (Y)(d − 1) implies conjectures. Take general strict normalcrossing divisors D1, ··· ,D2d on X excluding Y . Since X − Y is affine, let A denote Γ(X − Y,O), which is geometrically regular commutative kalgebra of finite type. Thus X − Y =SpecA.LetI1, ··· ,I2d be ideals of A such that V (I1)=D1, ···, V (I2d)=D1 ∩···∩D2d.ByN¨other’s nor- malization lemma, one obtains algebraically independent elements x1, ··· ,x2d of A such that (a) A is integral over k[x1, ··· ,x2d]. (b) Ii ∩ k[x1, ··· ,x2d]=(x1, ··· ,xi). Let K = k(x1, ··· ,x2d)andL = Q(A) the total fraction field of A.LetK ⊂ L be a finite extension and M the smallest quasi-Galois extension of L/K . Take the integral closure B of A over k[X]inM .LetG =Gal(M/K). Lemma 2 The action of G on M keeps B to be invariant. A Proof of the Hodge Conjecture 3 Proof. Since A is regular; hence normal, A coincides with the integral closure of k[x1, ··· ,x2d]inL .Leta ∈ A with the minimal polynomial hand let σ ∈ G. Then hσ = h. Take any b ∈ B , which satisfies the minimal polynomial g(b)=0 with coefficients in A .Thusgσ has its coefficients in A. Hence gσ(bσ)=0, which implies b is an integral element of M. Therefore bσ ∈ B . Let K bethefieldoftheinvariantsofM by G and C the integral closure of k[x1, ··· ,x2d]. Note that K is a radical extension of K of finite degree. Lemma 3 (i) BG = C (ii) Spec C → Spec k[x1, ··· , x2d] is a finite, surjective and radical morphism, which is a universal homeomorphism. G G Proof. Since M ⊃ B ⊃ A ⊃ C ⊃ k[x1, ··· ,x2d] , one has K = M ⊃ B ⊃ G G A ⊃ C = C.SinceB is a finite k[x1, ··· ,x2d]module and k[x1, ··· ,x2d]isa N¨otherian ring, every submodule of B is a finite k[x1, ··· ,x2d] module. Every element of BG is integral over C. Hence BG = C since C is integrally closed. φ :SpecC→ Spec k[x1, ··· , x2d] For every point x of Spec C one has κ(x)isa radical extension of κ(φ(x)); thus φ :SpecC→ Spec k[x1, ··· , x2d]isaradical morphism. It is clear that the mophism is finite and surjective; hence a universal homeomorphism. Let F be a suitable smooth sheaf on X. One has a trace map : trL/K : L → K , which naturally extends to a map trA/C :A→ C. If f ∈ C, one further has 1 1 a map trA[ 1 ]/C[ 1 ] : A[ ] → C[ ] and a cohomological map f f f f j 1 j 1 trA[ 1 ]/C[ 1 ] :H Spec A[ ], F → H Spec C[ ], F . f f f f 1 1 If f ∈ k[x1, ··· ,x2d], φ :SpecC[f ] → Spec k[x1, ··· , x2d, f ] is a finite, surjective and radical morphism; hence a universal homeomorphism. Thus one has an isomorphism j 1 j 1 H Spec C[ ], F → H Spec k[x1, ··· , x2d, ], F . f f Let H =Gal(M/L). Lemma 4 Assume that dimSpecA=n=2d≥ 4.Letf1, ··· ,fn be k-linear combinations of x1, ··· ,xn in k[x1, ··· ,xn]. Assume that the intersection locus V (f1, ··· ,fn) is void in Spec k[x1, ··· , xn], i.e., the hyperplanes intersect one n 1 point in infinity. One obtains Γ∗H Spec A[ ], F =0. f1···fn Proof. If n>2, by the affine vanishing theorem, one has H2n−1 (Spec A, F)=0, H2n−2 (Spec A, F)=0. 4 ACADEMIC REPORTS Fac. Eng. Tokyo Polytech. Univ. Vol. 29 No.1 (2006) One has a Cechˇ Spectral sequence: p,q q 1 p+q E1 = ⊕|I|=p+1H ∩i∈ISpec A[ ], F ⇒ H (Spec A − V(f1, ··· , fn), F) . fi pq Note that E1 =0forq>nby affine vanishing theorem and that p>n− 1by Cechˇ cohomology theory. One obtains n−1,n 2n−1 E∞ =H (Spec A − V(f1, ··· , fn), F)=0, n−2,n n−2,n n−2 2n−2 E2 = E∞ = Gr H ((Spec A, F)=0, n−4,n+1 n,n−1 since E2 = E2 = 0. Note that there exists the following exact se- quence: Γ 2d−2 F − −→ ∗H∗ (V(fik ), )(d 1) 2d 1 2d 1 Γ∗H (Spec A[ ], F)(d) −→ Γ∗H (Spec A[ ], F)(d) −→ 0. ∗ ··· ∗ ··· fi1 fik−1 fi1 fik Thus, n 1 n 1 ⊕|I|=n−2Γ∗H ( Spec A[ ], F →⊕|I|=n−1Γ∗H ( Spec A[ ], F i∈I fi i∈I fi and n 1 n 1 ⊕|I|=n−1Γ∗H ( Spec A[ ], F →⊕|I|=nΓ∗H ( Spec A[ ], F i∈I fi i∈I fi n−3,n+1 n+1,n−1 n,n are surjections, respectively. Using E2 = E2 = E1 = 0, one has n−1,n n−3,n+1 n−1,n n+1,n−1 E3 =H E2 → E2 → E2 = n−1,n n−2,n n−1,n n,n n−1,n E2 =H E1 → E1 → E1 =E∞ =0. Hence, the homomorphism n−2,n n−1,n E1 → E1 is a surjection. n−4,n+1 n,n−1 Applying E2 = E2 = 0, one has n−2,n n−4,n+1 n−2,n n,n−1 E3 =H E2 → E2 → E2 = A Proof of the Hodge Conjecture 5 n−2,n n−3,n n−2,n n−1,n n−2,n E2 =H E1 → E1 → E1 =E∞ =0. Moreover, since n−2,n n 1 E2 =H ⊕|I|=n−2H (Spec A[ ], F) −→ i∈I fi n 1 n 1 ⊕|I|=n−1H (Spec A[ ], F) →⊕|I|=nH (Spec A[ ], F) =0 i∈I fi i∈I fi and the homomorphism n 1 n 1 ⊕|I|=n−2Γ∗H (Spec A[ ], F) →⊕|I|=n−1Γ∗H (Spec A[ ], F) i∈I fi i∈I fi is surjective and a functor Γ∗ is exact, it implies that n 1 n 1 ⊕|I|=n−1Γ∗H (Spec A[ ], F) →⊕|I|=nΓ∗H (Spec A[ ], F) i∈I fi i∈I fi is a zero map. On the other hand, n 1 n 1 ⊕|I|=n−1Γ∗H (Spec A[ ], F) →⊕|I|=nΓ∗H (Spec A[ ], F) i∈I fi i∈I fi is a surjection. Thus one concludes that n 1 n 1 Γ∗H (Spec A[ ], F)=⊕|I|=nΓ∗H (Spec A[ ], F)=0. f1 ···fn i∈I fi Theorem 5 For 0 ≤ k ≤ n the images of d 1 2d 1 CH (Spec A[ ]) −→ Γ∗H (Spec A[ ])(d) f1 ···fk f1 ···fk generate the targets. In particular, the image of d 2d CH (Spec A) −→ Γ∗H∗ (Spec A)(d) generates the target.