By Lee,J M
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Example text
E5i , i = 1, 2, 3 ei e j , e2i e2j , . . 2) n=0 e−n i =0= 4 e¯ ni e¯ nj in R ⊂ End (A0 (Y )). 1) is trivial in R⊗Q. 2) show that R ⊗ Q is a quotient of the semi-simple ring F⊗Q F⊗Q F, so it will suffice to show that ¯ = algebraic closure of π∗ π∗ goes to zero under any homomorphism R ⊗ Q → Q ¯ Q. A homomorphism h : F ⊗ F ⊗ F → Q amounts to the choice of three nontrivial fifth roots of 1, ω1 , ω2 , ω3 . 2) force ωi ω j 1, i j. On the other hand, for the image of Zero-cycles on surfaces 13 π∗ π∗ to be non-trivial, one must have ω1 ω22 ω33 = 1.
Then there exist one-dimensional subschemes C , C ⊂ X and a 2-cycle Γ supported on (C × X) ∪ (X × C ) such that some non-zero multiple of the diagonal ∆ on X ×k X is rationally equivalent to Γ. Proof Let C → X be such that J(CΩ ) A0 (XΩ ), and let C ⊂ X be the image of C. Enlarging k, we may assume C defined over k. 3) Let k ⊂ K ⊂ K be extensions of fields. Then the kernel of CH2 (XK ) → CH2 (XK ) is torsion. Proof If [K : K] < ∞ this follows from the existence of a norm CH2 (XK ) → CH2 (XK ). The case K algebraic over K follows by a limit argument.
Xn ) where two or more of the xi coincide. On the complement (S n X)smooth of the singular set, ωn is well defined. There will be an open set ψ T 0 ⊂ T and a morphism T 0 → (S n X)smooth , and ωn,T will be a holomorphic extension of ψ∗ ωn . Using the definition of rational equivalence and the fact that there are no global holomorphic forms on projective space, Mumford shows that if the 14 Lecture 1 cycles in the family parameterized by T are all rationally equivalent, then ωn,T = 0. One next notices that if t ∈ (S n X)smooth is general, the two-form ωn will give a non-degenerate alternating pairing on the tangent space at t.