Show description

4. Let Ni be a direct summand of M (Qi ). Suppose for some odd number η and some ψ ∈ Hom(N1 |k , N2 |k ) we have η · ψ ∈ image Hom(N1 , N2 ) → Hom(N1 |k , N2 |k ) . Then ψ ∈ image Hom(N1 , N2 ) → Hom(N1 |k , N2 |k ) . Motives of Quadrics with Applications to the Theory of Quadratic Forms 45 Proof. Let F/k be a Galois extension of degree 2n such that Ni |F is a sum of Tate motives (for example, an extension which splits both quadrics completely). Then Hom(N1 |F , N2 |F ) → Hom(N1 |k , N2 |k ) is an isomorphism.

Then, for some d, ξ 2 is a projector. Proof. Let x := ξ 2 − ξ ∈ End M (Q) = CHm (Q × Q). Since ξ|k is a projector, x|k = 0. In particular, 2s · x = 0 for some s, since Q is hyperbolic over some Galois extension F/k of degree 2s , and TrF /k ◦jF /k (x) = [F : k] · x (here jF /k and TrF /k are the restriction and corestriction maps on Chow groups). 1, we have xt = 0 for some t. That means that for some large d 2d · xj = 0 j for all j > 0. From the equality ξ 2 = ξ + x (and the fact that ξ and x commute), we get ξ2 d+1 = 0≤j≤2d d 2d 2d −j j · x = ξ2 .

For any field extension E/k, we have that p|E is isotropic if and only if q|E is. In particular we get (2). So, we have rational (algebro-geometric) maps f: Q P , and g : P Q. Let α ∈ CHdim P (Q×P ) = Hom M (Q), M (P ) and β ∈ CHdim Q (P × Q) = Hom M (P ), M (Q) be the closures of the graphs of f and g, respectively. Clearly, α(l0 ) = l0 and β(l0 ) = l0 . So, the composition degQ ◦β ◦ α : CH0 (Q|k ) → Z/2 is nonzero. 8). 25. Let Q be an anisotropic quadric and L be an indecomposable direct summand of M (Q) with a(L) = 0.