Why is it true that the hom=num over finite fields implies the same statement over arbitrary fields? Harada uses this, and I have seen it before without understanding it.
Yes. I had to think a bit but it follows from the definitions:
Suppose you have a smooth and proper scheme over some base S = Spec R, where R is (say) an integral domain. Let y \in S be the generic point, s \in S a closed point. Then H^i(X_y) = H^i(X_s), so a cycle on X_y is hom^0 iff its reduction mod s is hom~0.
On the other hand, if a cycle Z on X_y is not num~0, there is a cycle Z’ with (Z.Z’) \ne 0 and then consider the reductions of Z, Z’ mod s.
Now suppose you have V smooth and proper over a field K. Find a finitely generated Z-algebra R with field of fractions F \subset K such that V comes by basechange from a smooth and proper scheme over Spec R - so the residue fields of closed points s of Spec R are finite.
Does this make sense? - Tony
I thought I would have a look at some of this to decide whether or not it makes sense (the author last published a paper in 1987 so if what he’s done were right, he would beat Wiles by a long short for persistence). Under such circumstances it is usually best to start at the end. On the last page before 6.4 he clearly misreads Milne’s paper (which only concerns abelian varieties). And nowhere in Andre’s paper does he claim that anything will prove the Hodge conjecture for abelian varieties. If he can’t even read simple statements of theorems correctly what chance does he have of proving the Tate conjecture? Both papers are half taken up by recalling elementary and well-known stuff. When you cut that out, the “proof” of Tate comes down to 2 or 3 pages…. I think there is little reason for optimism!
nLab page on Scholl about Harada