Homological equivalence for algebraic cycles

Question

How does one prove that for a smooth projective variety $X$ over an algebraically closed field $k$, an algebraic cycle that is homologically equivalent to zero is also numerically equivalent to zero? In all the papers I have seen, this is just stated without proof. Is there a simple proof or a good reference?

Answer 1

This just says cup product is well defined. Let $Z \subset X$ be an algebraic cycle; let $n=\dim X$ and $k=\dim Z$. $Z$ is homologically equivalent to zero if the image $[Z]$ of $Z$ in $H^{2n-2k}(X)$ is zero, and $Z$ is numerically zero if $[Z] \cup [Y]=0$ for all algebraic cycles $Y$ of dimension $n-k$.

It's possible you might have seen homological equivalence defined using the image in $H_{2k}(X)$ rather than $H^{2n-2k}(X)$. That's the same because the two maps are related by the Poincare duality isomorphism $\cap [X]: H^{2n-2k}(X) \to H_{2k}(X)$.

It's possible you might have seen numerical equivalence defined in the Chow ring rather than in cohomology. Then you need to use that $A^{\bullet}(X) \to H^{\bullet}(X)$ is a map of rings.