Cohomology of rationally equivalent cycles

Question

Let $X$ be a smooth projective variety over an algebraically closed field $k$ of characteristic zero. Let $Z_1, Z_2 \subset X$ be two rationally equivalent sub-varieties. Is it true that $$ H^\bullet (X, \mathcal{O}_{Z_1}) \cong H^\bullet (X, \mathcal{O}_{Z_2})? $$

Answer 1

No. Take a twisted cubic and an elliptic curve in $\mathbb{P}^3$. They have different $H^1$.

The condition of rational equivalence is much too weak to imply that sort of statement. Two equal-dimension projective varieties are rationally equivalent if and only if they have the same degree -- that's the only data remembered by the equivalence class when $X = \mathbb{P}^n$. Other $X$ will be more complicated, but the principle will be similar.

edit: They don't even need to have the same Euler characteristic, as the above example shows.