Rational equivalence

Remarks

When $W$ is normal along $V$, $\operatorname{ord}_V$ is the discrete valuation of $k(W)$ at $V$. Example: on $X = \mathbb{P}^1$ over a field $k$, take $W = X$ and $f = (x - a)/(x - b)$ with $a \ne b$ in $k$. Then $\operatorname{div}(f) = [a] - [b]$, so the points $a$ and $b$ are rationally equivalent. Non-example: on an elliptic curve $E$ over an algebraically closed field, two distinct points $P$ and $Q$ are not rationally equivalent. If they were, then $[P] - [Q] = \operatorname{div}(f)$ for some $f \in k(E)^{\times}$, since $W = E$ is the only choice on a curve. Then $f$ would be a rational function of degree $1$, hence an isomorphism $E \to \mathbb{P}^1$, which is impossible because $E$ has genus $1$. Equivalently, $Z - Z' = \sum_i \big([V_i(0)] - [V_i(\infty)]\big)$ for finitely many subvarieties $V_i \subset X \times \mathbb{P}^1$ of dimension $r + 1$ that map onto $\mathbb{P}^1$, where $V_i(a)$ is the fiber of $V_i$ over $a$, viewed in $X$.