Classical Chow relations and Bloch boundaries

Work on one integral component of $X$ of dimension $d$. Let $W\subset X\times {\mathbf{A}}_t^1$ be an admissible codimension-$r$ generator, so $\dim W=d-r+1$. Let $\bar{W}$ be its reduced closure in $X\times {\mathbf{P}}_t^1$ and let $Y\subset X$ be the reduced image under projection. The projection is proper, so $Y$ is closed and integral; its factorization through the reduced $Y$ is proper by base change along $Y\hookrightarrow X$. Its generic fiber is a closed integral subscheme of ${\mathbf{P}}_{k(Y)}^1$, and hence has dimension zero or one; these are also the possible transcendence degrees of $k(W)\mathrm{/}k(Y)$. The dimension formula in [F3] gives $\dim Y=\dim W$ or $\dim W-1$. If the latter holds, $W$ is an integral closed subset of $Y\times {\mathbf{A}}^1$ of its full dimension, hence equals that product; its two face cycles are both $[Y]$, and its boundary is zero. This also handles every possible generator when $r=0$, since then $\dim W=d+1$.

Conversely, for $r>0$ let $Y\subset X$ be integral of codimension $r-1$ and $f\in k(Y{)}^{\times }$. Form the graph closure $G\subset Y\times {\mathbf{P}}_u^1$ of [F2]. Remove the fiber $u=1$ and use $t=u\mathrm{/}(u-1)$ to identify ${\mathbf{P}}_u^1\setminus \{1\}$ with ${\mathbf{A}}_t^1$. If $f\mathrm{\ne }1$, this open part of $G$ is a nonempty integral closed codimension-$r$ subscheme of $X\times {\mathbf{A}}_t^1$. Its $t=0,1$ intersections are $G_0,G_{\mathrm{\infty }}$. Each is empty or an effective Cartier divisor of pure codimension one on the integral finite-type $G$, so the cycle is admissible. Its boundary equals ${\mathrm{div}}_Y(f)$ by the graph formula. If $f=1$, the divisor is zero and no generator is needed. Thus every principal-divisor relation is a boundary.

For the equal-dimensional case of step 1.1 $\bar{W}\to Y$ is proper dominant generically finite and $Y$ has codimension $r-1$. The rational function $h=t\mathrm{/}(t-1)$ on $\bar{W}$ is nonzero and defined as a rational function because admissibility excludes $W$ from either endpoint. Its divisor is the zero fiber minus the one fiber, with exactly the lengths in [F1]; it is invertible near infinity. By the norm formula, pushing this cycle to $X$ gives $\mathrm{\partial }[W]={\mathrm{div}}_Y({\mathrm{Nm}}_{k(W)\mathrm{/}k(Y)}h)$. Thus every boundary lies in $R^r(X)$, with zero as already shown in the product case.

The two inclusions prove $R^r(X)=B^r(X)$ for $r>0$, and step 1.1 proves it for $r=0$ because $R^0(X)=0$. The finite clopen decomposition of [F3] gives the equality on $X$ and the stated ordinary rational-equivalence quotient.