The Stacks project

Proof. We have to show that an element $\alpha$ of $Z_ k(X)$ which is rationally equivalent to zero, is mapped to zero in $K_0(\textit{Coh}_{\leq k + 1}(X)/\textit{Coh}_{\leq k - 1}(X))$. Write $\alpha = \sum (i_ j)_*\text{div}(f_ j)$ as in Definition 42.19.1. Observe that

is a finite morphism as each $i_ j : W_ j \to X$ is a closed immersion and the family of $W_ j$ is locally finite in $X$. Hence we may use Lemma 42.23.3 to reduce to the case of $W$. Since $W$ is a disjoint union of integral scheme, we reduce to the case discussed in the next paragraph.

Assume $X$ is integral of $\delta$-dimension $k + 1$. Let $f$ be a nonzero rational function on $X$. Let $\alpha = \text{div}(f)$. We have to show that $\alpha$ is mapped to zero in $K_0(\textit{Coh}_{\leq k + 1}(X)/\textit{Coh}_{\leq k - 1}(X))$. Let $\mathcal{I} \subset \mathcal{O}_ X$ be the ideal of denominators of $f$, see Divisors, Definition 31.23.10. Then we have short exact sequences

and

See Divisors, Lemma 31.23.9. We claim that

The claim implies the element $\alpha = \text{div}(f)$ is represented by $[\mathcal{O}_ X/\mathcal{I}] - [\mathcal{O}_ X/f\mathcal{I}]$ in $K_0(\textit{Coh}_{\leq k}(X)/\textit{Coh}_{\leq k - 1}(X))$. Then the short exact sequences show that this element maps to zero in $K_0(\textit{Coh}_{\leq k + 1}(X)/\textit{Coh}_{\leq k - 1}(X))$.

To prove the claim, let $Z \subset X$ be an integral closed subscheme of $\delta$-dimension $k$ and let $\xi \in Z$ be its generic point. Then $I = \mathcal{I}_\xi \subset A = \mathcal{O}_{X, \xi }$ is an ideal such that $fI \subset A$. Now the coefficient of $[Z]$ in $\text{div}(f)$ is $\text{ord}_ A(f)$. (Of course as usual we identify the function field of $X$ with the fraction field of $A$.) On the other hand, the coefficient of $[Z]$ in $[\mathcal{O}_ X/\mathcal{I}] - [\mathcal{O}_ X/f\mathcal{I}]$ is

Using the distance function of Algebra, Definition 10.121.5 we can rewrite this as

The equalities hold by Algebra, Lemmas 10.121.6 and 10.121.7. (Using these lemmas isn't necessary, but convenient.) $\square$