The Stacks project

Proposition 42.57.1. Let $(S, \delta )$ be as in Situation 42.7.1. Assume given a closed immersion $X \to Y$ of schemes locally of finite type over $S$ with $Y$ regular and quasi-compact. Then the composition

\[ K'_0(X) \to K_0(D_{X, perf}(\mathcal{O}_ Y)) \to A^*(X \to Y) \otimes \mathbf{Q} \to \mathop{\mathrm{CH}}\nolimits _*(X) \otimes \mathbf{Q} \]

of the map $\mathcal{F} \mapsto \mathcal{F}[0]$ from Remark 42.56.8, the map $ch(X \to Y, -)$ from Remark 42.56.11, and the map $c \mapsto c \cap [Y]$ induces an isomorphism

\[ K'_0(X) \otimes \mathbf{Q} \longrightarrow \mathop{\mathrm{CH}}\nolimits _*(X) \otimes \mathbf{Q} \]

which depends on the choice of $Y$. Moreover, the canonical map

\[ \mathop{\mathrm{CH}}\nolimits _ k(X) \otimes \mathbf{Q} \longrightarrow \text{gr}_ k K'_0(X) \otimes \mathbf{Q} \]

(see above) is an isomorphism of $\mathbf{Q}$-vector spaces for all $k \in \mathbf{Z}$.

Proof. Since $Y$ is regular, the construction in Remark 42.56.8 applies. Since $Y$ is quasi-compact, the construction in Remark 42.56.11 applies. We have that $Y$ is locally equidimensional (Lemma 42.42.1) and thus the “fundamental cycle” $[Y]$ is defined as an element of $\mathop{\mathrm{CH}}\nolimits _*(Y)$, see Remark 42.42.2. Combining this with the map $\mathop{\mathrm{CH}}\nolimits _ k(X) \to \text{gr}_ kK'_0(X)$ constructed above we see that it suffices to prove

  1. If $\mathcal{F}$ is a coherent $\mathcal{O}_ X$-module whose support has $\delta $-dimension $\leq k$, then the composition above sends $[\mathcal{F}]$ into $\bigoplus _{k' \leq k} \mathop{\mathrm{CH}}\nolimits _{k'}(X) \otimes \mathbf{Q}$.

  2. If $Z \subset X$ is an integral closed subscheme of $\delta $-dimension $k$, then the composition above sends $[\mathcal{O}_ Z]$ to an element whose degree $k$ part is the class of $[Z]$ in $\mathop{\mathrm{CH}}\nolimits _ k(X) \otimes \mathbf{Q}$.

Namely, if this holds, then our maps induce maps $\text{gr}_ kK'_0(X) \otimes \mathbf{Q} \to CH_ k(X) \otimes \mathbf{Q}$ which are inverse to the canonical maps $\mathop{\mathrm{CH}}\nolimits _ k(X) \otimes \mathbf{Q} \to \text{gr}_ k K'_0(X) \otimes \mathbf{Q}$ given above the proposition.

Given a coherent $\mathcal{O}_ X$-module $\mathcal{F}$ the composition above sends $[\mathcal{F}]$ to

\[ ch(X \to Y, \mathcal{F}[0]) \cap [Y] \in \mathop{\mathrm{CH}}\nolimits _*(X) \otimes \mathbf{Q} \]

If $\mathcal{F}$ is (set theoretically) supported on a closed subscheme $Z \subset X$, then we have

\[ ch(X \to Y, \mathcal{F}[0]) = (Z \to X)_* \circ ch(Z \to Y, \mathcal{F}[0]) \]

by Lemma 42.50.8. We conclude that in this case we end up in the image of $\mathop{\mathrm{CH}}\nolimits _*(Z) \to \mathop{\mathrm{CH}}\nolimits _*(X)$. Hence we get condition (1).

Let $Z \subset X$ be an integral closed subscheme of $\delta $-dimension $k$. The composition above sends $[\mathcal{O}_ Z]$ to the element

\[ ch(X \to Y, \mathcal{O}_ Z[0]) \cap [Y] = (Z \to X)_* ch(Z \to Y, \mathcal{O}_ Z[0]) \cap [Y] \]

by the same argument as above. Thus it suffices to prove that the degree $k$ part of $ch(Z \to Y, \mathcal{O}_ Z[0]) \cap [Y] \in \mathop{\mathrm{CH}}\nolimits _*(Z) \otimes \mathbf{Q}$ is $[Z]$. Since $\mathop{\mathrm{CH}}\nolimits _ k(Z) = \mathbf{Z}$, in order to prove this we may replace $Y$ by an open neighbourhood of the generic point $\xi $ of $Z$. Since the maximal ideal of the regular local ring $\mathcal{O}_{X, \xi }$ is generated by a regular sequence (Algebra, Lemma 10.106.3) we may assume the ideal of $Z$ is generated by a regular sequence, see Divisors, Lemma 31.20.8. Thus we deduce the result from Lemma 42.55.4. $\square$


Comments (3)

Comment #7997 by WhyAffineDiagonal on

In the proof, the condition that is affine diagonal is not used. Is it redundant?

Comment #7998 by on

WhyAffineDiagonal indeed. We upgraded our treatment of chern classes of complexes to but then we missed the slight improvement one gets here. Thanks! Unfortunately, in the only spot where we use the result (so far) we do need the affineness hypothesis I think (Lemma 42.58.1). See this commit.

Comment #9880 by Doug Liu on

What is the definition of "locally equidimensional" in the proof?


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