Lemma 42.14.2 (02RC)—The Stacks project

Table of contents
Part 3: Topics in Scheme Theory
Chapter 42: Chow Homology and Chern Classes
Section 42.14: Flat pullback
Lemma 42.14.2 (cite)

Lemma 42.14.2. Let $(S, \delta )$ be as in Situation 42.7.1. Let $X$ be locally of finite type over $S$. Let $U \subset X$ be an open subscheme, and denote $i : Y = X \setminus U \to X$ as a reduced closed subscheme of $X$. For every $k \in \mathbf{Z}$ the sequence

\[ \xymatrix{ Z_ k(Y) \ar[r]^{i_*} & Z_ k(X) \ar[r]^{j^*} & Z_ k(U) \ar[r] & 0 } \]

is an exact complex of abelian groups.

Proof. First assume $X$ is quasi-compact. Then $Z_ k(X)$ is a free $\mathbf{Z}$-module with basis given by the elements $[Z]$ where $Z \subset X$ is integral closed of $\delta $-dimension $k$. Such a basis element maps either to the basis element $[Z \cap U]$ or to zero if $Z \subset Y$. Hence the lemma is clear in this case. The general case is similar and the proof is omitted. $\square$

Comments (4)

Comment #177 by Adeel on March 20, 2013 at 18:58

Above the second arrow in the diagram, it should be j^* instead of lower star.

Comment #184 by Johan on March 27, 2013 at 18:06

Fixed, see here. Thanks!

Comment #2297 by Daniel on November 06, 2016 at 21:09

Why cannot there be to different cycles $[Z_1]$ and $[Z_2]$ in $Z_k(X)$ both mapping to $[Z_1\cap U]=[Z_2\cap U]$ ??

Comment #2323 by Johan on December 12, 2016 at 00:39

@Daniel: If $Z_1$ and $Z_2$ are integral closed subschemes of $X$ which both meet $U$ and if $Z_1 \cap U = Z_2 \cap U$ , then we have $Z_1 = Z_2$ . One argument to see this is if $\eta_i \in Z_i$ is the generic point, then $Z_1 \cap U = Z_2 \cap U$ nonempty implies $\eta_1 = \eta_2$ , hence $Z_1 = \overline{\{\eta_1\}} = \overline{\{\eta_2\}} = Z_2$ .

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