Lemma 42.62.6 (0FFE)—The Stacks project

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Table of contents
Part 3: Topics in Scheme Theory
Chapter 42: Chow Homology and Chern Classes
Section 42.62: Intersection products
Lemma 42.62.6 (cite)

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Lemma 42.62.6. Let $k$ be a field. Let $X$ be a scheme smooth over $k$ . Let $i : Y \to X$ be a regular closed immersion. Let $\alpha \in \mathop{\mathrm{CH}}\nolimits _*(X)$ . If $Y$ is equidimensional of dimension $e$ , then $\alpha \cdot [Y]_ e = i_*(i^!(\alpha ))$ in $\mathop{\mathrm{CH}}\nolimits _*(X)$ .

Proof. After decomposing $X$ into connected components we may and do assume $X$ is equidimensional of dimension $d$ . Write $\alpha = c \cap [X]_ n$ with $x \in A^*(X)$ , see Lemma 42.62.2. Then

$i_*(i^!(\alpha )) = i_*(i^!(c \cap [X]_ n)) = i_*(c \cap i^![X]_ n) = i_*(c \cap [Y]_ e) = c \cap i_*[Y]_ e = \alpha \cdot [Y]_ e$

The first equality by choice of $c$ . Then second equality by Lemma 42.59.7. The third because $i^![X]_ d = [Y]_ e$ in $\mathop{\mathrm{CH}}\nolimits _*(Y)$ (Lemma 42.59.8). The fourth because bivariant classes commute with proper pushforward. The last equality by Lemma 42.62.2. $\square$

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