Lemma 42.25.4. Let $(S, \delta )$ be as in Situation 42.7.1. Let $X$, $Y$ be locally of finite type over $S$. Let $p : X \to Y$ be a proper morphism. Let $\alpha \in Z_{k + 1}(X)$. Let $\mathcal{L}$ be an invertible sheaf on $Y$. Then

$p_*(c_1(p^*\mathcal{L}) \cap \alpha ) = c_1(\mathcal{L}) \cap p_*\alpha$

in $\mathop{\mathrm{CH}}\nolimits _ k(Y)$.

Proof. Suppose that $p$ has the property that for every integral closed subscheme $W \subset X$ the map $p|_ W : W \to Y$ is a closed immersion. Then, by definition of capping with $c_1(\mathcal{L})$ the lemma holds.

We will use this remark to reduce to a special case. Namely, write $\alpha = \sum n_ i[W_ i]$ with $n_ i \not= 0$ and $W_ i$ pairwise distinct. Let $W'_ i \subset Y$ be the image of $W_ i$ (as an integral closed subscheme). Consider the diagram

$\xymatrix{ X' = \coprod W_ i \ar[r]_-q \ar[d]_{p'} & X \ar[d]^ p \\ Y' = \coprod W'_ i \ar[r]^-{q'} & Y. }$

Since $\{ W_ i\}$ is locally finite on $X$, and $p$ is proper we see that $\{ W'_ i\}$ is locally finite on $Y$ and that $q, q', p'$ are also proper morphisms. We may think of $\sum n_ i[W_ i]$ also as a $k$-cycle $\alpha ' \in Z_ k(X')$. Clearly $q_*\alpha ' = \alpha$. We have $q_*(c_1(q^*p^*\mathcal{L}) \cap \alpha ') = c_1(p^*\mathcal{L}) \cap q_*\alpha '$ and $(q')_*(c_1((q')^*\mathcal{L}) \cap p'_*\alpha ') = c_1(\mathcal{L}) \cap q'_*p'_*\alpha '$ by the initial remark of the proof. Hence it suffices to prove the lemma for the morphism $p'$ and the cycle $\sum n_ i[W_ i]$. Clearly, this means we may assume $X$, $Y$ integral, $f : X \to Y$ dominant and $\alpha = [X]$. In this case the result follows from Lemma 42.25.3. $\square$

Comment #6642 by WhatJiaranEatsTonight on

I do not know why it need to construct $X'$ and $Y'$. Can we reduce the conclusion to $X,Y$ integral merely by the linearity of two sides of the equation?

Does the requirement of the cycles to be "locally finite" but not "finite" matter?

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