Lemma 42.26.4. 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 42.26.3. $\square$

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?

Comment #6867 by on

Yes, this is because the sums are only locally finite. Since the operations aren't linear for infinite sums (only for locally finite ones) we can't argue by linearity. This is why we construct $X'$ and $Y'$. But it is certainly a red herring if you are only interested in the case of varieties for example.

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