Lemma 42.29.5. Let $(S, \delta )$ be as in Situation 42.7.1. Let $X$ be locally of finite type over $S$. Let $(\mathcal{L}, s, i : D \to X)$ and $(\mathcal{L}', s', i' : D' \to X)$ be two triples as in Definition 42.28.1. Then the diagram

$\xymatrix{ \mathop{\mathrm{CH}}\nolimits _ k(X) \ar[r]_{i^*} \ar[d]_{(i')^*} & \mathop{\mathrm{CH}}\nolimits _{k - 1}(D) \ar[d]^{j^*} \\ \mathop{\mathrm{CH}}\nolimits _{k - 1}(D') \ar[r]^{(j')^*} & \mathop{\mathrm{CH}}\nolimits _{k - 2}(D \cap D') }$

commutes where each of the maps is a gysin map.

Proof. Denote $j : D \cap D' \to D$ and $j' : D \cap D' \to D'$ the closed immersions corresponding to $(\mathcal{L}|_{D'}, s|_{D'}$ and $(\mathcal{L}'_ D, s|_ D)$. We have to show that $(j')^*i^*\alpha = j^* (i')^*\alpha$ for all $\alpha \in \mathop{\mathrm{CH}}\nolimits _ k(X)$. Let $W \subset X$ be an integral closed subscheme of dimension $k$. Let us prove the equality in case $\alpha = [W]$. We will deduce it from the key formula.

We let $\sigma$ be a nonzero meromorphic section of $\mathcal{L}|_ W$ which we require to be equal to $s|_ W$ if $W \not\subset D$. We let $\sigma '$ be a nonzero meromorphic section of $\mathcal{L}'|_ W$ which we require to be equal to $s'|_ W$ if $W \not\subset D'$. Write

$\text{div}_{\mathcal{L}|_ W}(\sigma ) = \sum \text{ord}_{Z_ i, \mathcal{L}|_ W}(\sigma )[Z_ i] = \sum n_ i[Z_ i]$

and similarly

$\text{div}_{\mathcal{L}'|_ W}(\sigma ') = \sum \text{ord}_{Z_ i, \mathcal{L}'|_ W}(\sigma ')[Z_ i] = \sum n'_ i[Z_ i]$

as in the discussion in Section 42.26. Then we see that $Z_ i \subset D$ if $n_ i \not= 0$ and $Z'_ i \subset D'$ if $n'_ i \not= 0$. For each $i$, let $\xi _ i \in Z_ i$ be the generic point. As in Section 42.26 we choose for each $i$ an element $\sigma _ i \in \mathcal{L}_{\xi _ i}$, resp. $\sigma '_ i \in \mathcal{L}'_{\xi _ i}$ which generates over $B_ i = \mathcal{O}_{W, \xi _ i}$ and which is equal to the image of $s$, resp. $s'$ if $Z_ i \not\subset D$, resp. $Z_ i \not\subset D'$. Write $\sigma = f_ i \sigma _ i$ and $\sigma ' = f'_ i\sigma '_ i$ so that $n_ i = \text{ord}_{B_ i}(f_ i)$ and $n'_ i = \text{ord}_{B_ i}(f'_ i)$. From our definitions it follows that

$(j')^*i^*[W] = \sum \text{ord}_{B_ i}(f_ i) \text{div}_{\mathcal{L}'|_{Z_ i}}(\sigma '_ i|_{Z_ i})$

as cycles and

$j^*(i')^*[W] = \sum \text{ord}_{B_ i}(f'_ i) \text{div}_{\mathcal{L}|_{Z_ i}}(\sigma _ i|_{Z_ i})$

The key formula (Lemma 42.26.1) now gives the equality

$\sum \left( \text{ord}_{B_ i}(f_ i) \text{div}_{\mathcal{L}'|_{Z_ i}}(\sigma '_ i|_{Z_ i}) - \text{ord}_{B_ i}(f'_ i) \text{div}_{\mathcal{L}|_{Z_ i}}(\sigma _ i|_{Z_ i}) \right) = \sum \text{div}_{Z_ i}(\partial _{B_ i}(f_ i, f'_ i))$

of cycles. Note that $\text{div}_{Z_ i}(\partial _{B_ i}(f_ i, f'_ i)) = 0$ if $Z_ i \not\subset D \cap D'$ because in this case either $f_ i = 1$ or $f'_ i = 1$. Thus we get a rational equivalence between our specific cycles representing $(j')^*i^*[W]$ and $j^*(i')^*[W]$ on $D \cap D' \cap W$. By Remark 42.19.5 the result follows for general $\alpha$. $\square$

Comment #6644 by WhatJiaranEatsTonight on

In the last row of the statement, gysin map should be Gysin map.

There are also:

• 2 comment(s) on Section 42.29: Gysin homomorphisms and rational equivalence

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).