The Stacks project

Lemma 42.30.5. 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 42.29.1. 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 42.27. 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 42.27 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 42.27.1) 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 42.19.6 the result follows for general $\alpha $. $\square$


Comments (2)

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 42.30: Gysin homomorphisms and rational equivalence

Post a comment

Your email address will not be published. Required fields are marked.

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).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.




In order to prevent bots from posting comments, we would like you to prove that you are human. You can do this by filling in the name of the current tag in the following input field. As a reminder, this is tag 0B73. Beware of the difference between the letter 'O' and the digit '0'.