Lemma 42.29.2. Let $(S, \delta )$ be as in Situation 42.7.1. Let $X$ be locally of finite type over $S$. Let $(\mathcal{L}, s, i : D \to X)$ be as in Definition 42.28.1. The Gysin homomorphism factors through rational equivalence to give a map $i^* : \mathop{\mathrm{CH}}\nolimits _{k + 1}(X) \to \mathop{\mathrm{CH}}\nolimits _ k(D)$.

**Proof.**
Let $\alpha \in Z_{k + 1}(X)$ and assume that $\alpha \sim _{rat} 0$. This means there exists a locally finite collection of integral closed subschemes $W_ j \subset X$ of $\delta $-dimension $k + 2$ and $f_ j \in R(W_ j)^*$ such that $\alpha = \sum i_{j, *}\text{div}_{W_ j}(f_ j)$. Set $X' = \coprod W_ i$ and consider the diagram

of Remark 42.28.7. Since $X' \to X$ is proper we see that $i^*p_* = q_*(i')^*$ by Lemma 42.28.8. As we know that $q_*$ factors through rational equivalence (Lemma 42.20.3), it suffices to prove the result for $\alpha ' = \sum \text{div}_{W_ j}(f_ j)$ on $X'$. Clearly this reduces us to the case where $X$ is integral and $\alpha = \text{div}(f)$ for some $f \in R(X)^*$.

Assume $X$ is integral and $\alpha = \text{div}(f)$ for some $f \in R(X)^*$. If $X = D$, then we see that $i^*\alpha $ is equal to $c_1(\mathcal{L}) \cap \alpha $. This is rationally equivalent to zero by Lemma 42.27.2. If $D \not= X$, then we see that $i^*\text{div}_ X(f)$ is equal to $c_1(\mathcal{O}_ D) \cap [D]_{n - 1}$ in $\mathop{\mathrm{CH}}\nolimits _{n - 2}(D)$ by Lemma 42.29.1. Of course capping with $c_1(\mathcal{O}_ D)$ is the zero map (Lemma 42.24.2). $\square$

## 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Your email address will not be published.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.

## Comments (0)

There are also: