The Stacks project

Lemma 41.31.6. In the situation of Lemma 41.31.5 assume $Y$ is locally of finite type over $(S, \delta )$ as in Situation 41.7.1. Then we have $i_1^*p^*\alpha = p_1^*i^*\alpha $ in $\mathop{\mathrm{CH}}\nolimits _{k - 1}(D_1)$ for all $\alpha \in \mathop{\mathrm{CH}}\nolimits _ k(Y)$.

Proof. Let $W \subset Y$ be an integral closed subscheme of $\delta $-dimension $k$. We distinguish two cases.

Assume $W \subset D$. Then $i^*[W] = c_1(\mathcal{L}_1) \cap [W] + c_1(\mathcal{L}_2) \cap [W]$ in $\mathop{\mathrm{CH}}\nolimits _{k - 1}(D)$ by our definition of gysin homomorphisms and the additivity of Lemma 41.24.2. Hence $p_1^*i^*[W] = p_1^*(c_1(\mathcal{L}_1) \cap [W]) + p_1^*(c_1(\mathcal{L}_2) \cap [W])$. On the other hand, we have $g^*[W] = [g^{-1}(W)]_{k + 1}$ by construction of flat pullback. And $g^{-1}(W) = W_1 \cup W_2$ (scheme theoretically) where $W_ i = p_ i^{-1}(W)$ is a line bundle over $W$ by the lemma (since formation of the diagram commutes with base change). Then $[g^{-1}(W)]_{k + 1} = [W_1] + [W_2]$ as $W_ i$ are integral closed subschemes of $X$ of $\delta $-dimension $k + 1$. Hence

\begin{align*} i_1^*[g^{-1}(W)]_{k + 1} & = c_1(p_1^*\mathcal{L}_2) \cap [W_1] + [W_1 \cap W_2]_ k \\ & = c_1(p_1^*\mathcal{L}_2) \cap p_1^*[W] + [W_1 \cap W_2]_ k \\ & = p_1^*(c_1(\mathcal{L}_2) \cap [W]) + [W_1 \cap W_2]_ k \end{align*}

by construction of gysin homomorphisms, the definition of flat pullback (for the second equality), and compatibility of $c_1 \cap -$ with flat pullback (Lemma 41.25.2). Since $W_1 \cap W_2$ is the zero section of the line bundle $W_1 \to W$ we see from Lemma 41.31.4 that $[W_1 \cap W_2]_ k = p_1^*(c_1(\mathcal{L}_2) \cap [W])$. Note that here we use the fact that $D_1$ is the line bundle which is the relative spectrum of the inverse of $\mathcal{L}_2$. Thus we get the same thing as before.

Assume $W \not\subset D$. In this case, both $i_1^*p^*[W]$ and $p_1^*i^*[W]$ are represented by the $k - 1$ cycle associated to the scheme theoretic inverse image of $W$ in $D_1$. $\square$


Comments (1)

Comment #4640 by awllower on

What is the morphism in the proof? Is it supposed to refer to ?

Also, the end of the first paragraph seems to conclude that

which is different from .


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