Lemma 81.27.2 (Projective space bundle formula). Let $(S, \delta )$ be as in Situation 81.2.1. Let $X$ be locally of finite type over $S$. Let $\mathcal{E}$ be a finite locally free $\mathcal{O}_ X$-module $\mathcal{E}$ of rank $r$. Let $(\pi : P \to X, \mathcal{O}_ P(1))$ be the projective bundle associated to $\mathcal{E}$. The map

$\bigoplus \nolimits _{i = 0}^{r - 1} \mathop{\mathrm{CH}}\nolimits _{k + i}(X) \longrightarrow \mathop{\mathrm{CH}}\nolimits _{k + r - 1}(P),$
$(\alpha _0, \ldots , \alpha _{r-1}) \longmapsto \pi ^*\alpha _0 + c_1(\mathcal{O}_ P(1)) \cap \pi ^*\alpha _1 + \ldots + c_1(\mathcal{O}_ P(1))^{r - 1} \cap \pi ^*\alpha _{r-1}$

is an isomorphism.

Proof. Fix $k \in \mathbf{Z}$. We first show the map is injective. Suppose that $(\alpha _0, \ldots , \alpha _{r - 1})$ is an element of the left hand side that maps to zero. By Lemma 81.27.1 we see that

$0 = \pi _*(\pi ^*\alpha _0 + c_1(\mathcal{O}_ P(1)) \cap \pi ^*\alpha _1 + \ldots + c_1(\mathcal{O}_ P(1))^{r - 1} \cap \pi ^*\alpha _{r-1}) = \alpha _{r - 1}$

Next, we see that

$0 = \pi _*(c_1(\mathcal{O}_ P(1)) \cap (\pi ^*\alpha _0 + c_1(\mathcal{O}_ P(1)) \cap \pi ^*\alpha _1 + \ldots + c_1(\mathcal{O}_ P(1))^{r - 2} \cap \pi ^*\alpha _{r - 2})) = \alpha _{r - 2}$

and so on. Hence the map is injective.

To prove the map is surjective, we will argue exactly as in the proof of Lemma 81.25.1 to reduce to the case of schemes. We urge the reader to skip the proof.

Let $\beta \in \mathop{\mathrm{CH}}\nolimits _{k + r - 1}(P)$. Write $\beta = \sum m_ j[W_ j]$ with $m_ j \not= 0$ and $W_ j$ pairwise distinct integral closed subspaces of $\delta$-dimension $k + r$. Then the family $\{ W_ j\}$ is locally finite in $P$. Let $Z_ j \subset X$ be the “image” of $W_ j$ as in Lemma 81.7.1. For any quasi-compact open $U \subset X$ we see that $\pi ^{-1}(U) \cap W_ j$ is nonempty only for finitely many $j$. Hence the collection $Z_ j$ of images is a locally finite collection of integral closed subspaces of $X$.

Consider the fibre product diagrams

$\xymatrix{ P_ j \ar[r] \ar[d]_{\pi _ j} & P \ar[d]^\pi \\ Z_ j \ar[r] & X }$

Suppose that $[W_ j] \in Z_{k + r - 1}(P_ j)$ is rationally equivalent to

$\pi _ j^*\alpha _{j, 0} + c_1(\mathcal{O}(1)) \cap \pi _ j^*\alpha _{j, 1} + \ldots + c_1(\mathcal{O}(1))^{r - 1} \cap \pi _ j^*\alpha _{j, r - 1}$

for some $(k + i)$-cycle $\alpha _{j, i} \in \mathop{\mathrm{CH}}\nolimits _{k + i}(Z_ j)$. Then $\alpha _ i = \sum m_ j \beta _{j, i}$ will be a $(k + i)$-cycle on $X$ and

$\pi ^*\alpha _0 + c_1(\mathcal{O}(1)) \cap \pi ^*\alpha _1 + \ldots + c_1(\mathcal{O}(1))^{r - 1} \cap \pi ^*\alpha _{r - 1}$

will be rationally equivalent to $\beta$ (see Remark 81.15.3). This reduces us to the case $X$ integral, and $\alpha = [W]$ for some integral closed subscheme of $P$ dominating $X$. In particular we may assume that $d = \dim _\delta (X) < \infty$.

Hence we can use induction on $d = \dim _\delta (X)$. If $d < k$, then $\mathop{\mathrm{CH}}\nolimits _{k + r - 1}(X) = 0$ and the lemma holds; this is the base case of the induction. Consider a nonempty open $U \subset X$. Suppose that we can show that

$\beta |_{\pi ^{-1}(U)} = \pi ^*\alpha _0 + c_1(\mathcal{O}(1)) \cap \pi ^*\alpha _1 + \ldots + c_1(\mathcal{O}(1))^{r - 1} \cap \pi ^*\alpha _{r - 1}$

for some $\alpha _ i \in Z_{k + i}(U)$. By Lemma 81.10.2 we see that $\alpha _ i = \alpha '_ i|_ U$ for some $\alpha '_ i \in Z_{k + i}(X)$. By the exact sequences $\mathop{\mathrm{CH}}\nolimits _{k + i}(\pi ^{-1}(X \setminus U)) \to \mathop{\mathrm{CH}}\nolimits _{k + i}(P) \to \mathop{\mathrm{CH}}\nolimits _{k + i}(\pi ^{-1}(U))$ of Lemma 81.15.2 we see that

$\beta - \left(\pi ^*\alpha '_0 + c_1(\mathcal{O}(1)) \cap \pi ^*\alpha '_1 + \ldots + c_1(\mathcal{O}(1))^{r - 1} \cap \pi ^*\alpha '_{r - 1}\right)$

comes from a cycle $\beta ' \in \mathop{\mathrm{CH}}\nolimits _{k + r}(\pi ^{-1}(X \setminus U))$. Since $\dim _\delta (X \setminus U) < d$ we win by induction on $d$.

In particular, by replacing $X$ by a suitable open we may assume $X$ is a scheme and we have reduced our problem to Chow Homology, Lemma 42.36.2. $\square$

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