The Stacks project

Proof. Since $c_ r(\mathcal{E})$ is a bivariant class (Lemma 42.38.7) we may assume $X = X'$ and we have to show that $c_ r(\mathcal{E}) \cap [X]_ n = [Z(s)]_{n - r}$ in $\mathop{\mathrm{CH}}\nolimits _{n - r}(X)$. We will prove the lemma by induction on $r \geq 0$. (The case $r = 0$ is trivial.) The case $r = 1$ is handled by Lemma 42.25.4. Assume $r > 1$.

Let $\pi : P \to X$ be the projective space bundle associated to $\mathcal{E}$ and consider the short exact sequence

By the projective space bundle formula (Lemma 42.36.2) it suffices to prove the equality after pulling back by $\pi$. Observe that $\pi ^{-1}Z(s) = Z(\pi ^*s)$ has $\delta$-dimension $\leq n - 1$ and that the assumption on regular sequences at generic points of $\delta$-dimension $n - 1$ holds by flat pullback, see Algebra, Lemma 10.68.5. Let $t \in \Gamma (P, \mathcal{O}_ P(1))$ be the image of $\pi ^*s$. We claim

Assuming the claim we finish the proof as follows. The restriction $\pi ^*s|_{Z(t)}$ maps to zero in $\mathcal{O}_ P(1)|_{Z(t)}$ hence comes from a unique element $s' \in \Gamma (Z(t), \mathcal{E}'|_{Z(t)})$. Note that $Z(s') = Z(\pi ^*s)$ as closed subschemes of $P$. If $\xi \in Z(s')$ is a generic point with $\delta (\xi ) = n - 1$, then the ideal of $Z(s')$ in $\mathcal{O}_{Z(t), \xi }$ can be generated by a regular sequence of length $r - 1$: it is generated by $r - 1$ elements which are the images of $r - 1$ elements in $\mathcal{O}_{P, \xi }$ which together with a generator of the ideal of $Z(t)$ in $\mathcal{O}_{P, \xi }$ form a regular sequence of length $r$ in $\mathcal{O}_{P, \xi }$. Hence we can apply the induction hypothesis to $s'$ on $Z(t)$ to get $c_{r - 1}(\mathcal{E}') \cap [Z(t)]_{n + r - 2} = [Z(s')]_{n - 1}$. Combining all of the above we obtain

which is what we had to show.

Proof of the claim. This will follow from an application of the already used Lemma 42.25.4. We have $\pi ^{-1}(Z(s)) = Z(\pi ^*s) \subset Z(t)$. On the other hand, for $x \in X$ if $P_ x \subset Z(t)$, then $t|_{P_ x} = 0$ which implies that $s$ is zero in the fibre $\mathcal{E} \otimes \kappa (x)$, which implies $x \in Z(s)$. It follows that $\dim _\delta (Z(t)) \leq n + (r - 1) - 1$. Finally, let $\xi \in Z(t)$ be a generic point with $\delta (\xi ) = n + r - 2$. If $\xi$ is not the generic point of the fibre of $P \to X$ it is immediate that a local equation of $Z(t)$ is a nonzerodivisor in $\mathcal{O}_{P, \xi }$ (because we can check this on the fibre by Algebra, Lemma 10.99.2). If $\xi$ is the generic point of a fibre, then $x = \pi (\xi ) \in Z(s)$ and $\delta (x) = n + r - 2 - (r - 1) = n - 1$. This is a contradiction with $\dim _\delta (Z(s)) \leq n - r$ because $r > 1$ so this case doesn't happen. $\square$