Lemma 42.34.6. Let $(S, \delta )$ be as in Situation 42.7.1. Let $X$ be a finite type scheme over $S$ which has an ample invertible sheaf. Assume $d = \dim (X) < \infty$ (here we really mean dimension and not $\delta$-dimension). Then for any invertible sheaves $\mathcal{L}_1, \ldots , \mathcal{L}_{d + 1}$ on $X$ we have $c_1(\mathcal{L}_1) \circ \ldots \circ c_1(\mathcal{L}_{d + 1}) = 0$ in $A^{d + 1}(X)$.

Proof. We prove this by induction on $d$. The base case $d = 0$ is true because in this case $X$ is a finite set of closed points and hence every invertible module is trivial. Assume $d > 0$. By Divisors, Lemma 31.15.12 we can write $\mathcal{L}_{d + 1} \cong \mathcal{O}_ X(D) \otimes \mathcal{O}_ X(D')^{\otimes -1}$ for some effective Cartier divisors $D, D' \subset X$. Then $c_1(\mathcal{L}_{d + 1})$ is the difference of $c_1(\mathcal{O}_ X(D))$ and $c_1(\mathcal{O}_ X(D'))$ and hence we may assume $\mathcal{L}_{d + 1} = \mathcal{O}_ X(D)$ for some effective Cartier divisor.

Denote $i : D \to X$ the inclusion morphism and denote $i^* \in A^1(D \to X)$ the bivariant class given by the gysin hommomorphism as in Lemma 42.33.3. We have $i_* \circ i^* = c_1(\mathcal{L}_{d + 1})$ in $A^1(X)$ by Lemma 42.29.4 (and Lemma 42.33.4 to make sense of the left hand side). Since $c_1(\mathcal{L}_ i)$ commutes with both $i_*$ and $i^*$ (by definition of bivariant classes) we conclude that

$c_1(\mathcal{L}_1) \circ \ldots \circ c_1(\mathcal{L}_{d + 1}) = i_* \circ c_1(\mathcal{L}_1) \circ \ldots \circ c_1(\mathcal{L}_ d) \circ i^* = i_* \circ c_1(\mathcal{L}_1|_ D) \circ \ldots \circ c_1(\mathcal{L}_ d|_ D) \circ i^*$

Thus we conclude by induction on $d$. Namely, we have $\dim (D) < d$ as none of the generic points of $X$ are in $D$. $\square$

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