Lemma 42.40.2 (02UH)—The Stacks project

Table of contents
Part 3: Topics in Scheme Theory
Chapter 42: Chow Homology and Chern Classes
Section 42.40: Additivity of Chern classes
Lemma 42.40.2 (cite)

Lemma 42.40.2. Let $(S, \delta )$ be as in Situation 42.7.1. Let $X$ be locally of finite type over $S$. Let $\mathcal{E}$, $\mathcal{F}$ be finite locally free sheaves on $X$ of ranks $r$, $r - 1$ which fit into a short exact sequence

\[ 0 \to \mathcal{L} \to \mathcal{E} \to \mathcal{F} \to 0 \]

where $\mathcal{L}$ is an invertible sheaf. Then

\[ c(\mathcal{E}) = c(\mathcal{L}) c(\mathcal{F}) \]

in $A^*(X)$.

Proof. This relation really just says that $c_ i(\mathcal{E}) = c_ i(\mathcal{F}) + c_1(\mathcal{L})c_{i - 1}(\mathcal{F})$. By Lemma 42.40.1 we have $c_ j(\mathcal{E} \otimes \mathcal{L}^{\otimes -1}) = c_ j(\mathcal{F} \otimes \mathcal{L}^{\otimes -1})$ for $j = 0, \ldots , r$ were we set $c_ r(\mathcal{F} \otimes \mathcal{L}^{-1}) = 0$ by convention. Applying Lemma 42.39.1 we deduce

\[ \sum _{j = 0}^ i \binom {r - i + j}{j} (-1)^ j c_{i - j}({\mathcal E}) c_1({\mathcal L})^ j = \sum _{j = 0}^ i \binom {r - 1 - i + j}{j} (-1)^ j c_{i - j}({\mathcal F}) c_1({\mathcal L})^ j \]

Setting $c_ i(\mathcal{E}) = c_ i(\mathcal{F}) + c_1(\mathcal{L})c_{i - 1}(\mathcal{F})$ gives a “solution” of this equation. The lemma follows if we show that this is the only possible solution. We omit the verification. $\square$

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