Lemma 42.43.4. In Situation 42.7.1 let $X$ be locally of finite type over $S$. Let $\mathcal{E}$ and $\mathcal{F}$ be a finite locally free $\mathcal{O}_ X$-modules of ranks $r$ and $s$. Then we have

$c_1(\mathcal{E} \otimes \mathcal{F}) = r c_1(\mathcal{F}) + s c_1(\mathcal{E})$
$c_2(\mathcal{E} \otimes \mathcal{F}) = r c_2(\mathcal{F}) + s c_2(\mathcal{E}) + {r \choose 2} c_1(\mathcal{F})^2 + (rs - 1) c_1(\mathcal{F})c_1(\mathcal{E}) + {s \choose 2} c_1(\mathcal{E})^2$

and so on in $A^*(X)$.

Proof. Arguing exactly as in the proof of Lemma 42.43.3 we may assume we have invertible $\mathcal{O}_ X$-modules ${\mathcal L}_ i$, $i = 1, \ldots , r$ ${\mathcal N}_ i$, $i = 1, \ldots , s$ filtrations

$0 = \mathcal{E}_0 \subset \mathcal{E}_1 \subset \mathcal{E}_2 \subset \ldots \subset \mathcal{E}_ r = \mathcal{E} \quad \text{and}\quad 0 = \mathcal{F}_0 \subset \mathcal{F}_1 \subset \mathcal{F}_2 \subset \ldots \subset \mathcal{F}_ s = \mathcal{F}$

such that $\mathcal{E}_ i/\mathcal{E}_{i - 1} \cong \mathcal{L}_ i$ and such that $\mathcal{F}_ j/\mathcal{F}_{j - 1} \cong \mathcal{N}_ j$. Ordering pairs $(i, j)$ lexicographically we obtain a filtration

$0 \subset \ldots \subset \mathcal{E}_ i \otimes \mathcal{F}_ j + \mathcal{E}_{i - 1} \otimes \mathcal{F} \subset \ldots \subset \mathcal{E} \otimes \mathcal{F}$

with successive quotients

$\mathcal{L}_1 \otimes \mathcal{N}_1, \mathcal{L}_1 \otimes \mathcal{N}_2, \ldots , \mathcal{L}_1 \otimes \mathcal{N}_ s, \mathcal{L}_2 \otimes \mathcal{N}_1, \ldots , \mathcal{L}_ r \otimes \mathcal{N}_ s$

By Lemma 42.40.4 we have

$c(\mathcal{E}) = \prod (1 + x_ i), \quad c(\mathcal{F}) = \prod (1 + y_ j), \quad \text{and}\quad c(\mathcal{F}) = \prod (1 + x_ i + y_ j),$

in $A^*(X)$. The result follows from a formal computation which we omit. $\square$

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