Lemma 42.45.3. Let $(S, \delta )$ be as in Situation 42.7.1. Let $X$ be locally of finite type over $S$. Let $\mathcal{E}_1$ and $\mathcal{E}_2$ be finite locally free $\mathcal{O}_ X$-modules. Then we have the equality

\[ ch(\mathcal{E}_1 \otimes _{\mathcal{O}_ X} \mathcal{E}_2) = ch(\mathcal{E}_1) ch(\mathcal{E}_2) \]

More precisely, we have

\[ P_ p(\mathcal{E}_1 \otimes _{\mathcal{O}_ X} \mathcal{E}_2) = \sum \nolimits _{p_1 + p_2 = p} {p \choose p_1} P_{p_1}(\mathcal{E}_1) P_{p_2}(\mathcal{E}_2) \]

in $A^ p(X)$ where $P_ p$ is as in Example 42.43.6.

**Proof.**
It suffices to prove the more precise statement. By Section 42.43 this follows because if $x_{1, i}$, $i = 1, \ldots , r_1$ and $x_{2, i}$, $i = 1, \ldots , r_2$ are the Chern roots of $\mathcal{E}_1$ and $\mathcal{E}_2$, then $x_{1, i} + x_{2, j}$, $1 \leq i \leq r_1$, $1 \leq j \leq r_2$ are the Chern roots of $\mathcal{E}_1 \otimes \mathcal{E}_2$. Hence we get the result from the binomial formula for $(x_{1, i} + x_{2, j})^ p$ and the shape of our polynomials $P_ p$ in Example 42.43.6.
$\square$

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