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If $\mathscr{E}$ is a vector bundle of rank $k$ and $\mathcal{L}$ is a line bundle, Then: $$c_{k}(\mathscr{E} \otimes \mathcal{L}) = \displaystyle \sum_{i = 0}^{k}\binom{r-l}{k-l}c_{1}(\mathcal{l})^{k-l}c_{l}(\mathscr{E})$$ Is there a similar result for the tensor product of coherent sheaves?

For example, if $r_{1} = \mbox{rank}(E)$ and $r_{2} = \mbox{rank}(F)$, then: $$c_{1}(E \otimes F) = r_{2}c_{1}(E) + r_{1}c_{1}(F)$$ $$c_{2}(E \otimes F)?$$ $$\cdots$$ $$c_{k}(E \otimes F)?$$ where $E$ and $F$ are coherent sheaves (we can assume them in a smooth projective variety of dimension $n$).

Thank you very much.

Allan Ramos
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  • What is your definition of the rank of a coherent sheaf? – Ruben Jul 01 '19 at 22:49
  • Good quaestion, @Ruben du Burck. I'm a beginner in Algebraic Geometry. If an $\mathcal{O}{X}$-module $\mathscr{F}$ is ismorphic to $\mathcal{O}{X}^{n}$ as $\mathcal{O}_{X}$-modules, $\mathscr{F}$ is said to be a free module of rank $n$. – Allan Ramos Jul 01 '19 at 23:46
  • That is a definition on locally free sheaves. Locally free sheaves are coherent, but not all coherent sheaves are locally free. – Ruben Jul 08 '19 at 13:31
  • @Ruben. What is the correct definition of the rank of a coherent sheaf? Thank you. – Allan Ramos Jul 08 '19 at 15:53
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    I'm not sure if there is a "correct" definition. In certain cases a coherent sheaf $\mathcal F$ will admit a finite locally free resolution $\mathcal E_\bullet \to \mathcal F$. In that case you can define $\text{rank}; \mathcal F = \sum_i (-1)^i \text{rank}; \mathcal E_i$. Of course this requires proving that this is independent of the choice of resolution. – Ruben Jul 08 '19 at 15:57
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    @Ruben. Thank you very much. – Allan Ramos Jul 08 '19 at 16:01
  • You can find a formula for the first three Chern classes of a tensor product of locally free sheaves in this answer. – Michael Albanese Feb 18 '21 at 17:01
  • Ok. Thank you very much, Michael Albanese. – Allan Ramos Feb 19 '21 at 22:27

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