I'am try to work with Chern class of the coherent sheaves, in this sense. If I have a vector bundle $E$ of rank $r$ and $L$ a line bundle we have the Chern class property


$$c\_{r}(E\otimes L) = \sum\_{i = 0}^{r}c\_{i}(E)c\_{1}(L)^{r-i}.$$


I need a similar result for coherent sheaf.


If I have a coherent sheaf $\mathcal{F}$ over a smooth projective variety X we consider a locally free resolution


$$0 \longrightarrow E\_{n} \longrightarrow \cdots \longrightarrow E\_{0} \longrightarrow \mathcal{F} \longrightarrow 0.$$


We can define the total Chern class of $\mathcal{F}$ by


$$c(\mathcal{F}) = \displaystyle\Pi\_{i = 0}^{n} c(E\_{i})^{(-1)^{i}}.$$


**My question:** Is true that $c\_{n}(\mathcal{F}\otimes L) = \sum\_{i = 0}^{n}c\_{i}(\mathcal{F})c\_{1}(L)^{n-i},$ for the coherent sheaf $\mathcal{F}$? If this does not exist, we can have the similar relationship?


Thank you !!



How to calculate the Chern class of the tensor product of a torsion free sheaf with a line bundle
No, this is false. Take the simple example $\mathcal{F}=\mathcal{O}\_Y$, where $Y$ is a hypersurface in $X$. Using the resolution $0\rightarrow L(-Y)\rightarrow L\rightarrow L\otimes \mathcal{O}\_Y\rightarrow 0$, one gets $c\_1(L\otimes \mathcal{O}\_{Y})=c\_1(L)-c\_1(L(-Y))= [Y]$, so it is independent of $L$, while 
your formula gives $c\_1(L\otimes \mathcal{O}\_{Y})=c\_1(L)+ c\_1(\mathcal{O}\_{Y})=c\_1(L)+[Y]$. 


To get a general formula it is much easier to use the Chern character: we have $\operatorname{ch}(\mathcal{F}\otimes L)=\operatorname{ch}(\mathcal{F})\operatorname{ch}(L) $. Indeed the formula is classical if $\mathcal{F}$ is a vector bundle, and it extends readily to coherent sheaves using a finite locally free resolution.


