All varieties will be smooth when necessary.
Earlier i learned that the first Chern class of a line bundle on an algebraic variety does not determine the bundle up to algebraic isomorphism, i.e. the map $$ \alpha: \quad\operatorname{Pic}(X) \longrightarrow A^1(X) $$ sending a line bundle to its first Chern class can have a nontrivial kernel. (isn't this kernel nonzero whenever the albanese variety is nontrivial?)
(see the comments on Are vector bundles on $\mathbb{P}_{\mathbb{C}}^n$ of any rank completely classified? (main interest $n=3$))
However, i quote from Gathmann's notes "The top Chern class $c_r(F)$ has the additional geometric interpretation as the zero locus of a section of $F$". Here $F$ is a bundle of rank $r$. In the case of line bundles this is the first Chern class.
I found the same statement in Zach Teitlers notes: http://works.bepress.com/cgi/viewcontent.cgi?article=1001&context=zach_teitler, see fact 10.
The problem i am having is that these statements seem contradictory: it is known that a line bundle on a smooth variety is completely determined up to isomorphism by the divisor of zeroes of a section. But according to Gathmann, that is exactly its first Chern class, so Gathmann is basically saying that $c_1(\mathcal{L})$ does determine $\mathcal{L}$ up to isomorphism.
Question 1: Can someone clear this contradiction up? Do Gathmann and Teitler both make the assumption that $\alpha$ is injective? Or am i missing something? (i am new to this subject!)
Question 2 (probably related): Teitler explains the method of degeneracy loci to compute the Chern classes of a bundle. I have the feeling that this method needs special assumptions on the variety in order to work, that Teitler does not explain. Is this true? He does assumes projectivity, smoothness and irreducibility, are these necessary?
Thanks!