Let $X$ be a scheme. Let $mathcal E$ be a locally free sheaf of rank $r$ on $X$ and let $s$ be a section of $mathcal E$. Then the zero scheme of $s$ is defined as follows: Consider the homomorphism $mathcal O_X to mathcal E$ induced by $s$, taking duals, we obtain $mathcal E^vee to mathcal O_X$. Then $Z(s)$ is defined to be the scheme associated to the sheaf of ideals $mathop{mathrm{im}}(mathcal E^vee to mathcal O_X)$.

First question: is there some natural conditions of regularity defined on $s$. For example, when $mathcal E = mathcal L$ is an invertible $mathcal O_X$-module, then $s$ is said to be **regular** if and only if the induced homomorphism $mathcal O_X to mathcal L$ is injective, and in this case $Z(s)$ is an effective Cartier divisor on $X$. So I think the condition of regularity should satisfy that if $s$ is regular, then any generic point of irreducible components of $Z(s)$ has codimension $r$ in $X$.

Now suppose we have already defined some conditions of regularity. Assume now that $X$ is a smooth projective variety and let $s$ be a “regular” section. Consider the $r$-cycle associated to $Z(s)$. Prove the following statement: The class of the $r$-cycle associated to $Z(s)$ in the $r$-th Chow group $mathop{mathrm{CH}}^r(X)$ equals the $r$-th Chern class $c_r(mathcal E)$ of $mathcal E$.

Hence the linear equivalence class of $Z(s)$ is independent of the choice of $s$ and we get a well-defined map

$${text{locally free sheaves of rank } r text{ on } X} / {text{isomorphisms}} to mathop{mathrm{CH}}nolimits^r(X),$$

is this map bijective? (When $r = 1$, we obtain an isomorphism $mathop{mathrm{Pic}}(X) to mathop{mathrm{Cl}}(X)$.)

If the map above is bijective, then the group structure on $mathop{mathrm{CH}}nolimits^r(X)$ should induces a group structure on ${text{locally free sheaves of rank } r text{ on } X} / {text{isomorphisms}}$, and what is it? (When $r = 1$, it is tensor products of invertible $mathcal O_X$-modules. However, we cannot simply take the tensor product of two locally free sheaves of rank $r$).