Linear equivalence cartier divisors

Two Cartier divisors $D,D^{\mathrm{\prime }}$ on a scheme $X$ are linearly equivalent, written $D\sim D^{\mathrm{\prime }}$, if there is a global meromorphic unit $f\in \mathrm{\Gamma }(X,{\mathcal{K}}_X^{\times })$ such that $D-D^{\mathrm{\prime }}={\mathrm{div}}_C(f)\mathrm{.}$ Here subtraction is in the abelian group $\mathrm{CaDiv}(X)$ (Cartier divisor), and the right side is the principal divisor of Principal cartier divisor.

Equivalently, $D$ and $D^{\mathrm{\prime }}$ have the same class modulo the subgroup of principal Cartier divisors. Explicitly, reflexivity follows by taking $f=1$; if $D-D^{\mathrm{\prime }}={\mathrm{div}}_C(f)$, then $D^{\mathrm{\prime }}-D={\mathrm{div}}_C(f^{-1})$, giving symmetry; and if also $D^{\mathrm{\prime }}-D^{\mathrm{\prime }\mathrm{\prime }}={\mathrm{div}}_C(g)$, then $D-D^{\mathrm{\prime }\mathrm{\prime }}={\mathrm{div}}_C(fg)$, giving transitivity. Adding the same Cartier divisor to both sides preserves the relation because their difference is unchanged.

No assumption of effectiveness is imposed: either divisor may have positive or negative local equations in the sense of the Cartier group. On the empty scheme there is only the zero Cartier divisor and its single equivalence class.