Relative effective Cartier divisor

In algebraic geometry, a relative effective Cartier divisor is roughly a family of effective Cartier divisors. Precisely, an effective Cartier divisor in a scheme X over a ring R is a closed subscheme D of X that (1) is flat over R and (2) the ideal sheaf $I(D)$ of D is locally free of rank one (i.e., invertible sheaf). Equivalently, a closed subscheme D of X is an effective Cartier divisor if there is an open affine cover $U_{i}=\operatorname {Spec} A_{i}$ of X and nonzerodivisors $f_{i}\in A_{i}$ such that the intersection $D\cap U_{i}$ is given by the equation $f_{i}=0$ (called local equations) and $A/f_{i}A$ is flat over R and such that they are compatible.

An effective Cartier divisor as the zero-locus of a section of a line bundle[edit]

Let L be a line bundle on X and s a section of it such that $s:{\mathcal {O}}_{X}\hookrightarrow L$ (in other words, s is a ${\mathcal {O}}_{X}(U)$ -regular element for any open subset U.)

Choose some open cover $\{U_{i}\}$ of X such that $L|_{U_{i}}\simeq {\mathcal {O}}_{X}|_{U_{i}}$ . For each i, through the isomorphisms, the restriction $s|_{U_{i}}$ corresponds to a nonzerodivisor $f_{i}$ of ${\mathcal {O}}_{X}(U_{i})$ . Now, define the closed subscheme $\{s=0\}$ of X (called the zero-locus of the section s) by

\{s=0\}\cap U_{i}=\{f_{i}=0\},

where the right-hand side means the closed subscheme of $U_{i}$ given by the ideal sheaf generated by $f_{i}$ . This is well-defined (i.e., they agree on the overlaps) since $f_{i}/f_{j}|_{U_{i}\cap U_{j}}$ is a unit element. For the same reason, the closed subscheme $\{s=0\}$ is independent of the choice of local trivializations.

Equivalently, the zero locus of s can be constructed as a fiber of a morphism; namely, viewing L as the total space of it, the section s is a X-morphism of L: a morphism $s:X\to L$ such that s followed by $L\to X$ is the identity. Then $\{s=0\}$ may be constructed as the fiber product of s and the zero-section embedding $s_{0}:X\to L$ .

Finally, when $\{s=0\}$ is flat over the base scheme S, it is an effective Cartier divisor on X over S. Furthermore, this construction exhausts all effective Cartier divisors on X as follows. Let D be an effective Cartier divisor and $I(D)$ denote the ideal sheaf of D. Because of locally-freeness, taking $I(D)^{-1}\otimes _{{\mathcal {O}}_{X}}-$ of $0\to I(D)\to {\mathcal {O}}_{X}\to {\mathcal {O}}_{D}\to 0$ gives the exact sequence

0\to {\mathcal {O}}_{X}\to I(D)^{-1}\to I(D)^{-1}\otimes {\mathcal {O}}_{D}\to 0

In particular, 1 in $\Gamma (X,{\mathcal {O}}_{X})$ can be identified with a section in $\Gamma (X,I(D)^{-1})$ , which we denote by $s_{D}$ .

Now we can repeat the early argument with $L=I(D)^{-1}$ . Since D is an effective Cartier divisor, D is locally of the form $\{f=0\}$ on $U=\operatorname {Spec} (A)$ for some nonzerodivisor f in A. The trivialization $L|_{U}=Af^{-1}{\overset {\sim }{\to }}A$ is given by multiplication by f; in particular, 1 corresponds to f. Hence, the zero-locus of $s_{D}$ is D.

Properties[edit]

If D and D' are effective Cartier divisors, then the sum $D+D'$ is the effective Cartier divisor defined locally as $fg=0$ if f, g give local equations for D and D' .
If D is an effective Cartier divisor and $R\to R'$ is a ring homomorphism, then $D\times _{R}R'$ is an effective Cartier divisor in $X\times _{R}R'$ .
If D is an effective Cartier divisor and $f:X'\to X$ a flat morphism over R, then $D'=D\times _{X}X'$ is an effective Cartier divisor in X' with the ideal sheaf $I(D')=f^{*}(I(D))$ .

Examples[edit]

Hyperplane bundle[edit]

Effective Cartier divisors on a relative curve[edit]

From now on suppose X is a smooth curve (still over R). Let D be an effective Cartier divisor in X and assume it is proper over R (which is immediate if X is proper.) Then $\Gamma (D,{\mathcal {O}}_{D})$ is a locally free R-module of finite rank. This rank is called the degree of D and is denoted by $\deg D$ . It is a locally constant function on $\operatorname {Spec} R$ . If D and D' are proper effective Cartier divisors, then $D+D'$ is proper over R and $\deg(D+D')=\deg(D)+\deg(D')$ . Let $f:X'\to X$ be a finite flat morphism. Then $\deg(f^{*}D)=\deg(f)\deg(D)$ .^[1] On the other hand, a base change does not change degree: $\deg(D\times _{R}R')=\deg(D)$ .^[2]