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 $U_i = \operatorname A_i$ 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

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

Choose some open cover $\$ of ''X'' such that $L, _ \simeq \mathcal_X, _$. For each ''i'', through the isomorphisms, the restriction $s, _$ corresponds to a nonzerodivisor $f_i$ of $\mathcal_X(U_i)$. Now, define the closed subscheme $\$ of ''X'' (called the zero-locus of the section ''s'') by
:$\ \cap U_i = \,$
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 $f_i/f_j, _$ is a unit element. For the same reason, the closed subscheme $\$ 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 $\$ may be constructed as the fiber product of ''s'' and the zero-section embedding $s_0: X \to L$.

Finally, when $\$ 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 $I(D)^ \otimes_ -$ of $0 \to I(D) \to \mathcal_X \to \mathcal_D \to 0$ gives the exact sequence
:$0 \to \mathcal_X \to I(D)^ \to I(D)^ \otimes \mathcal_D \to 0$
In particular, 1 in $\Gamma(X, \mathcal_X)$ can be identified with a section in $\Gamma(X, I(D)^)$, which we denote by $s_D$.

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

Properties

*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

Hyperplane bundle

Effective Cartier divisors on a relative curve

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 $\Gamma(D, \mathcal_D)$ 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 $\operatorname R$. 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 $\deg(f^* D) = \deg(f) \deg(D)$. On the other hand, a base change does not change degree: $\deg(D \times_R R') = \deg(D)$.

A closed subscheme ''D'' of ''X'' is finite, flat and of finite presentation if and only if it is an effective Cartier divisor that is proper over ''R''.

Weil divisors associated to effective Cartier divisors

Given an effective Cartier divisor ''D'', there are two equivalent ways to associate Weil divisor /math> to it.

Notes

References

*{{cite book
, last1 = Katz
, first1 = Nicholas M
, authorlink1 = Nick Katz
, last2=Mazur
, first2=Barry
, authorlink2=Barry Mazur
, title = Arithmetic Moduli of Elliptic Curves
, publisher = Princeton University Press
, year = 1985
, location =
, pages =
, url =
, doi =
, id =
, isbn =0-691-08352-5

Algebraic geometry