Localization sequence for the divisor class group

Let $X$ be a Noetherian integral scheme that is regular in codimension one, and let $U=X\setminus {\bigcup }_{j=1}^r{Z}_j$, where the $Z_j$ are the codimension-one irreducible components of the complement. Restriction gives an exact sequence

If $A$ is a Noetherian normal domain and $S$ is a multiplicative set, then $\mathrm{Cl}(S^{-1}A)$ is the quotient of $\mathrm{Cl}(A)$ by the classes of height-one primes meeting $S$. Tracking the units of $S^{-1}A$ determines the relations among those prime divisors.

Let $A=k[x,y,z,w]/(xy-zw)$ over a field $k$. Its divisor class group is infinite cyclic. Inverting $x$ givesa unique factorization domain. The Nagata theorem for divisor class groups therefore generates $\mathrm{Cl}(A)$ by the two prime Weil divisors $D_1=V(x,z)$ and $D_2=V(x,w)$ above $x=0$.

At either generic point, the defining equation makes $x$ a uniformizer times a unit, so $\mathrm{div}(x)=D_1+D_2$. To prove that this is the only relation, a rational function with divisor supported on $D_1\cup D_2$ restricts to a unit of $A_x$. Every such unit is $c{x}^r$ with $c\in k^{\ast }$ and $r\in \mathbb{Z}$. Its divisor is $r(D_1+D_2)$, givingThus either ruling plane generates, with the other representing its negative. Computing the localized unit group rules out an additional torsion relation.

For a Noetherian integral scheme $X$ regular in codimension one and an integer $n\ge 1$,The first summand is the pullback of base Weil divisors; the second is the hyperplane divisor in the projective fibre.

Remove $H$ to obtain $X\times {\mathbb{A}}^n$. The localization sequence for the divisor class group, together with invariance under adjoining affine-space coordinates, shows that the base classes and $H$ generate. Restriction to the generic fibre ${\mathbb{P}}_{K(X)}^n$ sends a base class to zero and $H$ to its degree-one generator, so no nonzero multiple of $H$ can be a base class. This detects the extra summand geometrically by fibre degree.

The positive-dimension hypothesis is essential: ${\mathbb{P}}^0$ is a point and $\mathrm{Cl}(X\times {\mathbb{P}}^0)=\mathrm{Cl}(X)$, with no hyperplane summand.