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A Noetherian ring that is an integrally closed domain is a unique factorization domain exactly when its divisor class group vanishes. Indeed, vanishing says every prime Weil divisor, equivalently every height-one prime ideal, is principal.

If $C$ is an irreducible plane curve of degree $d$, the localization sequence for the divisor class group quotients $\mathrm{Cl}({\mathbb{P}}^2)=\mathbb{Z}[H]$ by the class $[C]=d[H]$. Nonsingularity of the removed curve is not required; irreducibility makes it a single removed prime Weil divisor.

Use the dense standard affine charts. Their product iswhich is also a dense affine chart of ${\mathbb{P}}^{n+m}$. The identification of these charts gives a birational isomorphism between the two projective varieties.

To distinguish them, calculate their divisor class groups explicitly. Let $H\subset {\mathbb{P}}^r$ be a coordinate hyperplane. Its complement is ${\mathbb{A}}^r$, whose coordinate ring is a unique factorization domain. Any Weil divisor on ${\mathbb{P}}^r$ restricts on this chart to a principal Weil divisor; subtract that principal Weil divisor on ${\mathbb{P}}^r$, leaving a Weil divisor supported on $H$. Thus $[H]$ generates $\mathrm{Cl}({\mathbb{P}}^r)$. If $aH=\mathrm{div}(q)$, then $q$ has zero Weil divisor on ${\mathbb{A}}^r$. A reduced fraction in a unique factorization domain with zero orders at every prime Weil divisor must be a unit, so $q\in k^{\ast }$. It follows that $a=0$. We have proved

In the product, let $H_1$ and $H_2$ be coordinate hyperplanes in the factors and putThe complement of $D_1\cup D_2$ is ${\mathbb{A}}^{n+m}$, again with a unique factorization domain as its coordinate ring. The same restriction-and-subtraction argument says that $[D_1],[D_2]$ generate the divisor class group. Ifthen on that affine chart $q$ has neither zeros nor poles along any prime Weil divisor. Unique factorization makes $q$ a unit of $k[s_1,\dots ,s_n,t_1,\dots ,t_m]$, hence a constant. It follows that $a=b=0$. ThereforeBoth spaces are smooth varieties, so their divisor class groups also equal their Picard groups. An isomorphism preserves the divisor class group; ${\mathbb{Z}}^2$ and $\mathbb{Z}$ are not isomorphic. Thus the spaces are birational but not isomorphic. The assumptions $n,m>0$ ensure that both boundary prime Weil divisors actually occur.

Write $a=t_1$, $b=t_2$, $c=t_3$, and $A=\mathbb{C}[a,b,c]/(ac-b^2)$. It is the quadratic cone invariant ringwhere the involution negates both $u$ and $v$. The invariant monomials have even total degree and are generated by $u^2,uv,v^2$; reduction by $ac=b^2$ proves the asserted presentation.

To verify the normal variety property, suppose $q\in \mathrm{Frac}(A)$ is integral over $A$. The same monic equation makes $q$ integral over $\mathbb{C}[u,v]$. This polynomial ring is an integrally closed domain, so $q\in \mathbb{C}[u,v]$. Since $q$ belongs to the fraction field of $A$, it is invariant under the involution, and hence belongs to $A$. Thus $X$ is normal. The gradient of its equation is $(c,-2b,a)$, so the Jacobian criterion shows that the origin $o$ is its only singular point of an algebraic variety. In particular $X$ is not smooth.

Take the prime Weil divisor $D=V_X(a,b)$ and $x=o$. We show that $D$ is not a Cartier divisor at $o$. In the local ringits prime ideal is $I=(a,b)R$. The quotient $I/\mathfrak{m}I$ has dimension of a vector space $2$ over $\mathbb{C}$: $a,b$ generate it, and they are independent since $\mathfrak{m}I\subset {\mathfrak{m}}^2$ while the relation $ac-b^2$ has no linear term. By Nakayama lemma, $I$ cannot be generated by one element.

On a normal variety, the ideal of an effective prime Weil divisor is the divisorial ideal of functions with order at least one along that divisor and order at least zero along every other prime Weil divisor. If $D$ were a Cartier divisor at $o$, a local defining rational function $h$ would identify this ideal with $hR$, because an integrally closed domain is the intersection of its height-one localizations at a prime ideal inside its fraction field. That would make $I$ principal, contradicting the calculation. Therefore $D$ has a nonzero class in the local divisor class group.

Now let $D^{\prime }=D+\mathrm{div}(q)$ be any linearly equivalent Weil divisor. If $o$ were outside its support of a Weil divisor, there would be an open subset containing $o$ on which $D^{\prime }$ is zero. On that neighbourhood $D=-\mathrm{div}(q)$ would be a principal Weil divisor, hence a Cartier divisor, which is impossible. Therefore this point is unavoidable in every representative:The argument does not assume that $D^{\prime }$ is effective. For comparison, $\mathrm{div}(a)=2D$ because on the chart $c\ne 0$, $a=b^2/c$; the obstruction has order two, as in the Divisor class group of an A-type surface singularity.

Let $U={\mathbb{P}}^2\setminus C$. The localization sequence for the divisor class group givesEvery prime Weil divisor of $U$ extends by closure to one of ${\mathbb{P}}^2$, and the only removed prime Weil divisor is $C$. Rational functions have the same function field on the two spaces, so the kernel on divisor class groups consists exactly of multiples of $[C]$.

The divisor class group of ${\mathbb{P}}^2$ is $\mathbb{Z}$, generated by the class $[H]$ of a line. To see the degree identification, if a plane curve $D$ has degree $e$ and homogeneous equation $F_D$, then $F_D/{\ell }^e$, for a line equation $\ell$, is a rational function with principal Weil divisor $D-eH$. Degrees of principal Weil divisors are zero, so $[H]$ has infinite order. In particular, $[C]=d[H]$. The localization sequence therefore yieldsThis is the divisor class group of a plane-curve complement. It includes $d=1$, when the group is zero, and does not require the removed curve to be nonsingular.

Put $R=k[x,y,z]/(xy-z^3)$ and let $X=\mathrm{Spec}R$. The ideal $P=(x,z)$ is a height-one prime ideal, because $R/P\cong k[y]$ and $\dim R=2$. It therefore defines a prime Weil divisor $D=V(x,z)$.

Localizing at $x$ eliminates $y$, givinga unique factorization domain. The Nagata theorem for divisor class groups says that $\mathrm{Cl}(X)$ is generated by the height-one primes containing $x$. Sincethe only such prime is $P$, and hence $[D]$ generates. At the generic point of $D$, $y$ is a unit in a ring and $x=z^3/y$, so the order of vanishing is three:Thus $3[D]=0$.

This relation has exact order three. Indeed, if a principal divisor were supported on $D$, its defining rational function would be a unit on $D(x)$. The units of $R_x=k[z][x,x^{-1}]$ are precisely $c{x}^m$ with $c\in k^{\times }$ and $m\in \mathbb{Z}$, whose divisors are $3mD$. Consequentlygenerated by $[V(x,z)]$. This is the $n=3$ case of the Divisor class group of an A-type surface singularity.

PutSinceis an integral domain, $(x,z)$ is a prime ideal. Moreover $\dim R=3$ and $\dim R/(x,z)=2$, so its height of an ideal is one. Therefore $D=V(x,z)$ is a prime Weil divisor.

Localizing at $x$ eliminates $y$ and givesa unique factorization domain. The Nagata theorem for divisor class groups says that $\mathrm{Cl}(R)$ is generated by the height-one primes containing $x$. Sincethese are $D=V(x,z)$ and $E=V(x,w)$. Both occur with multiplicity one in the principal divisorThe units of $R_x\cong k[z,w][x,x^{-1}]$ are exactly $c{x}^n$ with $c\in k^{\ast }$ and $n\in \mathbb{Z}$. Consequently the only relation supplied by localization is $[D]+[E]=0$, andThis is the divisor class group of the three-dimensional affine quadric cone.

The scheme ${\mathbb{P}}_k^3$ is Noetherian, integral, separated, and regular. Every open subscheme inherits these properties, so $U={\mathbb{P}}_k^3\setminus C$ satisfies $(\star )$. Because $C$ has dimension one, it has codimension two in ${\mathbb{P}}_k^3$ and contains no prime Weil divisor. The localization sequence for the divisor class group therefore makes restriction an isomorphismThe hyperplane divisor generates the class group of projective space, so

A Cartier divisor on an integral scheme is an open cover $(U_i)$ together with nonzero rational functions $f_i$ such that every ratio $f_i/f_j$ is a regular unit on $U_i\cap U_j$.

Every prime Weil divisor on ${\mathbb{P}}_k^n$ is cut out by an irreducible homogeneous polynomial because the polynomial ring $k[x_0,\dots ,x_n]$ is a unique factorization domain. Hence any Weil divisor can be represented by a homogeneous rational expressionof some total degree $d$. On the standard chart $U_i=D_{+}(x_i)$, put $f_i=F/x_i^d$. This is a degree-zero rational function, and on $U_i\cap U_j$,is a regular unit. These local equations form a Cartier divisor whose associated Weil divisor is the original one.

Now put $P=X\times {\mathbb{P}}^n$ and let $H=X\times {\mathbb{P}}^{n-1}$ be the hyperplane divisor at infinity. Its complement is $X\times {\mathbb{A}}^n$. Iterating the given invariance under multiplication by ${\mathbb{A}}^1$ givesThe localization sequence for the divisor class group shows that every class on $P$ is a pullback of a class on $X$ plus an integer multiple of $[H]$. Restriction to the generic fiber ${\mathbb{P}}_{K(X)}^n$ kills pullbacks from $X$ and sends $[H]$ to the generator of $\mathrm{Cl}({\mathbb{P}}_{K(X)}^n)\cong \mathbb{Z}$. Therefore the sum is direct, provingThis is the divisor class group of a projective-space bundle with trivial vector bundle.

A prime Weil divisor is an integral closed subscheme of codimension one. If $Y$ has generic point ${\eta }_Y$, regularity in codimension one makes the local ring ${\mathcal{O}}_{X,{\eta }_Y}$ a discrete valuation ring with fraction field $K(X)$. Its normalized discrete valuationis the order of vanishing along $Y$: writing $f=u{\pi }^m$ for a unit $u$ and uniformizer $\pi$ gives ${\nu }_Y(f)=m$. Additivity of exponents makes this a group homomorphism.