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For positive $n,m$, $\mathrm{Cl}({\mathbb{P}}^n\times {\mathbb{P}}^m)={\mathbb{Z}}^2$, generated freely by the pullbacks of coordinate hyperplanes. Restrict a divisor to the standard affine chart, where unique factorization makes it principal, and subtract this principal Weil divisor to leave only the two boundary hyperplanes. Any principal relation between them would be given by a rational function whose divisor vanishes on the chart; unique factorization then makes that function a constant, proving independence. The same argument gives $\mathrm{Cl}({\mathbb{P}}^r)=\mathbb{Z}$.

Two Weil divisors are linearly equivalent when their difference is a principal Weil divisor. This is the equivalence relation defining the divisor class group, and applies to normal varieties of any dimension, not just curves.

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.

If a Weil divisor $D$ on a normal variety is not Cartier at $x$, every linearly equivalent divisor has $x$ in its support of a Weil divisor. Otherwise that representative would vanish near $x$, expressing $D$ as a principal Weil divisor there. On $V(ac-b^2)$ the prime divisor $V(a,b)$ is not Cartier at the origin: its ideal has two independent generators modulo the maximal ideal times that ideal. This gives an unavoidable support point even for representatives with negative coefficients.