 Solution

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.