I cross-post a question that has not been answered on MSE, see here.
Consider the Milnor hypersurface $H_{ij}$, i.e., the smooth hypersufrace in $\mathbb CP^i \times \mathbb CP^j$ for fix pair of integers $j \ge i\ge 0$ defined by
$$H_{ij}= \left\{{z_0: \ldots :z_i)} \times {(\omega_0: \ldots :\omega_j) \in \mathbb CP^i \times \mathbb CP^j \; | \; z_0\omega_0+ \ldots +z_i\omega_i=0}\right\}. $$
Question. How can one prove that $H_{11}$ is homeomorphic to $\mathbb CP^1$?
Since the question was not answered on MSE, I will provide a short answer here.
The expression $H_{ij}$ is a bi-homogeneous polynomial of bi-degree $(1, \, 1)$, hence it defines an effective divisor in the complete linear system $|\mathcal{O}_{\mathbb{P}^i \times \mathbb{P}^j}(1, \, 1)|$. Moreover, a simple computation with derivatives shows that this divisor is smooth.
In particular, when $i=j=1$, we have a smooth divisor $D$ in the complete linear system $|\mathcal{O}_{\mathbb{P}^1 \times \mathbb{P}^1}(1, \, 1)|$. The rational map associated with this linear system is the usual embedding of $\mathbb{P}^1 \times \mathbb{P}^1$ as a smooth quadric in $\mathbb{P}^3$. Thus, the curve $D$ corresponds to a smooth hyperplane section of this quadric, namely, a smooth conic, which is isomorphic (not just homeomorphic) to $\mathbb{P}^1$.
