I cross-post a question that has not been answered on MSE, see [here](https://math.stackexchange.com/questions/4391605/question-about-milnor-hypersurface).


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


$$H\_{ij}= \left\{{z\_0: \ldots :z\_i)} \times {(\omega\_0: \ldots :\omega\_j) \in \mathbb CP^i \times \mathbb CP^j \; | \; z\_0\omega\_0+ \ldots +z\_i\omega\_i=0}\right\}. $$



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> **Question.** How can one prove that $H\_{11}$ is homeomorphic to $\mathbb CP^1$?
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Milnor hypersurface
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 $\mathbb{P}^1$.


