Let $Z:=\mathbb{P}^2_{\mathbb{C}}$, $\pi: Y\to Z$ be a double cover ramified on a smooth conic $E\subset Z$. Why $Y\cong \mathbb{P}^{1}\times\mathbb{P}^{1}$?
First I try to find linear system on $\mathbb{P}^{1}\times \mathbb{P}^{1}$ and write the map. I find $h^{0}(\mathbb{P}^{1}\times \mathbb{P}^{1},\mathcal{O}(2,0))=3$, so $|\mathcal{O}(2,0)|$ define a map to $\mathbb{P}^{2}$, but the map is actually $\mathbb{P}^{1}\times \mathbb{P}^{1}\to \mathbb{P}^{1}\hookrightarrow \mathbb{P}^{2}$. It is not $\pi$.
So I think the only way to get this is to write the expression of $\pi$: $$\mathbb{P}^{1}\times \mathbb{P}^{1} \to \mathbb{P}^{2}: ([x_{0}:y_{0}],[x_{1}:y_{1}])\to [x:y:z]$$ For example, if the equation $E$ is $xy-z^{2}=0$, then we can write what $\pi$ is. But I failed. I do not know how to the such a map. I think it should not be hard. Could you help me?