Let $\mathbb{P}^1(\mathbb{C})\times \mathbb{P}^1(\mathbb{C})\rightarrow \mathbb{P}^3(\mathbb{C})$, $[a:b],[c:d]\mapsto[ac:ad:bc:bd]=[z_0:z_1:z_2:z_3]$.
How to show that $Z_0=Z(z_0z_3-z_1z_2)$ is a subset of the image of this map?
I take a point $[z_0:z_1:z_2:z_3]\in Z_0$ and I need to find its preimage.
I tried to consider two cases: $z_0=0$ and $z_0\ne 0$. The former yields $5$ subcases:
- $(z_1,z_2,z_3)=(0,\ne 0, 0)$
- $(z_1,z_2,z_3)=(0, \ne 0, \ne 0)$
- $(z_1,z_2,z_3)=(\ne 0, 0, 0)$
- $(z_1,z_2,z_3)=(\ne 0, 0, \ne 0)$
- $(z_1,z_2,z_3)=(0,0,\ne 0)$
And the latter yields $4$ subcases:
- $(z_1,z_2,z_3)=(0,\ne 0, 0)$
- $(z_1,z_2,z_3)=(\ne 0, 0,0$
- $(z_1,z_2,z_3)=(0,0,0)$
- $(z_1,z_2,z_3)=(\ne 0, \ne 0, \ne 0)$
I concentrate on the first $5$ subcases. I managed to find preimages for cases $1$ and $3$ (they are respectively $[z_0:z_2],[1:0]$ and $[1:0],[z_0:z_1]$), but I don't know how to proceed because for example in case $2$ I need to find a preimage of $[z_0:0:z_2:z_3]$ with $z_i\ne 0$, but I think this is impossible.
Suppose that $z_0\not=0$ and satisfies $z_0z_3-z_1z_2=0$. Then, $z_3=\frac{z_1z_2}{z_0}$. Then, we can find $a,b,c,d$ as follows: \begin{align*} a&=1\\ c&=z_0\\ d&=z_1\\ b&=\frac{z_2}{z_0}. \end{align*} Prove that the image of $[a:b]\times[c:d]$ is the same as $[z_0:z_1:z_2:z_3]$. Continue similarly when $z_1\not=0$, $z_2\not=0$, and $z_3\not=0$ and you'll have all possible cases.
For example, if $z_1\not=0$, then $z_2=\frac{z_0z_3}{z_1}$ and we can find $a,b,c,d$ as \begin{align*} a&=1\\ c&=z_0\\ d&=z_1\\ b&=\frac{z_3}{z_1}. \end{align*}
When $z_2\not=0$ or $z_3\not=0$, then start with $b=1$.
There are other ways to get this, but, from your initial elementary approach, this might be the best approach.