I am a beginner at algebraic geometry. I am studying the Segre embedding $$\phi:\mathbb{P}^1\times \mathbb{P}^1\longrightarrow \mathbb{P}^3$$ sending $((x_0:x_1),(y_0,y_1))$ to $(x_0y_0:x_0y_1:x_1y_0:x_1y_1)$.
I have two questions:
(1) I would like to prove that $\phi$ is a regular map (is it?) and I have found this definition of what it means to be regular.
My doubt is whether this is the appropriate concept of regular map to work with, because in my case $X$ is a product $\mathbb{P}^1\times \mathbb{P}^1$, which I have not yet proved is a projective variety (I think, maybe wrongly?, that to prove that I first have to show that the map is regular and that the product is isomorphic to its image by this map, which is closed). Any clarification as to what is the adequate notion of regular map to work with here will be very useful.
(2) I have read in several books that the equation for this variety projective variety (the image of the map) is $z_{00}z_{11}-z_{01}z_{10}=0$ with $z_{ij}=x_iy_j$. If I plug the image of $\phi$ in this equation I can see that it verifies, and I also have read how to obtain a preimage of an element satisfying such equation, so I can see that it is correct. My question is how would one think of proposing such equation, is there any intuition or standard process to obtain the shape of the equations of a projective variety?
Thanks in advance
I think that the real proof coming for the Following Theorem: A variety $X$ is a projective variety if and only if there is an ample divisor on $X$. Note that if $X$ is projective, there is an immersion to a projective space $\mathbb{P}^N$, then the restriction of $\mathcal{O}_{\mathbb{P}^N}(1)$ to $X$ produces an ample divisor on $X$, and conversely, if $X$ posses an ample divisor, say $D$, some multiple will be very ample $mD$. The map $\phi_{mD}: X\to\mathbb{P}(|mD|)$ is an immersion. Following that idea, the Segre map is just the map defined by divisor $ \mathcal{O}_{\mathbb{P}^1}(1)\otimes \mathcal{O}_{\mathbb{P}^1}(1)$. Such map is given as follows: $H^0(\mathbb{P}^1\times \mathbb{P}^1,\mathcal{O}_{\mathbb{P}^1}(1)\otimes \mathcal{O}_{\mathbb{P}^1}(1))$ is generated by the tensor product of the basis of each $H^0(\mathbb{P}^1, \mathcal{O}_{\mathbb{P}^1}(1))$, If you choose variables $x,y$ for the first $\mathbb{P}^1$ and $z,w$ for the second, then the basis is $\{xz,xw,yz,yw\}$ and the map $\phi([x:y],[z:w])=[xz:xw:yz:yw]$.