This is related to the $P^1\times P^1$ image through segre embedding birational to $P^2$(Hartshorne I 4.5). I did a quite convoluted way by identifying the $A^1\times A^1\subset P^1\times P^1$ as quasi-projective variety isomorphic to $A^2\subset Im(Segre)$ where $A^2$ is treated as quasi-projective variety sitting inside $Im(Segre)$ and $Segre$ means the segre morphism$:P^1\times P^1\to P^3$.

Consider affine varieties $X\subset A^n\subset P^n,Y\subset A^m \subset P^m$. Through segre embedding, I have $X\times Y$ identified as quasi projective variety of $P^N$ for some $N$. Now $X\times Y$ has affine variety product structure and also has quasi projective variety structure in $P^n\times P^m$ through segre embedding. I have two different product variety structure. I would like to see that affine variety products can be embedded into product of projectives(i.e. Compatibility needs to be checked).

So I have $A^1\times A^1\subset P^1\times P^1$ open through segre embedding. Now I want to say $A^1\times A^1$ has inherent affine variety product structure (i.e. $A^1\times A^1\cong A^2\subset P^2$ where first iso requires product of affine varieties).

Q1: Is this generally true? There is no guarantee that segre embedding identifies affine varieties but projective/quasi-projectives.

Q2: Is the image of affine under segre image affine? The affines might be sitting inside some quasi-projectives and the image of affines might not be affine.

Q3: If image of affine is affine, then is this compatible with projective variety product structure through segre embedding?(i.e. $A^1\times A^1\cong A^2$ as affine product whereas I also have $A^1\times A^1\cong A^2\subset Im(Segre))$

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    $\begingroup$ For Q2, note that Segre is a closed immersion, hence an isomorphism onto its image. Since the affine property is about the isomorphism class of the variety, the image is still affine. $\endgroup$ Commented Aug 26, 2017 at 2:36

1 Answer 1


This is really a comment, but due to it's length, I'm writing it as an answer.

For simplicity, I will consider the case $\mathbb{A}^1 \subset \mathbb{P}^1.$ The Segre map $\Phi: \mathbb{P}^1 \times \mathbb{P}^1 \to \mathbb{P}^3$ is given by $$(x,w)\times (y,z) \mapsto (xy,xz,yw,zw).$$ Now assume that $\mathbb{A}^1 \subset \mathbb{P}^1$ is given by $w=1$ for the first component, and by $z=1$ for the second component. Then we have $$\Phi(\mathbb{A^1} \times \mathbb{A}^1)=\{(xy, x, y, 1) : x,y \in \mathbb{A}^1\}.$$ Now $\Phi(\mathbb{A}^1 \times \mathbb{A}^1) \subset \mathbb{A}^3 \subset \mathbb{P}^3$ (given by the last co-ordinate '1'). One can easily check that this is a closed subvariety of $\mathbb{A}^3$ (can you guess the ideal?), and it is isomorphic to $\mathbb{A}^2$ (we have a very obvious map here).

There is nothing special about $\mathbb{A}^1 \subset \mathbb{P}^1$. You can easily do it for $\mathbb{A}^n \subset \mathbb{P}^n$ and $\mathbb{A}^m \subset \mathbb{P}^m$. Roughly speaking, there are $n+m$ "independent variables" in the image $\Phi(\mathbb{A}^n \times \mathbb{A}^m)$, one component is $1$ (which says it is contained in $\mathbb{A}^{n+m+nm}$), and the rest of the components can be written using those "independent variables".


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