1
$\begingroup$

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 $A^1\times A^1\cong A^2\subset Im(Segre))$

$\endgroup$
  • 2
    $\begingroup$ 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. $\endgroup$ – Tabes Bridges Aug 26 '17 at 2:36
2
$\begingroup$

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".

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.