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We denote $\mathbb{P}^n$ to be the standard projective space over $\mathbb{C}$. Define $$\Gamma = \{ ([p_0 : ... : p_n] , [q_0 : ... : q_n]) \in \mathbb{P}^n \times \mathbb{P}^n | \sum_{i = 0}^n p_i q_i = 0\}.$$ The problem is to prove that $(\mathbb{P}^n \times \mathbb{P}^n) \setminus \Gamma$ is isomorphic to an affine variety.

Here is my understanding of the problem:

Let $\Sigma_{n,n} : \mathbb{P}^n \times \mathbb{P}^n \to \mathbb{P}^{(n+1)^2 - 1}$ be a the Segre map. Then with $\mathbb{P}^n \times \mathbb{P}^n$ in the problem, they mean $\Sigma_{n,n}(\mathbb{P}^n \times \mathbb{P}^n)$ as a subset of $\mathbb{P}^{(n+1)^2 - 1}$. I have seen a theorem stating that this is Zariski closed in $\mathbb{P}^{(n+1)^2 - 1}$.

We will denote the variables of polynomials as $Z_{i,j}$ for $0 \leq i,j \geq n$. For example if $n = 1$, we will work with polynomials in $\mathbb{C}[Z_{0,0} , Z_{0,1} , Z_{1,0} , Z_{1,1}]$. Then an observation I was able to make was that $$\Gamma = \mathbb{V}(\sum_{i = 0}^n Z_{i,i}) \cap \Sigma_{n,n}(\mathbb{P}^n \times \mathbb{P}^n),$$ so $\Gamma$ is closed in the Zariski topology.

Then we need to prove that $\Sigma_{n,n}(\mathbb{P}^n \times \mathbb{P}^n) \setminus \Gamma$ is isomorphic to an affine variety in $\mathbb{P}^{(n+1)^2 - 1}$. Note that $$\Sigma_{n,n}(\mathbb{P}^n \times \mathbb{P}^n) \setminus \Gamma = \Sigma_{n,n}(\mathbb{P}^n \times \mathbb{P}^n) \setminus \mathbb{V}(\sum_{i = 0}^n Z_{i,i}),$$ so it's definetly a quasi-projective variety.

I'm not entirely sure what is meant by an affine variety, but I'm guessing that they mean that we have to find a Zariski closed subset $V \subset \mathbb{P}^{(n+1)^2 - 1}$ such that $V \cap U_0$ is isomorphic to $\Sigma_{n,n}(\mathbb{P}^n \times \mathbb{P}^n) \setminus \Gamma$, where $$U_0 = \{[z_{0,0} : z_{0,1} : ... : z_{n,n}] \in \mathbb{P}^{(n+1)^2 - 1} | z_{0,0} \neq 0 \}.$$

Now since $\mathbb{V}(\sum_{i = 0}^n Z_{i,i})$ is a linear variety, we can take an automorphism $T : \mathbb{P}^{(n+1)^2 - 1} \to \mathbb{P}^{(n+1)^2 - 1}$ such that $T(\mathbb{V}(\sum_{i = 0}^n Z_{i,i})) = \mathbb{V}(Z_{0,0})$. Thus we can just take $V = T(\Sigma_{n,n}(\mathbb{P}^n \times \mathbb{P}^n))$ and I guess this solves the problem.

  1. Am I correct? Did I interpret this well?
  2. If I am, is there nothing missing in my solution?
  3. If I'm not, then how should I tackle this problem? Any Hints?
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    $\begingroup$ A projective variety (embedded in a specific projective space) minus a hyperplane section is an affine variety. $\endgroup$ Dec 13, 2023 at 23:16
  • $\begingroup$ So my proof is correct? $\endgroup$
    – Steve
    Dec 14, 2023 at 8:33
  • $\begingroup$ same question was asked here: math.stackexchange.com/questions/4825600/… $\endgroup$
    – Kenta S
    Dec 15, 2023 at 9:19

1 Answer 1

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Saying that $(\mathbb P^n\times\mathbb P^n)\backslash\Gamma$ means it is isomorphic to a Zariski closed subset of $\mathbb A^N$ for some $N\ge1$ as varieties (or schemes, if you like).

For example when $n=1$, there is an isomorphism $$ \begin{align*} \mathbb V(xz=y(y-1))&\xrightarrow\sim(\mathbb P^1\times\mathbb P^1)\backslash\Gamma\\ (x,y,z)&\mapsto \begin{cases} ([x:y],[z:-y])&y\ne0\\ ([y-1:z],[1-y:x])&y\ne1. \end{cases} \end{align*}$$

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