# $(\mathbb{P}^n \times \mathbb{P}^n) \setminus \Gamma$ is isomorphic to an affine variety

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?
• A projective variety (embedded in a specific projective space) minus a hyperplane section is an affine variety. Dec 13, 2023 at 23:16
• So my proof is correct? Dec 14, 2023 at 8:33
• same question was asked here: math.stackexchange.com/questions/4825600/… Dec 15, 2023 at 9:19

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*}