[Math] Affine Cover of $\mathbb{P}^n$

algebraic-geometryprojective-schemesprojective-spaceprojective-varieties

I have a question in understanding Hartshorne's Algebraic Geometry Corollary 2.3, Ch. 1. It says that any projective variety $Y\subset{\mathbb{P}}^n$ is covered by the open sets $Y\cap U_i$ (this part is clear) and that each $Y\cap U_i$ is homeomorphic to an affine variety via $\varphi_i$.

Definitions used: $U_i$ is defined as the set of all points in $\mathbb{P}^n$ with the i-th component non-zero and $\varphi_i$ is the (homeomorphic) map from $U_i$ to the affine space $\mathbb{A}^n$ given by $[a_0,…,a_n] \mapsto (a_1/a_i,…,a_n/a_i)$. A variety is always meant to be an irreducible algebraic subset.

Here are my questions concerning the second part of the corollary:

1) How is the intersection of $Y\cap U_i\cong Y\cap \mathbb{A}^n$ calculated? Does one use the (disjoint) open covering $\mathbb{P}^n=\bigcup_{i=0}^n \mathbb{A}^{n-i}$ and hence $Y\cap \mathbb{A}^n$ is the set of all $\mathbb{A}^n$-components of $Y\subset \mathbb{P}^n$?

2) How can one show that $Y\cap U_i\cong Y\cap \mathbb{A}^n$ is again irreducible? In which topology shall it be closed?

Thank you for your help.

Best Answer

  1. The ideal of the intersection $Y\cap U_i$ is calculated by "dehomogenizing," setting $x_i$ equal to $1$. The open cover you mention is not disjoint.

2.If $Y\cap U_i$ were reducible, then it would be the union of proper closed subsets. Taking closures in $\mathbb P^n$ would make $Y$ reducible. The topology is the Zariski topology on $U_i\cong \mathbb A^n$, which is also the subspace topology.

Related Question