Scheme-theoreic proof of the projective plane is quasi-compact

Question

The projective space is defined to be $\Bbb P^n(C)=\{\text{line through origin in $C^{n+1}$}\}=C^{n+1}\setminus \{0\}/\sim$ the precise definition coule be found here: https://en.wikipedia.org/wiki/Projective_space See "Definition of projective space".

I aim to prove that this space is quasi-compact. I think this space couele be covered by a finite collection of $U_i$ where $U_i$'s are defined by the subschemes of $\Bbb P^n$ which is the collection with the $i$-th coordinate non-zero. But I am not able to relat it to "Given any collection of open subschemes which covers $\Bbb P^n$, then it has a finite subcover".

Also I think that as $\Bbb P^n$ is not affine, it makes it difficult to talk about covers in terms of ideals. So do I need to find its Zariski closure to deal with the problem?

Any solution for that, please? Or any guidence or reference would be appreciate.

EDIT: I see, I was thinking about the wrong version of definition and misunderstood the question. For this problem, what I should think about is:

$\Bbb P^n_R(C)=\{\text {C-lines in $\Bbb A^{n+1}_C$}\}$

And the definition of $C$-lines should be:

A closed subscheme $L\subseteq \Bbb A^{m+1}_R$ is an $R$-line if it is principle locally on $R$.

But thinking about this version of definition. I could not even find an explicit finite cover of it...

Answer 1

If a topological space $X$ is a finite union of quasi-compact topological spaces, then $X$ is quasi-compact.

Answer 2

Here is a sort of zero-insight, very elementary argument.

Let $U_\lambda$, $\lambda \in \Lambda$, be an open cover of $\mathbb{P}^n$. Let $C_\lambda = \mathbb{P}^n - U_\lambda$, a Zariski closed set, and let $I_\lambda = I(C_\lambda)$, the homogeneous ideal of $C_\lambda$.

We will get a finite subcover by choosing some $\lambda_1,\dotsc,\lambda_m \in \Lambda$. Start with an empty list. Once we have $\lambda_1,\dotsc,\lambda_k$, we choose $\lambda_{k+1}$ as follows. (We start with $k=0$ and choose $\lambda_1$. If this seems weird, just choose $\lambda_1$ arbitrarily to start with, and start at $k=1$ to select $\lambda_2$...) First of all, if $U_{\lambda_1},\dotsc,U_{\lambda_k}$ is a cover of $\mathbb{P}^n$, then we're done. Otherwise it's not a cover, so the union $U_{\lambda_1} \cup \dotsb \cup U_{\lambda_k}$ has a nonempty complement. Pick a point $P$ in that complement. Since $\{U_\lambda, \lambda \in \Lambda\}$ is a cover, there is some $\lambda \in \Lambda$ such that $P \in U_\lambda$. Choose that $\lambda$ to be $\lambda_{k+1}$.

This terminates—at some point the $U_{\lambda_1},\dotsc,U_{\lambda_k}$ are a cover. Indeed, the complement at each step is $$ \mathbb{P}^n - (U_{\lambda_1}\cup\dotsb\cup U_{\lambda_k}) = C_{\lambda_1} \cap \dotsb \cap C_{\lambda_k} = V(I_{\lambda_1}+\dotsb+I_{\lambda_k}). $$ These ideals form an ascending chain: $$ I_{\lambda_1} \subseteq I_{\lambda_1} + I_{\lambda_2} \subseteq I_{\lambda_1} + I_{\lambda_2} + I_{\lambda_3} \subseteq \dotsb . $$ By the Ascending Chain Condition on the polynomial ring $k[\mathbb{P}^n]$ (or the Noetherian condition, or the Hilbert Basis Theorem, or whatever you call it), this chain stabilizes: there exists some $m$ such that $I_{\lambda_1}+\dotsb+I_{\lambda_k} = I_{\lambda_1}+\dotsb+I_{\lambda_m}$ for all $k \geq m$.

The $U_{\lambda_1},\dotsc,U_{\lambda_m}$ must be a cover of $\mathbb{P}^n$. Otherwise we'd pick a point $P$ in the complement and extend the cover to include $P$, then the complement would strictly decrease and the ideal of the complement would strictly increase.