# Affine charts are dense in projective space

Given a field $$k$$, we define the scheme-theoretic $$n$$-th affine space over $$k$$ by $$\mathbb{A}^n_k=\text{Spec}(k[X_1,\dots,X_n])$$ and the $$n$$-th projective space over $$k$$ by $$\mathbb{P}^n_k=\text{Proj}(k[X_0,\dots, X_n])$$. We know $$\mathbb{P}^n_k$$ is covered by $$n+1$$ affine charts given by $$D_+(X_i)=\mathbb{P}^n_k\smallsetminus V_+(X_i)$$ for $$i=0,\dots, n$$, each isomorphic to $$\mathbb{A}^n_k$$.

I was asking myself if - as it happens in the naive case - each of those charts is dense in $$\mathbb{P}^n_k$$.

Here is my attempt. Take e.g. $$D_+(X_0)$$. Then $$D_+(X_0)$$ being dense in $$\mathbb{P}^n_k$$ is equivalent to $$V_+(X_0)$$ having empty interior. Suppose there exists $$f\in (X_0,\dots, X_n)$$ s.t. $$D_+(f)\subset V_+(X_0)$$. Then every prime $$\mathfrak{p}\in \mathbb{P}^n_k$$ s.t. $$f\notin \mathfrak{p}$$ is s.t. $$X_0\in \mathfrak{p}$$, which means $$D_+(fX_0)=\emptyset$$, i.e. $$V_+(fX_0)=V_+(0)=\mathbb{P}^n_k$$, and thus $$f=0$$ since $$k[X_0,\dots, X_n]$$ is an integral domain.

Is this proof correct? Does anyone know a shorter way to prove it?

• You can also prove it by observing the fact that $D_+(X_0)$ is dense in every chart, i.e. $D_+(X_0)\cap D_+(X_i)$ is dense in $D_+(X_i)$. – Levent Jan 11 '19 at 14:57

Your argument is correct (you might just want to emphasize that the $$f$$ you introduce is homogeneous). The point is that the equality $$V_+(fX_0)=\mathbf{P}_k^n$$ is equivalent to $$(fX_0)\cap(X_0,\ldots,X_n)\subseteq\sqrt{0}$$; the lefthand side of this inclusion is just the ideal $$(fX_0)$$ since $$fX_0\in(X_0,\ldots,X_n)$$, while the righthand side is $$0$$ (the zero ideal) since $$k[X_0,\ldots,X_n]$$ is reduced.
This reasoning basically gets to the heart of what is going on, and I don't think it can really be shortened, although it can be modified slightly so as to be maximally general. What you are showing here is (equivalent to the assertion) that $$\mathbf{P}_k^n$$ is irreducible. This irreducibility is inherited from the irreducibility of the standard opens $$D_+(X_i)$$ and the manner in which they are glued together (more precisely, that $$D_+(X_i)\cap D_+(X_j)\neq\emptyset$$ for all $$i,j$$). In general, if you have a nonempty scheme $$X$$ that can be written as a union $$\bigcup_{i\in I}X_i$$ such that
(1) $$I$$ is nonempty,
(2) each $$X_i$$ is an irreducible open subset of $$X$$, and
(3) $$X_i\cap X_j\neq\emptyset$$ for all $$i,j\in I$$,
then $$X$$ is irreducible. So the reasoning you are using will apply to show the irreducibility of $$\mathrm{Proj}(S)$$ for a wider class of graded rings $$S$$ than just the class of polynomial rings over fields. For example, it can be applied to quotients of such rings by homogeneous prime ideals.