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?

  • $\begingroup$ 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 $D_+(X_i)$. $\endgroup$ – 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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.