So, I have the following definition of $\mathbb{P}_A^n$ for an arbitrary (commutative) ring $A$, from Hartshorne:

Set $S=A[x_0,\ldots,x_n]$, so that $S=\bigoplus_{d\geq 0}S_d$ as a graded ring, $S_+=\bigoplus_{d\geq 1}S_d$, and for convenience, let $S^\mathrm{H}=\bigcup_{d\geq 0} S_d$ denote the homogeneous elements. We define the set $\mathrm{Proj}\ S=\{\mathfrak{p}\subset S \mid \mathfrak{p} \mathrm{\ hmg.\ prime}, S_+ \nsubseteq\mathfrak{p}\}$ with closed sets $V(\mathfrak{a})=\{\mathfrak{p}\in\mathrm{Proj}\ S\mid \mathfrak{p}\supseteq\mathfrak{a}\}$ for all homogeneous ideals $\mathfrak{a}\subseteq S$.

Next, for all $\mathfrak{p}\in\mathrm{Proj}\ S$, we set $T_\mathfrak{p}=S^\mathrm{H}\setminus\mathfrak{p}$, $S_{(\mathfrak{p})}=\{\frac{f}{g}\in T_\mathfrak{p}^{-1} S \mid f\in S^\mathrm{H}, \deg f = \deg g\}$. Finally, for any open subset $U\subseteq\mathrm{Proj}\ S$, we define $$\mathcal{O}(U)=\{s:U\to\bigsqcup_\mathfrak{p} S_{(\mathfrak{p})} \mid \forall\ \mathfrak{p}\in U, s(\mathfrak{p})\in S_{(\mathfrak{p})}, \exists\ \mathfrak{p}\in V\subseteq U \mathrm{\ open}, a,f \in S^\mathrm{H} \mathrm{\ s.t.\ } \deg\frac{a}{f} = 0, \mathrm{\ and\ }\forall\ \mathfrak{q}\in V, f\notin\mathfrak{q}, s(\mathfrak{q})=\frac{a}{f}\in S_{(\mathfrak{q})}\}$$

When $A$ is an algebraically closed field, I can see the analogy with the projective space $\mathbb{P}^n$ of classical algebraic geometry. But for arbitrary rings, even for the simplest case of $A=\mathbb{Z}$, I struggle to make sense of this mess of symbols. Is there some good intuition to keep in mind when working with $\mathbb{P}_A^n$, or some simpler way of describing the ring of regular functions?

At the very least, I'd like to understand what's going on in $\mathbb{P}_\mathbb{Z}^n$, to gain some intuition for the more general case.

  • 2
    $\begingroup$ But you know what it looks like on $D(x_i)$, don't you? That's a copy of $\mathbb A^n_{\mathbb Z}$ and then these get glued together. The glueing maps are the same as in the field case. Probably the right interpretation is that this construction packages together projective $n$-spaces for all finite fields and for $\mathbb Q$. $\endgroup$
    – Hoot
    Jul 15, 2016 at 19:07

2 Answers 2


I agree that the point-set description is not very enlightening (the ratio of complexity of explicit description against complexity of abstract meaning can get much worse, e.g. Does the category of locally ringed spaces have products?); although one has to prove existence somehow, I think the right way to view the projective space is to look at the functor that the projective $n$-space represents, i.e. to describe how morphisms $X\to {\mathbb P}^n_A$ can be constructed:

For an $A$-scheme $X$, a morphism $X\to {\mathbb P}^n_A$ is equivalent to specifying a line bundle ${\mathcal L}$ on $X$ together with global sections $s_0,...s_n$ without simultaneous zero, up to isomorphism.

This is a very intuitive thing: Roughly, given a point $x$ of $X$, you pick some $s_i$ not vanishing at $x$, and map $x$ to $(\frac{s_j(x)}{s_i(x)})$. Choosing a different $s_{i^{\prime}}$ modifies the tuple by the constant $\frac{s_i(x)}{s_{i^{\prime}}(x)}$, hence defines the same point in projective space. I like this idea of 'measuring' the $s_j(x)$ against some chosen nonzero $s_i(x)$, and to remove the ambiguity by factoring out constants. Conversely, any morphism $p: X\to {\mathbb P}^n_A$ determines ${\mathcal L} := p^{-1}({\mathscr O}(1))$, and the $s_i$ are the pullbacks of the standard sections of ${\mathscr O}(1)$.

This description simplifies to the classical description of projective spaces when restricted to $X$ which does not admit any nontrivial line bundles: Then ${\mathcal L}\cong {\mathscr O}_X$ and a map $X\to {\mathbb P}^n_A$ is a tuple $(f_0,...,f_n)\in {\mathscr O}_X^n$ of generators of ${\mathscr O}_X^{n+1}$, modulo the action of ${\mathscr O}_X(X)^{\times}$. Taking $X=A=\text{Spec}(k)$ for example, this gives ${\mathbb P}^n_k(k)=(k^{n+1}\setminus\{0\})/k^{\times}$.

In the midway, you might consider $X=\text{Spec}(B)$ for some $A$-algebra $B$. Then specifying a map $X\to {\mathbb P}^n_A$ comes down to specifying a projective $B$-module $P$ together with a surjection $B^{n+1}\twoheadrightarrow P$, up to isomorphism. For $B=A={\mathbb Z}$ (over which every projective module is free) you get ${\mathbb P}^n_{\mathbb Z}({\mathbb Z})=\text{coprime}({\mathbb Z}^{n+1})/\{\pm 1\}$, where $\text{coprime}({\mathbb Z}^{n+1})$ is the set of $(a_0,...,a_n)$ which have no common prime divisor.

Note: You can also see the paving of ${\mathbb P}^n_A$ by ${\mathbb A}^n_A$ this way: For any $i\in\{0,1,...,n\}$ you have the subfunctor restricting to those $({\mathcal L},s_0,...,s_n)$ such that $s_i$ is nowhere vanishing. But if you have a nowhere vanishing section of a line bundle, this section trivializes the bundle, so the datum essentially comes down to specifying $n$ global functions on $X$, which is the same as a morphism $X\to {\mathbb A}^n_A$.


A typical way to picture this is to look at affine patches.

If, for example, $X,Y,Z$ are your standard coordinates on $\mathbb{P}^2$, then if we remove the subscheme $Z=0$, we are left with a copy of the affine plane $\mathbb{A}^2 \cong \mathrm{Spec}\left(\mathbb{Z}[\frac{X}{Z}, \frac{Y}{Z}] \right) \subseteq \mathbb{P}^2$.

The subscheme $Z=0$, incidentally, is isomorphic to the projective line $\mathbb{P}^1$. So you can think of the projective plane as the affine plane together with a projective line encircling it. This picture works in higher dimensions as well.

Similarly, if we remove $Y=0$, we get a different copy of the affine plane $\mathbb{A}^2 = \mathrm{Spec}\left(\mathbb{Z}[\frac{X}{Y}, \frac{Z}{Y}] \right)$.

These two copies of the affine plane overlap: their intersection is $$ \mathrm{Spec}\left(\mathbb{Z}\left[\frac{X}{Z}, \frac{Y}{Z}, \frac{X}{Y}, \frac{Z}{Y}\right] \right) = \mathrm{Spec}\left(\mathbb{Z}\left[\frac{X}{Z}, \frac{Y}{Z}, \left(\frac{Y}{Z}\right)^{-1}\right] \right) \cong \mathbb{A}^2 \setminus \mathbb{A}^1 $$

If we remove $X=0$, we get yet another copy of the affine plane $\mathbb{A}^2 = \mathrm{Spec}\left(\mathbb{Z}[\frac{Y}{X}, \frac{Z}{X}] \right)$.

These three copies of the affine plane are enough to cover the projective plane. So if you need to, you can do computations in these three affine patches, transferring from one to another as needed. Or to define things, define them on each patch and ensure they're consistent. (or other things, like define something on one affine patch and then extend it by continuity or whatever to the whole projective space)


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