Let $X=\mathbf{P}^n_A = \text{Proj} A[T_0,\ldots,T_n]$. If $A$ is a field, there is a simple classical description of $X(A)$. However, if $A$ is a more general ring, like $\mathbf{Z}$, I don't see an easy way to characterize rational points.
For example, we could try to characterize it as graded morphisms \begin{align*} A[T_0,\ldots,T_n] \to A[T] \end{align*} but it seems that two different morphisms can correspond to the same rational point.
What is the best way to see the set $X(A)$?
Intuitively, morphisms from some space $X$ into projective $n$-space are determined by line bundles ${\mathcal L}$ on $X$ equipped with $n+1$ global sections $s_0,...,s_n$ of ${\mathcal L}$ that are nowhere simultaneously vanishing. The idea is that, at any fixed $x\in X$, you can pick one $s_i$ with $s_i(x)\neq 0$, which then trivializes the fiber of ${\mathcal L}$ at $x$ and reduces all other sections $s_j(x)$ for $j\neq i$, to scalars $s_j(x)/s_i(x)$. You can then send $x$ to the point $\left[\frac{s_0(x)}{s_i(x)}:...:\frac{s_n(x)}{s_i(x)}\right]$ in ${\mathbb P}^n$, which is independent on the choice of $i$ such that $s_i(x)\neq 0$, since choosing instead any other $i^{\prime}$ with $s_{i^{\prime}}(x)\neq 0$ only results in a scaling of the homogeneous coordinates by $\frac{s_i(x)}{s_{i^{\prime}}(x)}$. Conversely, given any morphism $X\to {\mathbb P}^n$ induces a line bundle with $n+1$ nowhere simultaneously vanishing sections on $X$ by pulling back ${\mathcal O}(1)$ with its standard sections from ${\mathbb P}^n$.
Formally, for a scheme $X$ the $X$-rational points of ${\mathbb P}_{\mathbb Z}^n$ are in bijection with equivalence classes of pairs $({\mathcal L},(s_0,...,s_n))$ where ${\mathcal L}$ is rank $1$ locally free coherent sheaf on $X$ and $s_i\in {\mathcal L}(X)$ are global sections such that ${\mathcal L}_x = {\mathcal O}_{X,x}\cdot\{(s_i)_x\}$, i.e. for any $x\in X$ there is some $i$ with $(s_i)_x\notin{\mathfrak m}_{x}{\mathcal L}_x$. Note that given ${\mathcal L}$, the tuple $(s_0,...,s_n)$ with the above requirement is the same as a surjective morphism ${\mathcal O}_X^{n+1}\twoheadrightarrow{\mathcal L}$, and you identify two such morphisms ${\mathcal O}_X^{n+1}\to {\mathcal L}$ and ${\mathcal O}_X^{n+1}\to {\mathcal L}^{\prime}$ if they differ by an isomorphism between ${\mathcal L}$ and ${\mathcal L}^{\prime}$.
See Goertz-Wedhorn, Algebraic Geometry I, 8.5: Projective space as a Grassmannian
If $X=\text{Spec}(A)$, over which vector bundles correspond to projective $A$-modules, this means that $\text{Spec}(A)$-valued points of ${\mathbb P}_{\mathbb Z}^n$, i.e. $A$-rational points of ${\mathbb P}_A^n$, are in bijection with equivalence classes of rank $1$ projective $A$-modules $P$ equipped with $n+1$ elements $s_0,...,s_n\in P$ generating $P$. For example, if $P=A$, then the $s_i$ are ring elements that are to satisfy $(s_i)=A$, and you identify two such tuples $(s_i)$ and $(s^{\prime}_j)$ if they differ by a unit in $A$. In case $A$ is a field (over which any module is free), this gives you the familiar description of ${\mathbb P}_k^n(k)$ as $(k^{n+1}\setminus \{0\})/k^{\times}$, but you could also take $A={\mathbb Z}$ (over which any projective module is free) and see that ${\mathbb Z}$-points of ${\mathbb P}^n_{\mathbb Z}$ are in bijection with $(n+1)$-tuples $(s_0,...,s_n)\in{\mathbb Z}^{n+1}$ of coprime integers, up to overall multiplication by $\pm 1$.
Tracing, I don't know if we can apply the valuative criterion in this case (since $\mathbf{Z}$ is not a DVR).
– Dec 13 '14 at 18:16The algebraic description of $\mathbb P^n_A(A)$ is of biblical simplicity:
It consists of all submodules $M\subset A^{n+1}$ which are direct summands of rank one.
As the French poet Verlaine wrote: Et tout le reste est littérature ("And all the rest is literature")
Here however is some of that littérature:
1) To say that $M$ is a direct summand means that there exists a submodule $N\subset A$ such that $A^{n+1}=M\oplus N$.
2) To say that $M$ is of rank one means that for each maximal prime $\mathfrak m\subset A$ the automatically free $A_\mathfrak m$-module $M_\mathfrak m$ has dimension $1$.
3) Among these projective direct summands of rank are the free ones: they consist exactly of modules $M$ which can be written $M=A(a_0,a_1,\cdots,a_n)\subset A^{n+1}$ with the $a_i$'s satisfying $\sum_{i=0}^n Aa_i=A$.
We might then write for tradition's sake $M=[a_0:a_1:\cdots:a_n]$.
The module $M$ can also be written $M=A(b_0,b_1,\cdots,b_n)$ if and only if there exists an invertible element $u\in A^\times$ such that $(b_0,b_1,\cdots,b_n)=u(a_0,a_1,\cdots,a_n)\in A^{n+1}$.
4) For some rings $A$ all direct summands of rank one will be free [i.e. of the type described in 3)]: this is the case if $A$ is a PID or if $A$ is a local ring.
5) But this is not true for all rings. Here is an example of a non free direct summand of rank one:
The ideal $I=\langle 2,1+\sqrt -5\rangle\subset A=\mathbb Z[\sqrt -5]$ is known to be a non-free projective A-module of rank one.
There is a surjective morphism of $A$-modules $p:A^2\to I: (a,b)\mapsto a\cdot2+b\cdot\sqrt -5$.
Since $I$ is projective, $p$ has an $A$-linear section $s:I\to A^2$ (i.e. $p\circ s:I\to I$ is the identity).
Then the image $M:=s(I)$ is a direct summand of rank one of $A^2=M\oplus ker (p)$, and $M$ is thus an element of $\mathbb P^1_A(A)$ not of the form $A(a,b)$.
In other words $M\in \mathbb P^n_A(A)$ cannot be thought of as some $[a:b]$, in contrast to what happens over a field $k$, where elements of $\mathbb P^1_k(k)$ can always be written $[a:b]$.
