Points of $\mathbb{P}^1_\mathbb{Z}(\mathbb{Z})$ and reduction In $\mathbb{P}^1_\mathbb{Z}$ we have the different $\mathbb{P}^1_{\mathbb{F}_p}$ and $\mathbb{P}^1_\mathbb{Q},$ so we have $\mathbb{P}^1_{\mathbb{F}_p}\hookrightarrow\mathbb{P}^1_\mathbb{Z}$ and $\mathbb{P}^1_{\mathbb{Q}}\hookrightarrow\mathbb{P}^1_\mathbb{Z}$.
We know $\mathbb{P}^1_\mathbb{Q}(\mathbb{Q})$ is equal to $\mathbb{P}^1_\mathbb{Z}(\mathbb{Z})$ because in all $[a:b]$, $a,b\in\mathbb{Q}$ so we can simplify the denominators and have $a,b$ primes.
By reduction we have $\mathbb{P}^1_\mathbb{Z}(\mathbb{Z})\to\mathbb{P}^1_{\mathbb{F}_p}(\mathbb{F}_p)$.
I am very confused with the connection between all these arrows.
For example I know that the points $P$ of $\mathbb{P}^1_\mathbb{Q}(\mathbb{Q})=\mathbb{P}^1_\mathbb{Z}(\mathbb{Z})$ are not closed in $\mathbb{P}^1_\mathbb{Z}$ so their closure contains some points of the $\mathbb{P}^1_{\mathbb{F}_p}$.
Have these points something with the reduction $\overline{P}$ of $P$? For example take $[1:1/2]=[2:1]$. This is in fact the homogeneous prime ideal $(2X-Y)$ of $\mathbb{Z}[X,Y]$. Its closure contains $(2X-Y,3)\in\mathbb{P}^1_{\mathbb{F}_3}(\mathbb{F}_3)$ but what is the link with the point $[\overline{2}:\overline{1}]\in\mathbb{P}^1_{\mathbb{F}_3}(\mathbb{F}_3)$?
Please help me to be less confused.
 A: For the purposes of this question it is probably best to think of schemes as functors.
First, let me define $\mathbb{P}^1 (A)$ for a ring $A$.
Say a partial element of $\mathbb{P}^1 (A)$ is a tuple $(s, a_0, a_1)$ where $a_0$ and $a_1$ are elements of $A$ and $s$ is in the ideal generated by $a_0$ and $a_1$.
Say partial elements $(s, a_0, a_1)$ and $(t, b_0, b_1)$ are compatible if there is a natural number $r$ such that $s^r t^r a_0 b_1 = s^r t^r a_1 b_0$, i.e. if $a_0 b_1 = a_1 b_0$ in $A [s^{-1} t^{-1}]$.
For example, given a partial element $(s, a_0, a_1)$, if there exist $u, b_0, b_1$ such that $a_0 = b_0 u$, and $a_1 = b_1 u$, then $(s, a_0, a_1)$ is compatible with $(t, b_0, b_1)$ for any $t$ (in the ideal generated by $b_0$ and $b_1$).
Say a set of partial elements of $\mathbb{P}^1 (A)$ is consistent if they are pairwise compatible and say the set is total if the first elements generate the unit ideal of $A$.
A complete element of $\mathbb{P}^1 (A)$ is a set of partial elements that is consistent, total, and maximal with respect to inclusion among consistent total sets of partial elements.
Observe that if $A$ is a local ring, then any complete element of $\mathbb{P}^1 (A)$ contains a partial element $(s, a_0, a_1)$ where $s$ is a unit (and so $a_0$ and $a_1$ generate the unit ideal of $A$).
In general, you can use the data of a complete element of $\mathbb{P}^1 (A)$ to define a locally free submodule of $A^2$ of rank 1.
Thus, this is a generalisation of the usual definition of $\mathbb{P}^1 (A)$ in terms of homogeneous coordinates.
However, in general, it is not possible to represent complete elements of $\mathbb{P}^1 (A)$ by single partial elements – indeed, unlike equivalence classes, complete elements of $\mathbb{P}^1 (A)$ may fail to be disjoint without being equal.
Then, for a ring $k$, we define $\mathbb{P}^1_k (A) = \textrm{Hom}_\textbf{CRing}(k, A) \times \mathbb{P}^1 (A)$, i.e. the set of pairs $(\alpha, E)$ where $\alpha$ is a ring homomorphism $k \to A$ and $E$ is a complete element of $\mathbb{P}^1 (A)$.
It is clear that $\mathbb{P}^1_k (A)$ is covariantly functorial in $A$ and contravariantly functorial in $k$.
Since there is a unique ring homomorphism $\mathbb{Z} \to A$ for any ring $A$, the obvious natural map $\mathbb{P}^1_\mathbb{Z} (A) \to \mathbb{P}^1 (A)$ – the one that throws away $\alpha$ – is a bijection.
Similarly, there is at most one ring homomorphism $k \to A$ when $k$ is any quotient of any subring of $\mathbb{Q}$, so the obvious natural map $\mathbb{P}^1_k (A) \to \mathbb{P}^1 (A)$ is injective in that case.
Putting this together with the previous remark about $\mathbb{P}^1_\mathbb{Z}$, we find that the unique homomorphism $\mathbb{Z} \to k$ gives a natural injection $\mathbb{P}^1_k (A) \to \mathbb{P}^1_\mathbb{Z} (A)$ when $k$ is any quotient of any subring of $\mathbb{Q}$.
(Actually, this argument shows that either $\mathbb{P}^1_k (A)$ is empty or $\mathbb{P}^1_k (A) \to \mathbb{P}^1_\mathbb{Z} (A)$ is a bijection, but let's not dwell on that.)
On the other hand, we also get natural maps $\mathbb{P}^1_k (\mathbb{Z}) \to \mathbb{P}^1_k (A)$ for every ring $A$ and every ring $k$. Naturality means the following diagram commutes:
$$\require{AMScd}
\begin{CD}
\mathbb{P}^1_k (\mathbb{Z}) @>>> \mathbb{P}^1_\mathbb{Z} (\mathbb{Z}) \\
@VVV @VVV \\
\mathbb{P}^1_k (A) @>>> \mathbb{P}^1_\mathbb{Z} (A)
\end{CD}$$
In the case where $k = A = \mathbb{F}_p$, the vertical maps are reduction of the coordinates modulo $p$ but the horizontal maps do not do anything to coordinates.
In fact, $\mathbb{P}^1_{\mathbb{F}_p} (\mathbb{Z})$ is empty, because there are no homomorphisms $\mathbb{F}_p \to \mathbb{Z}$.
On the other hand, $\mathbb{P}^1_{\mathbb{F}_p} (\mathbb{F}_p) \to \mathbb{P}^1_\mathbb{Z} (\mathbb{F}_p)$ is a bijection – so while it is true that you can think of reduction as a map $\mathbb{P}^1_\mathbb{Z} (\mathbb{Z}) \to \mathbb{P}^1_{\mathbb{F}_p} (\mathbb{F}_p)$, in some sense it is going against the grain of naturality.
Similar remarks apply when $k = A = \mathbb{Q}$, regarding the map $\mathbb{P}^1_\mathbb{Z} (\mathbb{Z}) \to \mathbb{P}^1_\mathbb{Q} (\mathbb{Q})$.
Now suppose $F$ is a subfunctor of $\mathbb{P}^1_k$.
$F$ is closed if and only if, for every ring $A$ and every $(\alpha, E) \in \mathbb{P}^1_k (A)$, there is an ideal $I \trianglelefteq A$ such that, for every ring $B$ and every ring homomorphism $\phi : A \to B$,
$$(\phi \circ \alpha, E) \in F (B) \iff I \subseteq \ker \phi$$
We have seen that $\mathbb{P}^1_\mathbb{Q}$ is (isomorphic to) a subfunctor of $\mathbb{P}^1_\mathbb{Z}$.
It is not closed because the condition of the existence of a ring homomorphism $\mathbb{Q} \to B$ is not detectable by the kernel of the unique ring homomorphism $\mathbb{Z} \to B$.
On the other hand, each $\mathbb{P}^1_{\mathbb{F}_p}$ is (isomorphic to) a closed subfunctor of $\mathbb{P}^1_\mathbb{Z}$, because the existence of a ring homomorphism $\mathbb{F}_p \to B$ is detected by the kernel of the unique ring homomorphism $\mathbb{Z} \to B$.
The same argument shows that, for every complete element $G$ of  $\mathbb{P}^1 (\mathbb{Z})$, the subfunctor $P_k$ defined by
$$P_k (B) = \{ (\beta, F) \in \mathbb{P}^1_\mathbb{Z} (B) : G \subseteq F \text{ and } \exists \beta' \in \textrm{Hom}_\textbf{CRing} (k, B) \}$$
is not closed if $k = \mathbb{Q}$ and is closed if $k = \mathbb{F}_p$.
(Here, $G \subseteq F$ means, for every $(u, c_0, c_1) \in G$, we have $(u', c'_0, c'_1) \in F$, where $u', c'_0, c'_1$ are the integers $u, c_0, c_1$ considered as elements of $B$.)
You can think of $P_k$ as the part of $\mathbb{P}^1_\mathbb{Z}$ corresponding to $G$ considered as a $k$-valued point. The closure of $P_\mathbb{Q}$ is simply the subfunctor $P_\mathbb{Z}$, as you surmise. In particular, each $P_{\mathbb{F}_p} \subseteq P_\mathbb{Z}$, but from the functor point of view it is clearer that $P_\mathbb{Z}$ is not merely the union of $P_\mathbb{Q}$ and all the $P_{\mathbb{F}_p}$.
