In analogy with the function-field case we expect rational numbers $q \in \mathbb{Q}$ to define maps from the arithmetic curve $\mathsf{Spec}(\mathbb{Z})$ to the projective line $\mathbb{P}^1 / \mathbb{F}_1$. For all non-constant $q$ (that is, if $q \not= \pm 1$) we expect this map to be a cover.
the geometric picture
A non-constant rational function $f \in \overline{k}(C)$ defines a cover $C(\overline{k}) \mapsto \mathbb{P}^1(\overline{k})$ by sending a point $P$ to the point $[f(P):1]$ if $f$ is defined in $P$ or to $\infty = [1 : 0]$ otherwise.
If $f \in k(C)$, then this maps Galois orbits of points of $C(\overline{k})$ (that is, scheme-points of $C$) to Galois orbits of points of $\mathbb{P}^1(\overline{k})$, so it defines a scheme-map $f : C \mapsto \mathbb{P}^1$. We have seen before that the degree of this map equals the degree of the zero-divisor $div_0(f)$ of $f$ (or equivalently, of the pole-divisor $div_{\infty}(f)$). That is, if $div_0(f) = \sum n_P [P]$ then $deg(f)=\sum n_P deg(P)$.
the definition
The schematic points of $\mathbb{P}^1 / \mathbb{F}_1$ are $\{ 0,\infty \} \cup \{ [1],[2],[3],[4],\cdots \}$ with $[n]$ the set of all primitive $n$-th roots of unity.
For $q = \frac{a}{b} = \pm \frac{p_1^{e_1} \cdots p_r^{e_r}}{q_1^{f_1} \cdots q_s^{f_s}} \not= \pm 1 \in \mathbb{Q}$ we define a map $q~:~\mathsf{Spec}(\mathbb{Z}) \rightarrow \mathbb{P}^1 / \mathbb{F}_1$ by
- sending every $p_i$ to $0$
- sending every $q_j$ to $\infty$
- sending $\infty$ to $0$ if $log |q| < 0$ and to $\infty$ if $log |q| > 0$
- sending the prime number $p \notin \{ p_1,\cdots,p_r,q_1,\cdots,q_s \}$ to $[n]$ if $n$ is the order of the unit $\overline{q} = \overline{a}.\overline{b}^{-1} \in \mathbb{F}_p^*$.
To ‘explain’ the last line, it is equivalent to the existence of a primitive $n$-th root of unity $\epsilon$, of order prime to $p$, such that there is a prime ideal $P$ in $\mathbb{Z}[\epsilon]$ lying over $(p)$ such that $v_P(q-\epsilon) > 0$, or equivalently, that $q(P)=\epsilon(P)$.
the finite cover $q$
Assume that $q = \frac{a}{b}$ with $(a,b)=1$ and $0 < b < a$, then $div_0(q) = \sum_i e_i [p_i]$ and hence by the convention that $deg([p_i]) = log(p_i)$ we must define
$$deg(q) = deg(div_0(q)) = \sum_i e_i log(p_i) = log(a)$$
It is easy to see that the fibers of $q$ are all finite. For, if $q$ maps $p$ to $[n]$, then by definition we have
$a^n \equiv b^n~\text{mod}~p$ and $a^m \not\equiv b^m~\text{mod}~p$ for all $m < n$.
In other words, the fiber $q^{-1}([n])$ is the set of all prime-divisors of $a^n-b^n$ which do not divide $a^m-b^m$ for some $1 \leq m < n$.
It is a lot more challenging to prove that $q$ is indeed a cover, that is, that all fibers are non-empty. In fact, this is not always the case, but the exceptions are well understood.
This is the content of the so called Zsigmondy theorem after the Austrian mathematician Karl Zsigmondy who proved in 1892 that if $(a,b)=1$ and $1 \leq b < a$ then for any natural number $n > 1$ there is a prime number $p$ (called a primitive prime divisor) that divides $a^n-b^n$ and does not divide $a^k-b^k$ for any positive integer $k < n$, with the following exceptions:
- $a = 2, b = 1$, and $n = 6$ as $2^6-1=3^2.7$ and $7=2^3-1$, $3=2^2-1$, or
- $a+b$ is a power of two and $n=2$.
It is a pleasant exercise to draw part of the graph for specific maps $q~:~\mathsf{Spec}(\mathbb{Z}) \mapsto \mathbb{P}^1 / \mathbb{F}_1$ and to relate its geometric properties to deep open problems in number theory.
For example, a result of Schinzel’s from 1962 asserts that for all maps $q$ there are infinitely many points $[n]$ over which the fiber has at least two elements. However, nobody knows whether all such map have one fiber consisting of at least three elements…
