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$\mathbb{F}_1$ and the ABC-conjecture

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Finally, we’re closing in on Smirnov’s approach to the ABC-conjecture via geometry over the field with one element.

The geometric defect

Let $\phi : C_1 \mapsto C_2$ be a cover of curves defined over $k$, then the scheme-version of the Riemann-Hurwitz inequality is

$$2 g_{C_1} – 2 \geq deg(\phi) (2 g_{C_1} -2) + \sum^{scheme}_{P \in C_1} (e_{\phi}(P)-1) deg(P)$$

In the special case, when $f$ is a non-constant rational function in $k(C)$ and $f~:~C \mapsto \mathbb{P}^1_k$ is the corresponding cover, this reads

$$2g_C-2 \geq -2 deg(f) + \sum^{scheme}_{P \in C} (e_f(P)-1) deg(P)$$

which can be turned into the inequality

$$\sum^{scheme}_{P \in C} \frac{(e_f(P)-1) deg(P)}{deg(f)} \leq 2 – \frac{2-2g_C}{deg(f)}$$

We call the expression $\delta(P) = \tfrac{(e_f(P)-1) deg(P)}{deg(f)}$ the defect of $P$. Observe that $\delta(P) \geq 0$ and so this inequality only improves it we restrict the summation to some subset of schematic $C$-points.

The arithmetic defect

Take a positive rational number $q = \frac{m}{n}$ with $1 \leq n < m$ and $(m,n)=1$ and consider the cover

$$q~:~\mathsf{Spec}(\mathbb{Z}) \mapsto \mathbb{P}^1 / \mathbb{F}_1$$

Recall that the fiber over the point $[d] \in \mathbb{P}^1 / \mathbb{F}_1$ consists of all prime divisors of $m^d-n^d$ not dividing any $m^e-n^e$ for $e < d$. The fiber of $[0]$ (resp. of $[\infty]$) consists of all prime divisors of $m$ (resp. of $n$ together with $\infty$). Here's part of the cover for $q=\frac{104348}{33215}$ (a good rational approximation for $\pi$).

pi-map

It is tempting to define the ramification index $e_q(p)$ for the map $q$ in the prime $p$ lying in the fiber $q^{-1}([d])$ to be the largest power of $p$ dividing $m^d-n^d$. Likewise, for $p \in q^{-1}([0])$ (resp. in $q^{-1}([\infty])$) take for $e_q(p)$ the largest power of $p$ dividing $m$ (resp. dividing $n$). Finally, take $e_q(\infty) = log(q)$.

Combine this with our previous definitions for the degree of $p$ to be $log(p)$ and of the degree of the map $q$ to be $log(m)$, to define the arithmetic defect of $q$ in the prime $p$ to be

$$\delta(p) = \frac{(e_q(p)-1) log(p)}{log(m)}$$

We can now define the total defect of the cover $q$ over the point $[d] \in \mathbb{P}^1 / \mathbb{F}_1$ to be

$$\delta_{[d]} = \sum_{p \in q^{-1}([d])} \delta(p)$$

It is easy to work out these total defects for the four $\mathbb{F}_1$-rational points of $\mathbb{P}^1 / \mathbb{F}_1$ : $\{ [0],[1],[2],[\infty] \}$ (the primes lying on the blue lines in the graph).

For a natural number $a$ let $a_0$ be its square-free part and $a_1 = \tfrac{a}{a_0}$ the remaining part. Then

  • $\delta_{[0]} = \frac{log(m_1)}{log(m)}$
  • $\delta_{[\infty]} = \frac{log(n_1)+log(q)-1}{log(m)}$
  • $\delta_{[1]} = \frac{log((m-n)_1)}{log(m)}$
  • $\delta_{[2]}= \frac{log(k_1)}{log(m)}$

where $k$ is $m+n$ divided by the largest $2$-power it may contain.

Hurwitz-conjecture for $\mathbb{Q}$

If we sum the defects of $q$ in all primes over the points $\{ [0],[1],[\infty] \}$ we would get, in analogy with the Hurwitz-inequality in the function field case

$$\delta_{[0]}+\delta_{[1]}+\delta_{[\infty]} \leq 2 – \frac{2 – 2g_{\mathsf{Spec}(\mathbb{Z})}}{log(m)}$$

We do not know what the genus of the arithmetic curve $\mathsf{Spec}(\mathbb{Z})$ might be, but is sure is a constant not depending on the map $q$. If we could develop a geometry over $\mathbb{F}_1$ such that all wild guesses we made before would turn out to be the correct ones for an $\mathbb{F}_1$-version of the Hurwitz inequality, we would have the statement below :

For every $\epsilon > 0$ there exists a constant $C(\epsilon)$ such that the following inequality holds for every pair $1 \leq m < n$ with $(m,n)=1$

$$\frac{log(m_1) + log((m-n)_1) + log(n_1) + log(m)-log(n)-1}{log(m)} \leq 2 + \epsilon + \frac{C(\epsilon)}{log(m)}$$

‘Proof’ of the ABC-conjecture

The ABC-conjecture requires for every $\epsilon > 0$ a constant $D(\epsilon)$ such that for all coprime natural numbers $A$ and $B$ we have with $A+B=C$

$$C \leq D(\epsilon) (A_0B_0C_0)^{1+\epsilon}$$

Well, take $m=C$ and $n=min(A,B)$ then in the conjectural Hurwitz inequality for the cover corresponding to $q=\frac{m}{n}$ above we have that

  • $\frac{log(m_1)}{log(m)} = 1 – \frac{log(m_0)}{log(m)}$
  • $\frac{log(n_1)+log(m)-log(n)-1}{log(m)} = 1 – \frac{log(n_0)}{log(m)} – \frac{1}{log(m)}$
  • $\frac{log((m-n)_1)}{log(m)}=\frac{log(m-n)}{log(m)}-\frac{log((m-n)_0)}{log(m)} \geq 1 – \frac{log((m-n)_0)}{log(m)} – \frac{log(2)}{log(m)}$

(the latter inequality because $m-n \geq \frac{m}{2}$ and so $log(m-n) \geq log(m)-log(2)$). Plug this into the inequality above and get

$$3-\frac{log(n_0m_0(m-n)_0)}{log(m)} \leq 2 + \epsilon + \frac{C(\epsilon) + 1 + log(2)}{log(m)}$$

Take $log(C’(\epsilon))=C(\epsilon)+1+log(2)$ and reshuffle in order to get the inequality $m^{1-\epsilon} \leq C’(\epsilon)(n_0m_0(m-n)_0)$. But then, finally (finally!) with $D(\epsilon)=C’(\epsilon)^{1+\epsilon}$

$$C=m \leq D(\epsilon)(n_0m_0(m-n)_0)^{1+\epsilon} = D(\epsilon)(A_oB_0C_0)^{1+\epsilon}$$

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The exceptional map and Mersenne primes

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Last time we’ve seen that almost all rational numbers $q \in \mathbb{Q}$ determine a finite cover $q~:~\mathsf{Spec}(\mathbb{Z}) \mapsto \mathbb{P}^1 / \mathbb{F}_1$, the exceptional cases controlled by Zsigmondy’s theorem.

The prime exceptional case corresponds to $q=2$. Below,we sketch a small portion of the graph of this non-cover (the image does not contain the points $[1]$ and $[6]$, the red lines) where we use a logarithmic scale on $\mathsf{Spec}(\mathbb{Z})$.

Clearly, such images should be taken with a bucket of salt. The linear depiction of $\mathsf{Spec}(\mathbb{Z})$ suggests for example that the prime $3$ has more affinity with $2$ and $5$ than with say $2147483647$ or $524287$. However, the relevant topology on $\mathsf{Spec}(\mathbb{Z})$ is the cofinite topology, so one is always allowed to reshuffle finitely many primes in order to get a smoother covering map!

The fiber over $[n]$ consists of the primitive prime factors of $2^n-1$ (that is, those primes not dividing $2^d-1$ for $d$ a proper divisor of $n$). For small values of $n$, most fibers consist of just one prime (for $n \leq 35$ only $n=11,23,23,28$ and $35$ have two principal primes and only $n=29$ has three primes in its fiber).

A Mersenne prime is a prime number of the form $M_p = 2^p-1$ (from which it follows that $p$ must be prime, too). Only $47$ Mersenne primes are known, the smallest being

$M_2,M_3,M_5,M_7,M_{13},M_{17},M_{19}$ and $M_{31}$

corresponding to the points on ‘the diagonal’ in the graph of the exceptional map. Naturally, if $M_p$ is a Mersenne prime, the fiber $2^{-1}([p])$ has only one element.

If we believe a geometry over $\mathbb{F}_1$ can be developed such that the morphisms $q$ make sense (and hence their graphs define divisors in the Smirnov plane $\mathsf{Spec}(\mathbb{Z}) \times \mathbb{P}^1 / \mathbb{F}_1$) one might expect that the subsets of points $[n] \in \mathbb{P}^1 / \mathbb{F}_1$ with a fiber containing at least $k$ points, should be cofinite.

In particular, in view of Schinzel’s result (stating that all maps $q$ have infinitely many points with a fiber containing at least two elements) this would imply that the set of Mersenne primes has to be finite , contradicting the Lenstra-Pomerance-Wagstaff conjecture.

On the positive side, it would imply that infinitely many numbers of the form $2^p-1$ (with $p$ a prime number) are highly composite (as they must have at least two principal prime factors) which is another big open problem…

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Rational numbers and covers $\mathsf{Spec}(\mathbb{Z}) \mapsto \mathbb{P}^1 / \mathbb{F}_1$

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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…

About lieven le bruyn

Also blogs at NeverEndingBooks.

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