Water and cabbages in $\mathbb{F}_{1}$-geometry

Today we’ll try to grasp the meaning of Chebotarëv’s density theorem, and hint at how this may help us get a fix on $\mathbb{F}_{1}$-geometry, formerly known as recreational number theory.

Status quo

In our weekly seminar the problem at hand is to classify the $\mathbb{F}_{1}$-subschemes of $\mathbb{P}^{1}_{\mathbb{Z}}=\mathbb{P}^{1}_{\mathbb{F}_{1}} \  \times_{\mathbb{F}_{1}} \ \mathsf{Spec}(\mathbb{Z})$. This boils down to finding the ordinary $\mathbb{Z}$-subschemes that also carry a $\lambda$-structure. Solving this problem will help us formalize Smirnov’s approach to $\mathbb{F}_{1}$-geometry, as put forward in Hurwitz inequalities in number fields. To do so, crucial tools will be provided for by several density theorems in algebraic number theory. My experience in number theory is almost nonexistent, so it seems like a good idea to conjure up a post on the subject, with a little help from (once again) Hendrik Lenstra.

Chebo-who?

Nikolaï Chebotarëv (1894-1947) was a Russian number-theorist, famous for his proof of a conjecture by Frobenius. The conjecture had been open for 42 years before Cheb tamed it, and surprisingly (at least to me) it provided just the right ingredient for Artin to prove his well known reciprocity law, vastly generalizing the baby version put forward by Gauss. In fact, if Artin had waited a while longer to publish his paper, chances would have been in favour of the less groovy sounding Chebotarëv reciprocity. If you’re stuck for inspiration, let me just mention that Nikolaï (shown here with an exquisite bowtie) didn’t like working at a desk, but preferred a comfy bed to get the job done.

Densely packed

To explain the theorem, it’ll be worthwhile to check out the cases it generalizes: Dirichlet’s theorem on primes in arithmetic progressions, and Frobenius density. Very naturally, a set of primes $S$ is said to have density $\delta$ if the ratio of the number of primes in $S$, smaller than a number $n$, to the number of all primes smaller than $n$ converges to $\delta$, for $n \to \infty$. The Dirichlet theorem then says

Theorem. Fix $m \in \mathbb{Z}$. If an arbitary integer $a$ is coprime with $m$, then the set of primes $p$, with $p \equiv a \ \mathsf{mod} \ m$, has density $1/\phi(m)$, $\phi$ being Euler’s totient function.

Specializing to the case $m=10$, you get the intuitive result that there are “as many” primes ending in $1$, as there are in $3, 7$ and $9$ respectively.

Frobenius highest honour

Frobenius, the only mathematician I know of who’s also a noun, has his own density theorem. To formulate it correcty, we need two new concepts, decomposition type and cycle pattern. Given a monic integer polynomial of degree $d$ with non-zero discriminant $\Delta(f)$, and a prime $p$ that doesn’t divide $\Delta(f)$, you can factor $f \ \mathsf{mod} \ p$ into $n$ distinct irreducibles. The numbers $d_{1},…,d_{n}$, with $d_{i}$ the degree of the $i$-th factor in this decomposition, comprise the decomposition type of $f \ \mathsf{mod} \ p$. Notice that the $d_{i}$ form a partition of $d$. Since $f$ has $d$ different roots $\alpha_{1},…,\alpha_{d}$, the Galois group $G$ of $f$ consists of $\mathbb{Q}$-automorphisms of $\mathbb{Q}(\alpha_{1},…, \alpha_{d})$. Elementary Galois theory says this is a subgroup of $S_{d}$, so every element $\sigma \in G$ can be written as a product of disjoint cycles, including cycles of length $1$. The length of each of these cycles will again form a partition of the degree $d$, the cycle pattern. Frobenius connects the two.

Theorem. The set of primes $p$ for which $f$ has a given decomposition type $d_{1},…,d_{n}$ has a density which is equal to $1/\vert G \vert$ times the number of $\sigma \in G$ with cycle pattern $d_{1},…,d_{n}$.

Considering a special case once again, look at the partition $d=1+…+1$. Trivially, only the identity has this as a cycle pattern, and thus we see that the number of primes $p$ for which $f$ splits into linear factors has density $1/\vert G \vert$. It’s worth noting that by using a good algebra program like GAP, you can use this result to determine the order of the Galois group of a polynomial, and a fortiori the group itself. If your gut feeling is that the two theorems are connected, you’re absolutely right! To make this concrete, we’ll need Chebotarëv’s result.

Chebo-what?

In general, Dirichlet for a certain $m$ implies Frobenius for the polynomial $X^{m}-1$. To see this for the special decomposition type $1,1,…,1$, remember that the Galois group of $X^{m}-1$ is isomorphic to the unit group of $\mathbb{Z}/m\mathbb{Z}$, which has $\phi(m)$ elements. Taking $a$ to be $1$ in Dirichlet’s theorem, we see that the set of primes for which $p \equiv 1 \ \mathsf{mod} \ m$ is equal to the set of primes for which $X^{m}-1$ splits into linear factors, by Fermat’s little theorem. The general case is not much harder, but requires a couple of properties of finite fields I won’t go into. The problem rests in the converse. Given the statement of Frobenius density, a first natural question to ask is whether we can always find for a given prime $p$ that doesn’t divide $\Delta(f)$, an element $\sigma_{p} \in G$, such that the decomposition type of $f \ \mathsf{mod} \ p$ equals the cycle pattern of $\sigma_{p}$. This can be done up to conjugacy, which seems logical, since conjugate permutations also have the same cycle pattern. Chebotarëv’s theorem then deals with the density of the set of those primes $p$ for which $\sigma_{p}$ is equal to a given element of $G$.

It seems best to postpone the detailed formulation of Chebotarëv’s theorem, since I need some more technical stuff like places, which would make this post unnecessarily long. Let me finish by explaining the first part of the title with a quote from (tadaa) Lenstra:

So, Chebotarëv had to carry water and cabbages from the lower part of Odessa to the higher part of Odessa. This is, of course, the most difficult direction.

Inspite of its humorous simplicity, I can’t stop thinking about this quote which, like most of Lenstra’s debaucheries, says more than meets the eye.

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

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 $D(\epsilon)=C’(\epsilon)^{1+\epsilon}$

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

The exceptional map and Mersenne primes

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…