$\mathbb{F}_1$ 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 $D(\epsilon)=C’(\epsilon)^{1+\epsilon}$

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

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

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…

ABC-theorem for Curves

Here we give the promised proof of the ABC-conjecture for function fields.

As always, $k$ is a perfect (e.g. finite) field and $K=k(X)$ is the function field of a smooth projective curve $X$ defined over $k$. We take elements $u,v \in K^*$ satisfying $u+v=1$ and consider the cover $u : X \mapsto \mathbb{P}^1_k$ corresponding to the embedding $k(u) \hookrightarrow K$. We want to determine the (schematic) zero- and pole-divisors of $u$ and $v$ and call them $A=div_0(u), B=div_0(v)$ and $C=div_{\infty}(u)=div_{\infty}(v)$.

Let $R$ be the integral closure of $k[u]$ in $K$, then in $R$ we can write the ideals $u$ and $(v)=(1-u)$ as products of prime-ideals (which correspond to schematic points of $X$)

$(u) = P_1^{e_u(P_1)} \cdots P_r^{e_u(P_r)}$
$(v) = Q_1^{e_u(Q_1)} \cdots Q_s^{e_u(Q_s)}$

and so $A = \sum_i e_u(P_i) [P_i]$ and $B = \sum_j e_u(Q_j) [Q_j]$. If $S$ is the integral closure of $k[\frac{1}{u}]$ in $K$, then we have in $S$ a decomposition

$(\frac{1}{u}) = R_1^{e_u(R_1)} \cdots R_t^{e_u(R_t)}$

and therefore $C = \sum_l e_u(R_l)[R_l]$. We already know that $deg(A)=deg(B)=deg(C)=n=[K : k(u)]$.

Case 1 : Let us assume that the field extension $K/k(u)$ is separable. Then, by the Riemann-Hurwitz formula (or rather, the scheme-version of it) we get the inequality (use that the genus of $\mathbb{P}^1_k$ is zero) :

$2 g_K – 2 \geq -2n + \sum^{scheme}_{P \in C} (e_u(P)-1) deg(P)$

Because for all points $e_u(P)-1 \geq 0$, the inequality only becomes better if we restrict the sum to a subset of points, say to the support of $A+B+C$ (that is to ${ P_1,\cdots,P_r,Q_1,\cdots,Q_s,R_1,\cdots,R_t }$). Then we get

$2 g_K -2 \geq -2n + \sum_{P \in Supp(A+B+C)} e_u(P)deg(P) – \sum_{P \in Supp(A+B+C)}deg(P)$
$~\qquad = -2n+3n-\sum_{P \in Supp(A+B+C)} deg(P)$

which gives us the required form of the ABC-conjecture for curves

$n=deg(u)=deg_s(u) \leq 2g_K – 2 + \sum_{P \in Supp(A+B+C)} deg(P)$

Case 2 : If $K/k(u)$ is not separable, take a maximal separable subfield $k(u) \subset M \subset K$, then by definition of $deg_s(u)$ and case 1 we have

$deg_s(u) \leq 2g_M – 2 + \sum_{P’ \in Supp(A’+B’+C’)} deg(P’)$

where $A’$ (resp. $B’$,$C’$) are the schematic fibers of the cover $Y \mapsto \mathbb{P}^1_k$ over the $k$-rational points $0$ (resp. $1$, $\infty$) and where $Y$ is the curve with function field $M$. We are done if we can show that $g_K=g_M$ and that in the cover $X \mapsto Y$ there is a unique point $P$ lying over each point $P’$ with $deg(P)=deg(P’)$.

As $K/M$ is purely inseparable, we have a tower of subfields

$M=M_0 \subset M_1 \subset \cdots \subset M_z=K$

such that $M_i / M_{i-1}$ is purely inseparable of degree $p$ for all $i$. That is, raising to the $p$-th power gives a field-isomorphism $M_i \simeq M_{i-1}$. The genus is a field-invariant, so $g_{M_i}=g_{M_{i-1}}$ and there is a bijection between the dvr’s in $M_i$ and $M_{i-1}$. That is, a bijection between points $P_i \leftrightarrow P_{i-1}$ of the corresponding curves $Y_i \mapsto Y_{i-1}$. Finally, because $t_{P_i}^p = t_{P_{i-1}}$ it follows that $deg(P_i)=deg(P_{i-1})$, and we are done by induction on $i$.