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 $\pi$).

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

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