# $\mathbb{F}_1$ 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 $\pi$).

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 ABC-conjecture

In 1985 Joseph Oesterle (left) and David Masser (right) formulated the conjecture that for three relative prime integers satisfying $A+B=C$, the product of the prime divisors of $ABC$ is rarely much smaller than $C$.

More precisely, if $A,B,C \in \mathbb{Z}$ are such that $A+B=C$ and $\gcd(A,B,C)=1$, then their conjecture states that for each $\epsilon > 0$ there is a constant $M_{\epsilon}$ such that for all triples $(A,B,C)$ satisfying the conditions we have

$$\max( |A|, |B|, |C|) \leq M_{\epsilon}\left(\underset{p | ABC}{\prod} p\right)^{1+\epsilon}$$

The ABC-conjecture has several consequences, some obvious ones such as proving Fermat’s last theorem for large exponents, some less obvious such as Falting’s theorem. However, many people consider a proof the ABC-conjecture to be beyond the range of the available methods.

Since 2006 the ABC@Home project tries to find triples $(A,B,C)$ of large ‘quality’ meaning that the ratio

$$\operatorname{q}(A,B,C) = \frac{\log(C)}{\log(\operatorname{rad}(ABC))}$$

is as large as possible. To date, the champion-triple is $2+3^{10}109 = 23^5$ (discovered by Eric Reyssat) with a quality of $1.6299$.

If we write $u = \tfrac{A}{C}$ and $v=\tfrac{B}{C}$ then the ABC-conjecture can be recast as the statement that there is a constant $M_{\epsilon}$ such that when $u,v \in \mathbb{Q}^*$ satisfy $u+v=1$ we have

$$\max\left(\operatorname{ht}(u),\operatorname{ht}(v)\right) \leq M_{\epsilon} + (1+\epsilon)\left(\sum_{p | ABC} \log(p)\right)$$

where $A$ and $B$ are the numerators of $u$ and $v$ and $C$ is their common denominator, and where the ‘height’ $\operatorname{ht}(u)$ of a rational number $u=\tfrac{A}{C}$ with $(A,B)=1$ is $\max\left(\log|A|, \log|C|\right)$.

The latter formulation can be extended to the case of function fields of curves. So, let $K \in \mathsf{1Fields}$ with a perfect field of constants $k$ and suppose $u,v \in K^*$ are non-constants satisfying $u+v=1$. We need a substitute for the notion of height.

If $L$ is the maximal separable extension of $k(u)$ in $K$, then we call the dimension $[L : k(u)]$ the separability degree of $u$ and denote it with $\deg_s(u)$. Clearly, $\deg_s(u) \leq \deg(u) = [K : k(u)]$.

If $R$ is the integral closure of $k[u]$ in $K$, then there are maximal ideals $P_i$ in the Dedekind domain $R$ such that

$$(u) = P_1^{e_1} \cdots P_r^{e_r}$$

Because the local ring in $P_i$ is a discrete valuation ring in $K$ it determines a point in the curve $C$ with $K=k(C)$ (see here) also denoted $P_i$. But then, the zero-divisor of $u$ is $\operatorname{div}_0(u) = A = \sum_i e_i [P_i]$ with degree $\deg(A) = \sum_i e_i \deg(P_i)$.

Similarly, in the integral closure $S$ of $k[\tfrac{1}{u}]$ we have a decomposition

$$(\tfrac{1}{u}) = Q_1^{f_1} \cdots Q_s^{f_s}$$

and the pole-divisor of $u$ is $\operatorname{div}_{\infty}(u) = C = \sum_j f_j [Q_j]$ with degree $\deg(C) = \sum_j f_j \deg(Q_j)$. With these conventions, the ABC-conjecture for function fields can now be formulated as the following claim:

Let $K \in \mathsf{1Fields}/k$ and $u,v \in K^*$ with $u+v=1$, then

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

where $A=\operatorname{div}_0(u)$, $B=\operatorname{div}_0(v)$, $C=\operatorname{div}_{\infty}(u)=\operatorname{div}_{\infty}(v)$ and $g_K$ is the genus of $C$. Observe that there is no $\epsilon$ in this function field ABC-conjecture.

Perhaps surprisingly, the function-field ABC-conjecture can be proved fairly easily from the Riemann-Hurwitz genus formula. Details are in the book Number Theory in Function Fields by Michael Rosen (theorem 7.17) or in an upcoming prep-notes post.

# 0-geometry: Genus

In these rough prep-notes, we are working towards the proof of the Riemann-Hurwitz genus formula. “0-geometry” means we want to use only fields and their discrete valuations so that we can port some of this later to number fields.

Before we have seen that any field $K$ of transcendence degree $1$ over $k$ with $K \cap \overline{k} = k$ is really the function field $K = k(C)$ of a smooth projective curve $C$ defined over $k$.

A geometric point $P \in C$ is a discrete valuation ring $\mathcal{O}_P$ in the extended field $K^e = K \otimes \overline{k} = \overline{k}(C)$.

Aim: To determine the genus of $C$ from $K^e$ and the discrete valuation rings $\mathcal{O}_P$.

Divisors: For $f \in K^e$ and $P \in C$ we denote the valuation of $f$ in the discrete valuation ring $\mathcal{O}_P$ by $\operatorname{ord}_P(f)$ (that is, $f = u t^{\operatorname{ord}_P(f)}$ for $t$ is a uniformizer and $u$ a unit in $\mathcal{O}_P$).

We claim that there are only finitely many $P \in C$ such that $\operatorname{ord}_P(f) \not= 0$ and that $\sum_{P \in C} \operatorname{ord}_P(f) = 0$.

We can assume that $f \notin \overline{k}$ and so the subring $\overline{k}[f] \subset K^e$ is a polynomial ring. Let $R$ be the integral closure of $\overline{k}[f]$ in $K^e$ (which is a finite field extension of $\overline{k}(f)$ say of dimension $r$). Then $R$ is a Dedekind domain, projective of rank $r$ over $\overline{k}[f]$ and there are maximal ideals $\mathcal{P}_i$ in $R$ such that

$$(f) = \mathcal{P}_1^{e_1} \cdots \mathcal{P}_s^{e_s}$$

Because the localization of $R$ at $\mathcal{P}_i$ is a discrete valuation ring with residue field $\overline{k}$, each $\mathcal{P}_i$ defines a point $P_i \in C$ and we have $\sum_i e_i = r$.

Similarly, let $S$ be the integral closure of the polynomial algebra $\overline{k}[\frac{1}{f}]$ in $K^e$, then there are maximal ideals $\mathcal{Q}_j$ (corresponding to points $Q_j \in C$) such that

$$\left(\frac{1}{f}\right) = \mathcal{Q}_1^{f_1} \cdots \mathcal{Q}_t^{f_t}$$

and $\sum_j f_j = r$. But then the divisor of $f$ satisfies the claims

$$\operatorname{div}(f) = \sum_{P \in C} \operatorname{ord}_P(f) [P] = \sum_{i=1}^s e_i [P_i] – \sum_{j=1}^t f_j [Q_j]$$

Differentials forms: Consider the $K^e$-vectorspace $\Omega_C$ spanned by all ‘differential forms’ $\mathrm{d}f$ where $f \in K^e$, subject to the usual rules:

• $\mathrm{d}(f+g)=\mathrm{d}f+\mathrm{d}g$ for all $f,g \in K^e$.
• $\mathrm{d}(fg) = f\,\mathrm{d}g + g\,\mathrm{d}f$ for all $f,g \in K^e$.
• $\mathrm{d}a = 0$ for all $a \in \overline{k}$.

We claim that $\Omega_C$ has dimension one. More precisely, if $x \in K^e$ is transcendental over $\overline{k}$ such that $K^e$ is a finite separable field extension of the subfield $\overline{k}(x)$, then $\Omega_C = K^e \mathrm{d}x$.

The proof is a computation. Let $g \in K^e$ have a minimal polynomial over $\overline{k}(x)$ of the form

$$G(Y) = Y^n + f_1 Y^{n-1} + \cdots + f_{n-1} Y + f_n$$

with all $f_i \in \overline{k}(x)$. Now consider these two polynomials in $\overline{k}(x)[Y]$ :

$G_1(Y) = n Y^{n-1} + (n-1) f_1 Y^{n-2} + \cdots + f_{n-1}$, and

$G_2(Y) = Y^n + \frac{\partial f_{1}}{\partial x} Y^{n-1} + \cdots + \frac{\partial f_{n-1}}{\partial x} Y + \frac{\partial f_n}{\partial x}$.

By the above equations among differential forms we get

$$0 = \mathrm{d} G(g) = G_2(g)\,\mathrm{d}x + G_1(g)\, \mathrm{d}g$$

Because $G_1(g) \not= 0$ by separability, it follows that $\mathrm{d}g \in K^e \mathrm{d}x$. Done!

Genus: In particular, if $t$ is a uniformizing parameter of the discrete valuation ring $\mathcal{O}_P$, then for any differential form $\omega \in \Omega_C$ there is a unique $f \in K^e$ such that $\omega = f\,\mathrm{d}t$. We define $\operatorname{ord}_P(\omega) =\operatorname{ord}_P(f)$. Clearly, this number depends only on $\omega$ (and $P$), but not on the choice of uniformizer (check!).

Slightly more involved is the claim that $\operatorname{ord}_P(\omega) \not= 0$ for finitely many $P \in C$. Here’s the idea:

Take $x \in K^e$ such that $K^e$ is a finite separable extension of $\overline{k}(x)$ of dimension $r$, write $\omega = f\,\mathrm{d}x$ and consider the corresponding cover $x\colon C \rightarrow \mathbb{P}^1$. As before, there are at most $r$ points of $C$ lying over a point $Q \in \mathbb{P}^1$.

Now, write $K^e=\overline{k}(x)(\alpha)$ and let $D \in \overline{k}(x)$ be the discriminant of the minimal polynomial of $\alpha$ over $\overline{k}(x)$. Then, away from the finite number of poles and zeroes of $D$, there are precisely $r$ points of $C$ lying over any point $Q \in \mathbb{P}^1$.

So, removing a finite number of points from $C$, in the remaining $P \in C$ we have $f(P) \not= 0,\infty$, $x(P) \not= \infty$ and $x-x(P)$ is a uniformizer of $\mathcal{O}_P$. But in such points we have $\operatorname{ord}_P(\omega) =\operatorname{ord}_P(f\,\mathrm{d}(x-x(P))) = 0$.

The number $\sum_{P \in C} \operatorname{ord}_P(\omega)$ is thus well-defined and we claim that it doesn’t depend on the choice of differential form. For, any other form can be written as $\omega’ = f \omega$ for some $f \in K^e$ and then we have

$$\sum_{P \in C} \operatorname{ord}_P(\omega’) = \sum_{P \in C} (\operatorname{ord}_P(f) +\operatorname{ord}_P(\omega))$$

and we know already that $\sum_{P \in C} \operatorname{ord}_P(f)=0$. The genus $g_C$ of the curve $C$ is then determined from that number by $2g_C – 2 = \sum_{P \in C}\operatorname{ord}_P(\omega)$.

Example: Take the projective line $\mathbb{P}^1$ corresponding to the purely transcendental field $\overline{k}(x)$ and consider $\omega = \mathrm{d}x$. In a point $\alpha \not= \infty$ we know that $x-\alpha$ is a uniformizer, so

$$\operatorname{ord}_{\alpha}(\omega) = \operatorname{ord}_{\alpha}(\mathrm{d}x) = \operatorname{ord}_{\alpha}(\mathrm{d}(x-\alpha)) = 0$$

In $\infty$ the uniformizer is $\frac{1}{x}$, whence

$$\operatorname{ord}_{\infty}(\omega) =\operatorname{ord}_{\infty}(\mathrm{d}x) =\operatorname{ord}_{\infty}\left(-x^2\,\mathrm{d}\left(\frac{1}{x}\right)\right) = -2$$

Thus, $\sum_{P \in \mathbb{P}^1} \operatorname{ord}_P(\omega) = -2$ and so the genus of the projective line $g_{\mathbb{P}^1} = 0$.