Bounded height in pencils of subgroups of finite rank

Abstract

In a recent paper we proved some new bounded height results for equations involving varying integer exponents. Here we make a start on the problem of generalizing to rational exponents, which corresponds to the step from groups that are finitely generated to groups of finite rank. We discover two unexpected obstacles. The first is that bounded height may genuinely fail in the neighbourhood of certain exponents. The second concerns vanishing subsums, which seem to be much harder to deal with than in classical situations like S-unit equations. But for certain simple and natural equations we are able to clarify the first obstacle and eliminate the second. The proofs are partly based on our earlier work but there are also new considerations about successive minima over function fields.

Appendices

Appendix A: Beukers’ equation with a rational exponent

In this appendix we prove the following theorem which generalizes Theorem 1.2 stated in the introduction.

Theorem A.1

Let $\alpha$, $\beta$ be non-zero algebraic numbers and let $\lambda \in {\mathbb {Q}}$ be positive. Let $t_0\in {\overline{\mathbb {Q}}}\backslash \{0,1\}$. We fix determinations of $t_0^\lambda$ and of $(1-t_0)^\lambda$. Let us assume

$$\begin{aligned} \alpha t_0^\lambda +\beta (1-t_0)^\lambda =1 \end{aligned}$$

and

$$\begin{aligned} (\alpha t_0^{\lambda -1},\beta (1-t_0)^{\lambda -1})\ne (1,1). \end{aligned}$$

Then

$$\begin{aligned} h(t_0)\le 100\max (1,\lambda ^{-1})+121\lambda ^{-1}(h(\alpha )+h(\beta )). \end{aligned}$$

The special algebraic numbers $t_0$ such that $\alpha t_0^{\lambda -1}=\beta (1-t_0)^{\lambda -1}=1$ trivially satisfy the equation $\alpha t^\lambda +\beta (1-t)^\lambda =1$. They can be directly handled as we explain after the proof of the theorem. Thus Theorem 1.2 follows from Theorem A.1; see again the discussion after the proof for details.

Our theorem extends the following result of Beukers and Schlickewei which deals with even integers $\lambda$, and will be deduced from it.

Theorem A.2

([5, Lemma 2.3]) Let a, b, A, $B\in {\overline{\mathbb {Q}}}^*$ such that

$$\begin{aligned} A+B=1\quad \hbox {and}\quad aA^{2n}+bB^{2n}=1 \end{aligned}$$

for some integer $n\in {\mathbb {N}}$. Then⁴$H(A,B)\le 6\sqrt{3}\cdot 2^{1/n}H(a,b)^{1/n}$.

Proof of Theorem A.1

Let $\lambda >0$ be a rational parameter and let $t_0\in {\overline{\mathbb {Q}}}$ which satisfies $\alpha t_0^\lambda +\beta (1-t_0)^\lambda =1$ and such that

$$\begin{aligned} (\alpha t_0^{\lambda -1},\beta (1-t_0)^{\lambda -1})\ne (1,1). \end{aligned}$$

We shall prove:

$$\begin{aligned} h(t_0)\le {\left\{ \begin{array}{ll} 8\lambda ^{-1}+10\lambda ^{-1}(h(\alpha )+h(\beta )),&{}\quad \hbox { if }\lambda \le 1/6;\\[0.3cm] 100\lambda ^{-1}+121\lambda ^{-1}(h(\alpha )+h(\beta )),&{}\quad \hbox { if }1/6\le \lambda \le 1;\\[0.3cm] 100+20(h(\alpha )+h(\beta )),&{}\quad \hbox { if }1\le \lambda \le 6;\\[0.3cm] 8+9\lambda ^{-1}(h(\alpha )+h(\beta )),&{}\quad \hbox { if }\lambda \ge 6. \end{array}\right. } \end{aligned}$$

(A.1)

Theorem A.1 follows after a simple computation.

The strategy of the proof of (A.1) is the following. We distinguish three cases: for $\lambda \ge 6$ we apply Theorem A.2 choosing for n the integer part of $\lambda /2$. For $\lambda \in [1,6)$ we simply use the relations between roots and coefficients. Finally, we reduce the case $\lambda \in (0,1]$ to the previous ones by a duality argument involving exponent $1/\lambda$.

First case: $\lambda \ge 6$. Let n be the integer part of $\lambda /2$. By assumption $n\ge 3$. Let $a_0=t_0^{\lambda -2n}$, $b_0=(1-t_0)^{\lambda -2n}$ and $a=\alpha a_0$, $b=\beta b_0$. Thus $at_0^{2n}+b(1-t_0)^{2n}=1$. By Theorem A.2,

$$\begin{aligned} H(t_0,1-t_0)&\le 6\sqrt{3}\cdot 2^{1/n}H(a,b)^{1/n}\\&\le 6\sqrt{3}\cdot 2^{1/n}H(\alpha :\beta :1)^{1/n}H(a_0,b_0)^{1/n}. \end{aligned}$$

Since $\lambda -2n\ge 0$,

$$\begin{aligned} H(a_0,b_0)^{1/n}=H(t_0,1-t_0)^{(\lambda -2n)/n}. \end{aligned}$$

Since $\lambda <2(n+1)$ and $n\ge 3$ we have

$$\begin{aligned} 1-\frac{\lambda -2n}{n}>1-\frac{2}{n}\ge \frac{1}{3} \end{aligned}$$

and $2^{1/n}\le 2^{1/3}$. Thus

$$\begin{aligned} H(t_0,1-t_0)\le (6\sqrt{3}\cdot 2^{1/3})^3H(\alpha :\beta :1)^{3/n}. \end{aligned}$$

We have $3\log (6\sqrt{3}\cdot 2^{1/3})\le 8$ and $n>\lambda /2-1\ge \lambda /3$ (since $\lambda \ge 6$). Thus

$$\begin{aligned} h(t_0)\le {h^{\mathbb {A}}}(t_0,1-t_0)\le 8+9\lambda ^{-1}{h^{\mathbb {A}}}(\alpha ,\beta )\le 8+9\lambda ^{-1}(h(\alpha )+h(\beta )). \end{aligned}$$

Second case: $1\le \lambda \le 6$. Note that we can find a rational p/q with $\gcd (p,q)=1$ and $q=1$ or $q=2$ such that

$$\begin{aligned} {\varepsilon }:=\Big \vert \lambda -\frac{p}{q}\Big \vert \le \frac{2}{5q^2} \end{aligned}$$

(indeed, if the fractional part $\{\lambda \}$ is $\le 0.4$ or $\ge 0.6$, then the inequality holds with $q=1$; otherwise $\vert \{\lambda \}-1/2\vert \le 0.1$ and it holds with $q=2$). Since $\lambda \ge 1$ and ${\varepsilon }<1$ we have $p\ne 0$. Let $a_0=t_0^{\lambda -p/q}$, $b_0=(1-t_0)^{\lambda -p/q}$ and $a=\alpha a_0$, $b=\beta b_0$. Thus

$$\begin{aligned} b a t_0^{p/q}+b (1-t_0)^{p/q}=1 \end{aligned}$$

which means that there exists $s_0\in {\overline{\mathbb {Q}}}$ such that $s_0^q=t_0$ and $1-a s_0^p=b (1-s_0^q)^{p/q}$. Taking q powers, we see that $s_0$ is a root of the polynomial

$$\begin{aligned} f:=(1-a s^p)^q-b^q(1-s^q)^p\in {\overline{\mathbb {Q}}}[s]. \end{aligned}$$

Fact. $f\ne 0$.

Proof

We have

$$\begin{aligned} (1-a s^p)^q&=1-qas^p+\hbox {higher terms},\\ b^q(1-s^q)^p&=b^q-pb^qs^q+\hbox {higher terms.} \end{aligned}$$

Assume first $(p,q)\ne (1,1)$. Since $\gcd (p,q)=1$ we have $p\ne q$. We also recall that p, $q\ge 1$. Comparing the two expansions above, we deduce that $b^q=1$. Moreover, if $q<p$ then $b^q=0$, which is clearly impossible. But $a\ne 0$ since $\alpha \ne 0$, $a\ne 0$. Thus $(p,q)=(1,1)$ and $f=(1-b)+(b-a)s$ which is $=0$ if and only if $a=b=1$, that is

$$\begin{aligned} \alpha t_0^{\lambda -1}=\beta (1-t_0)^{\lambda -1}=1 \end{aligned}$$

which we have excluded in our assumption. $\square$

We now recall a classical relation between the height of the roots of a polynomial and the height of its coefficients. Let $K:={\mathbb {Q}}(a,b)$. Given a place v of K we denote by $M_v(f)$ the Mahler measure of $f^\sigma$, if v is archimedean and corresponds to the immersion $\sigma :K\hookrightarrow {\overline{\mathbb {Q}}}$. If v is non-archimedean, we let $M_v(f)$ be the maximum of the v-adic absolute values of the coefficients of f. We then define the normalized height of f as

$$\begin{aligned} \hat{h}(f)=\frac{1}{[K:{\mathbb {Q}}]}\sum _v d_v\log M_v(f) \end{aligned}$$

where v runs over the places of K and where $d_v$ denote the local degree. It is well known that $\hat{h}(f)=\sum _s h(s)$ for s running over the roots of f, counted with multiplicities. Thus ${h^{\mathbb {A}}}(s_0)\le \hat{h}(f)$ and we have to estimate this last quantity.

Given a non-archimedean place of K we have

$$\begin{aligned} M_v(f)\le \max \{\vert a\vert _v,\vert b\vert _v,1\}^q. \end{aligned}$$

Let now v be an archimedean place, corresponding to the immersion $\sigma :K\hookrightarrow {\overline{\mathbb {Q}}}$. Since the Mahler measure of a polynomial is bounded by the maximum of the absolute value of the polynomial on the disk of radius 1, we have:

$$\begin{aligned} M_v(f)&\le \max _{\vert s\vert =1}\vert (1-a^\sigma s^p)^q-(b^\sigma )^q (1-s^q)^p\vert \\&\le (1+\vert a\vert _v)^q+2^p\vert b\vert _v^q\\&\le (2^q+2^p)\max \{\vert a\vert _v,\vert b\vert _v,1\}^q. \end{aligned}$$

Thus

$$\begin{aligned} \hat{h}(f)\le q {h^{\mathbb {A}}}(a,b)+\log (2^q+2^p) \le q( {h^{\mathbb {A}}}(\alpha ,\beta )+ {h^{\mathbb {A}}}(a_0,b_0)+\log (2+2^{p/q})). \end{aligned}$$

By our choices of $a_0$, $b_0$ and ${\varepsilon }$,

$$\begin{aligned} {h^{\mathbb {A}}}(a_0,b_0)&\le h(a_0)+h(b_0)={\varepsilon }(h(t_0)+h(1-t_0))\le 2{\varepsilon }h(t_0)+{\varepsilon }\log 2\\&\le \frac{4h(t_0)}{5q^2}+\frac{2\log 2}{5q^2} \end{aligned}$$

and $p/q\le \lambda +\epsilon \le 7$. Thus, taking into account $q\le 2$,

$$\begin{aligned} h(t_0)&=q h(s_0)\le q\hat{h}(f)\\&\le q^2( {h^{\mathbb {A}}}(\alpha ,\beta )+ {h^{\mathbb {A}}}(a_0,b_0)+\log (2+2^{p/q}))\\&\le \frac{4h(t_0)}{5}+\frac{2\log 2}{5}+4(\log (2+2^7)+{h^{\mathbb {A}}}(\alpha ,\beta )) \end{aligned}$$

and so

$$\begin{aligned} h(t_0)&\le 2\log 2+5\cdot 4(\log (2+2^7)+{h^{\mathbb {A}}}(\alpha ,\beta ))\\&\le 100+20(h(\alpha )+h(\beta )). \end{aligned}$$

Third case: $\lambda \le 1$. We reduce to the previous two cases. Note that $t'_0=\alpha t_0^\lambda$ is a solution of

$$\begin{aligned} \alpha ^{-\lambda ^{-1}}(t'_0)^{\lambda ^{-1}}+\beta ^{-\lambda ^{-1}}(1-t'_0)^{\lambda ^{-1}}=1 \end{aligned}$$

and

$$\begin{aligned} h(t_0)=\lambda ^{-1}{h^{\mathbb {A}}}(\alpha ^{-1}t'_0)\le \lambda ^{-1}{h^{\mathbb {A}}}(t'_0)+\lambda ^{-1}h(\alpha ). \end{aligned}$$

Since $\lambda ^{-1}\ge 1$ the results of the previous cases apply. Suppose first $\lambda \le 1/6$. Then $\lambda ^{-1}\ge 6$ and

$$\begin{aligned} h(t_0)&\le \lambda ^{-1}h(t'_0)+\lambda ^{-1}h(\alpha )\\&\le \lambda ^{-1}\Big (8+9\lambda h(\alpha ^{-\lambda ^{-1}})+9\lambda h(\beta ^{-\lambda ^{-1}})\Big )+\lambda ^{-1}h(\alpha )\\&\le \lambda ^{-1}\big (8+9h(\alpha )+9h(\beta )\big )+\lambda ^{-1}h(\alpha )\\&\le 8\lambda ^{-1}+10\lambda ^{-1}(h(\alpha )+h(\beta )). \end{aligned}$$

Suppose now $1/6\le \lambda \le 1$. Then $1\le \lambda ^{-1}\le 6$ and

$$\begin{aligned} h(t_0)&\le \lambda ^{-1}h(t'_0)+\lambda ^{-1}h(\alpha )\\&\le \lambda ^{-1}\big (100+20h(\alpha ^{-\lambda ^{-1}})+20h(\beta ^{-\lambda ^{-1}})\big )+\lambda ^{-1}h(\alpha )\\&\le 100\lambda ^{-1}+(20\lambda ^{-1}+1)\lambda ^{-1}(h(\alpha )+h(\beta ))\\&\le 100\lambda ^{-1}+121\lambda ^{-1}(h(\alpha )+h(\beta )). \\ \end{aligned}$$

$\square$

We remark that the duality trick which we have used in the third case is only needed to get a better bound when $\lambda$ is close to zero. We could indeed modify the proof in the second case in order to get a result, depending on a fixed ${\varepsilon }\in (0,1]$, which holds for ${\varepsilon }\le \lambda \le 1$.

As promised, we now study the special algebraic numbers $t_0$ such that

$$\begin{aligned} \alpha t_0^{\lambda -1}=\beta (1-t_0)^{\lambda -1}=1. \end{aligned}$$

(A.2)

Then $t_0$ trivially satisfies the equation $\alpha t^\lambda +\beta (1-t)^\lambda =~1$.

Let us first assume $\lambda =1$. Then (A.2) is satisfied if and only if $\alpha =\beta =1$ and in this case the equation $\alpha t^\lambda +\beta (1-t)^\lambda =1$ becomes trivial. Thus we do not have bounded height.

Let now suppose $\lambda \ne 1$. Since

$$\begin{aligned} t_0=\alpha ^{-1/(\lambda -1)}\quad \hbox {and}\quad 1-t_0=\beta ^{-1/(\lambda -1)} \end{aligned}$$

(A.3)

we still have bounded height, but the bound seems to go to infinity as $\lambda \rightarrow 1$, unless $\alpha$, $\beta$ are both roots of unity. However we have the additional equation

$$\begin{aligned} \alpha ^{-1/(\lambda -1)}+\beta ^{-1/(\lambda -1)}=1 \end{aligned}$$

(A.4)

and it may be seen in several way that if $\alpha$, $\beta$ are not both roots of unity, this equation determines at most finitely many values of $\lambda \ne 1$ in terms of $\alpha$ and $\beta$.

For example, this is one of the few effective instances of the Skolem-Mahler-Lech Theorem, which follows from linear forms in logarithms combining [12, Theorem 1, p. 65] with Kummer theory. Or we could simply apply Liardet’s Theorem (as made effective for example by Bérczes, Evertse and Györy [3, Theorem 2.3]); this possibility was mentioned in [6, pp. 1121–1122].

In any case we get an effective bound $h(t_0)\le C(\alpha ,\beta )$.

Examples of (A.4) are

$$\begin{aligned} \alpha =2,\quad \beta ^q=\frac{2^q}{2^q-1},\quad \lambda =1+\frac{1}{q} \end{aligned}$$

for positive integer q with $t_0=2^{-q}$. Letting $q\rightarrow \infty$ shows that such a $C(\alpha ,\beta )$ cannot be bounded as any function of $h(\alpha )+h(\beta )$.

In the special case $\alpha =\beta =1$, the solutions $t_0$ of the Eq. (A.3) are roots of unity, and thus, by the previous remarks, any solution of $t^\lambda +(1-t)^\lambda =1$ with $\lambda >0$ and $\lambda \ne 1$ satisfies $h(t_0)\le 100\max (1,\lambda ^{-1})$. This completes the proof of Theorem 1.2 stated in the introduction.

Appendix B: Exceptional solutions of the Denz equation

We return to the system (6.1); this will finally settle the Denz equation as in Theorem 1.4.

Proposition B.1

Suppose $\delta$ is rational with $0<|\delta |\le 10^{-330}$. Then there is no $t_0 \ne 0,1,-1$ for which the determinations satisfy

$$\begin{aligned} t_0^\delta -2(1-t_0)^\delta =-1,~~~t_0^\delta +2(1+t_0)^\delta =1. \end{aligned}$$

(B.1)

We will need the following “two-circle” results, in which the adjectives refer to their mode of intersection.

Lemma B.2

For complex z, w with $|z|=|w|=1$ we have

$$\begin{aligned} |z^2-z+1| \le 3|z+w-1|. \end{aligned}$$

Proof

With $z=e^{i\theta },w=e^{i\phi }$ this is equivalent (after squaring) to $F \ge 0 ~(0 \le \theta ,\phi \le 2\pi )$ with

$$\begin{aligned} F=F(\theta ,\phi )=12-7\cos \theta -9\cos \phi +9\cos (\theta -\phi )-\cos 2\theta . \end{aligned}$$

We have

$$\begin{aligned} {\partial F \over \partial \phi }=9(\sin \phi +\sin (\theta -\phi )) \end{aligned}$$

which vanishes in the interior only for $\theta \equiv \pi +2\phi$ modulo $2\pi {\mathbb {Z}}$. At these points $F=G(\cos \phi )$ with

$$\begin{aligned} G(x)=2(2+x)(1-x)(1-2x)^2 \ge 0 \end{aligned}$$

for $-1 \le x \le 1$. And on the boundary $F(\theta ,0)=F(\theta ,2\pi )=H(\cos \theta )$ with

$$\begin{aligned} H(x)=2(1+x)(2-x) \ge 0 \end{aligned}$$

and $F(0,\phi )=4$. This completes the proof. $\square$

The constant 3 here is best possible, as the example $z=-1,w=1$ shows.

Lemma B.3

For complex z, w with $|z|=|w|=1$ we have

$$\begin{aligned} |z+1| \le 2|z+2w-1|^{1/2}. \end{aligned}$$

Proof

With $\epsilon =|z+2w-1|$ we have

$$\begin{aligned} \epsilon \ge |2w|-|z-1|=2-|z-1| \end{aligned}$$

so $|z-1| \ge 2-\epsilon$. If $\epsilon \le 2$ then drawing a picture shows that

$$\begin{aligned} |z+1| \le \sqrt{4-(2-\epsilon )^2}=\sqrt{4\epsilon -\epsilon ^2} \le 2\sqrt{\epsilon }. \end{aligned}$$

And if $\epsilon >2$ then even

$$\begin{aligned} |z+1| \le 2 < \sqrt{2}\sqrt{\epsilon } \end{aligned}$$

which completes the proof. $\square$

The exponent 1/2 comes from tangency. The multiplying constant 2 cannot be replaced by anything smaller, as the example

$$\begin{aligned} z=-1+{4\epsilon -\epsilon ^2 \over 2}+i{2-\epsilon \over 2}\sqrt{4\epsilon -\epsilon ^2},~~w=1-{\epsilon \over 2}-i{1 \over 2}\sqrt{4\epsilon -\epsilon ^2} \end{aligned}$$

with

$$\begin{aligned} |z+1| = \sqrt{4-\epsilon }|z+2w-1|^{1/2} \end{aligned}$$

and $\epsilon \rightarrow 0$ shows.

Proof of Proposition B.1

To keep better track of the determinations and their Galois conjugates, we write $\delta =r/q$ for integers $q>0,r \ne 0$ and consider the set $Y_{rq}$ in affine $\textbf{A}^4$ defined by the equations

$$\begin{aligned} y_1^q=y^r,~~y_2^{2q}=(1-y)^{2r},~~y_3^{2q}=(1+y)^{2r} \end{aligned}$$

$$\begin{aligned} y_1-2y_2=-1,~~~~y_1+2y_3=1 \end{aligned}$$

(B.2)

(and $y \ne 0,1,-1$). Note that this is a finite set in ${\overline{\mathbb {Q}}}^4$.

One checks that there is an involution on $Y_{rq}$ defined by sending $(y,y_1,y_2,y_3)$ to

$$\begin{aligned} \left( {1 \over y},{1 \over y_1},{y_2 \over y_1},-{y_3 \over y_1}\right) . \end{aligned}$$

(B.3)

Now a solution of (B.1) leads to a point

$$\begin{aligned} (y,y_1,y_2,y_3)=(t_0,t_0^\delta ,(1-t_0)^\delta ,(1+t_0)^\delta ) \end{aligned}$$

on $Y_{rq}$. So it suffices to show that $Y_{rq}$ is empty whenever

$$\begin{aligned} 0<|\delta |\le ~10^{-330}. \end{aligned}$$

(B.4)

Any element $\sigma$ of $\textrm{Gal}({\overline{\mathbb {Q}}}/{\mathbb {Q}})$ acts on points $(\eta ,\eta _1,\eta _2,\eta _3)$ of $Y_{rq}$.

Case 0. For some $\sigma$ we have

$$\begin{aligned} |\sigma (\eta )| \ge {1 \over 2},~~|1-\sigma (\eta )| \ge {1 \over 2},~~|1+\sigma (\eta )| \ge {1 \over 2},~~|\sigma (\eta )| \le 2. \end{aligned}$$

We note that for any z with $1/2 \le |z| \le 3$ we have

$$\begin{aligned} \left| |z|^\delta -1\right| =\left| \sum _{n=1}^\infty {\delta ^n(\log |z|)^n \over n!}\right| \le |\delta |\sum _{n=1}^\infty {|\log |z||^n \over n!}=|\delta |(\exp (|\log |z||)-1)\le 2|\delta |. \end{aligned}$$

Thus with $z=\sigma (\eta ),1-\sigma (\eta ),1+\sigma (\eta )$ and corresponding $s_1=\sigma (\eta _1),s_2=\sigma (\eta _2),s_3=\sigma (\eta _3)$, taking conjugates in (B.2) gives

$$\begin{aligned} s_1-2s_2=-1,~~~~s_1+2s_3=1 \end{aligned}$$

and

$$\begin{aligned} \left| |s_1|-1\right| \le 2|\delta |,~~\left| |s_2|-1\right| \le 2|\delta |,~~\left| |s_3|-1\right| \le 2|\delta | \end{aligned}$$

(note that $s_1^q=\sigma (\eta )^r$ so $|s_1|=|\sigma (\eta )|^\delta$, and similarly for $|s_2|,|s_3|$). As

$$\begin{aligned} \left| {w \over |w|}-w\right| =\left| |w|-1\right| ~~~~(w \ne 0) \end{aligned}$$

we get for $s_1'=s_1/|s_1|,s_2'=s_2/|s_2|,s_3'=s_3/|s_3|$ on the unit circle the inequalities

$$\begin{aligned} |s_1'-2s_2'+1| \le 6|\delta |,~~~~|s_1'+2s_3'-1| \le 6|\delta | \end{aligned}$$

(note that $s_1'-2s_2'+1=(s_1'-s_1)-2(s_2'-s_2)$ and so on). By Lemma B.3 with $z=-s_1',w=s_2'$ we get

$$\begin{aligned} |s_1'-1| \le 2(6|\delta |)^{1/2}; \end{aligned}$$

and then with $z=s_1',w=s_3'$

$$\begin{aligned} |s_1'+1| \le 2(6|\delta |)^{1/2}. \end{aligned}$$

These contradict each other provided $|\delta |<1/24$, certainly guaranteed by (B.4).

Thus for every $\sigma$ there are four possibilities coming from the failure of Case 0, and we consider each in turn.

Case 1a. For some $\sigma$ we have

$$\begin{aligned} |1-\sigma (\eta )| < {1 \over 2}. \end{aligned}$$

We write $\sigma (\eta )=1-u$ so that $|u| < 1/2$. Now

$$\begin{aligned} \sigma (\eta _1)^q=\sigma (\eta _1^q)=\sigma (\eta ^r)=\sigma (\eta )^r=(1-u)^r \end{aligned}$$

so that $\sigma (\eta _1)$ is a determination of $(1-u)^\delta$. There is a canonical such determination $b_1=1-\delta v$ for

$$\begin{aligned} v=u-{\delta -1 \over 2}u^2+{\delta -1 \over 2}{\delta -2 \over 3}u^3 -+\cdots , \end{aligned}$$

and because

$$\begin{aligned} \left| {\delta -k \over k+1}\right| \le 1~~~(k=1,2,\ldots ) \end{aligned}$$

we deduce

$$\begin{aligned} |b_1-1| \le |\delta |{|u| \over 1-|u|} \le |\delta |. \end{aligned}$$

(B.5)

Thus $\sigma (\eta _1)=b_1\zeta _1$ for a root of unity $\zeta _1$.

(Almost) similarly $\sigma (\eta _3)^{2q}=(2-u)^{2r}$ and so there is a root of unity $\zeta$ such that $\zeta \sigma (\eta _3)$ is a determination of $(2-u)^\delta$, and there is a canonical $b_3$ with

$$\begin{aligned} |b_3-2^\delta | \le 2^\delta |\delta | \end{aligned}$$

(B.6)

(with of course the natural determination of $2^\delta$). So $\sigma (\eta _3)=b_3\zeta _3$ for a root of unity $\zeta _3$.

The second equation in (B.2) now leads to

$$\begin{aligned} b_1\zeta _1+2b_3\zeta _3=1, \end{aligned}$$

(B.7)

so that by (B.5) and (B.6) we get first $|\zeta _1+2^{1+\delta }\zeta _3-1| \le (2^{1+\delta }+1)|\delta |$ and then

$$\begin{aligned} |\zeta _1+2\zeta _3-1| \le |2^\delta -1|+(2^{1+\delta }+1)|\delta | \le 6|\delta |. \end{aligned}$$

(B.8)

Now Lemma B.3 gives

$$\begin{aligned} |\zeta _1+1| \le 2(6|\delta |)^{1/2}. \end{aligned}$$

(B.9)

The first equation in (B.2) gives $2\sigma (\eta _2)=1+b_1\zeta _1=(b_1-1)\zeta _1+(\zeta _1+1)$ so we deduce

$$\begin{aligned} |\sigma (\eta _2)| \le {1 \over 2}(|\delta |+ 2(6|\delta |)^{1/2})\le 5|\delta |^{1/2}<1 \end{aligned}$$

(B.10)

by (B.5) and (B.9). Similarly $\sigma (\eta _1)+1=1+b_1\zeta _1$ leads to

$$\begin{aligned} |\sigma (\eta _1)+1| \le 10|\delta |^{1/2}. \end{aligned}$$

(B.11)

We may say that these $\sigma (\eta _1)$ cluster near $-1$. Also

$$\begin{aligned} |\sigma (\eta _1)-1| \le 2+10|\delta |^{1/2} <3. \end{aligned}$$

(B.12)

Now $\eta _2^{2q}=(1-\eta )^{2r}$ gives $|\sigma (\eta _2)|=|1-\sigma (\eta )|^\delta$. Thus

$$\begin{aligned} \delta ={\log |\sigma (\eta _2)| \over \log |1-\sigma (\eta )|}>0 \end{aligned}$$

(B.13)

by(B.10) and the basic assumption in this Case 1a.

We leave this case unresolved for the moment, and proceed straight to

Case 1b. For some $\sigma$ we have

$$\begin{aligned} |1+\sigma (\eta )| < {1 \over 2}. \end{aligned}$$

Here the same arguments with $-b_1\zeta _1+2b_2\zeta _2=1$ instead of (B.7) lead to

$$\begin{aligned} |\sigma (\eta _1)-1| \le 10|\delta |^{1/2} \end{aligned}$$

(B.14)

in place of (B.11), so clustering at 1, but also

$$\begin{aligned} |\sigma (\eta _1)+1| <3 \end{aligned}$$

(B.15)

as in (B.12), as well as

$$\begin{aligned} \delta > 0 \end{aligned}$$

(B.16)

as in (B.13). Again we jump to the next (somewhat critical)

Case 2a. For some $\sigma$ we have

$$\begin{aligned} |\sigma (\eta )| < {1 \over 2}. \end{aligned}$$

Now subtracting the two equations in (B.2) and using $\sigma (\eta _2)=b_2\zeta _2$ with $|b_2-1| \le |\delta |$ and $\sigma (\eta _3)=b_3\zeta _3$ with $|b_3-1| \le |\delta |$ leads to

$$\begin{aligned} |\zeta _2+\zeta _3-1| \le 2|\delta | \end{aligned}$$

as in (B.8). By Lemma B.2 we deduce $|\zeta _2^2-\zeta _2+1| \le 6|\delta |$ and then using $|\sigma (\eta _2)-\zeta _2| \le |\delta |$ also $|\sigma (\eta _2)^2-\sigma (\eta _2)+1| \le 12|\delta |$ and, multiplying by 4,

$$\begin{aligned} |\sigma (\eta _1)^2+3| \le 48|\delta |. \end{aligned}$$

(B.17)

Here we may say that these $\sigma (\eta _1)$ cluster near $\pm \sqrt{-3}$.

In particular

$$\begin{aligned} |\sigma (\eta _1)|^2 \ge 3-48|\delta | > 1 \end{aligned}$$

(B.18)

and also

$$\begin{aligned} |\sigma (\eta _1)^{-2}+3| < 4. \end{aligned}$$

(B.19)

Now we get

$$\begin{aligned} \delta ={\log |\sigma (\eta _1)| \over \log |\sigma (\eta )|}<0. \end{aligned}$$

(B.20)

Next we claim that $\eta _1$ is a unit. Write $p=-r>0$ so that $\delta =-p/q$. Pick any $\tau$ with $\tau ^p=\eta _1^{-1}$. Then $\tau ^{pq}=\eta _1^{-q}=\eta ^p$ so $\eta =\zeta \tau ^q$ for a root of unity $\zeta$. Now $2\eta _2=\eta _1+1$ so

$$\begin{aligned} 2^{2q}(1-\zeta \tau ^q)^{-2p}=2^{2q}(1-\eta )^{-2p}=(2\eta _2)^{2q}=(\eta _1+1)^{2q}=(\tau ^{-p}+1)^{2q}, \end{aligned}$$

or as a polynomial in $\tau$

$$\begin{aligned} (1-\zeta \tau ^q)^{2p}(1+\tau ^p)^{2q}=2^{2q}\tau ^{2pq}. \end{aligned}$$

Comparing highest and lowest coefficients we see that $\tau$ is a unit and therefore also $\eta _1=\tau ^{-p}$ as claimed (here it is crucial that $\delta <0$ and it is highly unlikely that it works when $\delta >0$).

Before proceeding further we jump to the final

Case 2b. For some $\sigma$ we have

$$\begin{aligned} |\sigma (\eta )| > 2. \end{aligned}$$

This is $|\sigma (\eta ^{-1})| < 1/2$, and according to (B.3) the point $(\eta ^{-1},\eta _1^{-1},\eta _2\eta _1^{-1},-\eta _3\eta _1^{-1})$ lies in $Y_{rq}$. By Case 2a we get

$$\begin{aligned} |\sigma (\eta _1^{-1})^2+3| \le 48|\delta | \end{aligned}$$

(B.21)

in place of (B.17), so clustering near $\pm 1/\sqrt{-3}$, as well as

$$\begin{aligned} |\sigma (\eta _1)|^{-2} \ge 3-48|\delta | > 1 \end{aligned}$$

as in (B.18), and also

$$\begin{aligned} |\sigma (\eta _1)^{2}+3| < 4 \end{aligned}$$

(B.22)

as in (B.19). Then

$$\begin{aligned} \delta <0 \end{aligned}$$

(B.23)

as in (B.20). Also $\eta _1^{-1}$, so $\eta _1$ too, is a unit.

Now let us sum up. As mentioned, we can ignore Case 0, so that every $\sigma$ falls into the other cases.

Suppose first that $\delta <0$.

Then Cases 1a and 1b are impossible by (B.13) and (B.16). We consider the number

$$\begin{aligned} \eta '=(\eta _1^2+3)(\eta _1^{-2}+3) \end{aligned}$$

which is an algebraic integer.

In Case 2a we use (B.17) and (B.19) to see that

$$\begin{aligned} |\sigma (\eta ')| < 192|\delta |. \end{aligned}$$

And in Case 2b we use (B.21) and (B.22) to get the same inequality.

Thus we see that $\eta '$ has norm of absolute value at most $(192|\delta |)^{d'}$ for its degree $d'$. So as soon as $|\delta | < {1 / 192}$ we deduce $\eta '=0$.

But this would imply that $\eta _1=i\sqrt{3}$ (say), so $\eta _2=(1+\eta _1)/2$ and $\eta _3=(1-\eta _1)/2$ are roots of unity so also $1-\eta ,1+\eta$ which is impossible as $\eta \ne 0$; and a similar argument works with say $\eta _1=1/(i\sqrt{3})$ using $\eta ^{-1}\ne 0$. This settles things when $\delta <0$.

It remains to deal with $\delta >0$.

Then Cases 2a and 2b are impossible by (B.20) and (B.23). But now we no longer know that $\eta _1$ is a unit (and probably it need not be), so we cannot use this method.

Instead we use Theorem A.1 on the equation $\eta _1-2\eta _2=-1$. We note that $\eta _1^q=\eta ^r$ so $\eta _1$ is a determination of $\eta ^\delta$. Also $\eta _2^{2q}=(1-\eta )^{2r}$ so there is a root of unity $\zeta$ such that $\eta _2/\zeta$ is a determination of $(1-\eta )^\delta$. Now

$$\begin{aligned} \alpha \eta _1+\beta (\eta _2/\zeta )=1 \end{aligned}$$

for $\alpha =-1,\beta =2\zeta$. It follows from Theorem A.1 with $\lambda =\delta >0$ that

$$\begin{aligned} h(\eta ) \le 100\delta ^{-1}+121\delta ^{-1}\log 2 \end{aligned}$$

(as long as we don’t have $\alpha \eta ^{\delta -1} = 1$, in which case $h(\eta )=0$ anyway). Therefore

$$\begin{aligned} h(\eta _1)=|\delta |h(\eta ) \le 100+121\log 2. \end{aligned}$$

Now we consider $\eta '=\eta _1^2-1$, so that

$$\begin{aligned} h(\eta ') \le 200+243\log 2. \end{aligned}$$

(B.24)

In Case 1a we use (B.11) and (B.12) to get

$$\begin{aligned} |\sigma (\eta ')| < 30|\delta |^{1/2}. \end{aligned}$$

And in Case 1b the same using (B.14) and (B.15).

If $\eta ' \ne 0$ the Product Formula gives $1 < (30|\delta |^{1/2})^{d'}e^{d'h(\eta ')}$ which by (B.24) is at most $(30|\delta |^{1/2}e^{200}2^{243})^{d'}$. So as soon as

$$\begin{aligned} |\delta | \le {1 \over 30^2e^{400}2^{486}}=10^{-322.972613\ldots } \end{aligned}$$

(accounting for (B.4) above) we conclude $\eta '=0$, now easily seen to be impossible. This settles things when $\delta >0$, thereby completing the proof of the Proposition. $\square$