Irrationality measures for cubic irrationals whose conjugates lie on a curve

Abstract

We describe an unexpected connection between bounded height in families of finitely generated subgroups in tori problems and irrationality. Our method allow us to recover effective irrationality measures for some values of algebraic functions

Appendix

The aim of this appendix is to provide a complete proof of the bound (5.13) announced in the last section.

Let g be a cubic algebraic function, regular at 0 with rational Taylor coefficients. We choose a determination of the conjugates $g_1=g$, $g_2$, $g_3$ of g over $\overline{{\mathbb {Q}}}(t)$, which we assume defined (and analytic) in the closed disk $\vert t\vert \le e^{-c_0}$, for some $c_0\ge 1$.

Proposition 6.1

There exists $C=C(g)\ge c_0\ge 1$ such that the following holds. Let ${\varepsilon }_1$, ${\varepsilon }_2$, ${\varepsilon }_3$, ${\varepsilon }_4\in (0,1)$ be such that ${\varepsilon }_2+{\varepsilon }_3+{\varepsilon }_4<1/2$. Let $a\in {{\mathbb {Z}}}$, $b\in {{\mathbb {N}}}$ coprime, with

$$\begin{aligned} \log b\ge {\left\{ \begin{array}{ll} \displaystyle \log \vert a\vert +\frac{C}{{\varepsilon }_2{\varepsilon }_3},&{}(a)\\ \displaystyle \frac{\log \vert a\vert +C/{\varepsilon }_3}{1/2-{\varepsilon }_2-{\varepsilon }_3-{\varepsilon }_4},&{}(b)\\ \displaystyle \frac{{\varepsilon }_2C^2}{{\varepsilon }_1^2{\varepsilon }_4^4}.&{}(c) \end{array}\right. } \end{aligned}$$

(6.1)

Let also

$$\begin{aligned} \kappa =\frac{(1-{\varepsilon }_3)\log (b/\vert a\vert )}{(1+{\varepsilon }_2)\log b +C/{\varepsilon }_3}. \end{aligned}$$

(6.2)

Then g(a/b) is irrational with effective irrationality measure

$$\begin{aligned} \le \frac{\kappa }{\kappa -1/2}+{\varepsilon }_1 \le 2+{\varepsilon }_1+\frac{1}{{\varepsilon }_4}\left( {\varepsilon }_2+{\varepsilon }_3+\frac{\log \vert a\vert +C/{\varepsilon }_3}{\log b}\right) . \end{aligned}$$

Proof

We first make some elementary remarks on the parameters. By definition (6.2) of $\kappa$ we have

$$\begin{aligned} \begin{aligned} 1-\kappa&=\frac{({\varepsilon }_2+{\varepsilon }_3)\log b+(1-{\varepsilon }_3)\log \vert a\vert +C/{\varepsilon }_3}{(1+{\varepsilon }_2)\log b +C/{\varepsilon }_3}\\&\le {\varepsilon }_2+{\varepsilon }_3+\frac{\log \vert a\vert +C/{\varepsilon }_3}{\log b}. \end{aligned} \end{aligned}$$

(6.3)

Thus $1-\kappa >0$ and $1-\kappa \le \frac{1}{2}-{\varepsilon }_4$ by (6.1b), i. e.

$$\begin{aligned} 1/2+{\varepsilon }_4\le \kappa <1 \end{aligned}$$

(6.4)

Let now $t_0=a/b$. Then, by (6.1a), $\log b\ge \log \vert a\vert + C$ and

$$\begin{aligned} \vert t_0\vert \le \frac{1}{2}e^{-c_0}, \end{aligned}$$

(6.5)

assuming $C\ge c_0+\log 2$. Remark also that $H(t_0)=b$. We denote by $c_1$, ..., $c_9$ positive constants depending only on g. By the functorial properties of the height,

$$\begin{aligned} H(\xi )\le e^{c_1}b^d \end{aligned}$$

(6.6)

where $\xi =g(t_0)$ and where where $d\ge 1$ is the geometric degree of the algebraic function g (i. e. the degree of its polar divisor).

Let N be a positive integer and $\delta \in [1/(N+1),3)$. We consider, in the terminology of [7], a $(N,\delta )$-Padé approximant at 0 for $(1,g,g^2)$, i. e. a non-zero vector of polynomials $(Q_0,Q_1,Q_2)\in {{\mathbb {Q}}}[t]$ of degree at most N such that

$$\begin{aligned} f=Q_0+Q_1g+Q_2g^2 \end{aligned}$$

vanishes at 0 with multiplicity $M\ge [(3-\delta )(N+1)]$. By Theorem 2 of [7] we can find a such vector of polynomials satisfying moreover

$$\begin{aligned} \max _jh(Q_j)\le \frac{c_2(3-\delta )^2(N+1)}{\delta }. \end{aligned}$$

We choose

$$\begin{aligned} N:=[6d/{\varepsilon }_2]+1\quad {{\mathrm{and}}}\quad \delta :=\frac{3({\varepsilon }_3 N+1)-1}{N+1}. \end{aligned}$$

Since ${\varepsilon }_3\le 1$ we have $\delta \in [1/(N+1),3)$ as needed. Moreover

$$\begin{aligned} M\ge (3-\delta )(N+1)-1=3(1-{\varepsilon }_3)N \end{aligned}$$

(6.7)

and

$$\begin{aligned} \delta \ge \frac{2({\varepsilon }_3 N+1)-1}{N+1}=\frac{2{\varepsilon }_3 N+1}{N+1}\ge {\varepsilon }_3. \end{aligned}$$

Thus

$$\begin{aligned} \max _jh(Q_j)\le \frac{c_3N}{{\varepsilon }_3}. \end{aligned}$$

(6.8)

Remark that this implies

$$\begin{aligned} \vert f\vert _R\le e^{c_4N/{\varepsilon }_3}\;\hbox {on the disk } \vert t\vert \le R:=e^{-c_0}\le 1. \end{aligned}$$

(6.9)

Let $\theta :=f(t_0)$. We want first show that $\theta$ is non-zero. Let us assume the contrary. Let $f(t)=\sum _{k=0}^{\infty } f_kt^k$ and $g(t)^j=\sum _{k=0}^\infty a_{j,k}t^k$ ($j=1$, 2) be the expansions of f and of g, $g^2$ around $t=0$. Let also $a_{0,k}=1$ if $k=0$ and $a_{0,k}=0$ otherwise. Writing $Q_j(t)=\sum _{l=0}^N q_{j,l}t^l$ for $j=0,1,2$ we have

$$\begin{aligned} f_M=\sum _{j=0}^2\sum _{l=0}^N q_{jl}a_{j,M-l}\ne 0 \end{aligned}$$

since f vanishes at 0 with multiplicity exactly M. Let v be a place of ${{\mathbb {Q}}}$. The Local Eisenstein Theorem [7, Corollary p.161] gives $\vert a_{j,k}\vert _v\le c'(v)R_v^{2k-1}$ for some $c'(v)$, $R_v$ depending on v and on g, with moreover $c'(v)=1$ if v is finite. Thus, by (6.8)

$$\begin{aligned} h(f_M)\le c_5M+\max _jh(Q_j)\le c_5M+c_3N/{\varepsilon }_3 \end{aligned}$$

and, by Liouville’s inequality,

$$\begin{aligned} \log \vert f_M\vert \ge -c_5M-c_3N/{\varepsilon }_3. \end{aligned}$$

On the other hand, $f_Mt_0^M=-\sum _{j>M}f_jt_0^j$ since we are assuming $f(t_0)=0$. Let as before $R:=e^{-c_0}\le 1$. Using Cauchy’s estimates $\vert f_j\vert \le \vert f\vert _R R^{-j}$, we get

$$\begin{aligned} \vert f_M\vert&\le \vert t_0\vert ^{-M}\sum _{j=M+1}^\infty \vert f\vert _R (\vert t_0\vert /R)^j =\vert t_0\vert ^{-M}\vert f\vert _R \frac{(\vert t_0\vert /R)^{M+1}}{1-\vert t_0\vert /R}\\&\le 2(1/R)^{M+1}e^{c_4N/{\varepsilon }_3}\vert t_0\vert \end{aligned}$$

by (6.9) and since $\vert t_0\vert \le \vert R\vert /2$ by (6.5). Comparing the lower and the upper bound for $\vert f_M\vert$ we find

$$\begin{aligned} \log (1/\vert t_0\vert )\le \log (2/R)+(\log (1/R)+c_5)M+(c_3+c_4)\frac{N}{{\varepsilon }_3}. \end{aligned}$$

We still need an (elementary) zero’s estimate to bound M. Since the $Q_j$’s are polynomials of degree $\le N$, the polar divisor of f has degree $\le c_6N$. Thus³$M\le c_6N$. Inserting this bound in the last displayed formula we get

$$\begin{aligned} \log (1/\vert t_0\vert )\le \log (2/R)+(\log (1/R)+c_5)c_6N +(c_3+c_4)\frac{N}{{\varepsilon }_3}<\frac{c_7}{{\varepsilon }_2{\varepsilon }_3} \end{aligned}$$

since $N\le 6d/{\varepsilon }_2+1$ by our choice. This contradicts (6.1a) provided that $C\ge c_7$. The proof of $\theta \ne 0$ is concluded.

We now prove an upper bound for ${H_0}(\theta )$ and for $\Vert \theta \Vert$. Inequality (6.8) implies

$$\begin{aligned} H(Q_0(t_0):Q_1(t_0):Q_2(t_0))\le e^{c_8N/{\varepsilon }_3} b^N. \end{aligned}$$

(6.10)

Using Schwarz lemma in the disk $\vert t\vert \le R:=e^{-c_0}\le 1$ and inequalities (6.9), (6.7), we get

$$\begin{aligned} \vert \theta \vert \le \vert f\vert _R(\vert t_0\vert /R)^M \le e^{c_4N/{\varepsilon }_3}(\vert t_0\vert /R)^{3(1-{\varepsilon }_3)N}. \end{aligned}$$

(6.11)

By (3.3) of Lemma 3.4, by (6.6), (6.10), (6.11), and since $N\ge 6d{\varepsilon }_2$, we have

$$\begin{aligned} {H_0}(\theta )\le 3H(\xi )^6H(Q_0(t_0):Q_1(t_0):Q_2(t_0))\le e^{c_9N/{\varepsilon }_3}b^{(1+{\varepsilon }_2)N}, \end{aligned}$$

(6.12)

and

$$\begin{aligned} \begin{aligned} \Vert \theta \Vert&\le H(\xi )^6H(Q_0(t_0):Q_1(t_0):Q_2(t_0))\vert \theta \vert \\&\le e^{c_9N/{\varepsilon }_3}b^{(1+{\varepsilon }_2)N}\vert t_0\vert ^{3(1-{\varepsilon }_3)N}. \end{aligned} \end{aligned}$$

(6.13)

We are quite in position to apply Theorem 5.2. Assuming $C\ge c_9$ and taking into account (6.13), (6.2), (6.4), we have

$$\begin{aligned} -\frac{1}{N}\log \Vert \theta \Vert&\ge 3(1-{\varepsilon }_3)\log (b/\vert a\vert )-(1+{\varepsilon }_2)\log b-C/{\varepsilon }_3\\&=((1+{\varepsilon }_2)\log b+C/{\varepsilon }_3)(3\kappa -1)>0. \end{aligned}$$

Liouville’s inequality (Lemma 3.3) implies that $\theta$ is irrational and thus $\xi$ is irrational as well, since $\theta \in {{\mathbb {Q}}}(\xi )$. If $\xi$ is a quadratic irrational our result is trivial, since its effective irrationality measure is 2 which is $\le \frac{\kappa }{\kappa -1/2}$ by (6.4). Thus we may assume that $\xi$ is a cubic irrational. Let ${{\mathbb {K}}}={{\mathbb {Q}}}(\xi )$, denote by $\sigma _1=\mathrm{Id},\sigma _2,\sigma _3$ the immersions ${{\mathbb {K}}}\hookrightarrow {{\mathbb {C}}}$ and put $\theta _j=f_j(t_0)$.

We apply Theorem 5.2 with ${{\mathcal {C}}}\subseteq {{\mathbb {P}}}_2$ be the projective curve parametrized by $(f_1:f_2:f_3)$, and ${\varepsilon }={\varepsilon }_1$. Let $c=c({{\mathcal {C}}})$ be the constant appearing in that theorem. Set $C:=\max (c_0+\log 2,c_7,c_9,4c)$,

$$\begin{aligned} \hbar :=((1+{\varepsilon }_2)\log b+C/{\varepsilon }_3)N,\qquad \lambda :=3\kappa -1. \end{aligned}$$

Inequality (6.4) implies

$$\begin{aligned} \frac{(\lambda -1/2)^2}{8(\lambda +1)}= & {} \frac{3(\kappa -1/2)^2}{8\kappa }\\\ge & {} \frac{{\varepsilon }_4^2}{4}. \end{aligned}$$

Thus, by (6.1c),

$$\begin{aligned} c^2\min \Big (1,\frac{(\lambda -1/2)^2}{8(\lambda +1)}{\varepsilon }_1\Big )^{-2} \le \frac{C^2}{{\varepsilon }_1^2{\varepsilon }_4^4}\le \frac{\log b}{{\varepsilon }_2}\le N\log b\le \hbar . \end{aligned}$$

Assumption (5.2) is satisfied. Moreover, by (6.12) we have $h_0(\theta )\le \hbar$, and, by (6.13) and (6.2),

$$\begin{aligned} -\frac{\log \Vert \theta \Vert }{\hbar } \ge \frac{3(1-{\varepsilon }_3)\log (b/\vert a\vert )-(1+{\varepsilon }_2)\log b-C/{\varepsilon }_3}{(1+{\varepsilon }_2)\log b+C/{\varepsilon }_3}=3\kappa -1=\lambda . \end{aligned}$$

Thus (5.3) is also satisfied. By Theorem 5.2, $\xi$ has irrationality measure

$$\begin{aligned} \le \frac{\lambda +1}{\lambda -1/2}+{\varepsilon }_1 =\frac{\kappa }{\kappa -1/2}+{\varepsilon }_1. \end{aligned}$$

We finally remark that, by (6.4) and (6.3),

$$\begin{aligned} \frac{\kappa }{\kappa -1/2}+{\varepsilon }_1 =2+\frac{1-\kappa }{\kappa -1/2}+{\varepsilon }_1 \le 2+{\varepsilon }_1+{\varepsilon }_4^{-1}\left( {\varepsilon }_2+{\varepsilon }_3+\frac{\log \vert a\vert +C/{\varepsilon }_3}{\log b}\right) . \end{aligned}$$

$\square$

From proposition (6.1), we easily deduce the bound (5.13) for the irrationality measure of the values of a cubic algebraic function.

Corollary 6.2

Let g be a cubic algebraic function, regular at 0 with rational Taylor coefficients. Then there exists $C=C(g)\ge 1$ such that the following holds. Let $a\in {{\mathbb {Z}}}$, $b\in {{\mathbb {N}}}$ coprime and ${\varepsilon }\in (0,4/7)$ such that

$$\begin{aligned} \vert a\vert \le b^{1/2-{\varepsilon }} \end{aligned}$$

(6.14)

and

$$\begin{aligned} \log b\ge 2^{10}C^3/{\varepsilon }^4. \end{aligned}$$

(6.15)

Then g(a/b) is irrational with effective irrationality measure

$$\begin{aligned} \le 2+\frac{1}{{\varepsilon }}\left( 9\sqrt{\frac{C}{\log b}}+2\frac{\log \vert a\vert }{\log b}\right) . \end{aligned}$$

Proof

Let $C=C(g)$ be the constant appearing in proposition (6.1). We let for short

$$\begin{aligned} {\varepsilon }'=\sqrt{\frac{C}{\log b}}. \end{aligned}$$

By (6.15) we have

$$\begin{aligned} {\varepsilon }'\le \sqrt{\frac{C}{2^{10}C^3/{\varepsilon }^4}}=\frac{{\varepsilon }^2}{32C}\le \frac{{\varepsilon }}{8}. \end{aligned}$$

(6.16)

We apply the proposition 6.1 choosing

$$\begin{aligned} {\varepsilon }_1={\varepsilon }'/{\varepsilon },\quad {\varepsilon }_2=2{\varepsilon }',\quad {\varepsilon }_3={\varepsilon }',\quad {\varepsilon }_4={\varepsilon }/2 \end{aligned}$$

which is an admissible choice since ${\varepsilon }_1\le 1/8<1$ by (6.16) and

$$\begin{aligned} {\varepsilon }_2+{\varepsilon }_3+{\varepsilon }_4=3{\varepsilon }'+{\varepsilon }/2\le 7{\varepsilon }'/8<1/2 \end{aligned}$$

again by (6.16) and by the assumption ${\varepsilon }<4/7$.

By the choice of ${\varepsilon }'$ and by (6.14) we have

$$\begin{aligned} \log \vert a\vert +\frac{C}{{\varepsilon }_2{\varepsilon }_3}=\log \vert a\vert +\frac{C}{2{\varepsilon }'^2}=\log \vert a\vert +\frac{1}{2}\log b\le \log b. \end{aligned}$$

Moreover, by (6.14) and since $1/2-{\varepsilon }+{\varepsilon }'\le 1/2-3{\varepsilon }'-{\varepsilon }/2$ by (6.16),

$$\begin{aligned} \frac{1}{\log b}\cdot \frac{\log \vert a\vert +C/{\varepsilon }_3}{1/2-{\varepsilon }_2-{\varepsilon }_3-{\varepsilon }_4} =\frac{\frac{\log \vert a\vert }{\log b}+{\varepsilon }'}{1/2-3{\varepsilon }'-{\varepsilon }/2} \le \frac{1/2-{\varepsilon }+{\varepsilon }'}{1/2-3{\varepsilon }'-{\varepsilon }/2}\le 1. \end{aligned}$$

Finally, by the choice of ${\varepsilon }'$ and by (6.15),

$$\begin{aligned} \frac{1}{\log b}\cdot \frac{{\varepsilon }_2C^2}{{\varepsilon }_1^2{\varepsilon }_4^4} =\frac{1}{\log b}\cdot \frac{2{\varepsilon }'C^2}{({\varepsilon }'/{\varepsilon })^2({\varepsilon }/2)^4} =\frac{32C^2}{{\varepsilon }'{\varepsilon }^2\log b} =\frac{32C^{3/2}}{{\varepsilon }^2\sqrt{\log b}} \le 1. \end{aligned}$$

The last three displayed lines show that (6.1) is satisfied. Proposition 6.1 asserts that g(a/b) is irrational with effective irrationality measure

$$\begin{aligned}&\le 2+{\varepsilon }_1+\frac{1}{{\varepsilon }_4}\left( {\varepsilon }_2+{\varepsilon }_3+\frac{\log \vert a\vert +C/{\varepsilon }_3}{\log b}\right) \\&= 2+\frac{{\varepsilon }'}{{\varepsilon }}+\frac{2}{{\varepsilon }}\left( 4{\varepsilon }'+\frac{\log \vert a\vert }{\log b}\right) \\&= 2+9\frac{{\varepsilon }'}{{\varepsilon }}+\frac{2}{{\varepsilon }}\frac{\log \vert a\vert }{\log b}\\&= 2+\frac{1}{{\varepsilon }}\left( 9\sqrt{\frac{C}{\log b}}+2\frac{\log \vert a\vert }{\log b}\right) . \end{aligned}$$

$\square$