2 Auxiliary results

In this section, we shall present some results that will be essential ingredients in the proof of our theorem. The first result is a generalisation of the usual Liouville criterion for transcendence [Reference Baker4, page 1], given by Adams [Reference Adams1].

Lemma 2.1. Let $P(X_1,\ldots , X_n)\in \mathbb {Z}[X_1,\ldots , X_n]$ be a nonzero polynomial with degree $d_i$ in the variable $X_i$ . Suppose that $P(\overline {\alpha }) = 0$ for some $\overline {\alpha } \in \mathbb {R}^n$ . Then, there exists a constant $c> 0$ depending on $\overline {\alpha }$ and P such that for any $\overline {\beta } = (a_1/b_1,\ldots , a_n/b_n) \in \mathbb {Q}^n,$ either

$$ \begin{align*}P(\overline{\beta}) = 0\end{align*} $$

or

$$ \begin{align*}{|\overline{\alpha} - \overline{\beta}|} \geq \frac{c}{b_1^{d_1}\cdots b_n^{d_n}}.\end{align*} $$

Here, $|{~\cdot ~}|$ is the standard Euclidean norm on $\mathbb {R}^n.$

For the sake of completeness, we shall establish the transcendence of Liouville series using the last lemma.

Proof. For all $n,$ consider the rational number $p_n/q_n=\sum _{m=1}^{n}a_mp^{-\omega _m}$ with $p_n=\sum _{m=1}^{n}a_mp^{\omega _n-\omega _m}$ and $q_n=p^{\omega _n}.$ Since $(a_n)_n$ is bounded with an infinity of entries different from zero, there exist an infinite number of distinct rationals in the sequence $(p_n/q_n)_n,$ and

(2.1) $$ \begin{align} \bigg|{\alpha-\frac{p_n}{q_n}}\bigg|\leq \frac{2M}{p^{\omega_{n+1}}}, \end{align} $$

where $M=\max \{|{a_n}|;~n\in \mathbb {N}\}$ (that is, $\alpha $ is a Liouville number). Assume, for the sake of contradiction, that $\alpha $ is algebraic. Thus, there exists a nonzero polynomial $P(X)$ with integer coefficients and degree D such that $P(\alpha )=0.$ Since $P(X)\neq 0,$ one has $P(p_n/q_n)\neq 0$ for an infinite number of $n.$ For each such n, by Lemma 2.1 and (2.1),

$$ \begin{align*} \frac{c}{p^{D\omega_n}}\leq{\left|\alpha - \frac{p_n}{q_n}\right|}\leq\frac{2M}{p^{\omega_{n+1}}}. \end{align*} $$

However, $\lim _{n\rightarrow {\infty }} {\omega _{n+1}}/{\omega _n}=\infty $ , so the last inequality is impossible.

To demonstrate our result concisely, we conclude this section with a simple definition. Given $(a_{n,1})_n,\ldots , (a_{n,k})_n$ sequences as in Theorem 1.2, one says that a k-tuple of integers $(b_1,\ldots ,b_k)$ is adherent if there exists an increasing sequence of natural numbers $(t_n)_n$ such that

$$ \begin{align*}b_1=a_{t_n,1}, \ldots, b_k=a_{t_n,k} \quad\text{for all } n\in \mathbb{N}.\end{align*} $$

Now, we are ready to deal with the proof of our result.

3 The proof of Theorem 1.2

Clearly, algebraic independence implies linear independence. So, we now focus on the other implication using an induction argument on $k.$ For $k=1,$ Proposition 2.2 implies that $\alpha _1$ is transcendental, so there is nothing to be done. Assume the result is false for $k>1,$ but true for $k-1$ , and let $P(X_1,\ldots ,X_k)$ be a nonzero polynomial with integer coefficients of minimal total degree such that $P(\alpha _1,\ldots ,\alpha _k)=0.$

Consider the Taylor expansion of P about $(\alpha _1,\ldots , \alpha _k)$ :

(3.1) $$ \begin{align} P(X_1,\ldots,X_k)=\sum_I C_I(X_1-\alpha_1)^{i_1}\cdots(X_k-\alpha_k)^{i_k}, \end{align} $$

where $I=(i_1,\ldots , i_k)$ runs over all k-tuples of nonnegative integers and the $C_I$ are real numbers which are zero for all but finitely many I. Set $C_i=C_{I_i}$ for $i=1,\ldots ,k,$ where $I_i$ has $1$ in the ith coordinate and 0 everywhere else. We have

$$ \begin{align*}C_i=\frac{\partial P}{\partial X_i}(\alpha_1, \ldots, \alpha_k).\end{align*} $$

Then, $C_i=0$ implies $\partial P/\partial X_i$ is a zero polynomial or else there would be a nonzero polynomial of smaller degree vanishing at $(\alpha _1,\ldots , \alpha _k).$ However, then $X_i$ did not occur in the polynomial P and we deduce that $\alpha _1,\ldots ,\alpha _{i-1}, \alpha _{i+1}, \cdots \alpha _k$ are algebraically dependent, violating our induction assumption.

The sequences $(a_{n,1})_n,\ldots ,(a_{n,k})_n$ take a finite number of values. Let $(b_{i,1},\ldots ,b_{i,k})$ for $i=1,\ldots , l,$ be all adherent points (see the definition at the end of Section 2) and suppose that

$$ \begin{align*}C_1b_{i,1}+\cdots+C_kb_{i,k}=0 \quad\text{for } i=1,\ldots,l.\end{align*} $$

Then, the matrix $B=(b_{i,j})_{1\leq i\leq l \atop 1\leq j \leq k}$ is not injective in $\mathbb {R}^k$ . Since the entries of B are integers, B is not injective in $\mathbb {Q}^k.$ Therefore, there exists $v=(\lambda _1,\ldots , \lambda _k)\in \mathbb {Q}^k\setminus (0,\ldots ,0)$ such that $B\cdot v=0$ and this implies that $\lambda _1a_{n,1}+\cdots +\lambda _ka_{n,k}=0$ for all n sufficiently large, so $\lambda _1\alpha _1+\cdots +\lambda _k\alpha _k$ is a rational number, which contradicts our hypothesis.

Choose now an adherent point $(b_1,\ldots ,b_k)$ with $C_1b_1+\cdots +C_kb_k\neq 0.$ For this point, fix an increasing sequence $(t_n)_n$ of natural numbers such that $b_1=a_{t_n,1}, \ldots , b_k=a_{t_n,k}$ for all $n\in \mathbb {N},$ and consider sequences $(R_{n,1})_n,\ldots ,(R_{n,k})_n$ of rational numbers given by

$$ \begin{align*}R_{n,j}=\sum_{m<t_n}a_{m,j}p^{-\omega_m} \quad\text{for } j=1,\ldots,k.\end{align*} $$

These satisfy

(3.2) $$ \begin{align} \alpha_j-R_{n,j}=\frac{b_j}{p^{\omega_{t_n}}}+O(p^{-\omega_{_{t_{n}+1}}}) \quad\text{for } j=1,\ldots,k. \end{align} $$

First, we establish that there are infinitely many n such that

$$ \begin{align*}P(R_{n,1},\ldots, R_{n,k})\neq 0.\end{align*} $$

In fact, by (3.1) and (3.2),

$$ \begin{align*} P(R_{n,1},\ldots, R_{n,k}) & = C_1(R_{n,1}-\alpha_1)+\cdots+C_k(R_{n,k}-\alpha_k)+o(p^{-\omega_{t_n}}) \\ &= p^{-\omega_{t_n}}[C_1p^{\omega_{t_n}}(R_{n,1}-\alpha_1)+\ldots+C_kp^{\omega_{t_n}}(R_{n,k}-\alpha_k)+o(1)]. \end{align*} $$

Now, by the inequalities in (3.2) and since $\lim _{n\rightarrow {\infty }}{\omega _{_{t_{n}+1}}}/{\omega _{t_n}}=\infty $ , one deduces that

$$ \begin{align*}\lim_{n\rightarrow{\infty}} p^{\omega_{t_n}}(R_{n,j}-\alpha_j)=b_j \quad\text{for } j=1,\ldots,k.\end{align*} $$

Since $C_1b_1+\cdots +C_kb_k\neq 0,$ choosing n large enough yields $P(R_{n,1}, \ldots , R_{n,k}) \neq 0$ for infinitely many n. For each such n, by Lemma 2.1,

$$ \begin{align*}|{(\alpha_1,\ldots,\alpha_k) - (R_{n,1},\ldots, R_{n,k})}| \geq \frac{c}{p^{D\omega_{_{t_{n}-1}}}} .\end{align*} $$

Here, $|{~\cdot ~}|$ is the standard Euclidean norm on $\mathbb {R}^k$ and D is the total degree of $P.$ However, (3.2) gives

$$ \begin{align*}|{(\alpha_1,\ldots,\alpha_k) - (R_{n,1},\ldots, R_{n,k})}| \leq \frac{c_0}{p^{\omega_{t_{n}}}}.\end{align*} $$

Since $\lim _{n\rightarrow {\infty }}{\omega _{t_n}}/{\omega _{_{t_n-1}}}=\infty $ , this contradicts the lower bound above.