Estimates of polynomials in a liouvillean polyadic integer
E. S. Krupitsyn Moscow State Pedagogical University
Abstract:
Let
$$
\alpha=\sum\limits_{n=0}^\infty a_kn_k!, \quad a_k\in\mathbb{Z}, \quad 0\leqslant a_k\leqslant n_k,
$$
with a rapidly growing sequence
$n_k$ of positive integers. This series converges in all
$p$-adic fields
$\mathbb{Q}_p$ so it is a polyadic number.
The ring of polyadic integers is a direct product of the rings
$\mathbb{Z}_p$ of
$p$-adic integers over all prime numbers
$p$.
So
$\alpha$ can be considered as the vector $\left(\alpha^{(1)}, \ldots, \alpha^{(n)}, \ldots\right)$ with coordinates equal to the sums
$\alpha^{(n)}$ of the series
$\alpha$ in the field
$\mathbb{Q}_{p_n}$ for the
$n$-th prime
$p_n$.
For any nonzero polynomial
$P(x)$ with integer coefficients one has
$$
P(\alpha)=\left(P\left(\alpha^{(1)}\right), \ldots, P\left(\alpha^{(n)}\right), \ldots \right).
$$
The polyadic integer
$\alpha$ is called transcendental, if for any nonzero polynomial
$P(x)$ with rational integer coefficients there exist a prime
$p^{(n)}$ with
$P\left(\alpha^{(n)}\right)\neq 0$ in
$p_n$.
The polyadic integer is infinitely transcendental if there exist infinitely many primes
$p_n$ such that
$P\left(\alpha^{(n)}\right)\neq 0$ in
$\mathbb{Q}_{p_n}$ and it is called globally transcendental, if
$P\left(\alpha^{(n)}\right)\neq 0$ for any
$n$.
The paper presents estimates from below of
$\left|P\left(\alpha^{(n)}\right)\right|_{p_n}$ in any
$\mathbb{Q}_{p_n}$. As a corollary we get the global transcendence of
$\alpha$.
Keywords:
polyadic integer, estimates of polynomials.
UDC:
517 Received: 14.09.2017
Accepted: 15.12.2017
DOI:
10.22405/2226-8383-2017-18-4-255-259