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JOURNALS // Chebyshevskii Sbornik // Archive

Chebyshevskii Sb., 2017 Volume 18, Issue 4, Pages 256–260 (Mi cheb609)

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



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