Galois groups of Taylor polynomials: prescriptions and arithmetic families

A mix of human-written and AI-generated textHuman understanding: some partsmath.GM — General Mathematicsmath.NT — Number Theory

Contributed by Khai-Hoan Nguyen-Dang ↗

Version 2 / Oct 26, 2026 / CC BY 4.0

Abstract

Let fn(x)=∑j=0najxjf_n(x)=\sum_{j=0}^n a_jx^j be the Taylor polynomials of a power series with nonzero rational coefficients. We study the proportions of degrees for which the Galois group over $\Q$ is symmetric, alternating, or neither. We prove that the set of degrees n≥3n\geq3 with symmetric group can be chosen arbitrarily, even for an entire function with positive rational coefficients whose Taylor polynomials have distinct negative real roots. More precisely, at each degree n≥3n\geq3 outside the chosen set, fnf_n has exactly one rational root, and the group on the other roots is the full symmetric group. Thus the proportions need not converge; when they do, the proportion of other groups can be any number in [0,1][0,1]. These conclusions answer negatively two questions of Katz and Rivin. Moreover, for fixed $s\in\Q\setminus\Z$, we also give a condition on the prime divisors of rational multipliers of the coefficients s/(s+j)s/(s+j) that forces symmetric groups in density one. For the Taylor polynomials of (1−x)−s(1-x)^{-s}, we determine the groups in all sufficiently large degrees. The number of exceptions to the symmetric group up to degree XX is csX+Os(1)c_s\sqrt X+O_s(1), with an explicit constant cs≥0c_s\geq0. We also determine the groups of the central-binomial Taylor polynomials in every degree and prove that every nonconstant Catalan Taylor polynomial has symmetric group. This proves Sun's Catalan group conjecture and his irreducible-reduction criterion for the monic reciprocals of the central-binomial polynomials.

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Version history

  1. v1Submitted by Khai-Hoan Nguyen-DangInitial depositSupersededOct 24, 2026
  2. v2Submitted by Khai-Hoan Nguyen-DangSubmitted as a new versionCurrentOct 26, 2026