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Galois group over \(\mathbb{Q}\) of some iterated polynomials. (English) Zbl 0758.11045
Let \(a\) be an integer such that \(-a\) is not a square in \(\mathbb{Q}\), \(f:=X^ 2+a\in\mathbb{Z}[X]\), and denote the iterates of \(f\) by \(f_ 0:=X\) and \(f_{n+1}:=f(f_ n)=f_ n^ 2+a\) for all \(n\geq 0\). Let \(c_ 1:=-a\) and \(c_{n+1}:=f(c_ n)=c_ n^ 2+a\) for \(n\geq 1\). There is an integer sequence \((b_ n)_{n\geq 1}\) with the \(b_ n\) coprime in pairs such that for all \(n\geq 1\), \(c_ n=\prod_{d\mid n} b_ d\). Let \(K_ n\) be the splitting field of \(f_ n\) over \(\mathbb{Q}\) and denote by \(\Omega_ n:=\text{Gal}(K_ n/\mathbb{Q})=\text{Gal}(f_ n/\mathbb{Q})\) its Galois group over the rational numbers. Let \([C_ 2]^ n\) denote the \(n\)-fold wreath product of the 2-element group. Then it is known that \(\Omega_ n\) always injects into \([C_ 2]^ n\). The following equivalence is shown in the paper: \(\Omega_ n\cong [C_ 2]\) if and only if \(c_ 1,c_ 2,\dots,c_ n\) are 2-independent in \(\mathbb{Q}\).
Here, “2-independent” means that no nonempty product of some of the \(c_ n\) is a square. This gives the following sufficient condition for \(\Omega_ n\cong[C_ 2]^ n\) to hold: \(| b_ m|\) is not a square for \(2\leq m\leq n\).
This condition is then verified for all \(n\) in the following cases: (\(a>0\) and \(a\equiv 1\) or \(2\bmod 4\)) or (\(a<0\) and \(a\equiv 0\bmod 4\) and \(-a\) not a square). So, for these \(a\), one always has \(\text{Gal}(f_ n/\mathbb{Q})\cong[C_ 2]^ n\) for all \(n\).

11R32 Galois theory
11R09 Polynomials (irreducibility, etc.)
11B37 Recurrences
Full Text: DOI
[1] J. E. Cremona, On the Galois groups of the iterates ofx 2+1. Mathematika36, 259-261 (1989). · Zbl 0699.12018 · doi:10.1112/S0025579300013127
[2] R. W. K. Odoni, Realising wreath products of cyclic groups as Galois groups. Mathematika35, 101-113 (1988). · Zbl 0662.12010 · doi:10.1112/S002557930000632X
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