## 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 $$n$$.
Reviewer: M.Stoll (München)

### MSC:

 11R32 Galois theory 11R09 Polynomials (irreducibility, etc.) 11B37 Recurrences

### Citations:

Zbl 0699.12018; Zbl 0622.12010
Full Text:

### References:

 [1] J. E. Cremona, On the Galois groups of the iterates ofx 2+1. Mathematika36, 259-261 (1989). · Zbl 0699.12018 [2] R. W. K. Odoni, Realising wreath products of cyclic groups as Galois groups. Mathematika35, 101-113 (1988). · Zbl 0662.12010
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