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Abstract

Let $\alpha $ be an algebraic integer of degree $n\geq 3$. Assume that the extension ${\mathbb Q}(\alpha )/{\mathbb Q}$ is Galois. Let ${\mathbb Z}[{\rm conj}(\alpha )]$ be the order of ${\mathbb Q}(\alpha )$ generated by the $n$ complex conjugates of $\alpha $. Apart from the case that ${\rm Gal}({\mathbb Q}(\alpha )/{\mathbb Q})$ is the symmetric group $\mathop {\mathfrak {S}}\nolimits _n$, only for the cyclic cubic case are an explicit ${\mathbb Z}$-basis and the discriminant of ${\mathbb Z}[{\rm conj}(\alpha )]$ known. Here, we prove that there always exists a ${\mathbb Z}$-basis of ${\mathbb Z}[{\rm conj}(\alpha )]$ containing $1$ and $\alpha $. We deduce a new proof of the cyclic cubic case. We hope that this new approach could be helpful to settle the unsolved Galois quartic case. Finally, for $\alpha $ an algebraic integer of any degree $n\geq 2$, it is known that the discriminants of the orders ${\mathbb Z}[\alpha ^k]$ go to infinity as $k$ goes to infinity (without assuming that ${\mathbb Q}(\alpha )/{\mathbb Q}$ is Galois). Then, in the Galois cubic and quartic cases, we propose several conjectures related to the apparent behavior of the orders ${\mathbb Z}[{\rm conj}(\alpha ^k)]$ as $k$ goes to infinity. In particular, the orders ${\mathbb Z}[{\rm conj}(\alpha ^k)]$ seem to behave completely differently from the orders ${\mathbb Z}[\alpha ^k]$, as $k\geq 1$ goes to infinity.