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1. CJM 2012 (vol 64 pp. 254)

Bell, Jason P.; Hare, Kevin G.
Corrigendum to ``On $\mathbb{Z}$-modules of Algebraic Integers''
We fix a mistake in the proof of Theorem 1.6 in the paper in the title.

Keywords:Pisot numbers, algebraic integers, number rings, Schmidt subspace theorem
Categories:11R04, 11R06

2. CJM 2009 (vol 61 pp. 264)

Bell, J. P.; Hare, K. G.
On $\BbZ$-Modules of Algebraic Integers
Let $q$ be an algebraic integer of degree $d \geq 2$. Consider the rank of the multiplicative subgroup of $\BbC^*$ generated by the conjugates of $q$. We say $q$ is of {\em full rank} if either the rank is $d-1$ and $q$ has norm $\pm 1$, or the rank is $d$. In this paper we study some properties of $\BbZ[q]$ where $q$ is an algebraic integer of full rank. The special cases of when $q$ is a Pisot number and when $q$ is a Pisot-cyclotomic number are also studied. There are four main results. \begin{compactenum}[\rm(1)] \item If $q$ is an algebraic integer of full rank and $n$ is a fixed positive integer, then there are only finitely many $m$ such that $\disc\left(\BbZ[q^m]\right)=\disc\left(\BbZ[q^n]\right)$. \item If $q$ and $r$ are algebraic integers of degree $d$ of full rank and $\BbZ[q^n] = \BbZ[r^n]$ for infinitely many $n$, then either $q = \omega r'$ or $q={\rm Norm}(r)^{2/d}\omega/r'$, where $r'$ is some conjugate of $r$ and $\omega$ is some root of unity. \item Let $r$ be an algebraic integer of degree at most $3$. Then there are at most $40$ Pisot numbers $q$ such that $\BbZ[q] = \BbZ[r]$. \item There are only finitely many Pisot-cyclotomic numbers of any fixed order. \end{compactenum}

Keywords:algebraic integers, Pisot numbers, full rank, discriminant
Categories:11R04, 11R06

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