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1. CJM 2009 (vol 61 pp. 264)
| 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 |

