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Orders with few rational monogenizations | Jan-Hendrik Evertse
; | Date: |
4 Jan 2023 | Abstract: | To an algebraic integer $alpha$, we can attach the order $mathbb{Z} [alpha
]$, and an order of this shape is called emph{monogenic}. For a not
necessarily integral algebraic number $alpha$, say of degree $n$, we define an
order $mathbb{Z}_{alpha}$ as follows: let $mathcal{M}_{alpha}$ be the
$mathbb{Z}$-module generated by $1,alpha ,ldots ,alpha^{n-1}$; then
$mathbb{Z}_{alpha}:={xiinmathbb{Q} (alpha ):,
ximathcal{M}_{alpha}subseteqmathcal{M}_{alpha}}$ is the ring of scalars
of $mathcal{M}_{alpha}$. We call an order of the shape $mathbb{Z}_{alpha}$
rationally monogenic. In fact, rationally monogenic orders are special types of
invariant orders of binary forms, and these have been intensively studied. If
$alpha ,eta$ are two $ ext{GL}_2(mathbb{Z})$-equivalent algebraic numbers,
i.e., $eta =(aalpha +b)/(calpha +d)$ for some
$ig(egin{smallmatrix}a&b\c&dend{smallmatrix}ig)in ext{GL}_2(mathbb{Z})$,
then $mathbb{Z}_{alpha}=mathbb{Z}_{eta}$. Given an order $mathcal{O}$ of
a number field, we call any $ ext{GL}_2(mathbb{Z})$-equivalence class of
$alpha$ with $mathbb{Z}_{alpha}=mathcal{O}$ a rational monogenization of
$mathcal{O}$.
We prove the following. If $K$ is a quartic number field, then $K$ has only
finitely many orders with more than two rational monogenizations. This is best
possible. Further, if $K$ is a number field of degree $geq 5$, the Galois
group of whose normal closure is five times transitive, then $K$ has only
finitely many orders with more than one rational monogenization. The proof uses
finiteness results for unit equations, which in turn were derived from
Schmidt’s Subspace Theorem.
We generalize the above results to rationally monogenic orders over rings of
$S$-integers of number fields. | Source: | arXiv, 2301.01552 | Services: | Forum | Review | PDF | Favorites |
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