I want to know whether number rings tend to the integers as the discriminant tends to infinity.
In detail, let $n$ be a natural number and let $C(n)$ be the set of all number fields $K$ of degree $n$. For $K\in C(n)$ let $K\_{\mathbb R}=K\otimes\_{\mathbb Q}\mathbb R$. The real vector space $K\_{\mathbb R}$ comes with a natural inner product and the ring of integers $\mathcal{O}\_K$ is a lattice in $K\_{\mathbb R}$.


Here's my question: Let $R>0$ and let $B\_R$ be the open ball of radius $R$ around zero in $K\_{\mathbb R}$. Is it true that there is $d>0$ such that for every $K\in C(n)$ of discriminant $|d\_K|\ge d$ one has
$$
B\_R\cap\mathcal{O}\_K=B\_R\cap{\mathbb Z}?
$$



Do number rings tend to Z?
The answer is yes if and only if $n$ is prime.


If $n$ is composite, fix $1<d<n$ a divisor of $d$, $L$ a number field of degree $n$. Then there are infinitely many $K$ a degree $n/d$ extension of $L$, with discriminants tending to infinity, and they satisfy $\mathcal O\_L \subset \mathcal O\_K$, so that $B\_{R \sqrt{d/n}} \cap \mathcal O\_L$ is contained in $B\_R \cap \mathcal O\_K$ for all such $K$.


If $n$ is prime, then any element in $\mathcal O\_K \setminus \mathbb Z$ generates $K$ as a field, and hence the discriminant of the field $K$ is bounded by the discriminant of the minimal polynomial of that element, which is bounded by the maximal absolute value of that element under any embedding, which is bounded by its norm in your sense. So only fields of bounded discriminant can contain an element of bounded norm.


