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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 $$ B_R\cap\mathcal{O}_K=B_R\cap{\mathbb Z}? $$

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  • $\begingroup$ What is the inner product on $K_{\mathbb{R}}$? $\endgroup$
    – Julian Rosen
    Commented Apr 30, 2020 at 18:12
  • $\begingroup$ The complex embeddings of $K$ identify $K_{\mathbb R}$ with a real subspace of ${\mathbb C}^n$. Then you take the natural inner product on ${\mathbb C}^n$. $\endgroup$
    – user130903
    Commented Apr 30, 2020 at 18:14
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    $\begingroup$ Isn't $B_R \cap \mathbb Z= \{0\}$ for $n$ large enough because $|1| = \sqrt{n}$ by this definition? When $n$ is sufficiently large, $B_R \cap \mathcal O_K$ will also be zero by the fact that the norm of a nonzero element of $\mathcal O_K$ is a nonzero integer and the arithmetic-geometric mean inequality. $\endgroup$
    – Will Sawin
    Commented Apr 30, 2020 at 18:20
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    $\begingroup$ @Will Sawin: The number n is fixed here. $\endgroup$
    – user130903
    Commented Apr 30, 2020 at 20:07

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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.

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