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The divisor bound asserts that for a large (rational) integer $n \in {\bf Z}$, the number of divisors of n $n$ is at most $n^{o(1)}$ as $n \to \infty$. It is not difficult to prove this bound using the fundamental theorem of arithmetic and some elementary analysis.

My question regards what happens if ${\bf Z}$ is replaced by the ring of integers in some other number field. For sake of concreteness let us work with the simple extension ${\bf Z}[\alpha]$, where $\alpha$ is some fixed algebraic integer. Of course, one may now have infinitely many units in this ring, but if we restrict the height then it appears that we have a meaningful question, namely:

Question: Let $n \in {\bf Z}[\alpha]$ be of height $O(H)$ (by which I mean that $n$ is a polynomial in $\alpha$ with rational integer coefficients of size $O(H)$ and degree $O(1)$). Is it true that the number of elements of ${\bf Z}[\alpha]$ of height $O(H)$ that divide $n$ is at most $H^{o(1)}$?

Here $o(1)$ denotes a quantity that goes to zero as $H \to \infty$, holding $\alpha$ fixed. (Actually, for my applications I would like $\alpha$ to not be fixed, but to have a minimal polynomial of bounded degree and coefficients of polynomial size in $H$, but for simplicity let me stick to the fixed $\alpha$ question first.)

It is tempting to take norms and apply the divisor bound to the norm, but then I end up needing to bound the number of elements in ${\bf Z}[\alpha]$ with a given norm and of controlled height, and I don't know how to do that except for quadratic extensions. A related problem comes up if one tries to exploit unique factorization of ideals to answer this problem. (On the other hand, it appears to me from the Dirichlet unit theorem that the number of units of height $O(H)$ is at most polylogarithmic in H, so the unit problem at least should go away.)

2 added 191 characters in body

The divisor bound asserts that for a large (rational) integer $n \in {\bf Z}$, the number of divisors of n is at most $n^{o(1)}$ as $n \to \infty$. It is not difficult to prove this bound using the fundamental theorem of arithmetic and some elementary analysis.

My question regards what happens if ${\bf Z}$ is replaced by the ring of integers in some other number field. For sake of concreteness let us work with the simple extension ${\bf Z}[\alpha]$, where $\alpha$ is some fixed algebraic integer. Of course, one may now have infinitely many units in this ring, but if we restrict the height then it appears that we have a meaningful question, namely:

Question: Let $n \in {\bf Z}[\alpha]$ be of height $O(H)$ (by which I mean that $n$ is a polynomial in $\alpha$ with rational integer coefficients of size $O(H)$ and degree $O(1)$). Is it true that the number of elements of ${\bf Z}[\alpha]$ of height $O(H)$ that divide $n$ is at most $H^{o(1)}$?

Here $o(1)$ denotes a quantity that goes to zero as $H \to \infty$, holding $\alpha$ fixed. (Actually, for my applications I would like $\alpha$ to not be fixed, but to have a minimal polynomial of bounded degree and coefficients of polynomial size in $H$, but for simplicity let me stick to the fixed $\alpha$ question first.)

It is tempting to take norms and apply the divisor bound to the norm, but then I end up needing to bound the number of elements in ${\bf Z}[\alpha]$ with a given norm and of controlled height, and I don't know how to do that except for quadratic extensions. A related problem comes up if one tries to exploit unique factorization of ideals to answer this problem. (On the other hand, it appears to me from the Dirichlet unit theorem that the number of units of height $O(H)$ is at most polylogarithmic in H, so the unit problem at least should go away.)

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# The divisor bound in number fields

The divisor bound asserts that for a large (rational) integer $n \in {\bf Z}$, the number of divisors of n is at most $n^{o(1)}$ as $n \to \infty$. It is not difficult to prove this bound using the fundamental theorem of arithmetic and some elementary analysis.

My question regards what happens if ${\bf Z}$ is replaced by the ring of integers in some other number field. For sake of concreteness let us work with the simple extension ${\bf Z}[\alpha]$, where $\alpha$ is some fixed algebraic integer. Of course, one may now have infinitely many units in this ring, but if we restrict the height then it appears that we have a meaningful question, namely:

Question: Let $n \in {\bf Z}[\alpha]$ be of height $O(H)$ (by which I mean that $n$ is a polynomial in $\alpha$ with rational integer coefficients of size $O(H)$ and degree $O(1)$). Is it true that the number of elements of ${\bf Z}[\alpha]$ of height $O(H)$ that divide $n$ is at most $H^{o(1)}$?

Here $o(1)$ denotes a quantity that goes to zero as $H \to \infty$, holding $\alpha$ fixed. (Actually, for my applications I would like $\alpha$ to not be fixed, but to have a minimal polynomial of bounded degree and coefficients of polynomial size in $H$, but for simplicity let me stick to the fixed $\alpha$ question first.)

It is tempting to take norms and apply the divisor bound to the norm, but then I end up needing to bound the number of elements in ${\bf Z}[\alpha]$ with a given norm and of controlled height, and I don't know how to do that except for quadratic extensions. A related problem comes up if one tries to exploit unique factorization of ideals to answer this problem.