Let $\alpha$ be an algebraic integer of degree $d$. Let $\tau(\alpha)$ be the number algebraic integers $\beta$ of degree $d$ such that $\alpha/\beta \in \mathbb{Z}$.

What is a good upper bound on $\tau(\alpha)$?

If $d = 1$, then $\alpha,\beta \in \mathbb{Z}$ and we have the divisor bound $$\tau(\alpha) \leq C_{\epsilon}|\alpha|^{\epsilon} \quad\forall \epsilon > 0.$$

Does something like this bound hold if $d > 1$?

EDIT: Let $R$ be the ring of integers of $\mathbb{Q}[\alpha]$. Any algebraic integer $\beta$ of degree $d$ such that $\alpha/\beta \in \mathbb{Z}$ must be in $R$. So one just need to count the number of $\beta$ of degree $d$ in $R$ such that $\alpha/\beta \in \mathbb{Z}$. An easy upper bound for this is the number of $\beta \in R$ such that $\alpha / \beta \in R$, i.e., the number of divisors of $\alpha \in R$. According to https://mathoverflow.net/questions/68437/the-divisor-bound-in-number-fields , the number of ideals in $R$ that divide $\alpha$ is $\leq C_{\epsilon}N(\alpha)^{\epsilon} \leq C_{\epsilon}H(\alpha)^{d\epsilon}$ (possibly with different $C_{\epsilon}$), where $N$ and $H$ denote the norm and height functions. If we impose the ADDITIONAL constraint that $\beta$ be of height $\leq M$, we are reduced to bounding the number of principal generators of height $\leq M$ of an arbitrary ideal in $R$. By an argument I don't fully understand from https://mathoverflow.net/questions/68437/the-divisor-bound-in-number-fields , the number of principal generators of an ideal in $R$ is $O(\log^r M)$, where $r$ is the rank of the unit group of $R$. So we get a bound $$ \tau_{M}(\alpha) \leq C_{\epsilon} N(\alpha)^{\epsilon} \log^r M \leq C_{\epsilon} H(\alpha)^{d\epsilon} \log^r M \quad \forall \epsilon > 0 $$ where the subscript $M$ on $\tau_{M}$ indicates we have restricted attention to $\beta$ with height $\leq M$.

The norm of $\alpha$, $N(\alpha)$, is the product of all images of $\alpha$ under embeddings from $\mathbb{Q}[\alpha]$ into $\mathbb{C}$. Can it be bounded by $|\alpha|^k$. for some $k$? If so, we get $$ \tau_{M}(\alpha) C_{\epsilon}|\alpha|^{d\epsilon} \log^r M \quad \forall \epsilon > 0 $$

Can I get from this or some other method the bound $$\tau(\alpha) \leq C_{\epsilon}|\alpha|^{\epsilon} \quad \forall \epsilon > 0?$$

Please correct me if I am wrong or sloppy about any of the above.

  • $\begingroup$ Try asking the question "What is a bound on the largest rational integer that can divide $\alpha$?" I think what you want to know will follow from the answer to that question, and this sounds much more tractable. $\endgroup$ – user14972 May 17 '14 at 18:00

The Euclidean absolute value $|\alpha|$ is a geometric quantity, whereas the norm $N(\alpha)$ is an arithmetic one. The property of $\beta$ and $\alpha/\beta$ being integral ($\alpha$ fixed) seems to be an arithmetic property purely (at least to my intuition), so would not correlate with $|\alpha|$. For illustration, it is possible to make the absolute value $|\alpha|$ tend to $0$ while keeping $|N(\alpha)|$ fixed: take $(\sqrt{2}-1)^k$ as $k\to\infty$.

Let $R$ be any number ring which has $0$ as an accumulation point. Let $g(\alpha)$ count the number of $\beta$ such that $\alpha/\beta\in\Bbb Z$. Since $g(\alpha)\ge2$ for all $\alpha\in R$, given any $\epsilon>0$ and constant $C$ it is not possible for $g(\alpha)\le C|\alpha|^\epsilon$ since the right-hand side can tend to $0$ while $g(\alpha)$ can't. The hypothesis is false.

Given an algebraic integer $\alpha$, there is a bijective correspondence

$$\{\beta~{\rm algebraic~ integer}:\alpha/\beta\in\Bbb Z\}\cong\{m\in\Bbb Z:\alpha/m~{\rm algebraic~integer}\}$$

given by $\beta\mapsto \alpha/\beta$. Therefore it suffices to count the latter.

For $\alpha/m$ to be integral, $N(\alpha/m)$ must be an integer, so $m^d\mid N(\alpha)$ is necessary (but not sufficient since $\alpha$ can be integral, $N(\alpha/m)$ an integer, but $\alpha/m$ not integral). Let $P_d(n)$ be the largest $d$th power dividing $n$. Then there is an upper bound $g(\alpha)\le 2\sigma_0(P_d(N(\alpha))^{1/d})$. In particular this gives us the bound $g(\alpha)\le2\sigma_0(N(\alpha)^{1/d})$. This is Hurkyl's answer. An explicit formula is possible too.

If the minimal polynomial of $\alpha$ over $\Bbb Q$ is $f(x)$ with $\deg f=d$, then the minimal polynomial of the ratio $\alpha/m$ is given by $m^{-d}f(mx)$. We have $\alpha/m$ integral iff the coefficients of $m^{-d}f(mx)$ are also integral. Suppose $f(x)=x^d+\sum_{k=0}^{d-1}c_kx^k$. Then $m^{-d}f(mx)=x^d+\sum_{k=0}^{d-1}m^{k-d}c_kx^k$. This means that $\alpha/m$ is integral iff $m^{d-k}\mid c_k$ for $0\le k<d$. Let $L(d)={\rm lcm}(1,2,\cdots,d)$. Then this condition can be rephrased as $m^L\mid c_k^{L/(d-k)}$ for $0\le k<d$, equivalently $m^L\mid{\rm lcm}(c_0^{L/d},c_1^{L/(d-1)},\cdots,c_{d-1}^L)$.

Every $m$ will be a divisor of the maximal such $m$. Therefore,

$$g(\alpha)=2\sigma_0\left(P_{L(d)}\left({\rm lcm}\left(c_0^{L(d)/d},c_1^{L(d)/(d-1)},\cdots,c_{d-1}^{L(d)}\right)\right)^{1/L(d)}\right).$$


The number of $\beta$ is completely determined by the largest rational integer $n$ dividing $\alpha$.

Your question should be formulated in terms of the norm, rather than the absolute value with respect to some complex embedding.

We can easily turn the question around, and instead ask "given an integer $n$, what is the smallest (measured by the norm) $\alpha$ of degree $d$ such that $n$ is the largest rational integer dividing $\alpha$?"

This last question is easy to answer: let $\beta$ be any algebraic integer of degree $d$ and norm $1$, and set $\alpha = \beta n$.

Therefore, given any bound $\tau(n) \leq f(n)$ on the number of divisors (including negative divisors) of a rational integer $n$, we have a corresponding bound $\tau(\alpha) \leq f(N(\alpha)^{1/d})$ on the number of $\beta$ such that $\alpha / \beta \in \mathbf{Z}$.

  • $\begingroup$ If you want to use height rather than norm, I imagine a similar argument can be made, but I am less comfortable working with height. $\endgroup$ – user14972 May 17 '14 at 18:46

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