I have read in the paper by Einsiedler, Lindenstrauss, Michel and Venkatesh on Duke's Theorem the following bound that I don't understand:

Let $d$ be a positive non-square interger and set let $K = \mathbb{Q}(\sqrt{d})$ and denote by $\mathcal{O}_K$ the ring of integers of $K$. We furthermore assume that $d$ is a discriminant, i.e. $d = 0,1 \mod 4$.

Then consider the order $$\mathcal{O}_d := \mathbb{Z}[\tfrac{d + \sqrt{d}}{2}] \subset \mathcal{O}_K.$$ We call an ideal $\mathfrak{a} \subset \mathcal{O}_d$ a proper $\mathcal{O}_d$-ideal if it is a $\mathbb{Z}$-module of rank 2 such that $$\mathcal{O}_d = \{ \lambda \in K \,:\, \lambda \mathfrak{a} \subset \mathfrak{a} \}.$$ The bound I am interested in is now the following: The number of proper $\mathcal{O}_d$-ideals $\mathfrak{a} \subset \mathcal{O}_K$ of norm equal to $n$ is bounded by the number of divisors of $n$, in formulas:

$$|\{\text{proper } \mathcal{O}_d \text{-ideals } \mathfrak{a} \subset \mathcal{O}_K \text{ with } N(\mathfrak{a}) = n \}| \leq |\{ \text{divisors of } n \}|.$$

  • 2
    $\begingroup$ Define "proper ideal" in the question since some may think it just means "proper subset of $\mathcal O_K$ that's an ideal," but it doesn't. On p. 220 of Borevich and Shafarevich's Number Theory is a proof that in all number fields $K$ (not just real quadratic fields), the number of ideals in $\mathcal O_K$ with norm at most $a$ -- which they write as $\psi(a)$ -- is $\leq \tau(a)^{[K:\mathbf Q]}$, where $\tau(a)$ is the number of divisors of $a$. For quadratic $K$ this upper bound on $\psi(a)$ is $\tau(a)^2$, not $\tau(a)$. In the proof, they make a very weak estimate near the end (cont). $\endgroup$
    – KConrad
    Mar 7, 2019 at 12:19
  • 1
    $\begingroup$ Perhaps by looking carefully at that proof with the goal of using a better bound at the end than B&S do (at least for quadratic fields) you can get the bound you're looking for. Consider checking the desired result first for the case when $\mathcal O_d = \mathcal O_K$. $\endgroup$
    – KConrad
    Mar 7, 2019 at 12:20
  • $\begingroup$ @KConrad: Thank you for the help – I have aded a more precise definition of proper $\mathcal{O}_d$-ideals. And yes, I agree that any bound as you indicate is sufficient for the argument. $\endgroup$ Mar 8, 2019 at 13:16

1 Answer 1


Here is a solution for counting ideals of a given norm in $\mathcal O_K$ with $K$ being a general number field. By relating the count of proper ideals with a given norm in a non-maximal order to counting ideals with a given norm in the maximal order containing it, I suspect the case of non-maximal orders could be deduced from the case of maximal orders, but I haven't thought that through.

For ${\rm Re}(s) > 1$, the Dirichlet series for $\zeta_K(s)$ can be written as $\sum_{n \geq 1} a_n/n^s$ where $a_n$ is the number of ideals in $\mathcal O_K$ with norm $n$. We want to get an upper bound on $a_n$. Write the Euler product for $\zeta_K(s)$ as a product where all prime ideals dividing a given rational prime appear together: $$ \zeta_K(s) = \prod_{p} \prod_{\mathfrak p \mid p} \frac{1}{1 - 1/{\rm N}(\mathfrak p)^s}. $$ For each $\mathfrak p$ dividing $p$, ${\rm N}(\mathfrak p) \geq p$ and there are at most $[K:\mathbf Q]$ prime ideals in $\mathcal O_K$ dividing $p$, so for real $s > 1$ we have $1/(1 - 1/{\rm N}(\mathfrak p)^s) \leq 1/(1 - 1/p^s)$ and therefore $$ \zeta_K(s) \leq \prod_{p} \left(\frac{1}{1 - 1/p^s}\right)^m = \zeta(s)^m, $$ where $m = [K:\mathbf Q]$. Thus the number $a_n = |\{\mathfrak a \subset \mathcal O_K : {\rm N}(\mathfrak a) = n\}|$ is at most the coefficient of $1/n^s$ in the Dirichlet series for $\zeta(s)^m$.

Multiplying Dirichlet series uses "Dirichlet convolution" on the coefficients: $$ \sum_{n \geq 1} \frac{b_n}{n^s} \sum_{n \geq 1} \frac{c_n}{n^s} = \sum_{n \geq 1} \left(\sum_{d \mid n} b_dc_{n/d}\right)\frac{1}{n^s}. $$ In particular, taking $b_n = c_n = 1$, the sum $\sum_{d \mid n} b_dc_{n/d}$ is the number of (positive) divisors of $n$. Writing that as $\tau(n)$, we get $\zeta(s)^2 = \sum_{n \geq 1} \tau(n)/n^s$, so $a_n \leq \tau(n)$ when $[K:\mathbf Q] = 2$. That tells you what you want when $\mathcal O_d = \mathcal O_K$ for all quadratic $K$ (not just real quadratic $K$).

The Euler product of $\zeta(s)^m$ is $\prod_{p} 1/(1-1/p^s)^m$, so by the power series $1/(1-x)^m = \sum_{k \geq 0} \binom{k+m-1}{k}x^k$ for $|x| < 1$ with $x = 1/p^s$, the coefficient of $1/n^s$ in $\zeta(s)^m$ can be expressed in terms of the prime factorization $n = p_1^{k_1}\cdots p_r^{k_r}$ as $$ \binom{k_1 + m-1}{k_1} \cdots \binom{k_r+m-1}{k_r}. $$ Thus in general, $$ [K:\mathbf Q] = m \text{ and } n = p_1^{k_1}\cdots p_r^{k_r} \Longrightarrow a_n \leq \binom{k_1 + m-1}{k_1} \cdots \binom{k_r+m-1}{k_r}. $$ When $m = 2$, the upper bound is $(k_1+1)\cdots (k_r+1)$, which is $\tau(n)$. For all $m$ we have $\binom{k+m-1}{k} \leq (k+1)^{m-1}$, so $a_n \leq \tau(n)^{m-1}$ for all $n$. That is a small savings in the exponent compared to the bound $a_n \leq \tau(n)^m$ that I mentioned from Borevich and Shafarevich's Number Theory in a comment to the question, although I think the same savings could probably be obtained from B&S's own proof.

Edit: The use of Dirichlet series is not strictly necessary, as the number of ideals in $\mathcal O_K$ with norm $n = p_1^{k_1}\cdots p_r^{k_r}$ is the product of the number of ideals of norm $p_i^{k_i}$ for each $i$, and the count of ideals with prime-power norm $p^k$ can be bounded by the count that could occur if $p$ split completely (to maximize the number of possibilities). Nevertheless, the convenience of using Dirichlet series (really, an Euler product) is the same reason that it's often convenient to use generating functions to answer combinatorial questions.

  • $\begingroup$ Another variant of this proof: Without passing to Euler products, one can see that the $n$th coefficient of $\zeta(s)^m$ is the number of ways of writing $n$ as an (ordered) product of $m$ positive integers. The first $m-1$ terms in such a product determine the last, and each of those $m-1$ terms must be a divisor of $n$. So this coefficient is bounded by $\tau(n)^{m-1}$. $\endgroup$ Mar 8, 2019 at 17:08
  • $\begingroup$ I just looked at Borevich & Shafarevich's book and see that at the end of the section where they prove the bound $a_n \leq \tau(n)^{[K:\mathbf Q]}$ (using my notation above), the very first exercise on page 231 asks the reader to prove a bound that, in the case of quadratic fields, is the same as $a_n \leq \tau(n)$. $\endgroup$
    – KConrad
    Mar 13, 2019 at 8:27

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