Let $K$ be a number field, and let $$\zeta_{K}(s):= \sum_{0 \neq I \text{ ideal of }O_K} \frac{1}{N_{K/\mathbb{Q}}(I)^s} = \sum_{n \ge 1} \frac{a_n}{n^s}$$ be the Dedekind zeta function of $K$. The quantity $s_K(x):=\sum_{n \le x} a_n$ counts ideals of $O_K$ of norm up to $x$.

$\zeta_K$ is analytic in $s\ge 1$ apart from a simple pole at $s=1$, with residue given by the class number formula. The Wiener-Ikehara theorem implies: $$s_K(x) \sim c_{K} x$$ as $x\to \infty$, where $c_K$ is given by the class number formula.

Let $E_K(x):=s_K(x) - c_K x$.

When $K=\mathbb{Q}(i)$, the problem of studying $E_K$ is known as the Gauss circle problem. Proving $E_K(x) = O(x^{1/2})$ is easy, but it is believed that $E_K(x) = O_{\varepsilon}(x^{1/4 + \varepsilon})$.

  1. What is known about $E_K(x)$ for general number fields (conditionally and unconditionally)? What is the heuristic for that?

  2. What distinguishes the case $K=\mathbb{Q}(i)$ from other cases? (Apart from the elementary geometric interpretation)

up vote 25 down vote accepted

There is nothing special about the circle problem (except of course that it goes back to Gauss)! The problem for number fields has been extensively investigated, and goes back to Landau. If the number field has degree $k$, then the problem is quite analogous to the error term in the $k$-divisor problem. One expects that the remainder term should be on the scale of $O(x^{\frac{k-1}{2k}+\epsilon})$, and an $\Omega$ result of this flavor is known. Regarding $O$-results, less is known and for large $k$ the best may still be $1-2/(k+1)$ (the analog of the $1/3$ exponent in the usual circle and divisor problems). You can look at Chapter XII in Titchmarsh's book (and Heath-Brown's notes) for the $k$-divisor problem, and also look at Vaughan or Nowak for more information and other references. For example, Vaughan notes that for cubic fields the exponent $43/96$ is known in the $O$-result, and is due to Muller.

  • 2
    Excellent references. Thank you for the answer. – Ofir Gorodetsky Dec 18 '15 at 10:24

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