The Chebotarev density theorem is one of the most celebrated and important results in number theory. We state the following version: for a number field $K$, Galois over $\mathbb{Q}$ with Galois group (isomorphic to) $G$ and a conjugacy class $C$ of $G$, put $\pi(x; C, K)$ to be the number of prime ideals $\mathfrak{p}$ in $\mathcal{O}_K$ which is unramified in $K$ and for which the Frobenius $\sigma_{\mathfrak{p}} = C$. Then $\pi(x; C, K)$ satisfies

$$\displaystyle \pi(x; C, K) = \frac{|C|}{|G|} \text{Li}(x)(1 + o(1)).$$

The original proof given by Chebotarev is ineffective, so that the determination of the error term $o(1)$ above is not tenable. Various effective improvements have been proved over the past several decades.

Conditioned on GRH, one can obtain the expression

$$\displaystyle \pi(x; C, K) = \frac{|C|}{|G|} \text{Li}(x) + O \left(x^{1/2} [K : \mathbb{Q}] \log(|\Delta_K x|)\right),$$

where $\Delta_K$ is the discriminant of $K$. Thus, assuming GRH and running over a family $\mathcal{F}$ of fields $K$ of equal degree and the same Galois group with the property that for any integer $n$ there is a uniformly bounded number of fields $K$ in $\mathcal{F}$ with discriminant $n$, one has


$$\displaystyle \sum_{\substack{n \leq Q \\ |\Delta_K| = n \\ K \in \mathcal{F}}} \sum_C \left \lvert \pi(x; C, K) - \frac{|C|}{|G|} \text{Li}(x) \right \rvert = O \left(Q x^{1/2} \log(xQ) \right).$$

Thus, the error term remains acceptable (i.e., $o(x)$) even when $Q$ is as large as $x^{1/2} (\log x)^{-A}$, for $A > 1$.

The latter statement is similar to a famous result of Bombieri and Vinogradov. For a positive integer $q$ and an integer $a$ such that $\gcd(a,q) = 1$, put $\pi(x; a, q)$ for the number of (rational) primes $p \leq x$ such that $p \equiv a \pmod{q}$. Then Bombieri-Vinogradov theorem asserts that

$$\displaystyle \sum_{q \leq Q} \max_{y < x} \max_{\substack{1 \leq a \leq q \\ \gcd(a,q) = 1}} \left \lvert \pi(y; a, q) - \frac{1}{\phi(q)} \text{Li}(y) \right \rvert = O \left(Q x^{1/2} (\log x)^5\right),$$

so we may take $Q$ to be as large as $x^{1/2} (\log x)^{-A}$ for $A > 5$.

The Elliott-Halberstam conjecture is the statement that one can take $Q$ as large as $x^{1 - \varepsilon}$ for any $\varepsilon > 0$ in the statement of Bombieri-Vinogradov.

Can one prove (1) without assuming GRH? Further, can one expect that the 'level of distribution' in (1) can in fact be taken as large as in the Elliott-Halberstam conjecture?

The most basic example of where this question may apply is the family of (real or imaginary) quadratic fields.

  • $\begingroup$ Have you seen Pierce, Turnage-Butterbaugh, and Wood's paper on Effective Chebotarev for families of number fields arxiv.org/abs/1709.09637 ? $\endgroup$ Aug 20, 2019 at 0:04
  • $\begingroup$ They're answering the harder problem of getting Chebotarev bounds for almost all fields in a family, but their answer should have implications for the average. $\endgroup$ Aug 20, 2019 at 0:08
  • 1
    $\begingroup$ @AlisonMiller I know of the paper (and quite impressed by their results), but I certainly can't claim that I understand the paper very well. It seems that the relevant result in that paper is their Theorem 1.1, which to me seems to be about improving on the effective Chebotarev result of Lagarias and Odlyzko; and it appears that the main point is removing the influence of the possible Siegel zero. The error term they produced is not conducive to summing, and would only lead to a very poor level of distribution. $\endgroup$ Aug 20, 2019 at 11:25
  • $\begingroup$ OK, in that case you probably know the paper better than I do! Wood also has work on averaging the behaviour of a single ideal class over a family of fields: cambridge.org/core/journals/compositio-mathematica/article/… but she doesn't say much about the size of the error term, so I presume they're not very good. $\endgroup$ Aug 20, 2019 at 11:37

1 Answer 1


Let us consider the related problem of finding a suitable $\delta>0$ such that

$\displaystyle\sum_{\substack{q\leq x^{\delta-\epsilon} \\ K\cap \mathbb{Q}(e^{2\pi i/q}) = \mathbb{Q}}}\max_{(a,q)=1}\Big|\sum_{\substack{p\leq x \\ p\equiv a\pmod{q} \\ [\frac{K/\mathbb{Q}}{p}]=C}}1 - \frac{|C|}{|G|}\frac{\mathrm{Li}(x)}{\varphi(q)}\Big|\ll \frac{x}{(\log x)^A}$.

Let $H$ be an abelian subgroup of $G$ such that $H\cap C$ is nonempty, and let $E$ be the fixed field of $H$. Ram Murty and Kumar Murty proved that this holds when $\delta = \frac{1}{\max\{2,[E:\mathbb{Q}]-2\}}$. The strong Artin conjecture for the irreducible representations $\rho$ of $G$ would imply that one can replace $[E:\mathbb{Q}]$ with $\max \rho(1)$.

Here, the averaging is much simpler than you propose in your initial problem. We are simply averaging over Dirichlet characters and have an optimal version of the large sieve. We have a level of distribution $\delta = 1/2$ when $[E:\mathbb{Q}]\leq 4$, which includes the cases where $K/\mathbb{Q}$ is abelian or dihedral.

If $K/\mathbb{Q}$ is "sufficiently nonabelian", then the level of distribution is quite small. Ultimately, this is related to the lack of strong bounds for Artin $L$-functions in the critical strip; we're using the convexity bound and Phragmen-Lindelof. So the level of distribution will be proportional to the reciprocal of the largest degree of all of the $L$-functions (when viewed over $\mathbb{Q}$) under consideration. I find it hard to believe that a proper analogue of Bombieri-Vinogradov (with a level of distribution equal to $1/2$) is possible without some serious advance toward Lindelof for Artin $L$-functions twisted by Dirichlet characters. If one cannot establish a proper BV in this setting, I don't see how one could do it in the more complicated setting you propose. This also ignores many subtleties in averaging over number fields with a fixed nonabelian Galois group, many of which are catalogued very nicely in the work of Pierce, Turnage-Butterbaugh, and Wood arxiv.org/abs/1709.09637 (as Alison already mentioned).

Also, a small point: You don't necessarily have that $|\{K\in \mathcal{F}\colon |\Delta_K|\leq Q\}|\ll Q$. So the RHS of your equation (1) might be off, depending on what $\mathcal{F}$ is.

  • $\begingroup$ Yes I am aware of the last comment you made... hence why I stated as a hypothesis that in $\mathcal{F}$ there are only a uniformly bounded number of fields having a given discriminant $\endgroup$ Aug 29, 2019 at 11:40
  • $\begingroup$ @StanleyYaoXiao sorry, I missed that. $\endgroup$
    – 2734364041
    Aug 29, 2019 at 13:42

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.