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Let $\alpha$ be irrational. A famous theorem of Vinogradov says that $\{ \alpha p\}$ is equidistributed in $[0,1]$ as $p$ runs over all primes.

Let $a,q$ be natural numbers with $\gcd(a,q) = 1$. Then is the sequence $\{ \alpha p\}$ equidistributed in $[0,1]$, as $p$ runs over primes with $p \equiv a \bmod q$?

Almost certainly this must be known. So I'm looking for a precise reference in the literature as I need it in a paper. Ideally, it would be nice to have an effective version which makes explicit the speed of convergence (via the Erdős-Turán inequality, say).

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I think the sought result follows from Vinogradov's theorem. By Weyl's criterion and the orthogonality of Dirichlet characters, the sought result can be reformulated as follows. For every nonzero integer $k$, and for every Dirichlet character $\chi$ modulo $q$, we have $$\sum_{p<x}\chi(p)e(k\alpha p)=o(\pi(x))\qquad\text{as}\qquad x\to\infty.$$ The function $n\mapsto\chi(n)$ can be written as a linear combination of additive characters $n\mapsto e((a/q)n)$ with $a\in\mathbb{Z}$, hence it suffices to show that $$\sum_{p<x}e((a/q+k\alpha)p)=o(\pi(x))\qquad\text{as}\qquad x\to\infty.$$ This in turn follows from the equidistribution of $\{(a/q+k\alpha)p\}$, because $a/q+k\alpha$ is an irrational number. QED

The error terms will be similar as in Vinogradov's theorem, but with additional constants depending on $q$. The rate of convergence will depend on how well $\alpha$ can be approximated by rational numbers. See Chapter XI and the subsequent notes in Vinogradov's "The method of trigonometrical sums in the theory of numbers". I am sure there is a modern reference, but I am no expert in this subject.

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    $\begingroup$ Great answer, thanks. $\endgroup$ Sep 2 '20 at 9:30
  • $\begingroup$ @DanielLoughran: Thank you. Actually, we don't even need to use Dirichlet characters. The condition $p\equiv a\pmod{q}$ can be detected directly by additive characters modulo $q$, so one arrives at the last display (with different $a$'s) directly. $\endgroup$
    – GH from MO
    Sep 2 '20 at 19:28
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    $\begingroup$ Yes I had realised this as well. It's a nice trick. $\endgroup$ Sep 2 '20 at 20:07
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    $\begingroup$ A more interesting problem seems to be the following. Let $K/\mathbb{Q}$ be a number field. Then is $\{\alpha p\}$ equidistributed as $p$ runs over all primes completely splti in $K$? (Or more general Chebotarev sets.) This doesn't seem to be immediately reducible to Vinogradov's result. $\endgroup$ Sep 3 '20 at 8:24
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    $\begingroup$ It is not difficult to extend Vinogradov's approach to $K/\mathbb Q$. Vaughan's identity is is a decomposition of $-\zeta'/\zeta$ and here you only have to replace zeta by the Dedekind zeta. $\endgroup$
    – Dr. Pi
    Sep 14 '20 at 16:53

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