# Equidistribution of $\{\alpha p\}$ for $p$ in an arithmetic progression

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).

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 $$\sum_{p The function $$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 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.
• @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. Sep 2 '20 at 19:28
• 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. Sep 3 '20 at 8:24
• 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. Sep 14 '20 at 16:53