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Sep 14, 2020 at 18:31 comment added GH from MO @DanielLoughran: The previous comment (by Captain Darling) was meant for you.
Sep 14, 2020 at 16:53 comment added Dr. Pi 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 3, 2020 at 8:24 comment added Daniel Loughran 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 2, 2020 at 20:07 comment added Daniel Loughran Yes I had realised this as well. It's a nice trick.
Sep 2, 2020 at 19:28 comment added GH from MO @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, 2020 at 9:30 comment added Daniel Loughran Great answer, thanks.
Sep 2, 2020 at 9:30 vote accept Daniel Loughran
Sep 2, 2020 at 3:26 history edited GH from MO CC BY-SA 4.0
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Sep 1, 2020 at 21:39 history answered GH from MO CC BY-SA 4.0