Let $\alpha$ an irrational real number. It is well known that the set $\{ \{n \alpha \}|\,\, n \in \mathbb{N} \}$ is dense in$[0,1]$. ($\{x\}$ denotes the fractional part of $x$)

But how to prove the set $\{ \{p \alpha \}|\,\, prime\,\,p\}$ is dense in $[0,1]$? Also, is it uniformly distributed?

It is somehow a strange generalization of Dirichlet's theorem of primes on arithmetic progressions, so I tried to emulate its proof, by considering sums of the form:

$S(k)=\sum\limits_{prime\,\,p} \frac{f(k p\alpha)}{p}$

Where $f$ would be a function analogous to a character (then we can "isolate" the $p$'s such that $\{p\alpha\}$ is in some some interval $[a,b]$ by considering an adequate linear combination of the $S(k)$'s). However, I fail to bound the sums when simply choosing $f(x)=e^{2i\pi x}$, because it is not multiplicative as characters are. But, a construction of such function seems hopeless. Any ideas?

Thanks in advance.

  • 5
    $\begingroup$ encyclopediaofmath.org/index.php/Vinogradov_method $\endgroup$ – alpoge Dec 18 '13 at 22:24
  • 9
    $\begingroup$ Already answered in the statement of mathoverflow.net/questions/131018 (Vinogradov proved in 1935 that the fractional parts are not just dense but equidistributed). $\endgroup$ – Noam D. Elkies Dec 18 '13 at 22:28
  • 3
    $\begingroup$ How is this a generalization of Dirichlet's theorem? $\endgroup$ – Qiaochu Yuan Dec 18 '13 at 22:28
  • 2
    $\begingroup$ Dirichlet's theorem modulo $n$ is about proving $\{p/n\}$ is "uniformly distributed". This is like looking at the primes modulo $1/\alpha$. It is a big abstraction, but it gave me the idea of trying to solve the problem like that. $\endgroup$ – Rodrigo Dec 18 '13 at 22:36

Exponential sums over primes can be reduced to exponential sums over arithmetic progressions, as discovered by Vinogradov. The basic idea is the same as in the sieve of Eratosthenes. Modern treatments of this reduction rely on particular decompositions of the von Mangoldt function into convolutions of "simpler" functions, e.g. by the Vaughan identity or the more general Heath-Brown identity.

Along these ideas, Vinogradov proved that $\{p\alpha\}$ is equidistributed in $[0,1]$ as $p$ runs through the prime numbers (in increasing order). Of course the rate of equidistribution depends on how well $\alpha$ can be approximated by rational numbers. For a newer version of this result see Corollary 2.2 in Vaughan: On the distribution of $\alpha p$ modulo 1, Mathematika 24 (1977), 135-141.

| cite | improve this answer | |

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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