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. 



Is $\{ p \alpha \}$ for prime $p$ dense in $[0,1]$?
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 For a newer version of this result see Corollary 2.2 in Vaughan: On the distribution of $\alpha p$ modulo 1, Mathematika 24 (1977), 135-141. 


