Let $a,q,N$ be integers such that $N/2 \leq q \leq N$ and $a/q \notin \mathbb{Z}$.


Is the following estimate true, and, if so, how can it be proved?
\[\left|\sum\_{1 \leq p \leq N} \exp(2\pi i p a/q) \right|\leq |a|^{o(1)} N^{o(1)},\]
where $f(x)=x^{o(1)}$ means $\lim\_{x \rightarrow \infty}\frac{\log f(x)}{\log x}=0$, or equivalently $f(x) = O\_{\epsilon} (x^{\epsilon})$ for all $\epsilon > 0$. Can it be obtained, for example, from Vinogradov-type estimates?


It would even be useful to know whether the estimate holds in the following average sense:
$$\sum\_{\substack{N/2 \leq q \leq N \\ a/q \notin \mathbb{Z}}} \left|\sum\_{1 \leq p \leq N} \exp(2\pi i p a/q) \right|\leq |a|^{o(1)} N^{1+o(1)}.$$


$N$ may be assumed to be as large as required.



An Exponential Sum Restricted to Primes
No, this is definitely not true. First of all, there's no way we should expect better than square-root cancellation in an exponential sum without an incredible amount of structure, which the primes do not possess. Second, it's known to be false for certain rational numbers, and conditionally known to be false for all rational numbers, because then the exponential sum essentially counts primes in arithmetic progressions.


For example, take $\frac aq=\frac14$; then
$$
\sum\_{p\le x} e^{2\pi i p a/q} = \sum\_{p\le x} e^{\pi i p/2} = -1 + i \big(\pi(x;4,1) - \pi(x;4,3) \big),
$$
where $\pi(x;4,b)$ denotes the number of primes up to $x$ that are congruent to $b$ (mod $4$). Littlewood proved that the quantity in parentheses is $\Omega(\sqrt x \log\log\log x/\log x)$, which is far larger than $x^{o(1)}$.


