8
$\begingroup$

Again in relation with some stuff I am currently reading, the authors make use of the following "standard argument in prime number theory":

Let $\chi$ be a non-principal Dirichlet-character. Then $$\sum_{y< p \leq x} \chi(p)\overline{\chi(p)}=\frac{x}{\log(x)}+ o\left(\frac{x}{\log(x)}\right),$$ when $x\to\infty$, where $p$ runs over prime numbers. This expression very much reminds of Polya's inequality plus some use of character orthogonality, but I don't see how to "restrict" the sum to only prime numbers.

I would be thankful if someone could point to the way how this is derived. As usual, references are most welcome!

$\endgroup$
3
  • $\begingroup$ You may find answers to following question of mine interesting... $\endgroup$
    – Sasha
    Commented Sep 26, 2011 at 4:45
  • 7
    $\begingroup$ Note that $\chi(p)\overline{\chi(p)}=1$ for almost all $p$. The above formula is then a restatement of the prime number theorem. $\endgroup$
    – Kevin Ventullo
    Commented Sep 26, 2011 at 5:28
  • $\begingroup$ @Kevin: thanks for your comment. This makes clear that the above some is simply counting the prime numbers. $\endgroup$
    – M.G.
    Commented Sep 26, 2011 at 17:54

1 Answer 1

Reset to default
6
$\begingroup$

Let $m$ be the conductor of $\chi$, $\omega(m)$ its number of prime divisors and $q$ the largest prime dividing it - then for $x\ge q$ the sum is precisely $\pi(x)-\omega(m)$, because $|\chi|^2$ is always either $1$ or $0$, and is only the latter for numbers that share common divisors with $m$. The only primes that share common divisors with $m$ are those that divide it, and there are $\omega(m)$ of those (a finite amount), so all other prime numbers will contribute exactly $1$ to the overall sum. This means that $\sum\sim x/\log x$ by the prime number theorem, which gives the identity.

$\endgroup$
1
  • $\begingroup$ Thank you for your answer! When I posted my question, I was thinking of anything else except the prime number theorem. Your answer made that relation clear. $\endgroup$
    – M.G.
    Commented Sep 26, 2011 at 17:53

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .