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I appreciate if you could help me to find a reference (and a proof).

Combining Dirichlet theorem on primes in arithmetic progression with Chebotarev densitiy theorem, we know that given two positive integers $a$ and $m$ with $\gcd(a,m)=1$ the proportion of primes $p$ with $p\cong a\mod m$ is $1/\varphi(m)$, where $\varphi$ is the Euler totient function.

Can you give me a reference (and possibly proof) of a quantitative version of this result? That is, a result giving an EXPLICIT function $\Phi(m)$ so that

$$\frac{|\{p\mid p\textrm{ prime with }p\leq n \textrm{ and }p\equiv a\mod m\}|}{|\{p\mid p\textrm{ prime with }p\leq n \}|}\geq \frac{n}{\log(n)\Phi(m)}.$$

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  • $\begingroup$ Are you willing to accept some dependence of $n$ and $m$? That is that $m$ is rather small relative to $n$ (if not the problem does not make all that much sense even.) $\endgroup$
    – user9072
    Aug 1, 2015 at 9:34
  • $\begingroup$ Further do you really want to divide by the number of primes? As you say yourself the proportion of primes is $1/\varphi(m)$ so the ratio is "constant." $\endgroup$
    – user9072
    Aug 1, 2015 at 10:57
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    $\begingroup$ yes, I am very much happy to have $n$ large compared to $m$ $\endgroup$ Aug 1, 2015 at 11:39

1 Answer 1

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If $n$ is sufficiently large compared to $m$, then we have $$\frac{|\{p\mid p\textrm{ prime with }p\leq n \textrm{ and }p\equiv a\mod m\}|}{|\{p\mid p\textrm{ prime with }p\leq n \}|}\geq \frac{1}{2\varphi(m)}.$$ This is, of course, a consequence of (the quantitative form of) Dirichlet's theorem on arithmetic progressions, whose proof can be found in many textbooks. You can replace the constant $2$ by any number bigger than $1$ (at the cost of increasing the necessary lower bound for $n$ in terms of $m$).

If you ask how large $n$ needs to be in terms of $m$, that is a much harder question. Even the easier question how large $n$ needs to be to have a positive number on the left hand side is very hard. Linnik proved in 1944 that, for $m$ sufficently large and for a suitable constant $L>1$, it suffices to take $n>m^L$ for positivity. The current record on the value of $L$ is due to Xylouris who established $L=5$. (Under GRH we could take any $L>2$.)

A quantitative version of Linnik's theorem, applicable to your question, is Corollary 18.8 in Iwaniec-Kowalski: Analytic number theory. The corollary implies that, for $m$ sufficently large and for suitable constants $L>1$ and $c>0$, we have, for $n>m^L$, $$\frac{|\{p\mid p\textrm{ prime with }p\leq n \textrm{ and }p\equiv a\mod m\}|}{|\{p\mid p\textrm{ prime with }p\leq n \}|}\geq \frac{c}{\varphi(m)\sqrt{m}}.$$ The proof allows to specify $L$ and $c$ explicitly, but I don't think this was carried out anywhere. At any rate, Iwaniec-Kowalski express $L$ in terms of four constants, contained in Principles 1-3 on page 428, for which explicit values can be found in the literature. That is, for the above quantitative version of Linnik's theorem, an explicit value of $L$ could be given with little work.

P.S. Your question has nothing to do with Chebotarev's density theorem, except that Chebotarev's theorem implies Dirichlet's theorem (in various forms) as a special case.

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