Someone could tell me if there is a proof of the Dirichlet's theorem on arithmetic progressions stated below using only the Riemann zeta function $\zeta(s)=\sum_{n=1}^\infty 1/n^s,\;\mbox{Re}(s)>1$? Someone reference?

Dirichlet's Theorem For any two positive coprime integers $a$ and $d$, there are infinitely many primes of the form $a + nd$, where $n \in \mathbb{N}$.

  • $\begingroup$ The comments here raise questions similar to those formulated (and left unanswered) there. $\endgroup$ – Did Jul 29 '13 at 16:03

Yes, there is, but it is quite technical (at least the one I know), and at the same time sooooo beautiful! I suggest you read the first 30-40 pages of Davenport's Multiplicative Number Theory. It illustrates the ideas of Dirichlet and if you go through the details, not only you will enjoy the proof, but learn a lot of ideas that emerged from number theory.

Hope that helps,

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    $\begingroup$ Which Theorem in Davenport proves Dirichlet's Theorem on primes in progressions using ONLY the Riemann zeta function? Specific citation please, because I'm unaware of such a proof. $\endgroup$ – stopple Jun 25 '13 at 15:16
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    $\begingroup$ I also am unaware and that'd be very nice for me. I know two different proofs, one using arithmetic functions galore and the other one, much more beautiful in my eyes, with lots of algebra, series, Dirichlet characters and stuff. One which "only" uses the zeta function would be really worth reading. $\endgroup$ – DonAntonio Jun 25 '13 at 17:15
  • $\begingroup$ I am pretty sure Patrick meant the usual proof with $L$-functions, but just didn't read the original question carefully. $\endgroup$ – KCd Jun 25 '13 at 22:04
  • $\begingroup$ I'll admit I didn't see the word 'only'. Why would you be interested in not using $L$-functions? Is it just because you've heard such a proof existed? I edited the question to make it more clear. $\endgroup$ – Patrick Da Silva Jun 26 '13 at 8:59
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    $\begingroup$ It looks like OP is not only forbidding $L$-functions, but the bit about the real part exceeding 1 suggests that even analytic continuation is out of the question. $\endgroup$ – Gerry Myerson Jun 26 '13 at 9:06

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