2

I'm am studying number theory and I came across this theorem by Dirichlet but the proof is not given and I do not know how to prove it

Can anyone help me?

$Theorem$ $of$ $Primes$ $in$ $Arithmetic$ $Progression:$ If $a$ and $b$ are integers, with $(a,b)=1$ so the arithmetic progression $an+b$, $n=1,2,3,...$ has infinite primes.

Arpan1729
  • 3,414
Gu Chiquetto
  • 157
  • 6
    many books have correct proofs. Not easy, and not something that would fit well in an answer box on this site. – Will Jagy Jul 03 '17 at 18:46
  • 1
    This is not an easy result! In certain cases, such as $4n+3$, there are ways to get at it but the general theorem requires a lot of machinery. – lulu Jul 03 '17 at 18:48
  • 1
    A classical reference is this book by Tom Apostol. There are very many references at MSE, e.g. here, here etc. Just look at the links on the right margin. – Dietrich Burde Jul 03 '17 at 18:48
  • 1
    This is mentioned at one of Dietrich's links in a comment, several special cases that do not require any analytic approach: http://www.math.uconn.edu/~kconrad/blurbs/gradnumthy/dirichleteuclid.pdf – Will Jagy Jul 03 '17 at 19:43
  • In general, the idea is that $F(s) =\displaystyle \sum_{p^k \text{ prime power }, p^k \equiv b \bmod a} p^{-sk}$ has the nice property that $F(s) =\frac{1}{\phi(a)} \sum_{\chi \bmod a} \overline{\chi(b)} \log L(s,\chi)$ where $L(s,\chi) = \sum_{n=1}^\infty \chi(n) n^{-s}$ are the Dirichlet L-functions. And hence to prove Dirichlet's theorem it is enough to prove $F(s)$ is not analytic at $s=1$. – reuns Jul 03 '17 at 20:16
  • Definitely agree with @WillJagy and @lulu; however, the traditional analytic approach is definitely not the only way to arrive at the elementary conclusion that in general there are infinitely many primes in any arithmetic progression if the residue and the modulus are coprime. – user56983 Jul 03 '17 at 23:06

0 Answers0