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The theorem can be found on Wikipedia.

In the subsection "Proof" Wikipedia says that there is a proof for the case $a=1$ which uses no calculus, instead splitting behavior of primes in cyclotomic extensions. Could you help me proving this?

Assumption. For every natural number $n$ there are infinitely many prime numbers $p\equiv 1 \pmod n$.

Proof: I assume there are only finitely many $p_1,...,p_i$, and let $P=p_1\cdot...\cdot p_i$. The cyclotomic polynomial

$$\phi_n(x):=\prod_{\gcd(k,n)=1,\ 1\le k<n}(x-\zeta_n^k)\;,\;\;\zeta_n=\cos\frac{2\pi}{n}+i\sin\frac{2\pi}{n}$$

The hint in Neukirch books states that not all numbers $\phi_n(xnP)$ for $x\in\mathbb Z$ can equal $1$. Why? Now let $p \mid \phi_n(xnP)$ for suitable $x$. How can a contradiction be followed from this?

DonAntonio
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Babla
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    You mean, for a fixed $x \in \mathbb{Z}$ and varying $n \in \mathbb{N}$, not all $\phi_n(x n P)$ can equal to 1? Otherwise, if $n$ is fixed and $x$ varies, if for all $x \in \mathbb{Z}$, $\phi_n(x n P) = 1$, then $\phi_n - 1$ has infinitely many zeros, hence it's a zero polynomial, a contradiction. – xyzzyz Apr 26 '13 at 10:53
  • Thanks, yes thats what I meant. Do you also have an idea for a contradiction for the last part, i.e why there have to be infinitely many prime numbers? – Babla Apr 26 '13 at 10:55
  • The contradiction is not so different from Euclid's proof for primes: any prime $p$ which divides $\Phi_n(y)$ must be relatively prime to $y$, and either divide $n$ or be congruent to $1$ mod $n$. – Erick Wong Apr 26 '13 at 13:18
  • Thanks, but where is the contradiction? – Babla Apr 26 '13 at 16:49

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That there are infinitely many primes $p \equiv 1 \pmod{n}$ for any fixed $n \in \mathbb{Z}^+$ is a nice application of some elementary properties of cyclotomic polynomials. A proof can be found in $\S$ 10.1 of my field theory notes, which begins by defining cyclotomic polynomials and establishing these properties.

Note that it is not necessary to construe the argument in terms of splitting of primes in cyclotomic fields (although I agree that it is natural to think of things in this way).

The proof of the general version of Dirichlet's Theorem is roughly an order of magnitude harder than this special case. Among many other places, a proof can be found in Chapter 17 of my under/graduate number theory notes. This proof is an adaptation of the one given in Serre's A Course in Arithmetic, with more attention given to certain details.

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Pete L. Clark
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