Dirichlet's Theorem on Arithmetic Progressions/Lemma 1

From ProofWiki
Jump to: navigation, search

Lemma for Dirichlet's Theorem on Arithmetic Progressions

Let $a, q$ be coprime integers.

Let $\mathcal P_{a, q}$ be the set of primes $p$ such that $p \equiv a \pmod q$.


Let $\chi$ be a Dirichlet character modulo $q$.

Let:

$\displaystyle f \left({s}\right) = \sum_p \chi \left({p}\right) p^{-s}$

If $\chi$ is non-trivial then $f \left({s}\right)$ is bounded as $s \to 1$.

If $\chi$ is the trivial character then:

$f \left({s}\right) \sim \ln \left({\dfrac 1 {s - 1} }\right)$

as $s \to 1$.


Proof

By Logarithm of Dirichlet L-Functions:

$(1):\quad \displaystyle \sum_p \chi \left({p}\right) p^{-s} = \ln L \left({s, \chi}\right) - \sum_p \sum_{n \ge 2} \frac{\chi \left({p}\right)^n} {n p^{n s}}$

If $\chi$ is non-trivial, then by L-Function does not Vanish at One, $\ln L \left({s, \chi}\right)$ is bounded as $s \to 1$.

If $\chi$ is trivial, then by Analytic Continuation of Dirichlet L-Function, $L \left({s, \chi}\right)$ has a simple pole at $s = 1$.

Therefore, in this case:

$L \left({s, \chi}\right) \sim \dfrac \lambda {s-1}$

where $\lambda$ is the residue of $L \left({s, \chi}\right)$ at $1$, and:

$\ln L \left({s, \chi}\right) \sim \ln \left({\dfrac \lambda {s-1}}\right) \sim \ln \left({\dfrac 1 {s-1}}\right)$

Thus if we can show that the second term of $(1)$ is bounded, the result holds.


On $\operatorname{Re} \left({s}\right) > 1$:

\(\displaystyle \left \vert \sum_p \sum_{n \mathop \ge 2} \frac {\chi \left({p}\right)^n} {n p^{n s} } \right\vert\) \(\le\) \(\displaystyle \sum_p \sum_{n \mathop \ge 2} \frac 1 {\left \vert{p^s}\right\vert^n}\)
\(\displaystyle \) \(=\) \(\displaystyle \sum_p \frac 1 {\left \vert{p^{2s} }\right\vert \ \left\vert{p^s - 1}\right\vert^n}\) Sum of Infinite Geometric Progression
\(\displaystyle \) \(\le\) \(\displaystyle \sum_p \frac 1 {p^2}\) because $\operatorname{Re} \left({s}\right) > 1$
\(\displaystyle \) \(\le\) \(\displaystyle \sum_n \frac 1 {n^2}\)

This last is $\zeta \left({2}\right)$ where $\zeta$ is the Riemann zeta function, so is finite.

$\blacksquare$