Dirichlet’s Theorem

We will prove Dirichlet’s Theorem on Primes in Arithmetic Progressions.

Euclid proved that there are infinitely many primes. This, however, says nothing about the distribution of those primes. Dirichlet proved a stronger statement: there are infinitely many primes in every arithmetic progression; in this sense, the primes are “uniformly distributed”.

Theorem 1 (Dirichlet’s Theorem) If {(a, q) = 1}, then there are infinitely many primes {p} satisfying {p \equiv a \pmod q}.

Many ideas of the proof of Dirichlet’s Theorem were first seen in Euler’s proof of the infinitude of primes. Here, Euler actually showed that {\sum_p 1/p} diverges, and that is the approach that we will take for Dirichlet’s Theorem.

1. Euler’s Proof of the Infinitude of Primes

Proposition 2 There are infinitely many primes.

Define the Riemann zeta function as

\displaystyle  \zeta(s) = \sum_{n = 1}^\infty \frac1{n^s} = \prod_p \left( 1 - \frac1{p^s} \right)^{-1}.

We begin with a lemma in the flavor of what we will consider in our proof.

Lemma 3

\displaystyle  \zeta(s) = \frac1{s-1} + O(1).

Proof: We can approximate this sum by integrals:

\displaystyle  \zeta(s) = \sum_{n=1}^\infty \frac1{n^s} 	\ge \int_{1}^\infty \frac{1}{t^s} \, dt 	= \left. \frac{-1}{(s-1)t^{s-1}} \right|_1^\infty = \frac{1}{s-1}

and similarly

\displaystyle  \zeta(s) = \sum_{n=1}^\infty \frac1{n^s} 	\le 1 + \int_{1}^\infty \frac{1}{t^s} \, dt 	= \left. \frac{-1}{(s-1)t^{s-1}} \right|_1^\infty = \frac{1}{s-1} + 1.

\Box

From the Taylor expansion of {\log (1 + x)}, we see that {- \log (1 - x) = x + O(x^2)}. This allows us to write

\displaystyle  \log \zeta(s) = \log \prod_p \left( 1 - \frac{1}{p^s} \right)^{-1} = - \sum_p \log \left( 1 - \frac{1}{p^s} \right).

Using {\log (1 - x) = - \sum_{k} \frac{x^k}k}, this means that

\displaystyle  \log \zeta(s) = \sum_p \sum_{k=1}^\infty \frac{1}{k p^{ks} } = \sum_p \left( \frac{1}{p^s} + O \left( \frac{1}{p^{2s}}\right) \right).

Now, since

\displaystyle  \sum_p \frac{1}{p^{2s}} \le \sum_{n=1}^\infty \frac{1}{n^{2s}} \le \sum_{n=1}^\infty \frac1{n^2}

converges, we have actually proven

Lemma 4

\displaystyle  \log \zeta(s) = \sum_p \frac{1}{p^s} + O(1).

Combining this with Lemma 3 shows the corollary

Corollary 5

\displaystyle  \sum_p \frac{1}{p^s} = \log \left( \frac{1}{s-1} \right) + O(1).

As {s \rightarrow 1^+}, the right hand side diverges, so {\sum_p \frac1p} diverges and hence there are an infinite number of primes.

2. Sketch of Proof of Dirichlet’s Theorem

We first sketch Dirichlet’s Theorem in a special case to illustrate and motivate the proof of the general theorem. Therefore, we will prove

Proposition 6 There are infinitely many primes {p \equiv 1 \pmod 4} and {p \equiv 3 \pmod 4}.

This proof will be in the same spirit as Euler’s proof of the infinitude of primes. However, since we are splitting the primes into two groups ({1 \mod 4} and {3 \mod 4}), we will need to construct a generalization of the zeta function. This generalization is the Dirichlet {L}-function.

Definition 7 Define

\displaystyle  \chi_{-4} (n) = \begin{cases} 1 & n \equiv 1 \pmod 4 \\ 0 & n \equiv 0,2 \pmod 4 \\ -1 & n \equiv 3 \pmod 4. \end{cases}

This is an example of a Dirichlet character. Note that it is multiplicative ({\chi_{-4}(mn) = \chi_{-4}(m) \chi_{-4}(n)}) and periodic ({\chi_{-4}(n) = \chi_{-4}(n+4)}).

Definition 8 A Dirichlet {L}-function is then

\displaystyle  L(s,\chi_{-4}) = \sum_{n=1}^\infty \frac{\chi_{-4}(s)}{n^s} = \frac{1}{1^s} - \frac{1}{3^s} + \cdots = \prod_p \sum_{k=1}^\infty \frac{\chi_{-4}(p)^k}{p^{ks}} = \prod_p \left( 1 - \frac{\chi_{-4}(p)}{p^s} \right)^{-1}.

Consider

\displaystyle  \log L(s, \chi_{-4}) 	= \sum_p - \log \left( 1 - \frac{\chi_{-4}(p)}{p^s}\right) 	= \sum_p 		\left( \frac{\chi_{-4}(p)}{p^s} + O \left( \frac{\chi_{-4}(p^2)}{p^{2s}} 			\right) \right) 	= \sum_p \frac{\chi_{-4}(p)}{p^s} + O(1).

Since we also know from Lemma 4 that

\displaystyle  \log \zeta(s) = \sum_p \frac{1}{p^s} + O(1),

we get the expressions

\displaystyle  \log \zeta(s) + \log L(s, \chi_{-4}) 	= \sum_{p \equiv 1 \bmod 4} \frac{2}{p^s} + O(1)

\displaystyle  \log \zeta(s) - \log L(s, \chi_{-4}) 	= \sum_{p \equiv 3 \bmod 4} \frac{2}{p^s} + O(1).

Now, let {s \rightarrow 1^+}, and consider

\displaystyle  \sum_{p \equiv 1 \bmod 4} \frac{2}{p^s} + O(1) = \log \zeta(s) + \log L(s, \chi_{-4}).

As {s \rightarrow 1^+}, we already know that {\log \zeta(s) \rightarrow \infty}. We want to show that {\log L(s, \chi_{-4})} does not go to {- \infty}, or equivalently, that {L(s, \chi_{-4})} does not go to zero. In fact, {L(s, \chi_{-4})} goes to {L(1, \chi_{-4})}, which in this case is actually

\displaystyle  L(1, \chi_{-4}) = \frac11 - \frac13 + \frac15 - \frac17 + \cdots = \frac\pi4.

Therefore, we see that {\log L(s, \chi_{-4})} is constant and hence the right hand side of

\displaystyle  \sum_{p \equiv 1 \bmod 4} \frac{2}{p^s} + O(1) = \log \zeta(s) + \log L(s, \chi_{-4}).

diverges as {s \rightarrow 1^+}, showing that {\sum_{p\equiv 1 \bmod 4} \frac2p} diverges and hence that there are an infinite number of primes congruent to {1 \bmod 4}. By precisely the same argument, there are also an infinitely number of primes congruent to {3 \bmod 4}.

Here, we used the character {\chi_{-4}} to produce the {L}-function {L(s, \chi_{-4})}, and we used this {L}-function together with the zeta function to pick out the arithmetic progression {1 \bmod 4}. This yielded a formula for {\sum_{p \equiv 1 \bmod 4} \frac1p} in terms of a linear combination of {\log \zeta(s)} and {\log L(s, \chi_{-4})}. We then showed that as {s \rightarrow 1^+}, {L(s,\chi_{-4})} converges to something nice, which means that it doesn’t contribute much to the sum. Therefore, {\log \zeta(s)} dominates, forcing {\sum \frac1p} to diverge, giving us our result.

To generalize this to arbitrary arithmetic progressions, we will need to find additional characters {\chi} and show how to pick out any arithmetic progression. We will also need to show that {L(s, \chi)} behaves nicely under the limit {s \rightarrow 1^+} and that {L(1, \chi) \ne 0}.

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