A simple note on Dirichlet's series (Not finished)

The note consists of four sections. The first sections introduces the Euler's proof for infiniteness of primes. The second section mentions some intuition about Dirichlet's series and Dirichlet's product, that leads to a loose proof of Moebius's inversion theorem. A more compact analytic theory of Section 2 will be mentioned in Section 3. And the fourth section includes a proof of Euler's product for Dirichlet's series, and some interesting factorizations.

As an application of Moebius' inversion theorem, one may refer to my thesis about counting irreducible polynomials over finite fields of a certain degree.

1. Euler's factorization.

We begin this section by recall some facts about series.

Theorem 1.1. Let $f$ be a one-to-one map from $\mathbb{Z}^+$ to $\mathbb{Z}^+$, and $\sum a_n$ is the absolutely convergent series. Then $\sum a_{f(n)}$ is also absolutely convergent. And $\sum a_n=\sum a_{f(n)}$.

Proof. Left as an exercise.

We can see that the series $\sum a_{f(n)}$ can be considered as a rearrangement of the series $\sum a_n$. That means, if $\sum a_n$ is absolutely convergent, then we can permute its elements, and the obtained series is also absolutely convergent, and they have the same sum.

We can use this to prove the following

Theorem 1.2. There exists infinitely many primes.

Proof. If there exists finitely many primes. Consider the following product $\prod_p (\frac{1}{1-p^-1})$. Using the identity $\frac{1}{1-x}=1+x+x^2+...$ for all $|x|<1$, our product is exactly $\prod_p (1+\frac{1}{p}+\frac{1}{p^2}+...)$. Because there exists finite primes, our product is in fact convergent. And hence, we can rearrange all terms in the product, and using the unique factorization of the integer, we obtain the sum $\sum_{n\ge 1}\frac{1}{n}$. But this sum diverges, and this leads the contradiction. Hence, there exists infinitely many primes. (Q.E.D)

For stronger result, we will prove

Theorem 1.3. The sum $\sum_p \frac{1}{p}$ that runs over all primes diverges.

Proof. If $\sum \frac{1}{p}$ converges, then $ \sum_p \frac{1}{p}+\sum_p (\frac{1}{p^2}+...+\frac{1}{p^n}+...)$ converges, since $\frac{1}{p^2}+\frac{1}{p^3}+...=\frac{1}{p^2}(1+\frac{1}{p}+...)=\frac{1}{p^2}\frac{1}{1-p^{-1}}<\frac{2}{p^2}$, and hence $\sum_p (\frac{1}{p^2}+...+\frac{1}{p^n}+...)<2(\sum_{n\ge 1}\frac{1}{n^2})$, which converges.

Because $\frac{1}{p^m}\ge \frac{1}{mp^m}$, we can see that the sum $\sum_p \frac{1}{p}+\sum_p (\frac{1}{2p^2}+...+\frac{1}{np^n}+...)$ converges. And we can rearrange the series so that it has the form $\sum_p (\frac{1}{p}+\frac{1}{2p^2}+\frac+...+\frac{1}{np^n}+...)$. Using the identity $\log(\frac{1}{1-x})=\sum_{n\ge 1}\frac{x^n}{n}$, we have $\sum_p (\frac{1}{p}+\frac{1}{2p^2}+\frac+...+\frac{1}{np^n}+...)=\sum_p \log(\frac{1}{1-p^-1})$ converges. That means, the sum $\sum_p \log(\frac{1}{1-p^{-1}})$ is bounded by a positive real number $M$. This implies $e^M > \prod_p (\frac{1}{1-p^{-1}})$. And hence, our product is convergent. This leads the contradiction to Theorem 1.2. And hence $\sum_p \frac{1}{p}$ is divergent. (Q.E.D)

From Theorem 1.1, one may think that Euler comes very closely to the identity $\sum_n \frac{1}{n}=\prod_p (\frac{1}{1-p^{-1}})$. But actually, this does not make sense, because the two sides of the identity go to infinity.  And it is natural to think about the convergent sum $\sum_n \frac{1}{n^s}$, where $Re(s)>1$. Can one obtain the Euler's factorization? That means, $\sum_n \frac{1}{n^s}=\prod_p (\frac{1}{1-p^{-s}})$? Amazingly, it is true! In the second section, we will give a more general context for the Euler's product.

2. Dirichlet's series, Dirichlet's product and Moebius inversion theorem.

To begin with this section, we recall about the product of two series.

Theorem 2.1. Let $\sum_m a_m$ and $\sum_n b_n$ be two absolutely convergent series, and their values are $A,B$ respectively. Then their Cauchy's product, defined by $\sum_k c_k$, where $c_k=\sum_{m+n=k}a_mb_n$ also converges absolutely, and $\sum_k c_k = AB$. Also, their Dirichlet's product, defined by $\sum_k d_k$, where $d_k = \sum_{d|k}a_db_{\frac{k}{d}}$, also converges absolutely, and $\sum_k d_k = AB$.

Proof. See T. M. Apostol "Mathematical Analysis", Chapter 8.

The Cauchy's product is very similar to the product of polynomials, and the Dirichlet's product appear naturally when we consider about product of two Dirichlet series, defined as follows.

Definition 2.2. Let $f:\mathbb{Z^+}\rightarrow \mathbb{C}$ be a map, then the Dirichlet series corresponds to $f$ is $\sum_n \frac{f(n)}{n^s}$.

Formally, one can take the product of two Dirichlet's series without focusing on their convergent information, and represent the result as another Dirichlet's series.

$$(\sum \frac{a_n}{n^s})(\sum \frac{b_n}{n^s})=\sum \frac{c_n}{n^s}$$

where $c_n=\sum_{d|n}a_db_{n/d}$. If $(\sum \frac{a_n}{n^s})$ and $(\sum \frac{b_n}{n^s})$ are absolutely convergent, then their Dirichlet's product will converge to their product, as Theorem 2.1 pointed out. The very important example will be constructed below.

Definition 2.3. The Moebius $\mu$ functions is defined as $\mu(1)=1, \mu(n)=(-1)^k$ where $n=\prod_{i=1}^k p_i$, where $p_i$'s are distinct primes. If $n$ is not square free, then $\mu(n)=0$.

We have an interesting property for Moebius's function.

Lemma 2.4. For all $n>1$, $\sum_{d|n}\mu(d)=0$.

Proof. Left as an exercise.

We recall that the series $\zeta(s)=\sum_{n\ge 1}\frac{1}{n^s}$ absolutely converges for all complex number $s$, such that $Re(s)>1$ (For $s=a+bi$, then $|n^s|=|e^{s\log n}|=|e^{(a+bi)\log n}|=e^{a\log n}=n^a$, and $\sum \frac{1}{n^a}$ converges for $a>1$). Let us consider next the series $M(s)=\sum_{n\ge1}\frac{\mu(n)}{n^s}$. Because the value of Moebius function does not exceed $1$, $M(s)$ is absolutely converge. And, their Dirichlet's product will be

$$\zeta(s) M(s) = \sum_n \sum_{d|n}(\frac{\mu(d)}{n^s})$$

By Lemma 2.4, we have $\zeta(s) M(s)$ is actually 1. And hence $\zeta(s)$ is the inverse of $M(s)$. And we have constructed the main ingredient for the Moebius' inversion theorem, which stated

Theorem 2.5. Let $F, f$ be two arithmetical functions (that means they go from $\mathbb{Z}^+$ to $\mathbb{C}$, and $F(n)=\sum_{d|n}f(d)$ for all $n\in \mathbb{Z}^+$, then $f(n)=\sum_{d|n}\mu(d)F(n/d)$ for all $n$.

Remark. Note that we will give here a loose proof, that may not make sense if we do not consider carefully about convergence of Dirichlet's series. Amazingly, though we will build a more compact theory latter, we also cannot prove the full theorem. Actually, the proof of the theorem is purely algebraic, when one considers the set of all arithmetical functions $f$, such that $f(1)\ne 0$, under the Dirichlet's product forms an abelian group.

Proof.  Let us consider $\sum \frac{F(n)}{n^s}, \sum \frac{f(n)}{n^s}$, the Dirichlet's series corresponding to $F$ and $f$, respectively. Then $F(n)=\sum_{d|n}f(d)$ implies that $\sum \frac{F(n)}{n^s}$ is the Dirichlet's product of two series $\sum \frac{f(n)}{n^s}$ and $\zeta(s)$. Hence, $M(s)(\sum \frac{F(n)}{n^s})=\sum \frac{f(n)}{n^s}$. And hence, when we consider all coefficients at each $n^s$, we obtain $f(n)=\sum_{d|n}\mu(d)F(n/d)$. (Q.E.D)

Look at the proof above, one may ask, if we omit all conditions about convergence, how can we make sure that if $\sum \frac{a(n)}{n^s}=\sum \frac{b^n}{n^s}$, then $a(n)$ is equal $b(n)$ for all $n$? Actually, it is the very important properties of Dirichlet's series, known under the name Uniqueness Theorem, as we will see later.

3. Some analytic properties of Dirichlet's series.



4. Euler's product.


5. An application of Moebius inversion theorem.

In this section, we will give an interesting application of Moebius inversion theorem to compute the $n^{th}$ cyclotomic polynomial. Let $\xi$ be the $n^{th}$ primitive root of unity in $\mathbb{C}$. The $n^{th}$ cyclotomic polynomial $\Phi_n(x)$ is defined

$$\Phi_n(x)=\prod_{k\le n,(k,n)=1}(x-\xi^k)$$

It can be seen that $\Phi_1(x)=(x-1), \Phi_2(x) = (x+1), \Phi_3(x) = (x^2+x+1)$. The natural question is that is $\Phi_n(x) \in \mathbb{Q}(x)$? And how to compute them? We first note that

Lemma 5.1. $(x^n-1)=\prod_{d|n}\Phi_d(x)$.

Proof. First, we can see that $x^n-1 = \prod_{k=1}^{n}(x-\xi^k)$, and The set $N = \{1,2,...,n\}$ can be partitioned into disjoint subsets $A(d) = \{k\in N|gcd(k,n)=d\}$, where $d|n$. Hence,

$$(x^n-1)=\prod_{d|n}\prod_{k\in A(d)}(x-\xi^k)$$

Furthermore, the product $\prod_{k\in A(d)}(x-\xi^k\big)$ is actually $\Phi_{n/d}(x)$, because $\xi^d$ is the $\frac{n}{d}^{th}$ primitive root of unity. This shows $(x^n-1)=\prod_{d|n}\Phi_{n/d}(x)=\prod_{d|n}\Phi_d(x)$ (Q.E.D).

From this, we can deduce $\log(\prod_{d|n}\Phi_d(x))=\log(x^n-1)$. Hence, $\sum_{d|n}\log(\Phi_d(x))=\log(x^n-1)$. Let us denote $F(n)=\log(x^n-1)$, and $f(d)=\log(\Phi_d(x))$. Then $F(n) = \sum_{d|n}f(d)$. Applying the Moebius inversion theorem, $f(d) = \sum_{d|n}\mu(d)F(n/d)$. That means,

$$\log(\Phi_n(x))=\sum_{d|n}\mu(d)\log(x^{n/d}-1)=\log(\prod_{d|n}(x^{n/d}-1)^{\mu(d)})$$

And hence,

$$\Phi_n(x)=\prod_{d|n}(x^{n/d}-1)^{\mu(d)}$$

Because the Moebius function $\mu(n)$ can receive only two values $1$ or $-1$. We can see that $\Phi_n(x)$ is a rational function in $\mathbb{Q}$. But it is actually a polynomial in $\mathbb{C}[x]$. We now prove the rationality of $\Phi_n(x)$.

Theorem 5.2. $\Phi_n(x)\in \mathbb{Q}[x]$.

Proof. Let $\Phi_n(x) = \frac{p(x)}{q(x)}$ where $p(x),q(x)\in\mathbb{Q}[x]$. And over $\mathbb{C}[x]$, we have $p(x) = q(x)\Phi_n(x)$. Assume that $\Phi_n(x)\notin \mathbb{Q}[x]$, and $\phi_l$ is the least coefficient that does not lie in $\mathbb{Q}$ of $\Phi_n(x)$.  Then the $l^{th}$ coefficient of $p(x)$ (denoted by $p_l$) will be $q_0\phi_l+q_1\phi_{l-1}+...+q_l\phi_0$, where $q_i$'s are coefficients of $q(x)$. From this, we can see that $p_l\notin \mathbb{Q}$, that leads to the contradiction. Hence $\Phi_n(x)\in\mathbb{Q}[x]$ (Q.E.D).

Via the Moebius inversion theorem, one can compute the $n^{th}$ cyclotomic polynomial. The rationality of it can also be obtained by Galois theory. We now discuss about it. First, it can be seen that $\mathbb{Q}(\xi)/\mathbb{Q}$ is Galois extension, because $\mathbb{Q}(\xi)$ is the splitting field of $(x^n-1)$, which is separable over $\mathbb{Q}$. And the Galois group can be computed by just looking at the action on $\xi$.

Let $\tau\in Gal(\mathbb{Q}(\xi)/\mathbb{Q})$, then $\tau(\xi)$ is another root of $(x^n-1)$, and because $\tau$ is a field automorphism, we must have $\tau(\xi)=\xi^k$, where $(k,n)=1$. And hence, $Gal(\mathbb{Q}(\xi)/\mathbb{Q})\cong (\mathbb{Z}/n\mathbb{Z})^\times$. And the minimum polynomial of $\xi$ over $\mathbb{Q}$ is

$$\prod_{\tau\in Gal(\mathbb{Q}(\xi)/\mathbb{Q})}(x-\xi^\tau)=\prod_{k\le n,(k,n)=1}(x-\xi^k)=\Phi_n(x)$$

And the rationality of $\Phi_n(x)$ now follows. We can deduce that $\Phi_n(x)\in\mathbb{Z}[x]$, by noting that $\xi$ is an algebraic integer, and then, so are all coefficients of $\Phi_n(x)$. This implies $\Phi_n(x)\in\mathbb{Z}[x]$.

Let us try with two interesting two exercises.

Problem 5.3. Prove that $p$ is a prime number iff $\Phi_p(x)=1+x+...+x^{p-1}$.

Problem 5.4. Prove that there exists infinitely many primes of the form $kn+1$.