| 1. | Lecture 1 |
| 2. | Lecture 2 |
| 3. | Lecture 3 |
| 4. | Lecture 4 |
| 5. | Lecture 5 |
| 6. | Lecture 6 |
| 7. | Lecture 7 |
| 8. | Lecture 8 |
| 9. | Lecture 9 |
| 10. | Lecture 10 |
| 11. | Lecture 11 |
| 12. | Lecture 12 |
| 13. | Lecture 13 |
Definition. Let \(q\in\mathbb Z_+\). A Dirichlet character modulo \(q\) is a map \(\chi: \mathbb{Z} \backslash\{0\} \to \mathbb{C}\) satisfying the following rules for all \(a, b \in \mathbb{Z} \backslash\{0\}\):
\(\chi(a)=\chi(a\pmod q)\);
\(\chi(a b)=\chi(a) \chi(b)\);
\(\chi(a)=0\) if \((a, q)>1\).
Remarks.
In fact, (i) and (ii) mean that Dirichlet character modulo \(q\) are homomorphisms of the multiplicative group \((\mathbb{Z} / q \mathbb{Z})^{\times}\) and property (iii) extends these maps to \(\mathbb{Z} \backslash\{0\}\).
On the other hand, we can readily see that corresponding to every multiplicative character \(\psi\in\widehat{(\mathbb{Z} / q \mathbb{Z})^{\times}}\) there is \(\chi: \mathbb{Z} \backslash\{0\} \to \mathbb{C}\), a Dirichlet character modulo \(q\), defined by \[\chi(a)= \begin{cases} \psi(a+q\mathbb Z) & \text{ if } (a, q)=1,\\ 0& \text{ if } (a, q)>1. \end{cases}\]
From this definition we immediately see that \(\chi\) satisfies (i)-(iii).
By the previous remark we see that the set of Dirichlet characters modulo \(q\) is a group isomorphic to the multiplicative group \((\mathbb{Z} / q \mathbb{Z})^{\times}\) of the units of the ring \(\mathbb{Z} / q \mathbb{Z}\).
In particular, there are \(\varphi(q)\) Dirichlet characters modulo \(q\).
The identity element of this group is called the principal, or trivial, character modulo \(q\) and is usually denoted by \(\chi_{0}\). Thus, \(\chi_{0}\) is defined for all \(a \in \mathbb{Z}\) by \[\chi_{0}(a)= \begin{cases} 1, & \text { if }(a, q)=1, \\ 0, & \text { otherwise }. \end{cases}\]
Let \(\chi\) be a Dirichlet character modulo \(q\). We define \(\overline{\chi}\) by \(\overline{\chi}(a)=\overline{\chi(a)}\). Clearly, \(\overline{\chi}\) is also a Dirichlet character modulo \(q\) called the conjugate character of \(\chi\).
It is also not difficult to see that, if \((a, q)=1\), then \(\chi(a)\) is a \(\varphi(q)\) th root of unity. Indeed, if \(a\in (\mathbb{Z} / q \mathbb{Z})^{\times}\), we have \[(\chi(a))^{\varphi(q)}=\chi\left(a^{\varphi(q)}\right)=\chi(1)=1.\]
Notation. Let \(m\in\mathbb Z_+\). We will use the following convenient notation:
\(\sum_{a\pmod m}\) will always denote the sum over a complete set of residue classes modulo \(m\).
\(\sum_{\chi\pmod m}\) will always denote the sum over the \(\varphi(m)\) Dirichlet characters modulo \(m\).
Applying the first orthogonality relation from Lecture 6 for \(\mathbb G=(\mathbb Z/m\mathbb Z)^{\times}\) we obtain the following orthogonality relation for Dirichlet characters:
If \(\chi\) is a Dirichlet character modulo \(m\), then \[\sum_{{a(\bmod m)}} \chi(a)= \begin{cases}\varphi(m) & \text { if } \chi=\chi_{0},\\ 0 & \text { if } \chi \neq \chi_{0}.\end{cases}\]
If \(a\in\mathbb Z\), then \[\sum_{\chi(\bmod m)} \chi(a)=\left\{\begin{array}{lll} \varphi(m) & \text { if } a \equiv 1 \quad(\bmod m), \\ 0 & \text { if } a \not \equiv 1 \quad(\bmod m). \end{array}\right.\]
Applying the second orthogonality relation from Lecture 6 for \(\mathbb G=(\mathbb Z/m\mathbb Z)^{\times}\) we obtain the following orthogonality relation for Dirichlet characters:
If \(\chi_{1}\) and \(\chi_{2}\) are Dirichlet characters modulo \(m\), then \[\sum_{{a(\bmod m)}} \chi_{1}(a) \overline{\chi_{2}}(a)= \begin{cases}\varphi(m) & \text { if } \chi_{1}=\chi_{2}, \\ 0 & \text { if } \chi_{1} \neq \chi_{2}.\end{cases}\]
If \(a,b\in \mathbb Z\), then \[\sum_{\chi(\bmod m)} \chi(a) \overline{\chi}(b)= \begin{cases}\varphi(m) & \text { if }(a, m)=(b, m)=1 \text { and } a \equiv b \pmod m, \\ 0 & \text { otherwise } .\end{cases}\]
For \(k=2\), the principal character is the only one.
For \(k=3\), there are two characters, namely \[\begin{aligned} \chi_{0}(n)=\begin{cases}1 & \text { if } n \equiv 1,2 \bmod 3, \\ 0 & \text { if } n \equiv 0 \quad \bmod 3,\end{cases} \quad \chi_{1}(n)= \begin{cases}1 & \text { if } n \equiv 1 \bmod 3, \\ -1 & \text { if } n \equiv-1 \bmod 3, \\ 0 & \text { if } n \equiv 0 \bmod 3 .\end{cases}. \end{aligned}\]
For \(k=4\), there are again two characters, namely \[\chi_{0}(n)=\begin{cases} 1 & \text { if } n \text { is odd }, \\ 0 & \text { if } n \text { is even, } \end{cases} \quad \chi_{1}(n)= \begin{cases}1 & \text { if } n \equiv 1 \bmod 4 \\ -1 & \text { if } n \equiv-1 \bmod 4, \\ 0 & \text { if } n \text { is even } .\end{cases}\]
For \(k=5\), there are four characters given in the following table: \[\begin{aligned} \begin{array}{|c||c|c|c|c|c|} \hline n\pmod 5 & 1 & 2 & 3 & 4 & 0 \\ \hline \hline\chi_{0}(n) & 1 & 1 & 1 & 1 & 0 \\ \chi_{1}(n) & 1 & i & -i & -1 & 0 \\ \chi_{2}(n) & 1 & -1 & -1 & 1 & 0 \\ \chi_{3}(n) & 1 & -i & i & -1 & 0 \\ \hline \end{array} \end{aligned}\]
Let \(\chi, \chi_1\) be Dirichlet characters modulo \(q\) and \(q_1\) respectively. Let \(q_1\) be a divisor of \(q\). We say that \(\chi\) is induced by \(\chi_1\) or \(\chi_1\) induces \(\chi\) if \[\chi(n)=\chi_{0}(n) \chi_1(n) \quad \text{ for all }\quad n\in\mathbb Z, \tag{*}\] where \(\chi_{0}\) is the principal character modulo \(q\).
A Dirichlet character \(\chi\) modulo \(q\) is said to be imprimitive if there exists a Dirichlet character \(\chi_1\) modulo \(q_1\) that induces \(\chi\) for some proper divisor \(q_1\) of \(q\).
The conductor of a Dirichlet character \(\chi\) modulo \(q\) is the smallest divisor \(q_1\) of \(q\) for which (*) holds.
A Dirichlet character \(\chi\) modulo \(q\) is said to be primitive if the conductor of \(\chi\) is simply equal to its modulus \(q\). In other words, \(\chi\) is primitive, if it is trivially induced by itself and \(\chi=\chi_{0}\chi\).
Remarks.
The principal character \(\chi_0\) is imprimitive for every integer \(q>1\).
Let \(q^\star\) be the conductor of a Dirichlet character \(\chi\) modulo \(q\). Then \(\chi=\chi_{0} \chi^\star\), where \(\chi^\star\) is a Dirichlet character modulo \(q^\star\), which is primitive and uniquely determined by \(\chi\). This follows from the fact that \(q^\star\) is the smallest divisor of \(q\) for which (*) holds.
Let \(m, d\in\mathbb Z_+\) be such that \(d\mid m\). If \(\gcd(a, d)=1\) for some \(a\in\mathbb Z\), then there exists \(b\in\mathbb Z\) such that \(b \equiv a \pmod d\) and \(\gcd(b, m)=1\).
Proof.
Let \(m=\prod_{i\in[k]} p_{i}^{r_{i}}\) and \(d=\prod_{i\in[k]} p_{i}^{s_{i}}\), where \(r_{i} \geq 1\) and \(0 \leq s_{i} \leq r_{i}\) for \(i\in[k]\). Let \(n\) be the product of the prime powers that divide \(m\) but not \(d\). Then \(n=\prod_{\substack{i\in[k] \\ s_{i}=0}} p_{i}^{r_{i}}\) and \(\gcd\left(n, d\right)=1\).
By the existence of solutions for linear congruences there is \(x\in\mathbb Z\) such that \(d x \equiv 1-a \pmod n\). Then \(b=a+d x \equiv 1 \pmod n\) and \(\gcd\left(b, n\right)=1\).
Also, \[b \equiv a \pmod d.\]
If \(\gcd\left(b, m\right) \neq 1\), there exists a prime \(p\in \mathbb P\) that divides both \(b\) and \(m\). However, \(p\) does not divide \(n\) since \(\gcd\left(b, n\right)=1\). It follows that \(p\mid d\), and so \(p\) divides \(b-d x=a\), which is impossible since \((a, d)=1\).
Therefore, \(\gcd\left(b, m\right)=1\) and we are done.$$\tag*{$\blacksquare$}$$
Definition. Let \(\chi\) be a Dirichlet character modulo \(q\). We say \(d\) is a quasiperiod of \(\chi\) if \(\chi(m) = \chi(n)\) whenever \(m\equiv n\pmod d\) and \((mn, q) = 1\). Obviously, every period is a quasiperiod.
Proposition. Let \(\chi\) be a Dirichlet character modulo \(q\). Then, \(\chi\) is imprimitive if and only if there is a proper divisor \(d\) of \(q\) which is a quasiperiod of \(\chi\). In particular, the conductor of \(\chi\) is its quasiperiod.
Proof.
Suppose that \(\chi\) is imprimitive, then it is induced by a Dirichlet character \(\chi_1\) modulo \(d\), that is \(\chi=\chi_0\chi_1\). Take \(m\equiv n\pmod d\) with \((mn, q) = 1\), then \((m, q)=(n, q)=1\) and \[\chi(m)=\chi_0(m)\chi_1(m)=\chi_1(m)=\chi_1(n)=\chi_0(n)\chi_1(n)=\chi(n),\] since \(\chi_1\) has period \(d\).
For the converse implication, we shall construct a Dirichlet character \(\chi^\star\) modulo \(d\) such that \(\chi=\chi_0\chi^\star\).
For this purpose we will use the following observation if \(d\mid q\) and \((n, d)=1\), then there is \(k\in\mathbb Z\) such that \((n+dk, q)=1\).
Then it suffices to set \[\begin{aligned} \chi^\star(n):= \begin{cases} \chi(n+dk) & \text{ if } (n, d)=1,\\ 0& \text{ if } (n, d)>1, \end{cases} \end{aligned}\] where \(k\in\mathbb Z\) is such that \((n+dk, q)=1\).
We note that although there are many possible choices of \(k\in\mathbb Z\), the value of \(\chi^\star(n)\) depends only on \(n\pmod d\). Indeed, if \(n\equiv m\pmod d\) and \((n,d)=(m, d)=1\) then there are integers \(k_n, k_m\in\mathbb Z\) such that \((n+dk_n, q)=(m+dk_m, q)=1\), which immediately implies \[\chi^\star(n)=\chi(n+dk_n, q)=\chi(m+dk_m, q)=\chi^\star(m),\] since \(d\) is the quasiperiod of \(\chi\) and \(n+dk_n\equiv m+dk_m\pmod d\).
It remains to prove that \(\chi^\star\) is a homomorphism. Let \(m, n\in\mathbb Z\) be such that \((mn, d)=1\), then \((m, d)=(n, d)=1\) and there are \(k_m, k_n, k_{mn}\in\mathbb Z\) such that \((m+dk_m, q)=(n+dk_n, q)=(mn+dk_{mn}, q)=1\). Thus we have \((m+dk_m)(n+dk_n)\equiv mn+dk_{mn}\pmod d\) and since \(\chi\) is homomorpfism and \(d\) is the quasiperiod of \(\chi\), we consequently have \[\begin{aligned} \chi^\star(mn)&=\chi(mn+dk_{mn})=\chi((m+dk_m)(n+dk_n))\\ &=\chi(m+dk_m)\chi(n+dk_n)=\chi^\star(m)\chi^\star(n). \quad\end{aligned}\qquad\blacksquare\]
Proposition 2. Let \(\chi\) be a Dirichlet character modulo \(q\) and let \(d\mid q\) and \(d<q\). Then, \(\chi\) has quasiperiod \(d\) if and only if \(\chi(n)=1\) for all \(n\equiv1\pmod d\) with \((n, q)=1\).
Proof.
Observe that if \(\chi\) has quasiperiod \(d\), then \(\chi(m) = \chi(n)\) whenever \(m\equiv n\pmod d\) and \((mn, q) = 1\), which implies the condition above by taking \(m=1\), since \(\chi(1)=1\). For the converse, assume that the above condition holds. Take \(m\equiv n\pmod d\) with \((m, q) = (n, q) =1\).
Since \((m, q) =1\) then we can find \(m'\in\mathbb Z\) such that \(mm'\equiv 1\pmod q\).
Thus \(\chi(mm')=\chi(1)=1\), since \(\chi\) has period \(q\).
We also obtain that \(mm'\equiv 1\pmod d\), since \(d\mid q\), and consequently \[nm'\equiv mm'\equiv 1\pmod d.\]
Since \((mm', q) = (nm', q) =1\), we obtain by the above condition that \[\chi(m)\chi(m')=\chi(mm')=1=\chi(nm')=\chi(n)\chi(m'),\] which yields \(\chi(n)=\chi(m)\) as desired. $$\tag*{$\blacksquare$}$$
Let \(\chi\) be a Dirichlet character modulo \(q\). Then, \(\chi\) is primitive if and only if for every proper divisor \(d\) of \(q\) there exists \(m\in\mathbb Z\) satisfying \(m\equiv1\pmod d\) and \((m, q)=1\) such that \(\chi(m)\neq1\).
Proof. The proof readily follows by applying the previous two propositions.$$\tag*{$\blacksquare$}$$
Let \(\chi\) be a Dirichlet character modulo \(q\). Then, the following are equivalent:
\(\chi\) is primitive;
if \(d\mid q\) and \(d<q\), then for any \(a\in\mathbb Z\), we have \[\sum_{\substack{n=1\\ n\equiv a\bmod d}}^q\chi(n)=0.\]
Assume that condition (i) is true. If \(\chi\) is primitive, then there is \(m\in\mathbb Z\) satisfying \(m\equiv1\pmod d\) and \((m, q)=1\) such that \(\chi(m)\neq1\). Then \[\begin{aligned} S=\sum_{\substack{n=1\\ n\equiv a\bmod d}}^q\chi(n)&=\sum_{\substack{n=1\\ n\equiv am\bmod d}}^q\chi(n) =\sum_{k=1}^{q/d}\chi(am+dkm)\\ &=\chi(m)\sum_{\substack{n=1\\ n\equiv a\bmod d}}^q\chi(n)=\chi(m)S. \end{aligned}\]
Since \(\chi(m)\neq1\), we deduce that \(S=0\), and we are done.
Assume that condition (ii) is true. Let \(d\mid q\) and \(d<q\), and \(a=1\). Then \[\sum_{\substack{n=1\\ n\equiv 1\bmod d}}^q\chi(n)=0,\] and since \(\chi(1)=1\), there must be \(2\le m\le q\) such that \(0\neq \chi(m)\neq1\).
But \(m\equiv 1\pmod d\) and \((m, q)=1\), hence by the previous theorem \(\chi\) must be primitive as desired. $$\tag*{$\blacksquare$}$$
Suppose that \((q_1, q_2) = 1\) and let \(\chi_i\) be a Dirichlet character \(\pmod {q_i}\) for \(i \in[2]\). Then \(\chi = \chi_1 \chi_2\) is primitive \(\pmod {q_1 q_2}\) if and only if \(\chi_1\) and \(\chi_2\) are both primitive.
Proof.
We first prove the implication \((\Longrightarrow)\). Let \(d_i\) be the conductor of \(\chi_i\). If \((mn, q_1q_2) = 1\) and \(m \equiv n \mod d_1 d_2\), then \(\chi_i(m) = \chi_i(n)\), hence \(d_1 d_2\) is a quasiperiod of \(\chi\). Thus \(d_1 d_2 = q_1q_2\) since \(\chi\) is primitive. Therefore, \(d_1 = q_1\) and \(d_2 = q_2\) since \(d_i \mid q_i\) for \(i \in [2]\).
We now prove the reverse implication \((\Longleftarrow)\). Let \(d\) be the conductor of \(\chi\). Set \(d_i = (d, q_i)\) for \(i\in[2]\). We show that \(d_1\) is a quasiperiod of \(\chi_1\). Suppose \((mn, q_1) = 1\) and \(m \equiv n \pmod {d_1}\). Choose \(m_0, n_0 \in \mathbb{Z}\) so that \[\begin{aligned} &m_0 \equiv m \pmod {q_1}, \quad \text{ and } \quad n_0 \equiv n \pmod {q_1},\\ &m_0 \equiv 1 \pmod {q_2}, \ \quad \text{ and } \quad n_0 \equiv 1 \pmod {q_2}. \end{aligned}\] Thus \(m_0 \equiv n_0 \pmod {d_1}\) and \(m_0 \equiv n_0 \pmod {d_2}\), hence \(m_0 \equiv n_0 \pmod {d_1d_2}\), but \(d_1d_2=(d, q_1q_2)=d\) since \(d\mid q_1q_2\). Consequently, \(m_0 \equiv n_0 \pmod d\) and \((m_0 n_0, q_1q_2) = 1\), yielding \(\chi(m_0) = \chi(n_0)\). Therefore, \(d_1\) is a quasiperiod of \(\chi_1\), since \(\chi_1(m)=\chi(m_0) = \chi(n_0) = \chi_1(n)\). Since \(\chi_1\) is primitive, we must have \(d_1 = q_1\). Similarly, \(d_2 = q_2\), and finally \(d = q_1q_2\).$$\tag*{$\blacksquare$}$$
By the Chinise reminder theorem, or more abstractly, by the structure theorem for finite abelian groups if \(q=\prod_{p\in\mathbb P}p^{\alpha_p(q)}\), then \[(\mathbb Z/q\mathbb Z)^{\times} \equiv \bigotimes_{p^{\alpha_p(q)}\mid q}(\mathbb Z/p^{\alpha_{p}(q)}\mathbb Z)^{\times}.\] In other words, any multiplicative group modulo \(q\) is isomorphic to the direct product of multiplicative groups modulo prime powers.
Therefore, understanding Dirichlet characters on \((\mathbb Z/q\mathbb Z)^{\times}\), is reduced to understand Dirichlet characters on \((\mathbb Z/p^{\alpha_{p}(q)}\mathbb Z)^\times\).
Let \(p\in\mathbb P\) be an odd prime number. Then the corresponding group \((\mathbb Z/p^{\alpha}\mathbb Z)^\times\) is cyclic. This means that there exists a primitive root \(g\) in \((\mathbb Z/p^{\alpha}\mathbb Z)^\times\). In fact, we can find a primitive root in \((\mathbb Z/p\mathbb Z)^\times\), which is also a primitive root in \((\mathbb Z/p^{\beta}\mathbb Z)^\times\) for all \(\beta\in\mathbb Z_+\).
If \((n, p)=1\) let \(\nu(n)=\operatorname{ind}_{g} n\pmod {p^{\alpha}}\), so that \(\nu(n)\) is the unique integer satisfying the conditions \[n \equiv g^{\nu(n)}\pmod {p^{\alpha}}, \quad \text{ where } \quad 0 \leq \nu(n)<\varphi\left(p^{\alpha}\right).\]
For \(h\in\mathbb N_{<\varphi(p^{\alpha})}\), define \(\chi_{h}\) by the relations \[\chi_{h}(n)=\chi_{h}(n; p^\alpha)= \begin{cases} e(h \nu(n) / \varphi(p^{\alpha})) & \text { if } p \nmid n, \\ 0 & \text { if } p \mid n.\end{cases}\]
Using the properties of indices \(\nu(n)=\operatorname{ind}_{g} n\pmod {p^{\alpha}}\) it is easy to verify that \(\chi_{h}\) is completely multiplicative and periodic with period \(p^{\alpha}\), so \(\chi_{h}\) is a Dirichlet character \(\bmod p^{\alpha}\), with \(\chi_{0}\) being the principal character. This verification is left as an exercise!
Since \[\chi_{h}(g)=e(h/\varphi (p^{\alpha}))\] the characters \(\chi_{0}, \chi_{1}, \ldots, \chi_{\varphi\left(p^{\alpha}\right)-1}\) are distinct because they take distinct values at \(g\). Therefore, since there are \(\varphi(p^{\alpha})\) such functions they represent all the Dirichlet characters \(\pmod {p^{\alpha}}\).
The same construction works for the modulus \(2^{\alpha}\) if \(\alpha=1\) or \(\alpha=2\), using \(g=3\) as the primitive root.
If \(\alpha \geq 3\) the modulus \(2^{\alpha}\) has no primitive root and a slightly different construction is needed to obtain the characters mod \(2^{\alpha}\).
We know that for every \(\alpha \geq 3\), and every odd integer \(n\in\mathbb Z\) there is a uniquely determined integer \(\nu(n)\) such that \[n \equiv(-1)^{(n-1) / 2} 5^{\nu(n)}\pmod {2^{\alpha}}, \quad \text { with } \quad 1 \leq \nu(n) \leq \varphi(2^{\alpha}) / 2.\]
With this knowledge we can construct all the characters \(\pmod {2^{\alpha}}\) if \(\alpha \geq 3\). Let \[f(n)= \begin{cases} (-1)^{(n-1) / 2} & \text { if } n \text { is odd }, \\ 0 & \text { if } n \text { is even }, \end{cases}\] and let \[g(n)= \begin{cases} e(\nu(n) / 2^{\alpha-2}) & \text { if } n \text { is odd }, \\ 0 & \text { if } n \text { is even }, \end{cases}\] where \(\nu(n)\) is the integer given in the previous item.
Then it is easy to verify that each of \(f\) and \(g\) is a character mod \(2^{\alpha}\). So is each product \[\begin{aligned} \chi_{a_1, a_2}(n)=\chi_{a_1, a_2}(n; 2^\alpha)=f(n)^{a_1} g(n)^{a_2}=e\bigg(\frac{a_1(n-1)}{4}+\frac{a_2\nu(n)}{2^{\alpha-2}}\bigg), \end{aligned}\] where \(a_1\in[2]\) and \(a_2\in[\varphi(2^{\alpha}) / 2]\). Moreover these \(\varphi(2^{\alpha})\) characters are distinct so they represent all the characters mod \(2^{\alpha}\).
If \(\chi\) is a real-valued Dirichlet character \(\pmod m\) and \((n, m)=1\), the number \(\chi(n)\) is both a root of unity and real, so \(\chi(n)= \pm 1\). From the construction from the previous slides we can determine all real Dirichlet characters \(\pmod {p^{\alpha}}\).
For an odd prime \(p\in\mathbb P\) and \(\alpha \in\mathbb Z_+\), consider the Dirichlet characters \[\chi_{h}(n)=\chi_{h}(n; p^\alpha)= \begin{cases} e(h \nu(n) / \varphi(p^{\alpha})) & \text { if } p \nmid n, \\ 0 & \text { if } p \mid n, \end{cases} \quad \text{ for } \quad h\in[\mathbb N_{<\varphi(p^{\alpha})}].\] Then \(\chi_{h}\) is real if, and only if, \(h=0\) or \(h=\varphi(p^{\alpha}) / 2\). Hence there are exactly two real characters \(\pmod {p^{\alpha}}\).
The next theorem describes the real characters \(\bmod 2^{\alpha}\) when \(\alpha \geq 3\).
If \(\alpha \geq 3\), consider the Dirichlet characters \[\begin{aligned} \chi_{a_1, a_2}(n)=\chi_{a_1, a_2}(n; 2^\alpha)=e\bigg(\frac{a_1(n-1)}{4}+\frac{a_2\nu(n)}{2^{\alpha-2}}\bigg), \end{aligned}\] where \(a_1\in[2]\) and \(a_2\in[\varphi(2^{\alpha}) / 2]\). Then \(\chi_{a_1, a_2}\) is real if, and only if, \(a_2=\varphi(2^{\alpha}) / 2\) or \(a_2=\varphi(2^{\alpha}) / 4\). Hence there are exactly four real characters \(\pmod 2^{\alpha}\) if \(\alpha \geq 3\).
For an odd prime \(p\in\mathbb P\) and \(\alpha \geq 2\), consider the Dirichlet characters \[\chi_{h}(n)=\chi_{h}(n; p^\alpha)= \begin{cases} e(h \nu(n) / \varphi(p^{\alpha})) & \text { if } p \nmid n, \\ 0 & \text { if } p \mid n, \end{cases} \quad \text{ for } \quad h\in[\mathbb N_{<\varphi(p^{\alpha})}].\] Then \(\chi_{h}\) is primitive \(\bmod p^{\alpha}\) if, and only if, \(p \nmid h\).
If \(\alpha \geq 3\), consider the Dirichlet characters \[\begin{aligned} \chi_{a_1, a_2}(n)=\chi_{a_1, a_2}(n; 2^\alpha)=e\bigg(\frac{a_1(n-1)}{4}+\frac{a_2\nu(n)}{2^{\alpha-2}}\bigg), \end{aligned}\] where \(a_1\in[2]\) and \(a_2\in[\varphi(2^{\alpha}) / 2]\). Then \(\chi_{a_1, a_2}\) is primitive \(\pmod {2^{\alpha}}\) if, and only if, \(a_2\) is odd.
The character corresponding to \(h=0\) is the principal character.
When \(\alpha=1\) the quadratic character \(\chi_p(n)=(n \mid p)\) is the only other real character \(\pmod p\).
For the moduli \(m=1,2\) and \(4\), all the Dirichlet characters are real.
There is only one primitive character modulo \(4\) defined for all odd positive integers \(n\) by \[\chi_{4}(n)=(-1)^{(n-1) / 2}.\]
There are two primitive characters modulo \(8\) defined for all odd positive integers \(n\) by \[\chi_{8}(n)=(-1)^{(n^{2}-1) / 8}, \quad \text { and } \quad \chi_{4} \chi_{8}(n)=(-1)^{(n-1) / 2+(n^{2}-1) / 8}.\]
If \(q=p^{\alpha}\) is a prime power, the only real primitive characters of conductor \(q\) are \(\chi_{4}, \chi_{8}, \chi_{4} \chi_{8}\) and \(\chi_{p}\). Every real primitive character can be obtained as the product of these characters.
This implies that the conductor of a real primitive character is of the form \(1, m, 4 m\) or \(8 m\) where \(m\) is a positive odd squarefree integer.
Let \(a, q\in\mathbb Z_+\) be coprime integers. Then there are infinitely many prime numbers \(p\in\mathbb P\) such that \(p \equiv a\pmod q\).
Around 1837, Dirichlet succeeded in using a generalization of Euler’s proof \(\sum_{p \in\mathbb P} \frac{1}{p}=\infty\) and some group-theoretic tools.
More precisely, Dirichlet proved the divergence of the series \[\sum_{p \equiv a (\bmod q)} \frac{1}{p}\] by discovering a clever expression for the characteristic function \[\mathbf{1}_{q, a}(n)= \begin{cases}1, & \text { if } n \equiv a \pmod q,\\ 0, & \text { otherwise }. \end{cases}\] and showing that \[\lim _{\sigma \rightarrow 1^{+}} \sum_{p\in\mathbb P} \frac{\mathbf{1}_{q, a}(p)}{p^{\sigma}}=\infty.\]
Proposition 3. Let \(a, q\in\mathbb Z_+\) be coprime integers and define \(\mathbf{1}_{q, a}(n)\) as \[\mathbf{1}_{q, a}(n)= \begin{cases}1, & \text { if } n \equiv a \pmod q, \\ 0, & \text { otherwise. } \end{cases}\] For all \(n\in\mathbb Z_+\) we have \[\mathbf{1}_{q, a}(n)=\frac{1}{\varphi(q)} \sum_{\chi(\bmod q)} \chi(n)\overline{\chi}(a)\] where the summation is taken over all Dirichlet characters modulo \(q\).
Proof.
This readily follows from the identity \(a,n\in \mathbb Z\), then \[\sum_{\chi(\bmod q)} \chi(n) \overline{\chi}(a)= \begin{cases}\varphi(q) & \text { if }(n, q)=(a, q)=1 \text { and } n \equiv a \pmod q, \\ 0 & \text { otherwise, } \end{cases}\] which is simply the orthogonality relation. $$\tag*{$\blacksquare$}$$
Proposition 4. Let \(a, q\in\mathbb Z_+\) be coprime integers and \(N>1\) be an integer. Then we have \[\sum_{\substack{p \le N \\ p \equiv a(\bmod q)}} \frac{1}{p} =\frac{1}{\varphi(q)} \sum_{\substack{p \leqslant N \\(p, q)=1}} \frac{1}{p} +\frac{1}{\varphi(q)} \sum_{\substack{\chi(\bmod q)\\\chi \neq \chi_{0}}} \overline{\chi}(a) \sum_{p \leqslant N} \frac{\chi(p)}{p}\]
Proof.
By the previous proposition we obtain \[\sum_{\substack{p \leqslant N \\ p \equiv a(\bmod q)}} \frac{1}{p} =\sum_{p \leqslant N} \frac{\mathbf{1}_{q, a}(p)}{p} =\frac{1}{\varphi(q)} \sum_{\chi(\bmod q)} \overline{\chi}(a) \sum_{p \leqslant N} \frac{\chi(p)}{p}.\]
We split the sum according to \(\chi=\chi_{0}\) or \(\chi \neq \chi_{0}\), leading to the claim.
This completes the proof.$$\tag*{$\blacksquare$}$$
Let \(a, q\in\mathbb Z_+\) be coprime integers and \(N>1\) be an integer. Then for any arithmetic function \(f:\mathbb N\to \mathbb C\) the following holds \[\sum_{\substack{n \leq N \\ n \equiv a(\bmod q)}} f(n)=\frac{1}{\varphi(q)} \sum_{\substack{n \leq N \\(n, q)=1}} f(n)+\frac{1}{\varphi(q)} \sum_{\substack{\chi(\bmod q)\\\chi \neq \chi_{0}}} \overline{\chi}(a) \sum_{n \leqslant N} \chi(n) f(n) .\]
Proposition 5. For all non-principal Dirichlet characters \(\chi\) modulo \(q\) and all non-negative integers \(M<N\), we have \[\left|\sum_{n=M+1}^{N} \chi(n)\right| \le \varphi(q).\]
Let \(K=q\lfloor(N-M-1) / q\rfloor\). By the orthognality relation, for all \(\chi \neq \chi_{0}\), we have \[\sum_{a(\bmod q)} \chi(a)=0.\]
Hence, by periodicity, we obtain \[\sum_{n=M+1}^{M+K} \chi(n)=\sum_{j=1}^{K / q} \sum_{n=M+1+(j-1) q}^{M+j q} \chi(n)=\sum_{j=1}^{K / q} \sum_{n=M+1}^{M+q} \chi(n)=0.\]
The interval \((M+K, N]\) contains at most \(q\) integers \(n_{1}, \ldots, n_{r}\) with \(r \in [q]\). Denoting by \(n_{i}\) the residue class of the integer \(n_{i}\) in \((\mathbb{Z} / q \mathbb{Z})^{\times}\), we obtain \[\bigg|\sum_{n=M+1}^{N} \chi(n)\bigg| \leqslant \sum_{\substack{i=1 \\\left(n_{i}, q\right)=1}}^{r}\left|\chi\left(n_{i}\right)\right| \leqslant \sum_{\substack{n \leqslant q \\(n, q)=1}} 1=\varphi(q)\] as asserted. $$\tag*{$\blacksquare$}$$
Proposition 6. Let \(F \in C^{1}((1,+\infty))\) be a decreasing function such that \(F>0\) and \(\lim_{x\to\infty}F(x)=0\). For all non-principal Dirichlet characters \(\chi\) modulo \(q\) and all real numbers \(x \geqslant 1\), we have \[\bigg|\sum_{k>x} \chi(k) F(k)\bigg| \leqslant 2 q F(x).\]
Proof.
If \((b_{k})_{k\in\mathbb Z}\subseteq \mathbb R_+\) is a monotone, then \[\Big|\sum_{k=m+1}^{n} a_{k} b_{k}\Big| \le 2 \max \{b_{m+1},b_{n}\} \max _{m \leqslant k \leqslant n}|s_k|,\] where \(s_k=\sum_{x<n\le k}\chi(n)\).
Applying this with \(a_k=\chi(k)\) and \(b_k=F(k)\) the result follows.
This completes the proof.$$\tag*{$\blacksquare$}$$
\(L\) -functions. Let \(\chi\) be a Dirichlet character modulo \(q \geqslant 2\). The \(L\)-function, or \(L\)-series, corresponding to \(\chi\) is the Dirichlet series of \(\chi\), given by \[L(s, \chi)=\sum_{n=1}^{\infty} \frac{\chi(n)}{n^{s}}\quad \text{ for all } \quad s=\sigma+i t \in \mathbb{C} \quad \text{ with } \sigma>1.\]
By the absolute convergence of \(L(s, \chi)\) for \(s\in \mathbb C\) with \(\sigma>1\) we have \[L(s, \chi)=\prod_{p\in\mathbb P}\left(1-\frac{\chi(p)}{p^{s}}\right)^{-1}.\]
If \(\chi=\chi_0\) we also have \[L\left(s, \chi_{0}\right)=\sum_{\substack{n=1 \\(n, q)=1}}^{\infty} \frac{1}{n^{s}}=\zeta(s) \prod_{p \mid q}\left(1-\frac{1}{p^{s}}\right).\]
If \(\chi\neq\chi_0\) then \(L(s, \chi)\) converges for all \(s\in \mathbb C\) with \(\sigma>0\), by the previous proposition.
If \(\chi \neq \chi_{0}\) is a non-principal Dirichlet character modulo \(q\) satisfying \[L(1, \chi) \neq 0,\] then the series \[\sum_{p\in\mathbb P} \frac{\chi(p)}{p}\] converges.
Proof. Let \(N \geqslant 2\) be an integer. We will estimate in two different ways the sum \[\sum_{1\le n \leqslant N} \frac{\chi(n) \log n}{n}.\]
Since \(\log n=\Lambda\star 1(n)\), we have \[\sum_{n \leqslant N} \frac{\chi(n) \log n}{n}=\sum_{n \leqslant N} \frac{\chi(n)}{n} \sum_{d \mid n} \Lambda(d).\]
Interchanging the order of summation and using multiplicativity of the Dirichlet characters we can write \[\begin{aligned} \sum_{n \leqslant N} \frac{\chi(n) \log n}{n} & =\sum_{d \leqslant N} \Lambda(d) \sum_{\substack{n \leqslant N \\ d \mid n}} \frac{\chi(n)}{n} \\ & =\sum_{d \leqslant N} \Lambda(d) \sum_{k \leqslant N / d} \frac{\chi(k d)}{k d} \\ & =\sum_{d \leqslant N} \frac{\chi(d) \Lambda(d)}{d} \sum_{k \leqslant N / d} \frac{\chi(k)}{k}\\ &=L(1, \chi) \sum_{d \leqslant N} \frac{\chi(d) \Lambda(d)}{d}-\sum_{d \leqslant N} \frac{\chi(d) \Lambda(d)}{d} \sum_{k>N / d} \frac{\chi(k)}{k}. \end{aligned}\]
Since \(L(1, \chi) \neq 0\), we infer that \[\sum_{d \leqslant N} \frac{\chi(d) \Lambda(d)}{d}=\frac{1}{L(1, \chi)}\bigg(\sum_{n \leqslant N} \frac{\chi(n) \log n}{n}+\sum_{d \leqslant N} \frac{\chi(d) \Lambda(d)}{d} \sum_{k>N / d} \frac{\chi(k)}{k}\bigg).\]
By the previous proposition, since \(|\sum_{k>N / d} \frac{\chi(k)}{k}|\le 2dq/N\), we obtain \[\bigg|\sum_{d \in [N]} \frac{\chi(d) \Lambda(d)}{d} \sum_{k>N / d} \frac{\chi(k)}{k}\bigg| \le \frac{2 q}{N} \sum_{d \in [N]} \Lambda(d)=\frac{2 q \Psi(N)}{N}.\]
By using \(\Psi(N)<2N\) we have \[\big|\sum_{d \in [N]} \frac{\chi(d) \Lambda(d)}{d} \sum_{k>N / d} \frac{\chi(k)}{k}\bigg|<4 q.\]
Inserting this bound to the last identity in the previous slide we have \[\begin{aligned} \bigg|\sum_{d \in [N]} \frac{\chi(d) \Lambda(d)}{d}\bigg| <\frac{1}{|L(1, \chi)|}\bigg(\bigg|\sum_{n \in [N]} \frac{\chi(n) \log n}{n}\bigg|+4 q\bigg). \end{aligned}\]
By partial summation we obtain \[\begin{aligned} &\sum_{n \leqslant N} \frac{\chi(n) \log n}{n}= \frac{\chi(2) \log 2}{2}+\sum_{3 \leqslant n \leqslant N} \frac{\chi(n) \log n}{n} \\ &\quad = \frac{\chi(2) \log 2}{2}+\frac{\log N}{N} \sum_{3 \leqslant n \leqslant N} \chi(n) +\int_{3}^{N} \frac{\log t-1}{t^{2}}\bigg(\sum_{3 \leqslant n \leqslant t} \chi(n)\bigg) d t. \end{aligned}\]
Using the fact that \(|\sum_{3 \leqslant n \leqslant N} \chi(n)|< q\), we obtain \[\begin{aligned} \bigg|\sum_{n \leqslant N} \frac{\chi(n) \log n}{n}\bigg| &\le \frac{\log 2}{2} +q\bigg(\frac{\log N}{N}+\int_{3}^{N} \frac{\log t-1}{t^{2}} dt\bigg)\\ &=\frac{\log 2}{2}+\frac{q \log 3}{3}<q. \end{aligned}\]
Inserting this bound to the last estimates \[\bigg|\sum_{d \leqslant N} \frac{\chi(d) \Lambda(d)}{d}\bigg|<\frac{5 q}{|L(1, \chi)|}.\]
We also have \[\sum_{p \leqslant N} \frac{\chi(p) \log p}{p}=\sum_{d \leqslant N} \frac{\chi(d) \Lambda(d)}{d}-\sum_{p \leqslant N} \log p \sum_{\alpha=2}^{\lfloor \log N / \log p\rfloor} \frac{\chi\left(p^{\alpha}\right)}{p^{\alpha}}.\]
The second sum is bounded since \[\bigg|\sum_{p \leqslant N} \log p \sum_{\alpha=2}^{\lfloor\log N / \log p\rfloor} \frac{\chi\left(p^{\alpha}\right)}{p^{\alpha}}\bigg| \leqslant \sum_{p \leqslant N} \log p \sum_{\alpha=2}^{\lfloor\log N / \log p\rfloor} \frac{1}{p^{\alpha}} \leqslant \sum_{p\in\mathbb P} \frac{\log p}{p(p-1)}<C,\] for some constant \(C\in\mathbb R_+\), which in fact we can take \(C=1\).
Now by partial summation we can write \[\sum_{p \leqslant N} \frac{\chi(p)}{p}=\frac{1}{\log N} \sum_{p \leqslant N} \frac{\chi(p) \log p}{p}+\int_{2}^{N}\bigg(\sum_{p \leqslant t} \frac{\chi(p) \log p}{p}\bigg) \frac{dt}{t(\log t)^{2}},\] so that by above we obtain \[\begin{aligned} \bigg|\sum_{p \leqslant N} \frac{\chi(p)}{p}\bigg| < & \frac{1}{\log N}\bigg(\bigg|\sum_{d \leqslant N} \frac{\chi(d) \Lambda(d)}{d}\bigg|+C\bigg) \\ & +\int_{2}^{N}\bigg(\bigg|\sum_{d \leqslant t} \frac{\chi(d) \Lambda(d)}{d}\bigg|+C\bigg) \frac{d t}{t(\log t)^{2}}. \end{aligned}\]
The estimate \(|\sum_{d \leqslant N} \frac{\chi(d) \Lambda(d)}{d}|<\frac{5 q}{|L(1, \chi)|}\) provides \[\bigg|\sum_{p \leqslant N} \frac{\chi(p)}{p}\bigg|<\frac{1}{\log 2}\bigg(\frac{5 q}{|L(1, \chi)|}+C\bigg),\] which completes the proof.$$\tag*{$\blacksquare$}$$
If \(\chi \neq \chi_{0}\) is a non-principal Dirichlet character modulo \(q\), then \[L(1, \chi) \neq 0.\]
Proof.
We form the product of all \(L(s, \chi)\): \[F(s)=\prod_{\chi(\bmod q)} L(s, \chi)=\prod_{p \nmid q} \prod_{\chi(\bmod q)} \frac{1}{1-\left(\chi(p) / p^{s}\right)} \quad \text { for all } \quad s>1.\]
If \(m\) is the smallest positive integer such that \(p^{m} \equiv 1\pmod q\), then \(\chi(p)\) is an \(m\)-th root of unity, say \(\varepsilon\). All such \(\varepsilon\) occur with the same multiplicity \(l=\phi(q) / m\) as \(\chi\) runs over all the characters modulo \(q\).
This means that \[\prod_{\chi(\bmod q)}\left(1-\frac{\chi(p)}{p^{s}}\right)=\prod_{\varepsilon}\left(1-\frac{\varepsilon}{p^{s}}\right)^{l},\] where \(\varepsilon\) runs over all the \(m\)-th roots of unity.
Now since \[\prod_{\varepsilon}(x-\varepsilon)=x^{m}-1,\] we have that \[\prod_{\varepsilon}\left(1-\frac{\varepsilon}{x}\right)=1-\frac{1}{x^{m}}.\]
Therefore \[\prod_{\varepsilon}\left(1-\frac{\varepsilon}{p^{s}}\right)=1-\frac{1}{p^{m s}},\] so that \[\prod_{\chi(\bmod q)}\left(1-\frac{\chi(p)}{p^{s}}\right)=\left(1-\frac{1}{p^{m s}}\right)^{l} \leq 1-\frac{1}{p^{lm s}}.\]
Here we used the inequality \((1-x)^{n} \leq 1-x^{n}\), which is clearly valid for all \(n \geq 1\) and \(x \in[0,1]\). Setting \(h=\phi(q)=lm\), we thus have \[F(s)=\prod_{\chi(\bmod q)} L(s, \chi) \geq \prod_{p \nmid q} \frac{1}{1-\left(1 / p^{h s}\right)} =\zeta(h s) \prod_{p \mid q}\left(1-\frac{1}{p^{h s}}\right).\]
This implies that for \(s>1\), that \[\begin{aligned} F(s)=\prod_{\chi(\bmod q)} L(s, \chi) \geq \zeta(h s) \prod_{p \mid q}\left(1-\frac{1}{p}\right)>\frac{\phi(q)}{q}. \qquad (*) \end{aligned}\]
We show that (*) precludes that two or more of the \(L(1, \chi)\) ’s vanish.
Indeed, assume that \(L\left(1, \chi_{1}\right)=L\left(1, \chi_{2}\right)=0\) for two characters \(\chi_{1}\) and \(\chi_{2}\). Clearly, \(\chi_{1}, \chi_{2} \neq \chi_{0}\). Then \(F(s)\) would contain, besides other factors that are continuous (thus bounded) at \(s=1\), the factor \[\begin{aligned} &L\left(s, \chi_{0}\right) L\left(s, \chi_{1}\right) L\left(s, \chi_{2}\right)\\ &\qquad =L\left(s, \chi_{0}\right)(s-1)^{2}\left(L^{\prime}\left(1, \chi_{1}\right)+\eta_{1}(s)\right)\left(L^{\prime}\left(1, \chi_{2}\right)+\eta_{2}(s)\right), \end{aligned}\] where \(\lim_{s\to 1}\eta_{1}(s)=\lim_{s\to 1}\eta_{2}(s)= 0\).
The Riemann zeta function has a simple pole at \(s=1\), and \(L\left(s, \chi_{0}\right)=\zeta(s) \prod_{p \mid k}\big(1-\frac{1}{p^{s}}\big)\), thus \[\lim_{s\to 1}(s-1) L\left(s, \chi_{0}\right) = \phi(q) / q,\] and we would get that \(\lim_{s\to 1}F(s) = 0\), which would contradict (*).
If now \(L(1, \chi)=0\) for some complex character \(\chi\) (that is, which assumes complex non-real values), then \(\overline{\chi}\) is also a character of modulus \(q\) which is distinct from \(\chi\), and clearly \(L(1, \overline{\chi})=\overline{L(1, \chi)}=0\). But we have just seen that this is impossible.
Thus, if \(L(s, \chi)=0\) for some \(\chi\), then \(\chi\) is unique and real (it assumes only the values \(\pm 1)\). In order to complete the proof, we will show that \(L(1, \chi) \neq 0\) for all real non-principal characters as well.
From now on, we assume that \(\chi\) is real non-principal character.
Note that \(\chi: \mathbb{N} \rightarrow\{0, \pm 1\}\) is, in particular, a multiplicative function. So, if we let \[f(n)=\sum_{d \mid n} \chi(d)=1 * \chi,\] then \(f\) is also multiplicative.
Note further that since \(\chi(p)= \pm 1\), we get that \[f\left(p^{l}\right)=\chi(1)+\chi(p)+\cdots+\chi\left(p^{l}\right) \geq 0\] for all \(l\in\mathbb Z_+\), and, in fact, \(f\left(p^{l}\right) \geq 1\) whenever \(2\mid l\).
Using the fact that \(f\) is multiplicative, we get that \(f(m^{2}) \geq 1\). Thus, \[\sum_{n=1}^{\infty} \frac{f(n)}{n^{1 / 2}} \geq \sum_{m \in\mathbb Z_+} \frac{f(m^{2})}{m} \geq \sum_{m \in\mathbb Z_+} \frac{1}{m}=\infty.\]
Let us take a closer look at this divergence. We have \[G(x)=\sum_{n \leq x} \frac{f(n)}{n^{1 / 2}}=\sum_{n \leq x} \frac{1}{n^{1 / 2}} \sum_{d \mid n} \chi(d)=\sum_{t d \leq x} \frac{\chi(d)}{(t d)^{1 / 2}}.\]
By using the Dirichlet hyperbola principle and splitting summation according to whether \(d \leq \sqrt{x}\) or \(d>\sqrt{x}\), we obtain \[\begin{aligned} G(x) & =\sum_{1 \leq d \leq \sqrt{x}} \frac{\chi(d)}{d^{1 / 2}} \sum_{1 \leq t \leq x / d} \frac{1}{t^{1 / 2}}+\sum_{t \leq \sqrt{x}} \frac{1}{t^{1 / 2}} \sum_{\sqrt{x+1} \leq d \leq x / t} \frac{\chi(d)}{d^{1 / 2}}\\ & =G_{1}(x)+G_{2}(x). \end{aligned}\]
By Abel’s summation formula, we have that \[\begin{aligned} \sum_{1 \leq t \leq y} \frac{1}{t^{1 / 2}}= & \frac{\lfloor y\rfloor}{y^{1 / 2}}-\int_{1}^{t}\lfloor t\rfloor \left(\frac{1}{t^{1 / 2}}\right)' d t=\frac{y-\{y\}}{y^{1 / 2}}+\frac{1}{2} \int_{1}^{y} \frac{t-\{t\}}{t^{3 / 2}} d t \\ = & y^{1 / 2}+O\left(\frac{1}{y^{1 / 2}}\right)+\frac{1}{2} \int_{1}^{y} \frac{d t}{t^{1 / 2}}-\frac{1}{2} \int_{1}^{y} \frac{\{t\}}{t^{3 / 2}} d t \\ = & 2y^{1 / 2}-1-\frac{1}{2}\left(\int_{1}^{\infty} \frac{\{t\}}{t^{3 / 2}} d t-\int_{y}^{\infty} \frac{\{t\}}{t^{3 / 2}} d t\right) +O\left(\frac{1}{y^{1 / 2}}\right) \\ = & 2 y^{1 / 2}+\left(-1-\frac{1}{2} \int_{1}^{\infty} \frac{\{t\}}{t^{3 / 2}} d t\right)+O\left(\frac{1}{y^{1 / 2}}+\int_{y}^{\infty} \frac{d t}{t^{3 / 2}}\right) \\ = & 2 y^{1 / 2}+C+O\left(\frac{1}{y^{1 / 2}}\right), \end{aligned}\] where \(C\) is the constant given by \[C=-1-\frac{1}{2} \int_{1}^{\infty} \frac{\{t\}}{t^{3 / 2}} d t.\]
Hence, we obtain \[\begin{aligned} G_{1}(x) & =\sum_{1 \leq d \leq \sqrt{x}} \frac{\chi(d)}{d^{1 / 2}}\left(2 \sqrt{\frac{x}{d}}+C+O\left(\sqrt{\frac{d}{x}}\right)\right) \\ & =2 \sqrt{x} \sum_{1 \leq d \leq \sqrt{x}} \frac{\chi(d)}{d}+C \sum_{1 \leq d \leq \sqrt{x}} \frac{\chi(d)}{d^{1 / 2}}+O\left(\frac{\sqrt{x}}{\sqrt{x}}\right) \\ & =2 \sqrt{x}\left(\sum_{d=1}^{\infty} \frac{\chi(d)}{d}-\sum_{d>\sqrt{x}} \frac{\chi(d)}{d}\right)+O(1), \end{aligned}\] where we used the fact that \[\begin{gathered} \sum_{1 \leq d \leq \sqrt{x}} \frac{\chi(d)}{d^{1 / 2}}=L(1 / 2, \chi)+o(1)=O(1),\\ \sum_{d>\sqrt{x}} \frac{\chi(d)}{d}=O\left(\frac{1}{\sqrt{x}}\right). \end{gathered}\]
Hence, we conclude that \[G_{1}(x)=2 \sqrt{x} L(1, \chi)+O(1).\]
We are left with examining the size of \[G_{2}(x)=\sum_{1 \leq t \leq \sqrt{x}} \frac{1}{t^{1 / 2}} \sum_{\sqrt{x+1} \leq d \leq x / t} \frac{\chi(d)}{d^{1 / 2}}.\]
The inner sums are bounded by \[\bigg|\sum_{\sqrt{x+1} \leq d \leq x / t} \frac{\chi(d)}{d^{1 / 2}}\bigg| =O \bigg(\frac{1}{x^{1 / 4}}\bigg).\] Therefore, \(G_{2}(x)=O(1)\), since \[\begin{aligned} G_{2}(x) & =O \bigg(\frac{1}{x^{1 / 4}} \sum_{1 \leq t \leq x^{1 / 2}} \frac{1}{t^{1 / 2}}\bigg) =O \bigg( \frac{1}{x^{1 / 4}}\bigg(1+\int_{1}^{\sqrt{x}} \frac{d t}{t^{1 / 2}}\bigg) \bigg)\\ & =\frac{1}{x^{1 / 4}}\big(1+2x^{1 / 4}\big) =O(1), \end{aligned}\]
Combining the above estimates, we obtain \[G(x)=G_{1}(x)+G_{2}(x)=2 \sqrt{x} L(1, \chi)+O(1)\]
Since we know that \(G(x)\) tends to infinity with \(x\) and, plainly, that this can happen only if \(L(1, \chi) \neq 0\). $$\tag*{$\blacksquare$}$$
Let \(a, q\in\mathbb Z_+\) be coprime integers. Then there are infinitely many prime numbers \(p\in\mathbb P\) such that \(p \equiv a\pmod q\).
Proof.
We know that \[\sum_{\substack{p \le N \\ p \equiv a(\bmod q)}} \frac{1}{p} =\frac{1}{\varphi(q)} \sum_{\substack{p \leqslant N \\(p, q)=1}} \frac{1}{p} +\frac{1}{\varphi(q)} \sum_{\substack{\chi(\bmod q)\\\chi \neq \chi_{0}}} \overline{\chi}(a) \sum_{p \leqslant N} \frac{\chi(p)}{p}.\tag{*}\]
It is easy to see that \[\lim_{N\to \infty}\sum_{\substack{p \leqslant N \\(p, q)=1}} \frac{1}{p}=\infty,\] since it only differs from \(\sum_{p \in \mathbb P_{\le N}} 1 / p\) by a finite number of terms.
On the other hand, we have shown \(\sum_{p \leqslant N} \frac{\chi(p)}{p}\) converges, thus we have \[\bigg|\sum_{\substack{\chi(\bmod q)\\\chi \neq \chi_{0}}} \overline{\chi}(a) \sum_{p \leqslant N} \frac{\chi(p)}{p}\bigg|=O(1).\]
Therefore, the series on the left-hand side of (*) must diverge, and consequently the Dirichlet’s theorem follows as desired.$$\tag*{$\blacksquare$}$$