Dirichlet character

Relation to group characters

The word "character" is used several ways in mathematics. In this section it refers to a homomorphism from a group ${\displaystyle G}$ (written multiplicatively) to the multiplicative group of the field of complex numbers:

${\displaystyle \eta :G\to {\mathbb{C}}^{\times },\eta (gh)=\eta (g)\eta (h),\eta (g^{-1})=\eta (g{)}^{-1}.}$

The set of characters is denoted ${\displaystyle \hat{G}.}$ If the product of two characters is defined by pointwise multiplication ${\displaystyle \eta \theta (a)=\eta (a)\theta (a),}$ the identity by the trivial character ${\displaystyle {\eta }_0(a)=1}$ and the inverse by complex inversion ${\displaystyle {\eta }^{-1}(a)=\eta (a{)}^{-1}}$ then ${\displaystyle \hat{G}}$ becomes an abelian group.

If ${\displaystyle A}$ is a finite abelian group then there is an isomorphism ${\displaystyle A\cong \hat{A}}$, and the orthogonality relations:

${\displaystyle \sum _{a\in A}\eta (a)=\{\begin{array}{ll}|A|& \text{ if }\eta ={\eta }_0\\ 0& \text{ if }\eta \ne {\eta }_0\end{array}}$ and ${\displaystyle \sum _{\eta \in \hat{A}}\eta (a)=\{\begin{array}{ll}|A|& \text{ if }a=1\\ 0& \text{ if }a\ne 1.\end{array}}$

The elements of the finite abelian group ${\displaystyle (\mathbb{Z}/m\mathbb{Z}{)}^{\times }}$ are the residue classes ${\displaystyle [a]=\{x:x\equiv a(\mathrm{mod}m)\}}$ where ${\displaystyle (a,m)=1.}$

A group character ${\displaystyle \rho :(\mathbb{Z}/m\mathbb{Z}{)}^{\times }\to {\mathbb{C}}^{\times }}$ can be extended to a Dirichlet character ${\displaystyle \chi :\mathbb{Z}\to \mathbb{C}}$ by defining

${\displaystyle \chi (a)=\{\begin{array}{lll}0& \text{if }[a]\notin (\mathbb{Z}/m\mathbb{Z}{)}^{\times }& \text{i.e. }(a,m)>1\\ \rho ([a])& \text{if }[a]\in (\mathbb{Z}/m\mathbb{Z}{)}^{\times }& \text{i.e. }(a,m)=1,\end{array}}$

and conversely, a Dirichlet character mod ${\displaystyle m}$ defines a group character on ${\displaystyle (\mathbb{Z}/m\mathbb{Z}{)}^{\times }.}$

Paraphrasing Davenport, Dirichlet characters can be regarded as a particular case of Abelian group characters. But this article follows Dirichlet in giving a direct and constructive account of them. This is partly for historical reasons, in that Dirichlet's work preceded by several decades the development of group theory, and partly for a mathematical reason, namely that the group in question has a simple and interesting structure which is obscured if one treats it as one treats the general Abelian group.

Elementary facts

4) Since ${\displaystyle gcd(1,m)=1,}$ property 2) says ${\displaystyle \chi (1)\ne 0}$ so it can be canceled from both sides of ${\displaystyle \chi (1)\chi (1)=\chi (1\times 1)=\chi (1)}$:

${\displaystyle \chi (1)=1.}$

5) Property 3) is equivalent to

if ${\displaystyle a\equiv b(\mathrm{mod}m)}$ then ${\displaystyle \chi (a)=\chi (b).}$

6) Property 1) implies that, for any positive integer ${\displaystyle n}$

${\displaystyle \chi (a^n)=\chi (a{)}^n.}$

7) Euler's theorem states that if ${\displaystyle gcd(a,m)=1}$ then ${\displaystyle a^{\varphi (m)}\equiv 1(\mathrm{mod}m).}$ Therefore,

${\displaystyle \chi (a{)}^{\varphi (m)}=\chi (a^{\varphi (m)})=\chi (1)=1.}$

That is, the nonzero values of ${\displaystyle \chi (a)}$ are ${\displaystyle \varphi (m)}$-th roots of unity:

${\displaystyle \chi (a)=\{\begin{array}{ll}0& \text{if }gcd(a,m)>1\\ {\zeta }_{\varphi (m)}^r& \text{if }gcd(a,m)=1\end{array}}$

for some integer ${\displaystyle r}$ which depends on ${\displaystyle \chi ,\zeta ,}$ and ${\displaystyle a}$. This implies there are only a finite number of characters for a given modulus.

8) If ${\displaystyle \chi }$ and ${\displaystyle {\chi }^{\prime }}$ are two characters for the same modulus so is their product ${\displaystyle \chi {\chi }^{\prime },}$ defined by pointwise multiplication:

${\displaystyle \chi {\chi }^{\prime }(a)=\chi (a){\chi }^{\prime }(a)}$ (${\displaystyle \chi {\chi }^{\prime }}$ obviously satisfies 1-3).

The principal character is an identity:

${\displaystyle \chi {\chi }_0(a)=\chi (a){\chi }_0(a)=\{\begin{array}{lll}0\times 0& =\chi (a)& \text{if }gcd(a,m)>1\\ \chi (a)\times 1& =\chi (a)& \text{if }gcd(a,m)=1.\end{array}}$

9) Let ${\displaystyle a^{-1}}$ denote the inverse of ${\displaystyle a}$ in ${\displaystyle (\mathbb{Z}/m\mathbb{Z}{)}^{\times }}$. Then

${\displaystyle \chi (a)\chi (a^{-1})=\chi (a{a}^{-1})=\chi (1)=1,}$ so ${\displaystyle \chi (a^{-1})=\chi (a{)}^{-1},}$ which extends 6) to all integers.

The complex conjugate of a root of unity is also its inverse (see here for details), so for ${\displaystyle (a,m)=1}$

${\displaystyle \overline{\chi }(a)=\chi (a{)}^{-1}=\chi (a^{-1}).}$ (${\displaystyle \overline{\chi }}$ also obviously satisfies 1-3).

Thus for all integers ${\displaystyle a}$

${\displaystyle \chi (a)\overline{\chi }(a)=\{\begin{array}{ll}0& \text{if }gcd(a,m)>1\\ 1& \text{if }gcd(a,m)=1\end{array};}$ in other words ${\displaystyle \chi \overline{\chi }={\chi }_0}$.

10) The multiplication and identity defined in 8) and the inversion defined in 9) turn the set of Dirichlet characters for a given modulus into a finite abelian group.

The group of characters

There are three different cases because the groups ${\displaystyle (\mathbb{Z}/m\mathbb{Z}{)}^{\times }}$ have different structures depending on whether ${\displaystyle m}$ is a power of 2, a power of an odd prime, or the product of prime powers.

Powers of odd primes

If ${\displaystyle q=p^k}$ is an odd number ${\displaystyle (\mathbb{Z}/q\mathbb{Z}{)}^{\times }}$ is cyclic of order ${\displaystyle \varphi (q)}$; a generator is called a primitive root mod ${\displaystyle q}$. Let ${\displaystyle g_q}$ be a primitive root and for ${\displaystyle (a,q)=1}$ define the function ${\displaystyle {\nu }_q(a)}$ (the index of ${\displaystyle a}$) by

${\displaystyle a\equiv g_q^{{\nu }_q(a)}(\mathrm{mod}q),}$

${\displaystyle 0\le {\nu }_q<\varphi (q).}$

For ${\displaystyle (ab,q)=1,a\equiv b(\mathrm{mod}q)}$ if and only if ${\displaystyle {\nu }_q(a)={\nu }_q(b).}$ Since

${\displaystyle \chi (a)=\chi (g_q^{{\nu }_q(a)})=\chi (g_q{)}^{{\nu }_q(a)},}$ ${\displaystyle \chi }$ is determined by its value at ${\displaystyle g_q.}$

Let ${\displaystyle {\omega }_q={\zeta }_{\varphi (q)}}$ be a primitive ${\displaystyle \varphi (q)}$-th root of unity. From property 7) above the possible values of ${\displaystyle \chi (g_q)}$ are ${\displaystyle {\omega }_q,{\omega }_q^2,...{\omega }_q^{\varphi (q)}=1.}$ These distinct values give rise to ${\displaystyle \varphi (q)}$ Dirichlet characters mod ${\displaystyle q.}$ For ${\displaystyle (r,q)=1}$ define ${\displaystyle {\chi }_{q,r}(a)}$ as

${\displaystyle {\chi }_{q,r}(a)=\{\begin{array}{ll}0& \text{if }gcd(a,q)>1\\ {\omega }_q^{{\nu }_q(r){\nu }_q(a)}& \text{if }gcd(a,q)=1.\end{array}}$

Then for ${\displaystyle (rs,q)=1}$ and all ${\displaystyle a}$ and ${\displaystyle b}$

${\displaystyle {\chi }_{q,r}(a){\chi }_{q,r}(b)={\chi }_{q,r}(ab),}$ showing that ${\displaystyle {\chi }_{q,r}}$ is a character and

${\displaystyle {\chi }_{q,r}(a){\chi }_{q,s}(a)={\chi }_{q,rs}(a),}$ which gives an explicit isomorphism ${\displaystyle \widehat{(\mathbb{Z}/p^k\mathbb{Z}{)}^{\times }}\cong (\mathbb{Z}/p^k\mathbb{Z}{)}^{\times }.}$

Examples m = 3, 5, 7, 9

2 is a primitive root mod 3. (${\displaystyle \varphi (3)=2}$)

${\displaystyle 2^1\equiv 2,2^2\equiv 2^0\equiv 1(\mathrm{mod}3),}$

so the values of ${\displaystyle {\nu }_3}$ are

${\displaystyle \overline{)\begin{array}{ccc}a& 1& 2\\ {\nu }_3(a)& 0& 1\end{array}}}$.

The nonzero values of the characters mod 3 are

${\displaystyle \overline{)\begin{array}{ccc}& 1& 2\\ {\chi }_{3,1}& 1& 1\\ {\chi }_{3,2}& 1& -1\end{array}}}$

2 is a primitive root mod 5. (${\displaystyle \varphi (5)=4}$)

${\displaystyle 2^1\equiv 2,2^2\equiv 4,2^3\equiv 3,2^4\equiv 2^0\equiv 1(\mathrm{mod}5),}$

so the values of ${\displaystyle {\nu }_5}$ are

${\displaystyle \overline{)\begin{array}{ccccc}a& 1& 2& 3& 4\\ {\nu }_5(a)& 0& 1& 3& 2\end{array}}}$.

The nonzero values of the characters mod 5 are

${\displaystyle \overline{)\begin{array}{ccccc}& 1& 2& 3& 4\\ {\chi }_{5,1}& 1& 1& 1& 1\\ {\chi }_{5,2}& 1& i& -i& -1\\ {\chi }_{5,3}& 1& -i& i& -1\\ {\chi }_{5,4}& 1& -1& -1& 1\end{array}}}$

3 is a primitive root mod 7. (${\displaystyle \varphi (7)=6}$)

${\displaystyle 3^1\equiv 3,3^2\equiv 2,3^3\equiv 6,3^4\equiv 4,3^5\equiv 5,3^6\equiv 3^0\equiv 1(\mathrm{mod}7),}$

so the values of ${\displaystyle {\nu }_7}$ are

${\displaystyle \overline{)\begin{array}{ccccccc}a& 1& 2& 3& 4& 5& 6\\ {\nu }_7(a)& 0& 2& 1& 4& 5& 3\end{array}}}$.

The nonzero values of the characters mod 7 are (${\displaystyle \omega ={\zeta }_6,{\omega }^3=-1}$)

${\displaystyle \overline{)\begin{array}{ccccccc}& 1& 2& 3& 4& 5& 6\\ {\chi }_{7,1}& 1& 1& 1& 1& 1& 1\\ {\chi }_{7,2}& 1& -\omega & {\omega }^2& {\omega }^2& -\omega & 1\\ {\chi }_{7,3}& 1& {\omega }^2& \omega & -\omega & -{\omega }^2& -1\\ {\chi }_{7,4}& 1& {\omega }^2& -\omega & -\omega & {\omega }^2& 1\\ {\chi }_{7,5}& 1& -\omega & -{\omega }^2& {\omega }^2& \omega & -1\\ {\chi }_{7,6}& 1& 1& -1& 1& -1& -1\end{array}}}$.

2 is a primitive root mod 9. (${\displaystyle \varphi (9)=6}$)

${\displaystyle 2^1\equiv 2,2^2\equiv 4,2^3\equiv 8,2^4\equiv 7,2^5\equiv 5,2^6\equiv 2^0\equiv 1(\mathrm{mod}9),}$

so the values of ${\displaystyle {\nu }_9}$ are

${\displaystyle \overline{)\begin{array}{ccccccc}a& 1& 2& 4& 5& 7& 8\\ {\nu }_9(a)& 0& 1& 2& 5& 4& 3\end{array}}}$.

The nonzero values of the characters mod 9 are (${\displaystyle \omega ={\zeta }_6,{\omega }^3=-1}$)

${\displaystyle \overline{)\begin{array}{ccccccc}& 1& 2& 4& 5& 7& 8\\ {\chi }_{9,1}& 1& 1& 1& 1& 1& 1\\ {\chi }_{9,2}& 1& \omega & {\omega }^2& -{\omega }^2& -\omega & -1\\ {\chi }_{9,4}& 1& {\omega }^2& -\omega & -\omega & {\omega }^2& 1\\ {\chi }_{9,5}& 1& -{\omega }^2& -\omega & \omega & {\omega }^2& -1\\ {\chi }_{9,7}& 1& -\omega & {\omega }^2& {\omega }^2& -\omega & 1\\ {\chi }_{9,8}& 1& -1& 1& -1& 1& -1\end{array}}}$.

Powers of 2

${\displaystyle (\mathbb{Z}/2\mathbb{Z}{)}^{\times }}$ is the trivial group with one element. ${\displaystyle (\mathbb{Z}/4\mathbb{Z}{)}^{\times }}$ is cyclic of order 2. For 8, 16, and higher powers of 2, there is no primitive root; the powers of 5 are the units ${\displaystyle \equiv 1(\mathrm{mod}4)}$ and their negatives are the units ${\displaystyle \equiv 3(\mathrm{mod}4).}$ For example

${\displaystyle 5^1\equiv 5,5^2\equiv 5^0\equiv 1(\mathrm{mod}8)}$

${\displaystyle 5^1\equiv 5,5^2\equiv 9,5^3\equiv 13,5^4\equiv 5^0\equiv 1(\mathrm{mod}16)}$

${\displaystyle 5^1\equiv 5,5^2\equiv 25,5^3\equiv 29,5^4\equiv 17,5^5\equiv 21,5^6\equiv 9,5^7\equiv 13,5^8\equiv 5^0\equiv 1(\mathrm{mod}32).}$

Let ${\displaystyle q=2^k,k\ge 3}$; then ${\displaystyle (\mathbb{Z}/q\mathbb{Z}{)}^{\times }}$ is the direct product of a cyclic group of order 2 (generated by −1) and a cyclic group of order ${\displaystyle \frac{\varphi (q)}{2}}$ (generated by 5). For odd numbers ${\displaystyle a}$ define the functions ${\displaystyle {\nu }_0}$ and ${\displaystyle {\nu }_q}$ by

${\displaystyle a\equiv (-1{)}^{{\nu }_0(a)}{5}^{{\nu }_q(a)}(\mathrm{mod}q),}$

${\displaystyle 0\le {\nu }_0<2,0\le {\nu }_q<\frac{\varphi (q)}{2}.}$

For odd ${\displaystyle a}$ and ${\displaystyle b,a\equiv b(\mathrm{mod}q)}$ if and only if ${\displaystyle {\nu }_0(a)={\nu }_0(b)}$ and ${\displaystyle {\nu }_q(a)={\nu }_q(b).}$ For odd ${\displaystyle a}$ the value of ${\displaystyle \chi (a)}$ is determined by the values of ${\displaystyle \chi (-1)}$ and ${\displaystyle \chi (5).}$

Let ${\displaystyle {\omega }_q={\zeta }_{\frac{\varphi (q)}{2}}}$ be a primitive ${\displaystyle \frac{\varphi (q)}{2}}$-th root of unity. The possible values of ${\displaystyle \chi ((-1{)}^{{\nu }_0(a)}{5}^{{\nu }_q(a)})}$ are ${\displaystyle \pm {\omega }_q,\pm {\omega }_q^2,...\pm {\omega }_q^{\frac{\varphi (q)}{2}}=\pm 1.}$ These distinct values give rise to ${\displaystyle \varphi (q)}$ Dirichlet characters mod ${\displaystyle q.}$ For odd ${\displaystyle r}$ define ${\displaystyle {\chi }_{q,r}(a)}$ by

${\displaystyle {\chi }_{q,r}(a)=\{\begin{array}{ll}0& \text{if }a\text{ is even}\\ (-1{)}^{{\nu }_0(r){\nu }_0(a)}{\omega }_q^{{\nu }_q(r){\nu }_q(a)}& \text{if }a\text{ is odd}.\end{array}}$

Then for odd ${\displaystyle r}$ and ${\displaystyle s}$ and all ${\displaystyle a}$ and ${\displaystyle b}$

${\displaystyle {\chi }_{q,r}(a){\chi }_{q,r}(b)={\chi }_{q,r}(ab)}$ showing that ${\displaystyle {\chi }_{q,r}}$ is a character and

${\displaystyle {\chi }_{q,r}(a){\chi }_{q,s}(a)={\chi }_{q,rs}(a)}$ showing that ${\displaystyle \widehat{(\mathbb{Z}/2^k\mathbb{Z}{)}^{\times }}\cong (\mathbb{Z}/2^k\mathbb{Z}{)}^{\times }.}$

Examples m = 2, 4, 8, 16

The only character mod 2 is the principal character ${\displaystyle {\chi }_{2,1}}$.

−1 is a primitive root mod 4 (${\displaystyle \varphi (4)=2}$)

${\displaystyle \overline{)\begin{array}{}a& 1& 3\\ {\nu }_0(a)& 0& 1\end{array}}}$

The nonzero values of the characters mod 4 are

${\displaystyle \overline{)\begin{array}{ccc}& 1& 3\\ {\chi }_{4,1}& 1& 1\\ {\chi }_{4,3}& 1& -1\end{array}}}$

−1 is and 5 generate the units mod 8 (${\displaystyle \varphi (8)=4}$)

${\displaystyle \overline{)\begin{array}{}a& 1& 3& 5& 7\\ {\nu }_0(a)& 0& 1& 0& 1\\ {\nu }_8(a)& 0& 1& 1& 0\end{array}}}$.

The nonzero values of the characters mod 8 are

${\displaystyle \overline{)\begin{array}{ccccc}& 1& 3& 5& 7\\ {\chi }_{8,1}& 1& 1& 1& 1\\ {\chi }_{8,3}& 1& 1& -1& -1\\ {\chi }_{8,5}& 1& -1& -1& 1\\ {\chi }_{8,7}& 1& -1& 1& -1\end{array}}}$

−1 and 5 generate the units mod 16 (${\displaystyle \varphi (16)=8}$)

${\displaystyle \overline{)\begin{array}{}a& 1& 3& 5& 7& 9& 11& 13& 15\\ {\nu }_0(a)& 0& 1& 0& 1& 0& 1& 0& 1\\ {\nu }_{16}(a)& 0& 3& 1& 2& 2& 1& 3& 0\end{array}}}$.

The nonzero values of the characters mod 16 are

${\displaystyle \overline{)\begin{array}{}& 1& 3& 5& 7& 9& 11& 13& 15\\ {\chi }_{16,1}& 1& 1& 1& 1& 1& 1& 1& 1\\ {\chi }_{16,3}& 1& -i& -i& 1& -1& i& i& -1\\ {\chi }_{16,5}& 1& -i& i& -1& -1& i& -i& 1\\ {\chi }_{16,7}& 1& 1& -1& -1& 1& 1& -1& -1\\ {\chi }_{16,9}& 1& -1& -1& 1& 1& -1& -1& 1\\ {\chi }_{16,11}& 1& i& i& 1& -1& -i& -i& -1\\ {\chi }_{16,13}& 1& i& -i& -1& -1& -i& i& 1\\ {\chi }_{16,15}& 1& -1& 1& -1& 1& -1& 1& -1\end{array}}}$.

Products of prime powers

Let ${\displaystyle m=p_1^{m_1}{p}_2^{m_2}\cdots p_k^{m_k}=q_1{q}_2\cdots q_k}$ where ${\displaystyle p_1<p_2<\cdots <p_k}$ be the factorization of ${\displaystyle m}$ into prime powers. The group of units mod ${\displaystyle m}$ is isomorphic to the direct product of the groups mod the ${\displaystyle q_i}$:

${\displaystyle (\mathbb{Z}/m\mathbb{Z}{)}^{\times }\cong (\mathbb{Z}/q_1\mathbb{Z}{)}^{\times }\times (\mathbb{Z}/q_2\mathbb{Z}{)}^{\times }\times \cdots \times (\mathbb{Z}/q_k\mathbb{Z}{)}^{\times }.}$

This means that 1) there is a one-to-one correspondence between ${\displaystyle a\in (\mathbb{Z}/m\mathbb{Z}{)}^{\times }}$ and ${\displaystyle k}$-tuples ${\displaystyle (a_1,a_2,\dots ,a_k)}$ where ${\displaystyle a_i\in (\mathbb{Z}/q_i\mathbb{Z}{)}^{\times }}$ and 2) multiplication mod ${\displaystyle m}$ corresponds to coordinate-wise multiplication of ${\displaystyle k}$-tuples:

${\displaystyle ab\equiv c(\mathrm{mod}m)}$ corresponds to

${\displaystyle (a_1,a_2,\dots ,a_k)\times (b_1,b_2,\dots ,b_k)=(c_1,c_2,\dots ,c_k)}$ where ${\displaystyle c_i\equiv a_i{b}_i(\mathrm{mod}{q}_i).}$

The Chinese remainder theorem (CRT) implies that the ${\displaystyle a_i}$ are simply ${\displaystyle a_i\equiv a(\mathrm{mod}{q}_i).}$

There are subgroups ${\displaystyle G_i<(\mathbb{Z}/m\mathbb{Z}{)}^{\times }}$ such that

${\displaystyle G_i\cong (\mathbb{Z}/q_i\mathbb{Z}{)}^{\times }}$ and

${\displaystyle G_i\equiv \{\begin{array}{ll}(\mathbb{Z}/q_i\mathbb{Z}{)}^{\times }& \mathrm{mod}{q}_i\\ \{1\}& \mathrm{mod}{q}_j,j\ne i.\end{array}}$

Then ${\displaystyle (\mathbb{Z}/m\mathbb{Z}{)}^{\times }\cong G_1\times G_2\times ...\times G_k}$ and every ${\displaystyle a\in (\mathbb{Z}/m\mathbb{Z}{)}^{\times }}$ corresponds to a ${\displaystyle k}$-tuple ${\displaystyle (a_1,a_2,...a_k)}$ where ${\displaystyle a_i\in G_i}$ and ${\displaystyle a_i\equiv a(\mathrm{mod}{q}_i).}$ Every ${\displaystyle a\in (\mathbb{Z}/m\mathbb{Z}{)}^{\times }}$ can be uniquely factored as ${\displaystyle a=a_1{a}_2...a_k.}$

If ${\displaystyle {\chi }_{m,\mathrm{\_}}}$ is a character mod ${\displaystyle m,}$ on the subgroup ${\displaystyle G_i}$ it must be identical to some ${\displaystyle {\chi }_{q_i,\mathrm{\_}}}$ mod ${\displaystyle q_i}$ Then

${\displaystyle {\chi }_{m,\mathrm{\_}}(a)={\chi }_{m,\mathrm{\_}}(a_1{a}_2...)={\chi }_{m,\mathrm{\_}}(a_1){\chi }_{m,\mathrm{\_}}(a_2)...={\chi }_{q_1,\mathrm{\_}}(a_1){\chi }_{q_2,\mathrm{\_}}(a_2)...,}$

showing that every character mod ${\displaystyle m}$ is the product of characters mod the ${\displaystyle q_i}$.

For ${\displaystyle (t,m)=1}$ define

${\displaystyle {\chi }_{m,t}={\chi }_{q_1,t}{\chi }_{q_2,t}...}$

Then for ${\displaystyle (rs,m)=1}$ and all ${\displaystyle a}$ and ${\displaystyle b}$

${\displaystyle {\chi }_{m,r}(a){\chi }_{m,r}(b)={\chi }_{m,r}(ab),}$ showing that ${\displaystyle {\chi }_{m,r}}$ is a character and

${\displaystyle {\chi }_{m,r}(a){\chi }_{m,s}(a)={\chi }_{m,rs}(a),}$ showing an isomorphism ${\displaystyle \widehat{(\mathbb{Z}/m\mathbb{Z}{)}^{\times }}\cong (\mathbb{Z}/m\mathbb{Z}{)}^{\times }.}$

Examples m = 15, 24, 40

${\displaystyle (\mathbb{Z}/15\mathbb{Z}{)}^{\times }\cong (\mathbb{Z}/3\mathbb{Z}{)}^{\times }\times (\mathbb{Z}/5\mathbb{Z}{)}^{\times }.}$

The factorization of the characters mod 15 is

${\displaystyle \overline{)\begin{array}{ccccc}& {\chi }_{5,1}& {\chi }_{5,2}& {\chi }_{5,3}& {\chi }_{5,4}\\ {\chi }_{3,1}& {\chi }_{15,1}& {\chi }_{15,7}& {\chi }_{15,13}& {\chi }_{15,4}\\ {\chi }_{3,2}& {\chi }_{15,11}& {\chi }_{15,2}& {\chi }_{15,8}& {\chi }_{15,14}\end{array}}}$

The nonzero values of the characters mod 15 are

${\displaystyle \overline{)\begin{array}{}& 1& 2& 4& 7& 8& 11& 13& 14\\ {\chi }_{15,1}& 1& 1& 1& 1& 1& 1& 1& 1\\ {\chi }_{15,2}& 1& -i& -1& i& i& -1& -i& 1\\ {\chi }_{15,4}& 1& -1& 1& -1& -1& 1& -1& 1\\ {\chi }_{15,7}& 1& i& -1& i& -i& 1& -i& -1\\ {\chi }_{15,8}& 1& i& -1& -i& -i& -1& i& 1\\ {\chi }_{15,11}& 1& -1& 1& 1& -1& -1& 1& -1\\ {\chi }_{15,13}& 1& -i& -1& -i& i& 1& i& -1\\ {\chi }_{15,14}& 1& 1& 1& -1& 1& -1& -1& -1\end{array}}}$.

${\displaystyle (\mathbb{Z}/24\mathbb{Z}{)}^{\times }\cong (\mathbb{Z}/8\mathbb{Z}{)}^{\times }\times (\mathbb{Z}/3\mathbb{Z}{)}^{\times }.}$ The factorization of the characters mod 24 is

${\displaystyle \overline{)\begin{array}{ccccc}& {\chi }_{8,1}& {\chi }_{8,3}& {\chi }_{8,5}& {\chi }_{8,7}\\ {\chi }_{3,1}& {\chi }_{24,1}& {\chi }_{24,19}& {\chi }_{24,13}& {\chi }_{24,7}\\ {\chi }_{3,2}& {\chi }_{24,17}& {\chi }_{24,11}& {\chi }_{24,5}& {\chi }_{24,23}\end{array}}}$

The nonzero values of the characters mod 24 are

${\displaystyle \overline{)\begin{array}{}& 1& 5& 7& 11& 13& 17& 19& 23\\ {\chi }_{24,1}& 1& 1& 1& 1& 1& 1& 1& 1\\ {\chi }_{24,5}& 1& 1& 1& 1& -1& -1& -1& -1\\ {\chi }_{24,7}& 1& 1& -1& -1& 1& 1& -1& -1\\ {\chi }_{24,11}& 1& 1& -1& -1& -1& -1& 1& 1\\ {\chi }_{24,13}& 1& -1& 1& -1& -1& 1& -1& 1\\ {\chi }_{24,17}& 1& -1& 1& -1& 1& -1& 1& -1\\ {\chi }_{24,19}& 1& -1& -1& 1& -1& 1& 1& -1\\ {\chi }_{24,23}& 1& -1& -1& 1& 1& -1& -1& 1\end{array}}}$.

${\displaystyle (\mathbb{Z}/40\mathbb{Z}{)}^{\times }\cong (\mathbb{Z}/8\mathbb{Z}{)}^{\times }\times (\mathbb{Z}/5\mathbb{Z}{)}^{\times }.}$ The factorization of the characters mod 40 is

${\displaystyle \overline{)\begin{array}{ccccc}& {\chi }_{8,1}& {\chi }_{8,3}& {\chi }_{8,5}& {\chi }_{8,7}\\ {\chi }_{5,1}& {\chi }_{40,1}& {\chi }_{40,11}& {\chi }_{40,21}& {\chi }_{40,31}\\ {\chi }_{5,2}& {\chi }_{40,17}& {\chi }_{40,27}& {\chi }_{40,37}& {\chi }_{40,7}\\ {\chi }_{5,3}& {\chi }_{40,33}& {\chi }_{40,3}& {\chi }_{40,13}& {\chi }_{40,23}\\ {\chi }_{5,4}& {\chi }_{40,9}& {\chi }_{40,19}& {\chi }_{40,29}& {\chi }_{40,39}\end{array}}}$

The nonzero values of the characters mod 40 are

${\displaystyle \overline{)\begin{array}{}& 1& 3& 7& 9& 11& 13& 17& 19& 21& 23& 27& 29& 31& 33& 37& 39\\ {\chi }_{40,1}& 1& 1& 1& 1& 1& 1& 1& 1& 1& 1& 1& 1& 1& 1& 1& 1\\ {\chi }_{40,3}& 1& i& i& -1& 1& -i& -i& -1& -1& -i& -i& 1& -1& i& i& 1\\ {\chi }_{40,7}& 1& i& -i& -1& -1& -i& i& 1& 1& i& -i& -1& -1& -i& i& 1\\ {\chi }_{40,9}& 1& -1& -1& 1& 1& -1& -1& 1& 1& -1& -1& 1& 1& -1& -1& 1\\ {\chi }_{40,11}& 1& 1& -1& 1& 1& -1& 1& 1& -1& -1& 1& -1& -1& 1& -1& -1\\ {\chi }_{40,13}& 1& -i& -i& -1& -1& -i& -i& 1& -1& i& i& 1& 1& i& i& -1\\ {\chi }_{40,17}& 1& -i& i& -1& 1& -i& i& -1& 1& -i& i& -1& 1& -i& i& -1\\ {\chi }_{40,19}& 1& -1& 1& 1& 1& 1& -1& 1& -1& 1& -1& -1& -1& -1& 1& -1\\ {\chi }_{40,21}& 1& -1& 1& 1& -1& -1& 1& -1& -1& 1& -1& -1& 1& 1& -1& 1\\ {\chi }_{40,23}& 1& -i& i& -1& -1& i& -i& 1& 1& -i& i& -1& -1& i& -i& 1\\ {\chi }_{40,27}& 1& -i& -i& -1& 1& i& i& -1& -1& i& i& 1& -1& -i& -i& 1\\ {\chi }_{40,29}& 1& 1& -1& 1& -1& 1& -1& -1& -1& -1& 1& -1& 1& -1& 1& 1\\ {\chi }_{40,31}& 1& -1& -1& 1& -1& 1& 1& -1& 1& -1& -1& 1& -1& 1& 1& -1\\ {\chi }_{40,33}& 1& i& -i& -1& 1& i& -i& -1& 1& i& -i& -1& 1& i& -i& -1\\ {\chi }_{40,37}& 1& i& i& -1& -1& i& i& 1& -1& -i& -i& 1& 1& -i& -i& -1\\ {\chi }_{40,39}& 1& 1& 1& 1& -1& -1& -1& -1& 1& 1& 1& 1& -1& -1& -1& -1\end{array}}}$.

Summary

Let ${\displaystyle m=p_1^{k_1}{p}_2^{k_2}\cdots =q_1{q}_2\cdots }$, ${\displaystyle p_1<p_2<\dots }$ be the factorization of ${\displaystyle m}$ and assume ${\displaystyle (rs,m)=1.}$

There are ${\displaystyle \varphi (m)}$ Dirichlet characters mod ${\displaystyle m.}$ They are denoted by ${\displaystyle {\chi }_{m,r},}$ where ${\displaystyle {\chi }_{m,r}={\chi }_{m,s}}$ is equivalent to ${\displaystyle r\equiv s(\mathrm{mod}m).}$ The identity ${\displaystyle {\chi }_{m,r}(a){\chi }_{m,s}(a)={\chi }_{m,rs}(a)}$ is an isomorphism ${\displaystyle \widehat{(\mathbb{Z}/m\mathbb{Z}{)}^{\times }}\cong (\mathbb{Z}/m\mathbb{Z}{)}^{\times }.}$

Each character mod ${\displaystyle m}$ has a unique factorization as the product of characters mod the prime powers dividing ${\displaystyle m}$:

${\displaystyle {\chi }_{m,r}={\chi }_{q_1,r}{\chi }_{q_2,r}...}$

If ${\displaystyle m=m_1{m}_2,(m_1,m_2)=1}$ the product ${\displaystyle {\chi }_{m_1,r}{\chi }_{m_2,s}}$ is a character ${\displaystyle {\chi }_{m,t}}$ where ${\displaystyle t}$ is given by ${\displaystyle t\equiv r(\mathrm{mod}{m}_1)}$ and ${\displaystyle t\equiv s(\mathrm{mod}{m}_2).}$

Also, ${\displaystyle {\chi }_{m,r}(s)={\chi }_{m,s}(r)}$

Orthogonality

The two orthogonality relations are

${\displaystyle \sum _{a\in (\mathbb{Z}/m\mathbb{Z}{)}^{\times }}\chi (a)=\{\begin{array}{ll}\varphi (m)& \text{ if }\chi ={\chi }_0\\ 0& \text{ if }\chi \ne {\chi }_0\end{array}}$ and ${\displaystyle \sum _{\chi \in \widehat{(\mathbb{Z}/m\mathbb{Z}{)}^{\times }}}\chi (a)=\{\begin{array}{ll}\varphi (m)& \text{ if }a\equiv 1(\mathrm{mod}m)\\ 0& \text{ if }a\not\equiv 1(\mathrm{mod}m).\end{array}}$

The relations can be written in the symmetric form

${\displaystyle \sum _{a\in (\mathbb{Z}/m\mathbb{Z}{)}^{\times }}{\chi }_{m,r}(a)=\{\begin{array}{ll}\varphi (m)& \text{ if }r\equiv 1\\ 0& \text{ if }r\not\equiv 1\end{array}}$ and ${\displaystyle \sum _{r\in (\mathbb{Z}/m\mathbb{Z}{)}^{\times }}{\chi }_{m,r}(a)=\{\begin{array}{ll}\varphi (m)& \text{ if }a\equiv 1\\ 0& \text{ if }a\not\equiv 1.\end{array}}$

The first relation is easy to prove: If ${\displaystyle \chi ={\chi }_0}$ there are ${\displaystyle \varphi (m)}$ non-zero summands each equal to 1. If ${\displaystyle \chi \ne {\chi }_0}$there is some ${\displaystyle a^{\ast },(a^{\ast },m)=1,\chi (a^{\ast })\ne 1.}$ Then

${\displaystyle \chi (a^{\ast })\sum _{a\in (\mathbb{Z}/m\mathbb{Z}{)}^{\times }}\chi (a)=\sum _a\chi (a^{\ast })\chi (a)=\sum _a\chi (a^{\ast }a)=\sum _a\chi (a),}$ implying

${\displaystyle (\chi (a^{\ast })-1)\sum _a\chi (a)=0.}$ Dividing by the first factor gives ${\displaystyle \sum _a\chi (a)=0,}$ QED. The identity ${\displaystyle {\chi }_{m,r}(s)={\chi }_{m,s}(r)}$ for ${\displaystyle (rs,m)=1}$ shows that the relations are equivalent to each other.