Editing Dirichlet character (section)

=== Summary ===
Let <math>m=p_1^{k_1}p_2^{k_2}\cdots = q_1q_2 \cdots</math>, <math>p_1<p_2< \dots </math> be the factorization of <math>m</math> and assume <math>(rs,m)=1.</math>

There are <math>\phi(m)</math> Dirichlet characters mod <math>m.</math> They are denoted by <math>\chi_{m,r},</math> where <math>\chi_{m,r}=\chi_{m,s}</math> is equivalent to <math>r\equiv s\pmod{m}.</math>
The identity <math>\chi_{m,r}(a)\chi_{m,s}(a)=\chi_{m,rs}(a)\;</math> is an isomorphism <math>\widehat{(\mathbb{Z}/m\mathbb{Z})^\times}\cong(\mathbb{Z}/m\mathbb{Z})^\times.</math><ref>This is true for all finite abelian groups: <math>A\cong\hat{A}</math>; See Ireland & Rosen pp. 253-254</ref>

Each character mod <math>m</math> has a unique factorization as the product of characters mod the prime powers dividing <math>m</math>:
:<math>\chi_{m,r}=\chi_{q_1,r}\chi_{q_2,r}...</math>

If <math>m=m_1m_2, (m_1,m_2)=1</math> the product <math>\chi_{m_1,r}\chi_{m_2,s}</math> is a character <math>\chi_{m,t}</math> where <math>t</math> is given by <math>t\equiv r\pmod{m_1}</math> and <math>t\equiv s\pmod{m_2}.</math>

Also,<ref>because the formulas for <math>\chi</math> mod prime powers are symmetric in <math>r</math> and <math>s</math> and the formula for products preserves this symmetry. See Davenport, p. 29.</ref><ref>This is the same thing as saying that the n-th column and the n-th row in the tables of nonzero values are the same.</ref>
<math> \chi_{m,r}(s)=\chi_{m,s}(r)</math>