# How to make Dirichlet character table modulo $5$

There are four reduced residue classes $$\mod 5$$, namely $$1, 2, 3, 4$$ and thus four Dirichlet characters $$\mod 5$$ since $$\phi(5)=4$$.

I understand how to deduce that the character can be $$1$$ or $$0$$(from defination):

$$\chi(n)=\begin{cases}1 &\text{ if } (n,5)=1\\0 &\text{ if } (n,5)>1\end{cases}$$

but i am not sure how to deduce for $$i$$ as shown here in $$\mod 5$$ table , using the following property we get:

$$\chi(mn)=\chi(m)\chi(n)$$

$$\chi(4)=\chi(2)\chi(2)=i\times i=i^2=-1$$

hence

$$\chi_2(4)=\chi_4(4)=-1$$

but,how is it known that

$$\chi_2(2)=\chi_4(3)=i$$

Every Dirichlet character $\chi$ (mod $5$) is determined by the value $\chi(2)$, since $(2,2^2,2^3,2^4) \equiv (2,4,3,1) \pmod 5$ and therefore $(\chi(1),\chi(2),\chi(3),\chi(4)) = (\chi(2)^4, \chi(2), \chi(2)^3, \chi(2)^2)$. (We're using the fact that $2$ is a "primitive root" modulo $5$.)
On the other hand, we already know that $\chi(1)=1$. Therefore $\chi(2)^4=1$, which means it's necessary that $\chi(2)$ is one of $\{i,-1,-i,1\}$.
• thanks alot,does the same approach work for other Dirichlet Character $\chi (\mod n)$, like $n=7,8,12$ Feb 24, 2014 at 19:16
• Definitely. What you need is a set of generators for $(\Bbb Z/n\Bbb Z)^\times$ (the multiplicative group of reduced residue classes modulo $n$), which we know how to find. For odd prime powers, primitive roots exist; for powers of $2$, take $5$ and/or $-1$; for more complicated numbers, use the Chinese remainder theorem. Figuring out what values the characters can assign to these generators then determines all possible Dirichlet characters. Feb 25, 2014 at 5:04