In my elementary number theory class we define the following:

Let $p$ be a prime, and let $\mathbb{Z}_p^*$ relatively prime residues modulo $p$. A Dirichlet character modulo $p$ is defined as a function $\chi:\mathbb{Z}_p^*\to\mathbb{C}\setminus\{0\}$ such that for all $m,n\in\mathbb{Z}_p^*$ $$ \chi(mn) = \chi(m)\chi(n)$$

I am asked to find all Dirichlet characters modulo $p$. I do not really understand how should I do that. I have found out that for all $x\in\mathbb{Z}_p^*$ $$ \chi(x) = e^{\frac{2\pi i k}{p-1}} $$

but I don't think that seals the deal. I am looking for leads to prove it with elementary tools without group theory.

  • $\begingroup$ All Dirichlet characters modulo $p$ form a group, which is isomorphic to $\mathbb{Z}_p^*$. So you have $\chi_1,\chi,\chi^2,\ldots, \chi^{p-1}$. $\endgroup$ – Dietrich Burde May 16 '16 at 19:14
  • $\begingroup$ @DietrichBurde In my elementary number theory class we did not talk about groups and isomorphisms, so I assume elementary way to show this. Are you familiar with one? $\endgroup$ – Joshhh May 16 '16 at 19:17
  • 1
    $\begingroup$ Yes, I am, see the answer. You need to talk about groups and group homomorphisms to be able to talk about $\mathbb{Z}_p^*$ and Dirichlet characters. $\endgroup$ – Dietrich Burde May 16 '16 at 19:24

Theorem $6.8$ in Tom Apostol's book "Introduction to Analytic Number Theory" says

Theorem: A finite abelian group $G$ of order $n$ has exactly $n$ distinct characters.

The proof is completely elementary, but too long to reproduce it here. It follows that there are exactly $\phi(k)$ different Dirichlet characters modulo $k$. For $k=p$ prime we have $\phi(p)=p-1$.

  • $\begingroup$ OK so I have established that there are $p-1$ characters, how do I find them all? $\endgroup$ – Joshhh May 16 '16 at 19:32
  • $\begingroup$ Just find a non-trivial one, say $\chi$. Then all others are powers of it, see the comment above. $\endgroup$ – Dietrich Burde May 16 '16 at 19:32
  • $\begingroup$ How can you compose one character on another? If $\chi(x) = i$, what is $\chi^2(x)$? $\endgroup$ – Joshhh May 16 '16 at 19:35
  • $\begingroup$ The composition $\chi\circ \psi$ is defined by $(\chi\circ \psi) (n)=\chi(n)\psi(n)$. $\endgroup$ – Dietrich Burde May 16 '16 at 19:40

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