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I started learning about Dirichlet Characters. Here is what I learned so far:

Definition: Let $m \in \mathbb{N}$. We call a function $\chi:\mathbb{Z} \rightarrow \mathbb{C}$ a Dirichlet Character mod $m$ if the following holds:

  1. $\chi(ab)=\chi(a)\chi(b)$
  2. $\chi (a+m)=\chi(a)$
  3. $\chi(a)=0$ iff $gcd(a,m)>1$

The set of all Dirichlet Characters mod $m$ is denoted as $\Gamma_m$.

Proposition: Let $m \in \mathbb{N}$. $\Gamma_m$ togheter with the pointwise multiplication is a abelian group. Further $\Psi :\Gamma_m \rightarrow (\mathbb{Z}/m\mathbb{Z})^{\times}$, $\chi \mapsto \phi:k+m \mathbb{Z} \mapsto \chi(k)$ for $gcd(k,m)=1$, is a Isomorphism.

As a result of that Proposition we get that $|\Gamma_m|=\Phi(m)$ , where $\Phi$ is the euler-function.

Proposition: Let (k,m)=1. Then $\sum_{\chi \in \Gamma_m}=\Phi(m)$ if $k \equiv 1 \mod m$ and $=0$ else.

Let $\chi \in \Gamma_m$. Dann gilt $\sum_{k=0}^m -1\chi(k)= \Phi(m)$ if $\chi$ is the principal character and $=0$ else.

Now I want to determine $\Gamma_m$ for $m=1,2,3,4,5$.

For $m=1$ and $m=2$, th values of the euler-function are $\Phi(1)=1$ and $\Phi(2)=1$. Thus $\Gamma_1,\Gamma_2$ do only have one element. Thus the only element can only be the principal character.

Let $m=3$: Since $\Phi(3)=2$, $|\Gamma_3|=2$. And since the principal character $\chi_0$ is always in $\Gamma_m$, we only need to find the second Dirichlet Character $\chi_1$. By the Proposition we have a Isomorphism between $\Gamma_3$ and $\mathbb{Z}/3\mathbb{Z}$. I want to somehow use this. That's how far I got.

As a hint I got:

In $\mathbb{Z}/3\mathbb{Z}$ one can easily find that $\overline{2}$ is a primitive element.

But I do not know how to use this

Edit:

As mentioned in the comments, it should have been: ... "Isomorphism between $\Gamma_3$ and $(\mathbb{Z}/3\mathbb{Z})^{\times}$" and "In $(\mathbb{Z}/3\mathbb{Z})^{\times}$ one can easily find that $\overline{2}$ is a primitive element".

NTc5
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    Regarding the case m=3: By the Proposition, you have an Isomorphism between $\Gamma_3$ and $(\mathbb{Z}/3 \mathbb{Z})^{\times}$ (and not between $\Gamma_3$ and $\mathbb{Z}/3 \mathbb{Z}$). Similarly the hint with the primitive element is supposed to be about $(\mathbb{Z}/3 \mathbb{Z})^{\times}$. Since $\overline{2}$ is primitive element of $(\mathbb{Z}/3 \mathbb{Z})^{\times}$, this mean that every element in $(\mathbb{Z}/3 \mathbb{Z})^{\times}$ can be "generated" by powers of $\overline{2}$. – Philip Jun 22 '24 at 07:35
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    Note that $(\mathbb{Z}/3 \mathbb{Z})^{\times}={\overline{1},\overline{2}}$. You get $\overline{2} \cdot \overline{2}= \overline{1} $. Now you can use your second Proposition to get $\chi_0(1)+\chi_1(1)=2$ (because $1\equiv 1 \mod 3$). This is equal to $1+\chi_1(1)=2$ and by writing $1=2^2$ you get $\chi_1(2^2)=1$. Using that $\chi$ is homomorphism, you can get $\chi_1(2)^2=1$. This implies that $\chi_(2)=-1$ or $\chi_1(2)=1$. With this information, you can now calculate $\chi_1(1)$. – Philip Jun 22 '24 at 07:35
  • I'm lost. You define $\Gamma_n,$ then write $\Gamma_m(\Bbb Z/m\Bbb Z)^×.$ Sorry I am still relatively new to characters. Could you explain what it means? – suckling pig Jun 22 '24 at 08:36
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    @icantry Sorry, that was a typo. I will correct it right away, there should have been a $\rightarrow$ between those two. – NTc5 Jun 22 '24 at 09:01
  • @Philip That was very detailed. Thank you very much! How would one do the cases for $m=4$ and $m=5$? – NTc5 Jun 22 '24 at 09:03
  • Just an Update, for $m=5$ it is the same argumentation as what @Philip wrote. But I am stuck on $m=4$. – NTc5 Jun 22 '24 at 09:16
  • Omg, then since they're isomorphic, the character group $\Gamma_m$ is just my old friend the group of units (of the cyclic group). – suckling pig Jun 22 '24 at 14:45
  • I mean ring.... – suckling pig Jun 22 '24 at 20:06
  • We should have $\Gamma_3=\Gamma_4.$ Or do you want to go beyond the isomorphism? – suckling pig Jun 22 '24 at 20:21

1 Answers1

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For $m=3$, as you mentioned, the group of characters is isomorphic to $(\mathbb{Z}/3\mathbb{Z})^\times$, which is a group of order $\varphi(3)=2$.

We know that $(\mathbb{Z}/3\mathbb{Z})^\times=\{1,2\} $, where $1$ and $2$ are really equivalence classes mod $3 $ (but I am a bit lazy). In particular, this group is cyclic, generated by the element $2$.

We have $1\equiv2^2$ mod $3$ and clearly $2\equiv 2$ mod $3$.

So $\chi(1)=\chi(2^2)=\chi(2)^2$ by the first and second properties. Obviously, $\chi(2)=\chi(2)$.

Again by the second property, the character is entirely determined by $\chi(1)$ and $\chi(2)$. But $\chi(1)=\chi(2)^2$, so the character is actually determined entirely by the value of $\chi(2)$.

In general, when $(\mathbb{Z}/m\mathbb{Z})^\times$ is cyclic, the characters are entirely determined by where they send the generator of this group.

Returning to the $m=3$ case, the first property pretty easily shows that $\chi(1)=1$, so that $\chi(2)^2 = 1$. Thus $\chi(2)=1$ or $\chi(2)=-1$. These give our two characters mod $3$.

In general, if $a$ is the generator for the cyclic group $(\mathbb{Z}/m\mathbb{Z})^\times$, then $\chi(a)^{\varphi(m)}=1$ so that $\chi(a)=\zeta$ where $\zeta$ is a $\varphi(m)^{th}$ root of unity.

Let’s look at the case $m=4$. We know that $\varphi(4)=2$ so the group of characters is cyclic (every group of order $2$ is cyclic). You can check that $(\mathbb{Z}/4\mathbb{Z})^\times=\{1,3\}$ is generated by (the equivalence class) $3$.

So we need to find $\chi(3)$. But as above, $\chi(3)^2=1$ so that our two characters are determined by the two cases $\chi(3)=1$ or $\chi(3)=-1$.

For $m=5$, the group $(\mathbb{Z}/5\mathbb{Z})^\times=\{1,2,3,4\}$ is cyclic and generated by $2$ (or $3$).

As $\varphi(5)=4$, we have $\chi(2)^4 = 1$, so that $\chi(2)=\pm 1$ or $\chi(2)=\pm i$. These four cases determine our four characters mod $5$.

ljfirth
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