- #1

Math100

- 768

- 214

- Homework Statement
- Let ## G ## be the group of reduced residue classes modulo ## 21 ##. Construct a table showing the values of all the real-valued, and one of the complex-valued, Dirichlet characters modulo ## 21 ##. (In your table, use ## \omega ## for ## e^{\pi i/3} ##.)

- Relevant Equations
- Definition: Dirichlet characters. Let ## G ## be the group of reduced residue classes modulo ## k ##. Corresponding to each character ## f ## of ## G ##, we define an arithmetical function ## \chi=\chi_{f} ## as follows:

## \chi(n)=f(\hat{n}) ## if ## (n, k)=1 ##,

## \chi(n)=0 ## if ## (n, k)>1 ##.

The function ## \chi ## is called a Dirichlet character modulo ## k ##. The principal character ## \chi_{1} ## has the properties ## \chi_{1}(n)=1 ## if ## (n, k)=1 ##, and ## \chi_{1}(n)=0 ## if ## (n, k)>1 ##.

Corresponding to each character ## f_{i} ## of ## G ##, we define an arithmetical function ## \chi_{i} ## as follows:

## \chi_{i}(n)=f_{i}(\hat{n}) ## if ## (n, k)=1 ##,

## \chi_{i}(n)=0 ## if ## (n, k)>1 ##.

\begin{array}{|c|c|c|c|c|c|c|c|c|c|c|c|c|}

\hline n & 1 & 2 & 4 & 5 & 8 & 10 & 11 & 13 & 16 & 17 & 19 & 20 \\

\hline \chi_{1}(n) & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 \\

\hline \chi_{2}(n) & 1 & -1 & 1 & 1 & -1 & -1 & -1 & -1 & 1 & 1 & -1 & 1 \\

\hline \chi_{3}(n) & 1 & 1 & 1 & -1 & 1 & -1 & 1 & -1 & 1 & -1 & -1 & -1 \\

\hline \chi_{4}(n) & 1 & -1 & 1 & -1 & -1 & 1 & -1 & 1 & 1 & -1 & 1 & -1 \\

\hline \chi_{5}(n) & 1 & \omega & \omega^{2} & -\omega & -1 & -\omega^{2} & -\omega^{2} & -1 & -\omega & \omega^{2} & \omega & 1 \\

\hline \chi_{6}(n) & 1 & \omega^{2} & -\omega & \omega^{2} & 1 & -\omega & -\omega & 1 & \omega^{2} & -\omega & \omega^{2} & 1 \\

\hline \chi_{7}(n) & 1 & -\omega & \omega^{2} & -\omega & 1 & \omega^{2} & \omega^{2} & 1 & -\omega & \omega^{2} & -\omega & 1 \\

\hline \chi_{8}(n) & 1 & -\omega^{2} & -\omega & \omega^{2} & -1 & \omega & \omega & -1 & \omega^{2} & -\omega & -\omega^{2} & 1 \\

\hline \chi_{9}(n) & 1 & \omega & \omega^{2} & \omega & -1 & \omega^{2} & -\omega^{2} & 1 & -\omega & -\omega^{2} & -\omega & -1 \\

\hline \chi_{10}(n) & 1 & \omega^{2} & -\omega & -\omega^{2} & 1 & \omega & -\omega & -1 & \omega^{2} & \omega & -\omega^{2} & -1 \\

\hline \chi_{11}(n) & 1 & -\omega & \omega^{2} & \omega & 1 & -\omega^{2} & \omega^{2} & -1 & -\omega & -\omega^{2} & \omega & -1 \\

\hline \chi_{12}(n) & 1 & -\omega^{2} & -\omega & -\omega^{2} & -1 & -\omega & \omega & 1 & \omega^{2} & \omega & \omega^{2} & -1 \\

\hline

\end{array}

The table above is the answer/solution for this problem.

I know that ## \varphi(21)=\varphi(3)\varphi(7)=2\cdot 6=12 ##, which means there are ## 12 ## elements such that ## G=\{1, 2, 4, 5, 8, 10, 11, 13, 16, 17, 19, 20\} ##. But I do not know how to get ## \chi_{2}(n)... ## except ## \chi_{1}(n) ##. May anyone please tell me how to get these values?

\hline n & 1 & 2 & 4 & 5 & 8 & 10 & 11 & 13 & 16 & 17 & 19 & 20 \\

\hline \chi_{1}(n) & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 \\

\hline \chi_{2}(n) & 1 & -1 & 1 & 1 & -1 & -1 & -1 & -1 & 1 & 1 & -1 & 1 \\

\hline \chi_{3}(n) & 1 & 1 & 1 & -1 & 1 & -1 & 1 & -1 & 1 & -1 & -1 & -1 \\

\hline \chi_{4}(n) & 1 & -1 & 1 & -1 & -1 & 1 & -1 & 1 & 1 & -1 & 1 & -1 \\

\hline \chi_{5}(n) & 1 & \omega & \omega^{2} & -\omega & -1 & -\omega^{2} & -\omega^{2} & -1 & -\omega & \omega^{2} & \omega & 1 \\

\hline \chi_{6}(n) & 1 & \omega^{2} & -\omega & \omega^{2} & 1 & -\omega & -\omega & 1 & \omega^{2} & -\omega & \omega^{2} & 1 \\

\hline \chi_{7}(n) & 1 & -\omega & \omega^{2} & -\omega & 1 & \omega^{2} & \omega^{2} & 1 & -\omega & \omega^{2} & -\omega & 1 \\

\hline \chi_{8}(n) & 1 & -\omega^{2} & -\omega & \omega^{2} & -1 & \omega & \omega & -1 & \omega^{2} & -\omega & -\omega^{2} & 1 \\

\hline \chi_{9}(n) & 1 & \omega & \omega^{2} & \omega & -1 & \omega^{2} & -\omega^{2} & 1 & -\omega & -\omega^{2} & -\omega & -1 \\

\hline \chi_{10}(n) & 1 & \omega^{2} & -\omega & -\omega^{2} & 1 & \omega & -\omega & -1 & \omega^{2} & \omega & -\omega^{2} & -1 \\

\hline \chi_{11}(n) & 1 & -\omega & \omega^{2} & \omega & 1 & -\omega^{2} & \omega^{2} & -1 & -\omega & -\omega^{2} & \omega & -1 \\

\hline \chi_{12}(n) & 1 & -\omega^{2} & -\omega & -\omega^{2} & -1 & -\omega & \omega & 1 & \omega^{2} & \omega & \omega^{2} & -1 \\

\hline

\end{array}

The table above is the answer/solution for this problem.

I know that ## \varphi(21)=\varphi(3)\varphi(7)=2\cdot 6=12 ##, which means there are ## 12 ## elements such that ## G=\{1, 2, 4, 5, 8, 10, 11, 13, 16, 17, 19, 20\} ##. But I do not know how to get ## \chi_{2}(n)... ## except ## \chi_{1}(n) ##. May anyone please tell me how to get these values?