Dirichlet character tables up to mod 21

dI have written about Dirichlet characters in previous notes (see Advanced Number Theory Note #13 and the references therein). For reference purposes, I will set out the first twenty Dirichlet character tables in the present note and demonstrate in detail the calculation of the first ten tables. Recall that there are \varphi(k) distinct Dirichlet characters \chi modulo k, each of which is completely multiplicative and periodic with period k. Each character value \chi(n) is a (complex) root of unity if gcd(n, k) = 1 whereas \chi(n) = 0 whenever gcd(n, k) > 1. We also have \chi(1) = 1 for all Dirichlet characters. These facts uniquely determine each Dirichlet character table.

k = 2

mod 02

We have \varphi(2) = 1 so there is only one Dirichlet character in this case (the principal one), with values \chi_1(1) = 1 and \chi_1(2) = 0.

k = 3

mod 03

We have \varphi(3) = 2 so there are two Dirichlet characters in this case. One of them will be the principal character which takes the values \chi_1(1) = 1, \chi_1(2) = 1 and \chi_1(3) = 0. To work out the second Dirichlet character we consider the two roots of unity

\omega = e^{2 \pi i/2} = e^{\pi i} = -1

and

\omega^2 = 1

Note that the set of least positive residues mod 3 is generated by 2:

2 \equiv 2 mod(3)

2^2 \equiv 1 mod(3)

Therefore the non-principal Dirichlet character will be completely determined by the values of \chi(2). If we set

\chi_2(2) = \omega = -1

then

\chi_2(1) = \chi_2(2^2) = \chi_2^2(2) = \omega^2 = 1

(though this calculation is superfluous here since \chi_2(1) = 1 anyway. This is a fundamental property of Dirichlet characters arising from the fact that they are completely multiplicative). We also have \chi_2(3) = 0. This completes the second character. (From now on we will omit the statements of the zero values of the Dirichlet characters, which as stated earlier arise whenever gcd(n, k) > 1).

k = 4

mod 04

We have \varphi(4) = 2 so there are two Dirichlet characters in this case. One of them will be the principal character. (From now on we will always denote the principal character by \chi_1). To work out the second Dirichlet character we again consider the two roots of unity

\omega = e^{2 \pi i/2} = e^{\pi i} = -1

and

\omega^2 = 1

Note that the set of least positive residues mod 4 is generated by 3:

3 \equiv 3 mod(4)

3^2 \equiv 1 mod(4)

Therefore the non-principal Dirichlet character will be completely determined by the values of \chi(3). If we set

\chi_2(3) = \omega = -1

then

\chi_2(1) = \chi_2(3^2) = \chi_2^2(3) = \omega^2 = 1

(though again this second calculation is superfluous since \chi_2(1) = 1 anyway). This completes the second character.

k = 5

mod 05

We have \varphi(5) = 4 so there are four Dirichlet characters in this case. We consider the four roots of unity

\omega = e^{2 \pi i/4} = e^{\pi i/2} = i

\omega^2 = e^{4 \pi i/4} = e^{\pi i} = -1

\omega^3 = e^{6 \pi i/4} = e^{3 \pi i/2} = -i

\omega^4 = e^{8 \pi i/4} = e^{2 \pi i} = 1

Note that the set of least positive residues mod 5 is generated by 2:

2 \equiv 2 mod(5)

2^2 \equiv 4 mod(5)

2^3 \equiv 3 mod(5)

2^4 \equiv 1 mod(5)

Therefore the non-principal Dirichlet characters will be completely determined by the values of \chi(2). If we set

\chi_2(2) = \omega = i

then

\chi_2(3) = \chi_2(2^3) = \chi_2^3(2) = -i

\chi_2(4) = \chi_2(2^2) = \chi_2^2(2) = -1

(and we have \chi_2(1) = 1). This completes the second character.

To compute the third character we can set

\chi_3(2) = \omega^2 = -1

then

\chi_3(3) = \chi_3(2^3) = \chi_3^3(2) = -1

\chi_3(4) = \chi_3(2^2) = \chi_3^2(2) = 1

(and we have \chi_3(1) = 1). This completes the third character.

To compute the fourth character we set

\chi_4(2) = \omega^3 = -i

then

\chi_4(3) = \chi_4(2^3) = \chi_4^3(2) = i

\chi_4(4) = \chi_4(2^2) = \chi_4^2(2) = -1

(and we have \chi_4(1) = 1). This completes the fourth character.

k = 6

mod 06

We have \varphi(6) = 2 so there are two Dirichlet characters in this case. We consider the two roots of unity

\omega = e^{2 \pi i/2} = e^{\pi i} = -1

and

\omega^2 = 1

Note that the set of least positive residues mod 6 is generated by 5:

5 \equiv 5 mod(6)

5^2 \equiv 1 mod(6)

Therefore the non-principal Dirichlet character will be completely determined by the values of \chi(5). If we set

\chi_2(5) = \omega = -1

then

\chi_2(1) = \chi_2(5^2) = \chi_2^2(5) = \omega^2 = 1

(though again this second calculation is superfluous since \chi_2(1) = 1 anyway). This completes the second character.

k = 7

mod 07

We have \varphi(7) = 6 so there are six Dirichlet characters in this case. We consider the six roots of unity

\omega = e^{2 \pi i/6} = e^{\pi i/3}

\omega^2 = e^{2 \pi i/3}

\omega^3 = e^{3 \pi i/3} = e^{\pi i} = -1

\omega^4 = \omega \cdot \omega^3 = -\omega

\omega^5 = \omega \cdot \omega^4 = -\omega^2

\omega^6 = e^{6 \pi i/3} = e^{2\pi i} = 1

Note that the set of least positive residues mod 7 is generated by 3:

3 \equiv 3 mod(7)

3^2 \equiv 2 mod(7)

3^3 \equiv 6 mod(7)

3^4 \equiv 4 mod(7)

3^5 \equiv 5 mod(7)

3^6 \equiv 1 mod(7)

Therefore the non-principal Dirichlet characters will be completely determined by the values of \chi(3). If we set

\chi_2(3) = \omega

then

\chi_2(2) = \chi_2(3^2) = \chi_2^2(3) = \omega^2

\chi_2(4) = \chi_2(3^4) = \chi_2^4(3) = \omega^4 = -\omega

\chi_2(5) = \chi_2(3^5) = \chi_2^5(3) = \omega^5 = -\omega^2

\chi_2(6) = \chi_2(3^3) = \chi_2^3(3) = \omega^3 = -1

(and we have \chi_2(1) = 1). This completes the second character.

To compute the third character we can set

\chi_3(3) = \omega^2

then

\chi_3(2) = \chi_3(3^2) = \chi_3^2(3) = \omega^4 = -\omega

\chi_3(4) = \chi_3(3^4) = \chi_3^4(3) = \omega^8 = \omega^2

\chi_3(5) = \chi_3(3^5) = \chi_3^5(3) = \omega^{10} = -\omega

\chi_3(6) = \chi_3(3^3) = \chi_3^3(3) = \omega^6 = 1

(and we have \chi_3(1) = 1). This completes the third character.

To compute the fourth character we can set

\chi_4(3) = \omega^3 = -1

then

\chi_4(2) = \chi_4(3^2) = \chi_4^2(3) = 1

\chi_4(4) = \chi_4(3^4) = \chi_4^4(3) = 1

\chi_4(5) = \chi_4(3^5) = \chi_4^5(3) = -1

\chi_4(6) = \chi_4(3^3) = \chi_4^3(3) = -1

(and we have \chi_4(1) = 1). This completes the fourth character.

To compute the fifth character we can set

\chi_5(3) = \omega^4 = -\omega

then

\chi_5(2) = \chi_5(3^2) = \chi_5^2(3) = \omega^2

\chi_5(4) = \chi_5(3^4) = \chi_5^4(3) = \omega^4 = -\omega

\chi_5(5) = \chi_5(3^5) = \chi_5^5(3) = \chi_5^4(3) \cdot \chi_5(3) = \omega^2

\chi_5(6) = \chi_5(3^3) = \chi_5^3(3) = -\omega^3 = 1

(and we have \chi_5(1) = 1). This completes the fifth character.

Finally, to compute the sixth character we set

\chi_6(3) = \omega^5 = -\omega^2

then

\chi_6(2) = \chi_6(3^2) = \chi_6^2(3) = \omega^4 = -\omega

\chi_6(4) = \chi_6(3^4) = \chi_6^4(3) = \omega^8 = \omega^2

\chi_6(5) = \chi_6(3^5) = \chi_6^5(3) = -\omega^{10} = \omega

\chi_6(6) = \chi_6(3^3) = \chi_6^3(3) = -\omega^6 = -1

(and we have \chi_6(1) = 1). This completes the sixth character.

k = 8

mod 08

We have \varphi(8) = 4 so there are four Dirichlet characters in this case. We consider the four roots of unity

\omega = e^{2 \pi i/4} = e^{\pi i/2} = i

\omega^2 = e^{4 \pi i/4} = e^{\pi i} = -1

\omega^3 = e^{6 \pi i/4} = e^{3 \pi i/2} = -i

\omega^4 = e^{8 \pi i/4} = e^{2 \pi i} = 1

In this case, none of the four elements of the set of least positive residues mod 8 generates the entire set. However, the characters must satisfy the following relations, which restrict the choices:

\chi(3) \cdot \chi(5) = \chi(15) = \chi(7)

\chi(3) \cdot \chi(7) = \chi(21) = \chi(5)

\chi(5) \cdot \chi(7) = \chi(35) = \chi(3)

Each character’s values must be chosen in such a way that these three relations hold.

To compute the second character, suppose we begin by trying to set

\chi_2(3) = \omega = i

and

\chi_2(5) = \omega^2 = -1

Then we must have

\chi_2(7) = \chi_2(3) \cdot \chi_2(5) = -i

but then

\chi_2(3) \cdot \chi_2(7) = 1 \neq \chi_2(5)

so this does not work. If instead we try to set

\chi_2(5) = -i

then we must have

\chi_2(7) = \chi_2(3) \cdot \chi_2(5) = 1

but then

\chi_2(3) \cdot \chi_2(7) = i \neq \chi_2(5)

so this does not work either. Computations like these show that \pm i cannot appear in any of the characters mod 8. All the characters must be formed from \pm 1. (Fundamentally, this is due to the fact that the group of least positive residues mod 8 can be subdivided into four cyclic subgroups of order 2, each of which has characters whose values are the two roots of unity, 1 and -1).

To compute the second character we can set

\chi_2(3) = 1

and

\chi_2(5) = -1

then we must have

\chi_2(7) = -1

and this works.

To compute the third character we can set

\chi_3(3) = -1

and

\chi_3(5) = -1

then we must have

\chi_3(7) = 1

and this works too.

Finally, to compute the fourth character we can set

\chi_4(3) = -1

and

\chi_4(5) = 1

then we must have

\chi_4(7) = -1

and this works too.

k = 9

mod 09

We have \varphi(9) = 6 so there are six Dirichlet characters in this case. We consider the six roots of unity

\omega = e^{2 \pi i/6} = e^{\pi i/3}

\omega^2 = e^{2 \pi i/3}

\omega^3 = e^{3 \pi i/3} = e^{\pi i} = -1

\omega^4 = \omega \cdot \omega^3 = -\omega

\omega^5 = \omega \cdot \omega^4 = -\omega^2

\omega^6 = e^{6 \pi i/3} = e^{2\pi i} = 1

Note that the set of least positive residues mod 9 is generated by 2:

2 \equiv 2 mod(9)

2^2 \equiv 4 mod(9)

2^3 \equiv 8 mod(9)

2^4 \equiv 7 mod(9)

2^5 \equiv 5 mod(9)

2^6 \equiv 1 mod(9)

Therefore the non-principal Dirichlet characters will be completely determined by the values of \chi(2). If we set

\chi_2(2) = \omega

then

\chi_2(4) = \chi_2(2^2) = \chi_2^2(2) = \omega^2

\chi_2(5) = \chi_2(2^5) = \chi_2^5(2) = \omega^5 = -\omega^2

\chi_2(7) = \chi_2(2^4) = \chi_2^4(2) = \omega^4 = -\omega

\chi_2(8) = \chi_2(2^3) = \chi_2^3(2) = \omega^3 = -1

(and we have \chi_2(1) = 1). This completes the second character.

To compute the third character we can set

\chi_3(2) = \omega^2

then

\chi_3(4) = \chi_3(2^2) = \chi_3^2(2) = \omega^4 = -\omega

\chi_3(5) = \chi_3(2^5) = \chi_3^5(2) = \omega^{10} = \omega^6 \cdot \omega^4 = -\omega

\chi_3(7) = \chi_3(2^4) = \chi_3^4(2) = \omega^8 = \omega^6 \cdot \omega^2 = \omega^2

\chi_3(8) = \chi_3(2^3) = \chi_3^3(2) = \omega^6 = 1

(and we have \chi_3(1) = 1). This completes the third character.

To compute the fourth character we can set

\chi_4(2) = \omega^3 = -1

then

\chi_4(4) = \chi_4(2^2) = \chi_4^2(2) = 1

\chi_4(5) = \chi_4(2^5) = \chi_4^5(2) = -1

\chi_4(7) = \chi_4(2^4) = \chi_4^4(2) = 1

\chi_4(8) = \chi_4(2^3) = \chi_4^3(2) = -1

(and we have \chi_4(1) = 1). This completes the fourth character.

To compute the fifth character we can set

\chi_5(2) = \omega^4 = -\omega

then

\chi_5(4) = \chi_5(2^2) = \chi_5^2(2) = \omega^2

\chi_5(5) = \chi_5(2^5) = \chi_5^5(2) = -\omega^5 = -\omega^3 \cdot \omega^2 = \omega^2

\chi_5(7) = \chi_5(2^4) = \chi_5^4(2) = \omega^4 = \omega^3 \cdot \omega = -\omega

\chi_5(8) = \chi_5(2^3) = \chi_5^3(2) = -\omega^3 = 1

(and we have \chi_5(1) = 1). This completes the fifth character.

Finally, to compute the sixth character we can set

\chi_6(2) = \omega^5 = -\omega^2

then

\chi_6(4) = \chi_6(2^2) = \chi_6^2(2) = \omega^4 = -\omega

\chi_6(5) = \chi_6(2^5) = \chi_6^5(2) = -\omega^{10} = -\omega^6 \cdot \omega^4 = \omega

\chi_6(7) = \chi_6(2^4) = \chi_6^4(2) = \omega^8 = \omega^6 \cdot \omega^2 = \omega^2

\chi_6(8) = \chi_6(2^3) = \chi_6^3(2) = -\omega^6 = -1

(and we have \chi_6(1) = 1). This completes the sixth character.

k = 10

mod 10

We have \varphi(10) = 4 so there are four Dirichlet characters in this case. We consider the four roots of unity

\omega = e^{2 \pi i/4} = e^{\pi i/2} = i

\omega^2 = e^{4 \pi i/4} = e^{\pi i} = -1

\omega^3 = e^{6 \pi i/4} = e^{3 \pi i/2} = -i

\omega^4 = e^{8 \pi i/4} = e^{2 \pi i} = 1

Note that the set of least positive residues mod 10 is generated by 3:

3 \equiv 3 mod(10)

3^2 \equiv 9 mod(10)

3^3 \equiv 7 mod(10)

3^4 \equiv 1 mod(10)

Therefore the non-principal Dirichlet characters will be completely determined by the values of \chi(3). If we set

\chi_2(3) = \omega = i

then

\chi_2(7) = \chi_2(3^3) = \chi_2^3(3) = -i

\chi_2(9) = \chi_2(3^2) = \chi_2^2(3) = -1

(and we have \chi_2(1) = 1). This completes the second character.

To compute the third character we can set

\chi_3(3) = \omega^2 = -1

then

\chi_3(7) = \chi_3(3^3) = \chi_3^3(3) = -1

\chi_3(9) = \chi_3(3^2) = \chi_3^2(3) = 1

(and we have \chi_3(1) = 1). This completes the third character.

Finally, to compute the fourth character we set

\chi_4(3) = \omega^3 = -i

then

\chi_4(7) = \chi_4(3^3) = \chi_4^3(3) = i

\chi_4(9) = \chi_4(3^2) = \chi_4^2(3) = -1

(and we have \chi_4(1) = 1). This completes the fourth character.

k = 11

mod 11

We have \varphi(11) = 10 so there are ten Dirichlet characters in this case. We consider the ten roots of unity

\omega = e^{2 \pi i/10} = e^{\pi i/5}

\omega^2 = e^{2 \pi i/5}

\omega^3 = e^{3 \pi i/5}

\omega^4 = e^{4 \pi i/5}

\omega^5 = e^{5 \pi i/5} = e^{\pi i} = -1

\omega^6 = -\omega

\omega^7 = -\omega^2

\omega^8 = -\omega^3

\omega^9 = -\omega^4

\omega^{10} = -\omega^5 = 1

Note that the set of least positive residues mod 11 is generated by 2:

2 \equiv 2 mod(11)

2^2 \equiv 4 mod(11)

2^3 \equiv 8 mod(11)

2^4 \equiv 5 mod(11)

2^5 \equiv 10 mod(11)

2^6 \equiv 9 mod(11)

2^7 \equiv 7 mod(11)

2^8 \equiv 3 mod(11)

2^9 \equiv 6 mod(11)

2^{10} \equiv 1 mod(11)

Therefore the non-principal Dirichlet characters will be completely determined by the values of \chi(2). If we set

\chi_2(2) = \omega

then

\chi_2(3) = \chi_2(2^8) = \chi_2^8(2) = \omega^8 = -\omega^3

\chi_2(4) = \chi_2(2^2) = \chi_2^2(2) = \omega^2

\chi_2(5) = \chi_2(2^4) = \chi_2^4(2) = \omega^4

\chi_2(6) = \chi_2(2^9) = \chi_2^9(2) = \omega^9 = -\omega^4

\chi_2(7) = \chi_2(2^7) = \chi_2^7(2) = \omega^7 = -\omega^2

\chi_2(8) = \chi_2(2^3) = \chi_2^3(2) = \omega^3

\chi_2(9) = \chi_2(2^6) = \chi_2^6(2) = \omega^6 = -\omega

\chi_2(10) = \chi_2(2^5) = \chi_2^5(2) = \omega^5 = -1

(and we have \chi_2(1) = 1). This completes the second character.

To compute the third character we can set

\chi_3(2) = \omega^2

then

\chi_3(3) = \chi_3(2^8) = \chi_3^8(2) = \omega^{16} = -\omega

\chi_3(4) = \chi_3(2^2) = \chi_3^2(2) = \omega^4

\chi_3(5) = \chi_3(2^4) = \chi_3^4(2) = \omega^8 = -\omega^3

\chi_3(6) = \chi_3(2^9) = \chi_3^9(2) = \omega^{18} = -\omega^3

\chi_3(7) = \chi_3(2^7) = \chi_3^7(2) = \omega^{14} = \omega^4

\chi_3(8) = \chi_3(2^3) = \chi_3^3(2) = \omega^6 = -\omega

\chi_3(9) = \chi_3(2^6) = \chi_3^6(2) = \omega^{12} = \omega^2

\chi_3(10) = \chi_3(2^5) = \chi_3^5(2) = \omega^{10} = 1

(and we have \chi_3(1) = 1). This completes the third character.

To compute the fourth character we can set

\chi_4(2) = \omega^3

then

\chi_4(3) = \chi_4(2^8) = \chi_4^8(2) = \omega^{24} = \omega^4

\chi_4(4) = \chi_4(2^2) = \chi_4^2(2) = \omega^6 = -\omega

\chi_4(5) = \chi_4(2^4) = \chi_4^4(2) = \omega^{12} = \omega^2

\chi_4(6) = \chi_4(2^9) = \chi_4^9(2) = \omega^{27} = -\omega^2

\chi_4(7) = \chi_4(2^7) = \chi_4^7(2) = \omega^{21} = \omega

\chi_4(8) = \chi_4(2^3) = \chi_4^3(2) = \omega^9 = -\omega^4

\chi_4(9) = \chi_4(2^6) = \chi_4^6(2) = \omega^{18} = -\omega^3

\chi_4(10) = \chi_4(2^5) = \chi_4^5(2) = \omega^{15} = -1

(and we have \chi_4(1) = 1). This completes the fourth character.

To compute the fifth character we can set

\chi_5(2) = \omega^4

then

\chi_5(3) = \chi_5(2^8) = \chi_5^8(2) = \omega^{32} = \omega^2

\chi_5(4) = \chi_5(2^2) = \chi_5^2(2) = \omega^8 = -\omega^3

\chi_5(5) = \chi_5(2^4) = \chi_5^4(2) = \omega^{16} = -\omega

\chi_5(6) = \chi_5(2^9) = \chi_5^9(2) = \omega^{36} = -\omega

\chi_5(7) = \chi_5(2^7) = \chi_5^7(2) = \omega^{28} = -\omega^3

\chi_5(8) = \chi_5(2^3) = \chi_5^3(2) = \omega^{12} = \omega^2

\chi_5(9) = \chi_5(2^6) = \chi_5^6(2) = \omega^{24} = \omega^4

\chi_5(10) = \chi_5(2^5) = \chi_5^5(2) = \omega^{20} = 1

(and we have \chi_5(1) = 1). This completes the fifth character.

To compute the sixth character we can set

\chi_6(2) = \omega^5 = -1

then

\chi_6(3) = \chi_6(2^8) = \chi_6^8(2) = 1

\chi_6(4) = \chi_6(2^2) = \chi_6^2(2) = 1

\chi_6(5) = \chi_6(2^4) = \chi_6^4(2) = 1

\chi_6(6) = \chi_6(2^9) = \chi_6^9(2) = -1

\chi_6(7) = \chi_6(2^7) = \chi_6^7(2) = -1

\chi_6(8) = \chi_6(2^3) = \chi_6^3(2) = -1

\chi_6(9) = \chi_6(2^6) = \chi_6^6(2) = 1

\chi_6(10) = \chi_6(2^5) = \chi_6^5(2) = -1

(and we have \chi_6(1) = 1). This completes the sixth character.

To compute the seventh character we can set

\chi_7(2) = \omega^6 = -\omega

then

\chi_7(3) = \chi_7(2^8) = \chi_7^8(2) = \omega^8 = -\omega^3

\chi_7(4) = \chi_7(2^2) = \chi_7^2(2) = \omega^2

\chi_7(5) = \chi_7(2^4) = \chi_7^4(2) = \omega^4

\chi_7(6) = \chi_7(2^9) = \chi_7^9(2) = -\omega^9 = \omega^4

\chi_7(7) = \chi_7(2^7) = \chi_7^7(2) = -\omega^7 = \omega^2

\chi_7(8) = \chi_7(2^3) = \chi_7^3(2) = -\omega^3

\chi_7(9) = \chi_7(2^6) = \chi_7^6(2) = \omega^6 = -\omega

\chi_7(10) = \chi_7(2^5) = \chi_7^5(2) = -\omega^5 = 1

(and we have \chi_7(1) = 1). This completes the seventh character.

To compute the eighth character we can set

\chi_8(2) = \omega^7 = -\omega^2

then

\chi_8(3) = \chi_8(2^8) = \chi_8^8(2) = \omega^{16} = -\omega

\chi_8(4) = \chi_8(2^2) = \chi_8^2(2) = \omega^4

\chi_8(5) = \chi_8(2^4) = \chi_8^4(2) = \omega^8 = -\omega^3

\chi_8(6) = \chi_8(2^9) = \chi_8^9(2) = -\omega^{18} = \omega^3

\chi_8(7) = \chi_8(2^7) = \chi_8^7(2) = -\omega^{14} = -\omega^4

\chi_8(8) = \chi_8(2^3) = \chi_8^3(2) = -\omega^6 = \omega

\chi_8(9) = \chi_8(2^6) = \chi_8^6(2) = \omega^{12} = \omega^2

\chi_8(10) = \chi_8(2^5) = \chi_8^5(2) = -\omega^{10} = -1

(and we have \chi_8(1) = 1). This completes the eighth character.

To compute the ninth character we can set

\chi_9(2) = \omega^8 = -\omega^3

then

\chi_9(3) = \chi_9(2^8) = \chi_9^8(2) = \omega^{24} = \omega^4

\chi_9(4) = \chi_9(2^2) = \chi_9^2(2) = \omega^6 = -\omega

\chi_9(5) = \chi_9(2^4) = \chi_9^4(2) = \omega^{12} = \omega^2

\chi_9(6) = \chi_9(2^9) = \chi_9^9(2) = -\omega^{27} = \omega^2

\chi_9(7) = \chi_9(2^7) = \chi_9^7(2) = -\omega^{21} = -\omega

\chi_9(8) = \chi_9(2^3) = \chi_9^3(2) = -\omega^9 = \omega^4

\chi_9(9) = \chi_9(2^6) = \chi_9^6(2) = \omega^{18} = -\omega^3

\chi_9(10) = \chi_9(2^5) = \chi_9^5(2) = -\omega^{15} = 1

(and we have \chi_9(1) = 1). This completes the ninth character.

Finally, to compute the tenth character we set

\chi_{10}(2) = \omega^9 = -\omega^4

then

\chi_{10}(3) = \chi_{10}(2^8) = \chi_{10}^8(2) = \omega^{32} = \omega^2

\chi_{10}(4) = \chi_{10}(2^2) = \chi_{10}^2(2) = \omega^8 = -\omega^3

\chi_{10}(5) = \chi_{10}(2^4) = \chi_{10}^4(2) = \omega^{16} = -\omega

\chi_{10}(6) = \chi_{10}(2^9) = \chi_{10}^9(2) = -\omega^{36} = \omega

\chi_{10}(7) = \chi_{10}(2^7) = \chi_{10}^7(2) = -\omega^{28} = \omega^3

\chi_{10}(8) = \chi_{10}(2^3) = \chi_{10}^3(2) = -\omega^{12} = -\omega^2

\chi_{10}(9) = \chi_{10}(2^6) = \chi_{10}^6(2) = \omega^{24} = \omega^4

\chi_{10}(10) = \chi_{10}(2^5) = \chi_{10}^5(2) = -\omega^{20} = -1

(and we have \chi_{10}(1) = 1). This completes the tenth character.

The remaining ten Dirichlet character tables which follow in this note can be calculated analogously (the details are not shown).

k = 12

mod 12

k = 13

mod 13

k = 14

mod 14

k = 15

mod 15

k = 16

mod 16

k = 17

mod 17

k = 18

mod 18

k = 19

mod 19

k = 20

mod 20

k = 21

mod 21