# Advanced Number Theory Note #13: A study of primitive Dirichlet characters

In this (rather long) note, intended mainly as a technical memo for myself, I want to explore in detail the concept of a primitive Dirichlet character mod k, and the role of induced moduli in determining primitive and non-primitive Dirichlet characters. (I plan to expand further on some aspects of primitive characters in a couple of subsequent notes). As usual I will try herein to illustrate key ideas with concrete examples.

Dirichlet characters form a group and satisfy other strong properties, such as being completely multiplicative, which make them mathematically interesting in their own right as well as useful. (See my Note about them). For example, one of their most important roles is in Dirichlet L-functions which in turn are used in the proof of the prime number theorem, something I will explore in great detail in future notes. Primitive Dirichlet characters in particular are important because all Dirichlet characters can be viewed as extensions of primitive ones. The primitive Dirichlet characters are the irreducible ones from which the others are made, somewhat analogously to how prime numbers are the irreducible integers from which other integers are made.

Primitive Dirichlet characters are defined using the concept of an induced modulus which I have already used in a couple of earlier notes. In one previous note I tried to formulate an intuitively simple proof of a result concerning Dirichlet characters with two induced moduli. The concept of an induced modulus was also central to the separability of Gauss sums which I explored in Advanced Number Theory Note #12. I briefly mentioned its relevance in the preamble to Theorem 4 there, and the basic definition of an induced modulus is actually contained in equation (3) in Theorem 4 of that note.

Definition of induced modulus. Let $\chi$ be a Dirichlet character mod $k$ and let $d$ be any positive divisor of $k$. The number $d$ is called an induced modulus for $\chi$ if we have

$\chi(a) = 1$    whenever    $gcd(a, k) = 1$    and    $a \equiv 1$   (mod $d$)            (1)

Example: Suppose $\chi$ is the nonprincipal character mod 6. The reduced residue classes mod 6 are $\{1, 5\}$ so there are two Dirichlet characters mod 6. One is the principal character taking the value 1 for both 1 and 5. The other is the nonprincipal character $\chi$ which must have the square roots of unity as its values, so $\chi(1) = 1$, $\chi(5) = -1$. The proper divisors of $k = 6$ are $1$, $2$, $3$. Now consider any $a$ satisfying $gcd(a, 6) = 1$ and $a \equiv 1$ (mod 3). Then we must necessarily have $a \equiv 1$ (mod 6), so $\chi(a) = \chi(1) = 1$. Thus, $d = 3$ is an induced modulus of $\chi$. But note that neither $d = 1$ nor $d = 2$ are induced moduli. A counterexample for both is the case $a = 5$. We have $5 \equiv 1$ (mod 1) and $gcd(5, 6) = 1$, but $\chi(5) = -1 \neq 1$. Similarly, we have $5 \equiv 1$ (mod 2) and $gcd(5, 6) = 1$, but $\chi(5) = -1 \neq 1$. Thus, the nonprincipal character mod 6 has only one induced modulus smaller than 6. END

In the above example the divisor 1 of 6 was found not to be an induced modulus of the nonprincipal character mod 6. The following theorem shows that there is only one character for each mod k such that 1 is an induced modulus, and that is the principal character.

Theorem 1. Let $\chi$ be a Dirichlet character mod $k$. Then 1 is an induced modulus for $\chi$ if and only if $\chi$ is the principal character $\chi_1$.

Proof: Suppose first that $\chi = \chi_1$. Then $\chi(a) = 1$ for all $a$ which are relatively prime to $k$. But since all such integers $a$ also satisfy $a \equiv 1$ (mod 1), the number 1 is an induced modulus. Conversely, suppose that the number 1 is an induced modulus. Then $\chi(a) = 1$ whenever $gcd(a, k) = 1$ and so it must be the case that $\chi = \chi_1$ since both $\chi$ and $\chi_1$ vanish when the argument is not coprime with $k$. QED

Observe that for any Dirichlet character mod $k$ the modulus $k$ itself is an induced modulus, because $k$ is a divisor of $k$ and $\chi(a) = 1$ whenever $gcd(a, k) = 1$ and $a \equiv 1$ (mod $k$). The Dirichlet character is called primitive when it has no other induced moduli. Formally:

Definition of primitive characters. A Dirichlet character $\chi$ mod $k$ is said to be primitive mod $k$ if it has no induced modulus $d < k$. Thus, a primitive Dirichlet character mod $k$ will be such that, for every proper divisor $d$ of $k$, there is some integer $a \equiv 1$ (mod $d$) such that $gcd(a, k) = 1$ and $\chi(a) \neq 1$.

Note that for any modulus $k > 1$, the principal character $\chi_1$ will not be primitive since it will have $1 < k$ as an induced modulus. However, as the next theorem shows, if the modulus $k$ is a prime number, then every nonprincipal character mod $k$ is primitive.

Theorem 2. Every nonprincipal character $\chi$ modulo a prime $p$ is a primitive character mod $p$.

Proof: The only proper divisor of $p$ is 1, but since $\chi$ is not a principal character the number 1 cannot be an induced modulus, so $\chi$ must be primitive since it has no induced modulus $< p$. QED

The next theorem establishes an important and useful property of Dirichlet characters, namely that $d$ is an induced modulus for $\chi$ mod $k$ if and only if $\chi$ is periodic mod $d$ on those integers relatively prime to $k$. This can sometimes be used to identify induced moduli very quickly by looking at the pattern of a Dirichlet character’s values (see examples below).

Theorem 3. Let $\chi$ be a Dirichlet character mod $k$ and assume $d|k$ with $d > 0$. Then $d$ is an induced modulus for $\chi$ if and only if

$\chi(a) = \chi(b)$    whenever    $gcd(a, k) = gcd(b, k) = 1$    and    $a \equiv b$     (mod $d$)       (2)

Proof: Observe first that if (2) holds, then simply setting $b = 1$ gives equation (1), so $d$ must be an induced modulus if (2) holds. Going the other way, suppose that $d$ is an induced modulus. We will prove that then (2) must hold. Choose $a$ and $b$ such that

$gcd(a, k) = gcd(b, k) = 1$    and    $a \equiv b$   (mod $d$)

The following argument then shows $\chi(a) = \chi(b)$. Let $a\prime$ be the reciprocal of $a$ mod $k$, so that

$aa\prime \equiv 1$   (mod $k$)

This reciprocal exists because $gcd(a, k) = 1$. And note that also $gcd(a\prime, k) = 1$, so then we must have $gcd(aa\prime, k) = 1$. Since $d|k$, we can deduce immediately that

$aa\prime \equiv 1$ (mod $d$)

and therefore it must be the case that $\chi(aa\prime) = 1$ since $d$ is an induced modulus. But since $a \equiv b$ (mod $d$) we deduce that

$aa\prime \equiv ba\prime = 1$ (mod $d$)

and therefore

$\chi(aa\prime) = \chi(ba\prime)$

so

$\chi(a)\chi(a\prime) = \chi(b)\chi(a\prime)$

But $\chi(a\prime) \neq 0$ since $\chi(a)\chi(a\prime) = 1$, so we can cancel $\chi(a\prime)$ to deduce that $\chi(a) = \chi(b)$ which is what we needed to prove. QED

The following examples illustrate the situation.

Example: The following table shows the values of one of the Dirichlet characters $\chi$ mod 9.

We can see straight away that this table is periodic modulo 3 so by Theorem 3 it must be the case that 3 is an induced modulus for $\chi$. In fact, $\chi$ acts like the following Dirichlet character $\psi$

in the sense that $\chi(n) = \psi(n)$ for all $n$. Here, $\psi$ is a primitive Dirichlet character and $\chi$ is a non-primitive character called an extension of $\psi$.

Whenever $\chi$ is an extension of a character $\psi$ modulo $d$, then $d$ will be an induced modulus for $\chi$ because we will necessarily have $\chi(a) = \chi(b)$ whenever $gcd(a, k) = gcd(b, k) = 1$ and $a = b$ (mod $d$). In the above case we have for example $\chi(8) = \chi(2)$ with $gcd(8, 9) = gcd(2, 9) = 1$ and $8 = 2$ (mod $3$). END

Example: Here is an example showing that it is possible for a character $\chi$ mod $k$ to have an induced modulus $d < k$ without $\chi$ being an extension of any character $\psi$ mod $d$. The case is the one considered in the first example above, where $\chi$ is the (only) nonprincipal character mod 6.

As seen earlier, in this case the number 3 is an induced modulus because $\chi(n) = 1$ whenever $gcd(n, 6) = 1$ and $n \equiv 1$ (mod $3$). (There is only one such $n$ in this case, namely $n = 1$). However, $\chi$ is not an extension of any character $\psi$ modulo 3, because the only characters modulo 3 are the nonprincipal one shown above and the principal character

Since $\chi(2) = 0$, we see immediately that $\chi$ cannot be an extension of either $\psi$ or $\psi_1$. END

Example: There are four reduced residue classes mod 8, namely 1, 3, 5, 7, and thus four Dirichlet characters mod 8 as follows.

We look for induced moduli among the proper divisors of 8 other than 1, namely 2, 4. We see immediately that $\chi_2(5) \neq 1$, $\chi_3(7) \neq 1$ and $\chi_4(5) \neq 1$ are counterexamples for 2 being an induced modulus in the cases $\chi_2$, $\chi_3$ and $\chi_4$. Also $\chi_2(5) \neq 1$ and $\chi_4(5) \neq 1$ are counterexamples for 4 being an induced modulus in the cases $\chi_2$ and $\chi_4$. However, it is immediately apparent that $\chi_3$ is periodic mod 4, so 4 must be an induced modulus for $\chi_3$. END

Example: There are four reduced residue classes mod 12, namely 1, 5, 7, 11, and thus four Dirichlet characters mod 12 as follows.

Again we look for induced moduli among the proper divisors of 12 other than 1, namely 2, 3, 4, 6. We see immediately that $\chi_2(5) \neq 1$, $\chi_3(7) \neq 1$ and $\chi_4(5) \neq 1$ are counterexamples for 2 being an induced modulus in the cases $\chi_2$, $\chi_3$ and $\chi_4$.
Also $\chi_2(5) \neq 1$ and $\chi_4(5) \neq 1$ are counterexamples for 4 being an induced modulus in the cases $\chi_2$ and $\chi_4$. However, whenever $gcd(a, 12) = 1$ and $a \equiv 1$ (mod $4$) we have $\chi_3(a) = 1$ so 4 is an induced modulus for $\chi_3$.
$\chi_2(7) \neq 1$ and $\chi_3(7) \neq 1$ are counterexamples for 3 and 6 being induced moduli in the cases $\chi_2$, $\chi_3$. However, whenever $gcd(a, 12) = 1$ and $a \equiv 1$ (mod $3$) we have $\chi_4(a) = 1$ so 3 is an induced modulus for $\chi_4$. We also see that $\chi_4$ is periodic mod 6, so it follows from Theorem 3 that 6 must be an induced modulus for $\chi_4$. END

The above examples also help to clarify the following theorem.

Theorem 4. Let $\chi$ be a Dirichlet character modulo $k$, and assume $d|k$ and $d > 0$. Then the following two statements are equivalent:
(a). $d$ is an induced modulus for $\chi$
(b). There is a character $\psi$ modulo $d$ such that

$\chi(n) = \psi(n)\chi_1(n)$    for all   $n$                (3)

where $\chi_1$ is the principal character modulo $k$.

Example: In examples above we saw that 4 is an induced modulus for $\chi_3$ mod 8 as well as $\chi_3$ mod 12. Therefore there must be a character $\psi$ modulo 4 such that

$\chi_3(n) = \psi(n)\chi_1(n)$      for all   $n$

in both cases. This character is the only nonprincipal character mod 4, which takes the values $\psi(1) = 1$ and $\psi(3) = -1$.

Similarly, we saw that 3 is an induced modulus for $\chi_4$ mod 12, so there must be a character $\psi$ modulo 3 such that

$\chi_4(n) = \psi(n)\chi_1(n)$      for all   $n$

This character is the only nonprincipal character mod 3, which takes the values $\psi(1) = 1$ and $\psi(2) = -1$.

Finally, we saw that 6 is an induced modulus for $\chi_4$ mod 12, so there must be a character $\psi$ modulo 6 such that

$\chi_4(n) = \psi(n)\chi_1(n)$      for all   $n$

This character is the only nonprincipal character mod 6, which takes the values $\psi(1) = 1$ and $\psi(5) = -1$. END

Proof of Theorem 4: First, we assume (b) holds and then show that (a) must be true. If (b) holds we can choose any $n$ satisfying $gcd(n, k) = 1$ and $n \equiv 1$ (mod $d$) and for this $n$ we will then have $\chi_1(n) = 1$ and also $\psi(n) = 1$ (because by assumption $\psi$ is a character mod $d$, and $n \equiv 1$ (mod $d$)). Therefore we will have $\chi(n) = 1$ for any such $n$ and so $d$ is an induced modulus, which proves (a).

Going the other way, we now assume that (a) holds and try to find a character $\psi$ modulo $d$ for which equation (3) above holds. We can obtain the desired character by defining $\psi(n)$ as follows. If $gcd(n, d) > 1$, let $\psi(n) = 0$. In this case we also have $gcd(n, k) > 1$ (because $d|k$) so equation (3) holds because both sides are zero.

If $gcd(n, d) = 1$ then there exists an integer $m$ such that

$m \equiv n$ (mod $d$) and $gcd(m, k) = 1$

[This can be proved immediately with Dirichlet’s theorem on primes in arithmetic progressions. (See my Facebook Note for a detailed discussion of Dirichlet’s theorem and its proof). The arithmetic progression $xd + n$ contains infinitely many primes. Simply choose one that does not divide $k$ and call it $m$].

Having chosen $m$, which is unique modulo $d$, we then define

$\psi(n) = \chi(m) = \chi(n + xd)$

The number $\psi$ is then a well defined character mod $d$. It takes equal nonzero values at numbers which are congruent mod $d$ and coprime with $d$, and takes the value zero by definition at numbers which are not coprime with $d$.

We now verify that equation (3) holds for all $n$. If $gcd(n, k) = 1$, then it must also be the case that $gcd(n, d) = 1$, so $\psi(n) = \chi(m)$ for some $m \equiv n$ (mod $d$). So we have $gcd(n, k) = 1$, $gcd(m, k) = 1$, and $m \equiv n$ (mod $d$) and since $d$ is an induced modulus by (a) it then follows from Theorem 3 above that

$\chi(n) = \chi(m) = \psi(n) = \psi(n)\chi_1(n)$

where the last equality holds since $\chi_1(n) = 1$ for all $n$. This confirms that (b) holds in the case $gcd(n, k) = 1$.

If $gcd(n, k) > 1$, then $\chi(n) = \chi_1(n) = 0$ and both sides of (3) are zero. This confirms that (b) holds in the case $gcd(n, k) > 1$, so (b) holds for all $n$ if (a) is true. QED

It is useful to introduce the following terminology at this stage.

Definition. Let $\chi$ be a Dirichlet character mod $k$. The smallest induced modulus $d$ for $\chi$ is called the conductor of $\chi$.

The following theorem finally establishes the role that primitive Dirichlet characters play as the building blocks of Dirichlet characters generally.

Theorem 5. Every Dirichlet character $\chi$ mod $k$ can be expressed as a product

$\chi(n) = \psi(n)\chi_1(n)$      for all    $n$                      (4)

where $\chi_1$ is the principal character mod $k$ and $\psi$ is a primitive character modulo the conductor of $\chi$.

Proof: Let $d$ be the conductor of $\chi$. We already know from Theorem 4 that $\chi$ can be expressed as a product of the form (4), where $\psi$ is a character mod $d$. To prove Theorem 5 we therefore only need to prove the additional claim that $\psi$ is primitive mod $d$.

The proof will be by contradiction, assuming that $\psi$ is not primitive mod $d$. If $\psi$ is not primitive mod $d$ there is a proper divisor $q$ of $d$ which is an induced modulus for $\psi$. In equation (4) choose any $n$ such that

$gcd(n, k) = 1$      and      $n \equiv 1$   (mod $q$)

Then we will also have $gcd(n, q) = 1$ (since $q|k$) and since $q$ is an induced modulus for $\psi$ we must have $\psi(n) = 1$. But then equation (4) says

$\chi(n) = \psi(n)\chi_1(n) = \psi(n) = 1$      for all     $n$

so $q$ is also an induced modulus for $\chi$ which is smaller than $d$. This is a contradiction since $d$ is supposed to be the conductor of $\chi$. Therefore the assumption that $\psi$ is not primitive mod $d$ in (4) must be false. QED