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

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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.

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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

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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.

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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

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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.

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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.

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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