18.785: Analytic Number Theory, MIT, spring 2007 (K.S. Kedlaya)
Dirichlet characters and Dirichlet L-series
In this unit, we introduce some special multiplicative functions, the Dirichlet characters,and study their corresponding Dirichlet series. We will use these in a subsequent unit to prove
Dirichlet’s theorem on primes in arithmetic progressions, and the prime number theorem inarithmetic progressions.

1

Dirichlet characters

For N a positive integer, a Dirichlet character of level N is an arithmetic function χ which
factors through a homomorphism (Z/N Z)∗ → C on integers n ∈ N coprime to N , and iszero on integers not coprime to N ; such a function is completely multiplicative. Note thatthe nonzero values must all be N -th roots of unity, and that the characters of level N forma group under termwise multiplication.
For each level N , there is a Dirichlet character taking the value 1 at all n coprime to N ;it is called the principal (or trivial ) character of level N . A non-principal Dirichlet characterof level N is given by the Legendre-Jacobi symboln.
χ(n) =NLemma 1. If χ is nonprincipal of level N , thenχ(1) + · · · + χ(N ) = 0.
Proof. The sum is invariant under multiplication by χ(m) for any m ∈ N coprime to N , but
if χ is nonprincipal, then we can choose m with χ(m) 6= 1.
Sometimes a Dirichlet character of level N can be written as the termwise product ofthe principal character of level N with a character of some level N 0 < N (of course N 0 must
divide N ). We say the character is imprimitive in this case and primitive otherwise.

2L-series

The Dirichlet series associated to a Dirichlet character χ of level N is called a Dirichlet Lseries (or Dirichlet L-function) of level N , denoted L(s, χ). (It may also be denoted L χ (s),so that one can refer to Lχ as a function without explicitly naming the variable.) Since χ iscompletely multiplicative, L(s, χ) formally factors asY(1 − χ(p)p−s )−1 .
(1)p1

In particular, if χ is imprimitive corresponding to the character χ0 of level N 0 , thenYL(s, χ) = L(s, χ0 )(1 − χ0 (p)p−s ).

(2)

p|N,p6|N 0(Note that (2) reduces to L(s, χ) = L(s, χ0 ) if N and N 0 have the same prime factors, e.g.,
if N 0 is prime and N = (N 0 )2 .) In particular, the abscissa of absolute convergence of theprincipal character of level N , and hence of each of the characters of level N , is 1, and theproduct representation (1) is valid for Re(s) > 1. In particular, L(s, χ) 6= 0 for Re(s) > 1.
Theorem 2. Let χ be a Dirichlet character of level N . Then L(s, χ) extends to a meromorphic function on Re(s) > 0 with no poles
Q away from s = 1. If χ is principal, then L(s, χ)
has a simple pole at s = 1 of residue p|N (1 − p−1 ); otherwise, L(s, χ) is holomorphic also
at s = 1.
Proof. If χ is principal, then by (2),L(s, χ) = ζ(s)Y
(1 − p−s ),p|N

and the claims about L(s, χ) follow from what we already know about ζ. So assume hereafterthat χ is nonprincipal. By partial summation, we can writeL(s, χ) =∞X(χ(1) + · · · + χ(n))(n−s − (n + 1)−s ).

(3)n=1Since χ(1) + · · · + χ(N ) = 0 by Lemma 1, the quantities χ(1) + · · · + χ(n) are bounded for
all n. Meanwhile,n−s − (n + 1)−s = n−s (1 − (1 + 1/n)−s )= sn−s−1 + O(n−s−2 ),where the implied constant in the big O can be taken uniform over s in a compact set. Consequently, the sum representation for L(s, χ) given by (3) converges uniformly for Re(s) ≥ for any  > 0. This yields the claim.

3Nonvanishing of L-functions on Re(s) = 1

Much as we used nonvanishing of ζ on the line Re(s) = 1 to study the prime number theorem,
we will use nonvanishing of L-functions on that line to study the prime number theorem inarithmetic progressions. An additional wrinkle, though, is that we have to do some extra
work to understand what is going on at s = 1 itself; see next section.
Lemma 3. Let f (s) be a meromorphic function on a neighborhood of Re(s) ≥ L, with atworst a simple pole at s = L and no other poles. Suppose that log f (s) is represented by aDirichlet series with abscissa of convergence ≤ L and nonnegative real coefficients. Thenf (s) 6= 0 for Re(s) ≥ L.
2

Proof. See exercises.
Theorem 4. Let N be a positive integer. Let fN (s) be the product of all of the Dirichlet
L-series of level N . Then fN (s) 6= 0 for s ∈ C with Re(s) = 1.
Proof. Note that for Re(s) > 1, we havelog fN (s) =X∞XX

p:(p,N )=1 n=1χ!

χ(pn ) p−ns ,(4)which is a Dirichlet series with nonnegative real coefficients. (The sum over χ is invariant
under multiplication by χ(pn ) for any single χ, so either the sum is zero or all of the summands
are equal to 1.) We may thus apply Lemma 3.
This tells us a lot about nonvanishing of individual L-functions, but not quite everything.
Theorem 5. For any Dirichlet character χ, L(s, χ) 6= 0 when Re(s) = 1 and s 6= 1.
Proof. Let N be the level of χ. Then fN (s) is a product of functions, one of which is L(s, χ),
all of which are holomorphic at s. By Theorem 4, fN (s) has no zero at s, so none of thefactors can either.
It will take a bit more work to deal with s = 1; see next section.

4

Nonvanishing for L-functions at s = 1

At s = 1 (the so-called critical point for Dirichlet L-functions), life is a bit more complicated;to deduce that none of the L(1, χ) vanish, I would need to know that the function fN (s) inTheorem 4 has a simple pole, rather than being holomorphic, at s = 1.
We say a Dirichlet character is real if it takes values in ±1, and nonreal (or complex )otherwise.
Theorem 6. For any nonreal Dirichlet character χ, L(1, χ) 6= 0.
Proof. Let N be the level of χ. If L(1, χ) = 0, then also L(1, χ) = 0, where χ denotes thecomplex conjugate character. But then fN (s) is the product of one factor with a simple poleat s = 1 (coming from the principal character), two factors with zeroes at s = 1 (comingfrom χ and χ, and a bunch of factors which are holomorphic at s = 1. This would forcefN (s) to have a zero at s = 1, contradicting Theorem 4.
For the real characters, the above argument fails because χ and χ are the same character,so they don’t give two different contributions to fN (s). Instead, we use a different trick.
(There are a number of proofs of this result; see exercises for a second approach.)
Theorem 7. For any real nonprincipal Dirichlet character χ, L(1, χ) 6= 0.
3

Proof. Assume on the contrary that L(1, χ) = 0. Defineψ(s) =L(s, χ)L(s, χ0 ),L(2s, χ0 )where χ0 is the principal character of level N . Then the numerator of ψ is holomorphic forRe(s) > 0, because L(s, χ) counterbalances the simple pole of L(s, χ0 ) at s = 1. On theother hand, the denominator of ψ is holomorphic and nonzero for Re(s) > 1/2; moreover,it extends meromorphically to a neighborhood of s = 1/2 with a simple pole at s = 1/2.
Therefore ψ is holomorphic for Re(s) > 1/2, and extends holomorphically to a neighborhoodof 1/2 with a simple zero at s = 1/2.
However, ψ(s) admits the formal factorization
Y  1 + p−s ψ(s) =1 − p−sp:χ(p)=1and so expands as a Dirichlet series with nonnegative real coefficients and constant coefficient
1. The product factorization converges absolutely for Re(s) > 1, so the Dirichlet series does
too. But ψ is holomorphic for Re(s) > 1/2, so Landau’s theorem implies that the Dirichlet
series converges absolutely on Re(s) > 1/2.
This yields ψ(s) ≥ 1 for s > 1/2, whereas lims→(1/2)+ ψ(s) = 0, contradiction.
The proofs above have the merit that one could rewrite them without using complex
analysis, in order to obtain a complex analysis-free proof of Dirichlet’s theorem. (Dirichlet
was working before the properties of complex analytic functions were completely understood,
so his proofs tend to only involve real s.) However, Dani and Sawyer pointed out that you canalso argue more directly as follows. Suppose any of the L(s, χ) had a zero at s = 1; then f N (s)would be holomorphic on Re(s) > 0. Since log fN (s) has nonnegative real coefficients, sodoes fN (s) by formal exponentiation. Landau’s theorem would then imply that the Dirichlet
series for fN (s) converges absolutely for Re(s) > 0. However, since the Dirichlet series forlog fN (s) diverges for s = 1 − 1/φ(N ) (exercise), so does the series for fN (s), contradiction.

5

Historical aside: Dirichlet’s class number formula

Dirichlet introduced the Dirichlet L-series before Riemann had introduced complex function
theory into the picture, and so did not have access to such simple arguments in order toprove L(1, χ) 6= 0. However, the workaround he found is quite important in its own right; hewas able to express the value L(1, χ) in terms of a important numerical invariant, the classnumber of binary quadratic forms of a given discriminant. That number evidently being
positive, he obtained nonvanishing of L(1, χ), and even determined its sign.
Nowadays, one typically expresses this in the language of algebraic number theory. (Ifyou are not familiar with this language, feel free to ignore the rest of this section.) Let K

4

be a quadratic number field, and let χK be the character such thatp is unramified and split in K
1
χK (p) = −1 p is unramified and inert in K
0p is ramified in K.

One then proves that L(1, χK ) is equal to the class number hK of K times the regulator
RK . (The latter equals 1 if K is imaginary quadratic, and otherwise is equal to a fixed
normalization factor times the logarithm of the fundamental unit of K.)
The point here is that L(1, χ0 )L(1, χK ) is (up to multiplication by Euler factors for theramified primes) equal to the Dedekind zeta function ζK of K, defined byXζK (s) =Norm(a)−s ,afor a running over nonzero ideals of the ring of integers oK . For a general number field K,
ζK has a simple pole at 1, whose residue is the class number of K times the regulator of K
times a normalization factor (determined by the number of real and complex places of K);
the point is that each factor in this product is visibly nonzero.
One sometimes turns this around and tries to use analytic information about ζK to getinformation about the product hK RK . It is quite difficult to separate the two factors in thisexpression; indeed, one can make a good case that they really are simply two separate factors
in the computation of the volume of a certain compact topological group, the Arakelov classgroup of R, whose group of components is isomorphic to the usual class group.
Notable exception: there is no regulator for an imaginary quadratic field, so you can get
good bounds in this case. For instance, the Brauer-Siegel theorem says that the class numberof an imaginary quadratic field of discriminant D is at least c D 1/2− for any  > 0, though
unfortunately the constant c cannot be effectively determined from . The best effective
results are due to Goldfeld, who proves an effective lower bound which is polynomial inlog(D); this is a far cry from the truth, but is for instance enough to solve Gauss’s classnumber 1 problem (there are exactly nine imaginary quadratic fields of class number 1).

Exercises
1. Prove Lemma 3. (Hint: recall how you proved the special case f = ζ earlier.)
2. Let f be a meromorphic function on some neighborhood of Re(s) ≥ L, with a pole oforder e > 0 at s = L and no other poles. Suppose that log f (s) is represented by aDirichlet series with nonnegative real coefficients and abscissa of absolute convergence≤ L. Prove that every zero of f on the line Re(s) = L has multiplicity ≤ e/2. (Fore = 1, this implies Lemma 3.)
3. Prove directly that the Dirichlet series in (4) does not converge for s = 1 − 1/φ(N ).
(Hint: for every Dirichlet character χ of order N , χφ(N ) is principal.)5

4. Let χ be a real nonprincipal character. Use Dirichlet’s hyperbola method (from a priorhomework) to show thatXf (n)n−1/2 = 2L(1, χ)x1/2 + O(1),n≤xfor f (n) =Pd|n χ(d).

5. Use the previous exercise to show that L(1, χ) > 0, giving a second proof of Theorem 7.
(Hint: prove that f (n) ≥ 1 if n is a perfect square, and f (n) ≥ 0 otherwise.)6