Introduction to Dirichlet Series and the Dedekind Zeta Function

Home , Cauchy sequence, General Dirichlet series

Introduction to Dirichlet series and the Dedekind zeta function

Gil Moss November 11th, 2008

These notes are meant to give an introduction to the basic analysis behind Dirichlet series and the Dedekind zeta function. They include a description of the conditions which make Dirichlet series converge uniformly to continuous functions, using Abel’s summation formula, computing a residue at x = 1 of the Riemann zeta function, deﬁning an Euler product of Dirichlet series, and concluding with a deﬁnition of the Dedekind zeta function, proving its convergence and Euler product formula, and stating a theorem describing its residue at x = 1.

Dirichlet series and Abel summation P∞ −x A general Dirichlet series is deﬁned for a sequence {ai} of complex numbers to be i=1 aii . We can prove the following proposition regarding its convergence:

P∞ −x Proposition. If i=1 aii converges absolutely at x0, then it converges uniformly on [x0, ∞).

Proof. Take x > x0, then a i−x i x0−x −x = n aii 0 P∞ −x converges to zero as n goes to inﬁnity. Since the absolute value of the ratio of the coeﬃcients of i=1 aii P∞ −x0 P∞ −x and i=1 aii converges, the comparison test tells us that i=1 aii converges absolutely. To show uniform convergence, notice −x −x0 0 ≤ |aii | ≤ |aii |, −x P so that each term |aii | is bounded by a constant Mi such that Mi converges. This demonstrates uniform Pn −x convergence on the interval [x0, ∞), since the convergence of the partial sums i=1 |aii | is thus controlled independently of x. This is also called the Weierstrass M-test. Because a uniform limit of continuous functions is continuous, the Dirichlet series on this interval is thus continuous (it is a uniform limit of its partial sums, each of which is continuous). Now we introduce Abel summation, which will be useful for Dirichlet series. Pn Abel summation. Let {ai}, {bi} be sequences, let An = i=m+1 ai. Then

n n X X aibi = Anbn+1 − Ambm+1 − Ai(bi+1 − bi) m+1 m+1 .

1 Proof.

n n X X aibi = (Ai − Ai−1)bi i=m+1 i=m+1 n n X X = Aibi − Ai−1bi i=m+1 i=m+1 n n X X = Aibi − ( Aibi+1 + Ambm+1 − Anbn+1) i=m+1 i=m+1 n X = Anbn+1 − Ambm+1 − Ai(bi+1 − bi) m+1

Before applying Abel summation to Dirichlet series, we prove the following lemma.

Lemma. Given {ai}, {bi} sequences, assume |An| is bounded and {bi} is monotonically decreasing, con- P∞ verging to zero. Then i=1 aibi converges. Proof. Use Abel summation and the absolute value inequalities:

n n X X | aibi| ≤ |An|bn+1 + |Am|bm+1 + |Ai|(bi − bi+1), i=m+1 i=m+1 then if M is the bound for |An|, and for m, n sufficiently large, this expression is less than n X M(bn+1) + M(bm+1) + M (bi − bi+1) = M(bn+1 + bm+1 + bm+1 − bn+1) i=m+1 ≤ M(/2M + /2M) = Pn Hence the partial sums | i=m+1 aibi| form a Cauchy sequence and hence converge as desired. We now apply the Abel business to Dirichlet series. Pn Proposition. Assume the partial sums of the sequence {ai}, | i=1 ai| are bounded for each n. Then P∞ −x i=1 aii converges uniformly on [y, ∞) for all y > 0, and hence defines a continuous function on (0, ∞). −y P∞ −y Proof. Letting bi = i for y > 0 we note that i=1 aii converges on y > 0 since the conditions of the Lemma are fulfilled. 0 −y 0 −x 0 Pn 0 Now take x ≥ 0 and let ai = aii , bi = i , Ai = i=m+1 ai. Then we have n n X −(x+y) 0 −x 0 −x X 0 −x −x | aii | ≤ |An|(n + 1) + |Am|(m + 1) + |Ai|(i − (i + 1) ) i=m+1 i=m+1 n X ≤ (/2)(n + 1)−x + (/2)(m + 1)−x + (/2)(i−x − (i + 1)−x) i=m+1 = (/2)((n + 1)−x + (m + 1)−x + (m + 1)−x − (n + 1)−x) ≤ 2/2 = ,

0 P∞ 0 where the second inequality assumes m, n suﬃciently large that |An| < /2, which is valid since i=1 ai converges. The last inequality is due to the fact that (m+1)−x ≤ 1 when x ≥ 0. This shows the convergence on [y, ∞) is uniform since it does not rely on y. The continuity of the Dirichlet series again follows from the uniformity of convergence.

2 The Riemann zeta function at x = 1 P∞ i −x The previous proposition then shows that i=1(−1) i converges to a continuous function on (0, ∞). P∞ −x However the Riemann zeta function i=1 i does not fulﬁll the hypotheses of the proposition, but is convergent and continuous on (1, ∞) by our very ﬁrst proposition. We will now compute the residue of the Riemann zeta function at x = 1.

1 Proposition. Let φ(x) = ζ(x) − x−1 , then φ is continuous on (1, ∞) and 0 ≤ φ(x) ≤ 1 for x in this range. Proof. Continuity follows from the continuity of the Riemann zeta function on (1, ∞). To show the second part of the theorem, take x > 1, then for n ≥ 1, we have

Z n+1 (n + 1)−x ≤ y−xdy ≤ n−x, n since for n ≤ y ≤ n + 1 we have (n + 1)−x ≤ y−x ≤ n−x, and the interval of integration has length 1. Then summing over all n we have

R ∞ −x ζ(x) − 1 ≤ 1 y dy ≤ ζ(x) 1 ζ(x) − 1 ≤ x−1 ≤ ζ(x), which shows 0 ≤ φ(x) ≤ 1. Corollary. lim (x − 1)ζ(x) = 1 x→1+

1 Proof. Since φ takes values in [0, 1], limx→1+ (x − 1)φ(x) = 0, so limx→1+ (x − 1)(ζ(x) − x−1 ) = 0 and thus limx→1+ (x − 1)ζ(x) = 1.

Euler products for Dirichlet series Q∞ First we recall some facts about infinite products. i=1(1 + bi) for bi 6= −1 is convergent if the partial Qn products i=1(1 + bi) are convergent. Similarly to sums, an infinite product converges absolutely when Q∞ i=1(1 + |bi|) is convergent. Re-ordering of an absolutely convergent product does not change its value and an absolutely convergent product is convergent. Q∞ P∞ Lemma. i=1(1 + |bi|) converges iff i=1 |bi| converges. Qn Pn Proof. Expanding partial products we have the inequality i=1(1 + |bi|) ≥ i=1 |bi|, proving one direction of the lemma. To prove the other direction, note that for x ≥ 0, ex ≥ 1 + x by its power series expansion. Then Pn n i=1 |bi| Q e ≥ i=1(1 + |bi|) and if the partial sums converge, then so do the partial products.

A multiplicative sequence is a sequence {ai} such that aiaj = aij when i, j are relatively prime. Theorem. The following two conditions are equivalent:

P∞ −ix Q∞ P∞ −ix 1. i=1 api p converges absolutely and i=1(1 + i=1 |api |p ) converges P∞ −x 2. 1 + i=2 aii converges absolutely. P∞ −ix P∞ −x Proof. (1 ⇐ 2) Since i=1 api p is a subseries of 1 + i=2 aii , it converges absolutely. Now let S be a ﬁnite set of primes, then we have the equality

∞ Y X −ix X −x (1 + |api |p ) = 1 + aii p∈S i=1 i≥2,S3p|i

3 (where the sum on the right hand side is taken over all i such that p|i for some p ∈ S), since expanding the product will give us products of inﬁnite series which, for two diﬀerent primes p 6= p0 look like

∞ ∞ ∞ X −ix X 0−ix X X −dx 0−(i−d)x ( |api |p )( |ap0i |p ) = |apd ap0i−d |p p , i=1 i=1 i=1 d≤i

and since the sequence is multiplicative, this is

∞ X X d 0(i−d) −x |apdp0i−d |(p p ) , i=1 d≤i

and expanding the ﬁnite product will give all combinations of powers of primes in S, yielding the sum on the P −x right. But this sum 1 + i≥2,S3p|i aii is bounded above by assumption, so that all the partial products Q P∞ −ix are bounded, and thus the product p(1 + i=1 |api |p ) converges. (1 ⇒ 2) We use a similar argument to prove the opposite direction. Let Sm be the set of primes dividing an integer m. Then we have

m Y X −ix X −x (1 + |api |p ) ≤ 1 + aii

p i=1 i≥2,Sm3p|i ∞ Y X −ix = (1 + |api |p )

p∈Sm i=1 ∞ Y X −ix ≤ (1 + |api |p ), p i=1

which converges by assumption. Using the equality from the previous proof,

∞ X −x Y X −ix 1 + aii = (1 + |api |p ).

i≥2,Sm3p|i p∈Sm i=1

Then assuming the two equivalent convergence conditions of the previous theorem hold, these two partial sums converge to the same thing as m → ∞.

The Dedekind zeta function and its Euler product

If K is a number field of finite degree over Q, we defined the Dedekind zeta function of K to be

X −x ζK (x) = NI ,

I⊂OK where the sum ranges over all nonzero integral ideals I. We consider when this series converges and prove its Euler product representation.

Theorem. ζK (x) converges on (1, ∞) and on this interval,

Y −x −1 ζK (x) = (1 − NP ) , P

where P denotes a prime ideal of OK .

4 Proof. First note that 1 − (NP −x)2 < 1 and that 1 + NP −x − 2NP −2x ≥ 1 (the second fact is true since −x −2x −x −x 1 −x NP ≤ 1/2, so 2NP ≤ NP ). This gives us the inequalities 1N P < 1−NP −x ≤ 1 + 2NP . −x −1 −x −1 −x −x Now let (1 − NP ) = 1 + bp, for P |(p), so that bp = (1 − NP ) − 1 ≤ 1 + 2NP − 1 = 2NP = 2p−fx where f is the residual degree of P . Then

X X −x X −x bp ≤ 2[K : Q] p ≤ 2[K : Q] n , p p n

Pg since [K : Q] is the number of primes P above (p) (this is because [K : Q] = i=1 ef). Since x > 1 this −x series converges. And in fact, bn > 1+NP −1 > 0, so that |bn| = bn so the series is absolutely convergent. 1 P −ix Now using a geometric series expansion, 1−NP −x = 1 + P NP . Then it is left to show that, if Sm denotes the ﬁnite set of prime ideals P with NP ≤ m, Y X X (1 + NP −ix) = 1 + NI−x.

P ∈Sm P I⊂OK ,Sm3P |I

But this is true by the exact same reasoning as in proof of the Euler product theorem for Dirichlet series in the previous subsection, where the coeﬃcients ai are 1 in this case, and we deal with multiplication of prime ideals rather than prime integers. Unique factorization of ideals and multiplicativity of the norm make the argument the same, giving this equality for all m. Since both sides converge as m → ∞ by the same lemma in the previous subsection, the two limits of the two sides are equal. The Riemann zeta function can be considered a special case of the Dedekind zeta function, in particular

it is ζQ(x) since the norm of a rational prime ideal (p) is p. In considering the residue of the general Dedekind zeta function at x = 1, we discover it contains arithmetic information about the number ﬁeld K (which in the case of Q is uninteresting). To conclude these notes, we state, but do not prove, the following theorem: Theorem. 2s+tπtR h lim (x − 1)ζ (x) = K K . K 1/2 x→1+ WK |dK |

In this theorem, s and 2t are the number of real and imaginary imbeddings of K, respectively, hK is the order of the ideal class group of K, RK is the regulator of K, WK is the number of roots of unity contained in K, and dK is the ﬁeld discriminant of K.