Exercises Analytic Number Theory (Part 5) Date of Delivery: April 28

Exercises Analytic Number Theory (Part 5) Date of delivery: April 28.

1 2 1 25.a) Prove that x−log(1+x) ≥ 3 x if 0 ≤ x ≤ 2 . Prove also that −x+log(1−x) ≤ 1 2 1 − 3 x for 0 ≤ x ≤ 2 . b) Complete the proof of Theorem 1 of Chapter 5 on page 5.19. That is, deﬁne 1 δ(x) by A(x) = x(1+δ(x)). Let x be such that δ(x) ≤ 0. Put ε := min( 2 , |δ(x)|) R x A(y)−y 1 2 and x1 := (1 − ε)x. Prove that 2 dy ≤ − ε . x1 y 3

P −c(log x)1/10 26. Prove that for some constant c > 0 we have n≤x µ(n) = O xe as x → ∞. Hint. Prove that f(s) := ζ(s)+(ζ(s))−1 ∈ D(u, v) for certain suitable functions u, v by using the results from §7 of Chapter 3, and then apply Theorem 5 of Chapter 5.

27. In general, one deduces a version of the prime number theorem with error term using a zero-free region of ζ(s). In this exercise you are asked to do the converse, i.e., to prove the existence of a zero-free region for ζ(s) assuming a version of the Prime Number Theorem with an error term. Assume that for some ε > 0 we have

1 +ε π(x) = li(x) + O x 2 as x → ∞.

1 Deduce from this that ζ(s) 6= 0 for all s ∈ C with Re s > 2 + ε. Work out the P P following steps (as usual, θ(x) = p≤x log x and ψ(x) = n≤x Λ(n)):

R x π(t) a) Prove that θ(x) = π(x) log x− 2 t dt (using partial summation), li(x) log x = R x li(t) 2 t dt + x − 2 (using integration by parts) and

Z x π(t) − li(t) θ(x) − x = (π(x) − li(x)) log x − 2 − dt . 2 t

1 +ε 1 +ε b) Prove that θ(x) = x+O x 2 log x and ψ(x) = x+O x 2 log x as x → ∞.

1 ζ0(s) c) Compute an upper bound for the abscissa of convergence of f(s) := − ζ(s) −ζ(s) using Theorem 5 of Chapter 2 and give a region on which f(s) is holomorphic. d) Complete the proof.

28. Let G be a ﬁnite abelian group with basis {g1, . . . , gr}. Put mi := ord gi (i = 1, . . . , r). ∼ a) Prove that < g1 > × · · · × < gr >= G.

b) Prove that χ 7→ (χ(g1), . . . , χ(gr)) deﬁnes an isomorphism from Gˆ to m µm1 × · · · × µmr where µm denotes the group of roots of unity ρ with ρ = 1. c) Prove that there is an isomorphism from G to Gˆ.

In the next two exercises you have to use the Chinese Remainder Theorem: Let q1, q2 be two coprime integers and q = q1q2. Then for each pair of integers m1, m2, there exists a unique integer m such that

m ≡ m1(mod q1), m ≡ m2(mod q2), 0 ≤ m < q .

This integer m is given by

0 0 m ≡ q2q2m1 + q1q1m2(mod q), 0 ≤ m < q ,

0 0 0 0 where q1, q2 are integers such that q2q2 ≡ 1(mod q1) and q1q1 ≡ 1(mod q2). Moreover, if mi runs through 0, 1, , . . . , qi − 1 for i = 1, 2, then m runs through 0, 1, . . . , q − 1.

29.a) Let q = 2d where d is an odd positive integer. Let χ be a character modulo q. Prove that χ is induced by a character modulo d. Hint. Given any integer a, put χ0(a) = 0 if gcd(a, d) > 1 and χ0(a) := χ(b) if gcd(a, d) = 1, where b is any integer with b ≡ a(mod d) and gcd(b, q) = 1. Show that such integers b exist, that χ0(a) is well-deﬁned, i.e., independent of the choice of b and that χ0 is a character. b) Let now d, q be arbitrary positive integers with d a divisor of q. Prove that for every integer a with gcd(a, d) = 1 there is an integer b with b ≡ a(mod d) and gcd(b, q) = 1. Hint. Write q = q1q2 such that q1 is composed of primes dividing d and q2 is coprime with d. Use the Chinese Remainder Theorem.

2 c) Let χ be a character modulo q and d a divisor of q. Prove that the following three statements are equivalent: (i) χ is induced by a character modulo d; (ii) for all integers a, b with a ≡ b(mod d) and gcd(a, q) = gcd(b, q) = 1 we have χ(a) = χ(b). (iii) for all integers a with a ≡ 1(mod d) and gcd(a, q) = 1 we have χ(a) = 1.

d) Let χ be a character modulo q. Assume that χ is induced by characters χ1 modulo d1 and χ2 modulo d2, where d1, d2 are divisors of q. Prove that χ is induced by a character modulo d := gcd(d1, d2) and that this character induces χ1 and χ2.

30. Let q1, q2 be positive integers with gcd(q1, q2) = 1. and put q := q1q2. Let χ be ∗ ∗ a character modulo q. Deﬁne maps χ1 : Z → C , χ2 : Z → C as follows: given m ∈ Z, deﬁne a, b by

a ≡ m(mod q1), a ≡ 1(mod q2),

b ≡ 1(mod q1), b ≡ m(mod q2).

Then put χ1(m) := χ(a), χ2(m) := χ(b). The maps χ1, χ2 are called the projections of χ modulo q1, q2, respectively.

a) Prove that χ1 is a character modulo q1 (then one proves similarly that χ2 is a character modulo q2). Further prove that if m ≡ m1(mod q1), m ≡ m2(mod q2), then χ1(m1)χ2(m2) = χ(m).

b) Prove that τ(1, χ) = χ1(q2)χ2(q1)τ(1, χ1)τ(1, χ2). Hint. Use the above remarks concerning the Chinese Remainder Theorem.

Remark. By repeatedly applying the formula in b), one may express a Gauss sum for a character modulo q in terms of Gauss sums for characters modulo the prime powers in the factorization of q.

Points for the exercises: 25 a) 5, b) 5; 26) 10; 27 a) 3, b) 2, c) 3, d) 2; 28 a) 3, b) 4, c) 3; 29 a) 5, b) 5, c) 5, d) 5; 30 a) 5, b) 5.

Total: 70; Mark: (number of points)/7.