# ELEMENTS OF ANALYTIC NUMBER THEORY

Save this PDF as:

Size: px
Start display at page:

## Transcription

1 ELEMENTS OF ANALYTIC NUMBER THEORY P. S. Kolesnikov, E. P. Vdovin Lecture course Novosibirsk, Russia 2013

2 Contents Chapter 1. Algebraic and transcendental numbers Field of algebraic numbers. Ring of algebraic integers 4 1. Preliminary information 4 2. Minimal polynomial 6 3. Algebraic complex numbers 9 4. Algebraic integers Diophantine approximations of algebraic numbers Diophantine approximation of degree ν Dirichlet approximation theorem Liouville theorem on Diophantine approximation of algebraic numbers Transcendentality of e and π Hermite identity Transcendentality of e Symmetrized n-tuples Transcendentality of π Problems 29 Chapter 2. Asymptotic law of distribution of prime numbers Chebyshev functions Definition and estimates Equivalence of the asymptotic behavior of Chebyshev functions and of the prime-counting function Von Mangoldt function Riemann function: Elementary properties Riemann function in Re z > Distribution of the Dirichlet series of a multiplicative function Convolution product and the Möbius inversion formula Euler identity Logarithmic derivative of the Riemann function 39 2

3 Contents 3 6. Expression of the integral Chebyshev function via the Riemann function Riemann function: Analytic properties Analytic extension of the Riemann function Zeros of the Riemann function Estimates of the logarithmic derivative Proof of the Prime Number Theorem Problems 54 Chapter 3. Dirichlet Theorem Finite abelian groups and groups of characters Finite abelian groups Characters Characters modulo m Dirichlet series Convergence of L-series Landau Theorem Proof of the Dirichlet Theorem 68 Chapter 4. p-adic numbers Valuation fields Basic properties Valuations over rationals The replenishment of a valuation field Construction and properties of p-adic fields Ring of p-adic integers and its properties The field of p-adic rationals is the replenishment of rationals in p-adic metric Applications Problems 86 Bibliography 87 Glossary 88 Index 89

4 CHAPTER 1 Algebraic and transcendental numbers 1.1. Field of algebraic numbers. Ring of algebraic integers 1. Preliminary information. Let us recall some basic notions from Abstract Algebra. Throughout we use the following notations: P is the set of all prime numbers; N is the set of positive integers (the set of natural numbers); Z is the set of all integers; Q is the set of all rational numbers; R is the set of all real numbers; C is the set of all complex numbers, C = R + ιr, ι 2 = 1. Given a field F, symbol F [x] denotes the ring of polynomials in variable x with coefficients in F. If f(x) = a 0 + a 1 x + + a n x n F [x], a i F, is chosen so that a n 0, then n is called the degree of f(x), it is denoted by deg f, while a n F is called the leading coefficient of f(x), and if a n = 1 then f(x) is called monic. If f(x) = 0 (all coefficients are equal to zero) then the degree of f(x) is said to be. If f(x), g(x) F [x], g(x) 0, then there exist unique q(x), r(x) F [x] such that f(x) = g(x)q(x) + r(x), deg r < deg g. (1.1) These polynomials (quotient q(x) and remainder r(x)) can be found by the well-known division algorithm. If r(x) = 0 then we write g f (g divides f). One may easily note the similarity between division algorithms in the ring of integers Z and in the ring of polynomials F [x]. Indeed, these are particular examples of Euclidean rings, and there are many common features and problems that can be solved in similar ways for integers and polynomials. In particular, the greatest common divisor (gcd) d of two polynomials f, g F [x] is defined as a monic common divisor which is divided by every 4

5 1.1. Field of algebraic numbers 5 other common divisor, i.e., d = gcd(f, g) if and only if d f, d g, and for every h F [x] with h f and h g it follows that h divides d. To find gcd of f and g, one may use the Euclidean algorithm based on the following observation: If f and g are related by (1.1) then gcd(f, g) = gcd(g, r). Moreover, if d = gcd(f, g) then there exist p(x), s(x) F [x] such that f(x)p(x) + g(x)s(x) = d(x). Exercise 1.1. Let f 1,..., f n F [x] be a finite family of polynomials over a field F. Prove that there exists a unique monic greatest common divisor of f 1,..., f n. Suppose R is a commutative ring with an identity (e.g., R = Z or R = F [x] as above). A subset I R is called an ideal of R if a ± b I for every a, b I, and ax I for every a I, x R. For example, the set of all even integers is an ideal of Z; the set {f(x) F [x] f(α) = 0} is an ideal of F [x], where α is an element of some extension field of F. Since an intersection of any family of ideals is again an ideal, for every set M R there exists minimal ideal of R which contains M, it is denoted by (M). It is easy to note that { } (M) = x i a i x i R, a i M. i An ideal I of R is said to be principal if there exists a R such that I = (a), where (a) stands for ({a}). Recall that a commutative ring R is called an integral domain (or simply a domain) if ab = 0 implies a = 0 or b = 0 for all a, b R. In particular, Z and F [x] are integral domains. An integral domain R such that every ideal of R is principal is called a principal ideal domain. Exercise 1.2. Prove that Z and F [x] (where F is a field) are principal ideal domains. In particular, if f(x), g(x) F [x] then ({f, g}) = (gcd(f, g)). If R is a domain, then we can consider the field of fractions of R. In order to construct it we start with the Cartesian product R (R \ {0}) of R (here each pair (a, b) corresponds to fraction a b ). Now define an equivalence relation (a 1, b 1 ) (a 2, b 2 ) a 1 b 2 = a 2 b 1. Let Q be the set of equivalence classes of R (R \ {0}) under this equivalence. Define the addition and multiplication on representatives by (a 1, b 1 ) + (a 2, b 2 ) = (a 1 b 2 + a 2 b 1, b 1 b 2 ), (a 1, b 1 ) (a 2, b 2 ) = (a 1 a 2, b 1 b 2 ).

6 1.1. Field of algebraic numbers 6 We leave for the reader to prove that all operations defined are correct and that Q is a field under these operations. Let I be an ideal of R. Then R is split into a disjoint union of congruence classes a+i = {a+x x I}, a R, and the set of all these classes (denoted by R/I) is a ring with respect to natural operations (a + I) + (b + I) = (a + b) + I, (a + I)(b + I) = ab + I. The ring R/I obtained is called a factor ring of R over I. For example, Z/(n) = Z n, the ring of remainders modulo n. A proper ideal I of R is maximal if there are no proper ideals J of R such that I J. For example, if R = Z then (n) is maximal if and only if n = ±p, where p is a prime natural number; if R = F [x] then (f) is maximal if and only if the polynomial f is irreducible over F. Note that if I is a maximal ideal of a commutative ring R then R/I is a field. Indeed, if a + I 0 (i.e., a / I) then J = {xa + b x R, b I} is an ideal of R such that I J. Therefore, J = R, and thus all equations of the form (a + I)X = c + I, c R, have solutions in R/I. Exercise 1.3. Prove that R[x]/(x 2 +x+1) is a field isomorphic to the field C of complex numbers. 2. Minimal polynomial. A complex number α C is algebraic if there exists a nonzero polynomial f(x) Q[x] such that f(α) = 0. A non-algebraic complex number is said to be transcendental. An algebraic number α is called an algebraic integer if there exists a monic polynomial f(x) Z[x] such that f(α) = 0. Every rational number α Q C is obviously an algebraic one. Moreover, as we will see later, a rational number is an algebraic integer if and only if it is an integer. Exercise 1.4. (1) Prove that 2 and ι are algebraic integers. 2 (2) Prove that if α C is an algebraic number then Re α and Im α are algebraic numbers. Whether the same statements are true for an algebraic integer α? (3) Show that the cardinality of the set of all algebraic numbers is countable. Since the cardinality of the entire set of complex numbers is uncountable (continuum), transcendental numbers do exist. However, it is not so easy to show an example of such a number accompanied with reasonable proof.

7 1.1. Field of algebraic numbers 7 Let α be an algebraic number. Denote by x a formal variable, and let I(α, x) = {f(x) Q[x] f(α) = 0}. It is clear that I(α, x) is an ideal of the ring Q[x]. Since Q[x] is a principal ideal domain, the ideal I(α, x) is generated by a single polynomial. Namely, the monic polynomial of minimal positive degree from h(x) I(α, x) is a generator of the ideal I(α, x), it is called the minimal polynomial for α. Given an algebraic number α, denote its minimal polynomial by h α (x) and say deg h α to be the degree of α. Lemma 1.5. Let α be an algebraic number, and let h(x) be a monic polynomial from Q[x]. Then the following conditions are equivalent: (1) h(x) = h α (x); (2) h(α) = 0, and h divides every f I(α, x); (3) h(α) = 0, and h(x) is irreducible over Q. Proof. (1) (2) It is obvious by definition. (2) (3) Assume h(x) is reducible over Q, i.e., it can be decomposed into nonscalar factors as follows: h(x) = h 1 (x)h 2 (x), where h i (x) Q[x], deg h i 1. Then h(α) = h 1 (α)h 2 (α) = 0, so for either of i = 1, 2 we have h i (α) = 0. Therefore, the corresponding polynomial h i (x) belongs to I(α, x), hence, h i (x) is a multiple of h(x), which is impossible due to deg h i < deg h. (3) (1) Let h α (x) be the minimal polynomial for α. Then h(x) I(α, x) = (h α (x)), so h α h. Since h(x) is irreducible and deg h α > 0, we have h(x) = h α (x). For example, if α Q then h α = x α. For α = 2, h α = x 2 2. The number α = ι 3 2 satisfies the equation α3 + 1 = 0, but its minimal polynomial is h α (x) = x 2 x + 1. Exercise 1.6. Prove that a minimal polynomial does not have multiple roots. If α is an algebraic number, and β C is a root of h α (x) then β is said to be conjugate to α. Therefore, every algebraic number α of degree n has exactly n pairwise different conjugate complex numbers α 1,..., α n,

8 1.1. Field of algebraic numbers 8 including α itself. Moreover, h α (x) = n (x α i ) Q[x]. i=1 Given an algebraic number α, the minimal polynomial h α Q[x] is uniquely defined. If α is an algebraic integer then there also exists a monic polynomial f Z[x] of minimal degree such that h α f. In order to prove that these f and h α coincide, we need the following observation. Let h(x) Q[x] be a monic polynomial with rational coefficients, h(x) = p 0 q 0 + p 1 q 1 x + + p n 1 q n 1 x n 1 + x n, gcd(p i, q i ) = 1. Denote by q(h) the least common multiple of the coefficients denominators: Then for q = q(h) we have q(h) = lcm (q 0,..., q n 1 ). (1.2) qh(x) = a 0 + a 1 x + + a n 1 x n 1 + a n x n Z[x], where the gcd of all coefficients is equal to the identity: (a 0,..., a n ) = 1. Indeed, assume a 0,..., a n have a common divisor d > 1. Then d q = a n, and for b = q/d Z we have bh(x) Z[x]. Therefore, q i bp i for all i = 0,..., n 1, so q i b, and, finally, q b = q/d, which is impossible for d > 1. Polynomials with relatively prime integral coefficients are studied in Abstract Algebra, they are called primitive. The following statement is well-known. Exercise 1.7 (Hauss Lemma ). Prove that the product of primitive polynomials is also a primitive polynomial. Proposition 1.8. Let α be an algebraic integer. Then the minimal polynomial h α (x) has integral coefficients. Proof. Suppose f(x) Z[x] be a monic polynomial with integral coefficients such that f(α) = 0. Then Lemma 1.5 implies h α f, i.e., f(x) = h α (x)g(x), g(x) Q[x]. Since both f and h α are monic polynomials, so is g. Denote q = q(h α ), q = q(g), where the function q( ) is given by (1.2). According to the remark above, qh α (x) and q g(x) are primitive polynomials in Z[x]. However, qq f(x) = (qh α (x))(q g(x)).

9 1.1. Field of algebraic numbers 9 The Hauss Lemma implies qq f(x) to be primitive, while f(x) has integral coefficients itself. Therefore, qq = 1, i.e., all coefficients of h α and g are integral. Thus, there is no need to define separately integral minimal polynomial and degree for algebraic integers. Let us also note that for every algebraic number α there exists c N such that cα is an algebraic integer: It is enough to consider c = q(h α ) deg hα. 3. Algebraic complex numbers. Let us denote the set of all algebraic numbers by A. Recall that the Fundamental Theorem of Algebra states C to be an algebraically closed field, i.e., every non-constant polynomial over C has a root in C. In this section, we will prove that A C is the minimal algebraically closed subfield of C, i.e., A is the algebraic closure of Q. Lemma 1.9. The following statements are equivalent for α C: (1) α A; (2) Q[α] := {f(α) f(x) Q[x]} is a finite-dimensional vector space over Q; (3) Q[α] is a subfield of C. Proof. (1) (2). Note that f(x) = q(x)h α (x) + r(x) by the division algorithm, deg r < deg h α. Hence, f(α) = r(α), and the latter is a linear combination over Q of 1, α,..., α n 1, where n = deg h α. Therefore, dim Q[α] n. Moreover, 1, α,..., α n 1 are linearly independent since n is the minimal possible degree of a polynomial over Q annihilating α, so dim Q[α] = deg h α. (2) (1). It is enough to note that 1, α, α 2,... are linearly dependent, so there exist a 0, a 1,..., a n Q such that a a 1 α + + a n α n = 0, end at least one of a i is nonzero. Hence, h(x) = a 0 +a 1 x+ +a n x n Q[x] is a nonzero polynomial annihilating α. (1) (3) Obviously, Q[α] is a subring of C. Let 0 f(α) Q[α], then f Q[x] \ I(α, x). Therefore, h α does not divide f. Since h α is irreducible, we have gcd(f, h α ) = 1. Hence, there exist polynomials u(x), v(x) Q[x] such that u(x)f(x) + v(x)h α (x) = 1.

10 1.1. Field of algebraic numbers 10 Then for x = α we obtain u(α)f(α) = 1, i.e., f(α) 1 = u(α) Q[α], i.e., Q[α] is a field. (3) (1) Suppose Q[α] is a subfield of C. It is enough to consider the case when α 0. Since α 1 Q[α], there exists f(x) Q[x] such that α 1 = f(α), i.e., h(α) = 0 for h(x) = xf(x) 1, deg h 1. Corollary If α 1,..., α n A then Q[α 1,..., α n ] := {f(α 1,..., α n ) f(x 1,..., x n ) Q[x 1,..., x n ]} is a finite-dimensional vector space over Q. Proof. For n = 1, it follows from Lemma 1.9. By induction, since dim Q[α 1,..., α n 1 ] < and dim Q[α n ] <, dim Q[α 1,..., α n ] dim Q[α 1,..., α n 1 ] dim Q[α n ] < (all dimensions are over Q). Exercise Prove that Q[α 1,..., α n ] is a subfield of C provided that α 1,..., α n A. Theorem The set of all algebraic numbers is a subfield of C. Proof. Since 1, 0 C are obviously algebraic, it is enough to prove the following two statements: (1) If α and β are algebraic numbers then α ± β and αβ are also algebraic numbers; (2) If β 0 is an algebraic number then 1/β is also an algebraic number. (1) Since Q[α ± β], Q[α β] Q[α, β] C, we have dim Q[α ± β] <, dim Q[α β] < by Corollary Thus by Lemma 1.9 α ± β, αβ A. (2) By Lemma 1.9, β 1 Q[β], so Q[β 1 ] Q[β] which is finitedimensional over Q. Hence, dim Q[β 1 ] < and thus β 1 A. Theorem The field A is algebraically closed. Proof. Suppose α 0,..., α n A, n 1, α n 0, ϕ(x) = α 0 + α 1 x + + α n x n A[x]. It is enough to show that ϕ(x) has a root in A. Without loss of generality, assume α n = 1. Indeed, by Theorem 1.12 we may divide ϕ(x) by α n, and the result is still in A[x]. By the Fundamental Theorem of Algebra, ϕ(x) has a root β C. Note that Q[β] Q[α 0,..., α n 1, β].

11 1.1. Field of algebraic numbers 11 Since β n can be expressed as a linear combination of β k, k = 0,..., n 1, with coefficients depending on α i, i = 0,..., n 1, as we have β n = α 0 α 1 β α n 1 β n 1, dim Q[α 0,..., α n 1, β] n dim Q[α 0,..., α n 1 ] <. Hence, Q[β] is a finite-dimensional vector space over Q, and by Lemma 1.9 β A Theorems 1.12 and 1.13 imply that A is the algebraic closure of Q, which is often denoted by Q. 4. Algebraic integers. Suppose K[x 1,..., x n ] is the ring of polynomials in several variables over a ring K. For f K[x 1,..., x n ] denote by deg f the maximal sum of the degrees of the variables that appear in a term of f with a nonzero coefficient. Namely, f may be uniquely written as f = f 0 + f 1 x n + + f m x m n, where f i K[x 1,..., x n 1 ]. Assuming deg f i are defined by induction, set deg f = max (i + deg f i). i=0,...,m Theorem The set of all algebraic integers is a subring of the field C. Proof. Ii is enough to show that if α, β are algebraic integers then α ± β and αβ are algebraic integers as well. We will prove a more general fact: Every number of the form γ = k,l c kl α k β l, c kl Z, is an algebraic integer. Let h α (x) = a 0 + a 1 x + + a n 1 x n 1 + x n, h β (x) = b 0 + b 1 x + + b m 1 x m 1 + x m. By Proposition 1.8, a i, b j Z. Lemma 1.5 implies that h α and h β are irreducible polynomials over Q, thus they have no multiple roots. Denote by α 1,..., α n all complex roots of h α (x), and let β 1,..., β m stand for all complex roots of h β (x). To be more precise, set α 1 = α, β 1 = β. Consider the polynomial n m p(x) = c kl αi k βj l C[x]. i=1 j=1 x k,l

12 1.1. Field of algebraic numbers 12 It is clear that p(γ) = 0 and the leading coefficient of p(x) is equal to the identity. It remains to show that p(x) Z[x]. It follows from the definition of p(x) that p(x) = f(x, α 1,..., α n, β 1,..., β m ), where f Z[x, y 1,..., y n, z 1,..., z m ]. Namely, n m f(x) = x c kl yi k z l j Z[x, y 1,..., y n, z 1,..., z m ]. i=1 j=1 k,l Moreover, every permutation of the variables y 1,..., y n or z 1,..., z m does not change the polynomial f, i.e., it is symmetric with respect to y i and with respect to z i. Hence, f(x, y 1,..., y n, z 1,..., z m ) = nm a=0 g a (y 1,..., y n, z 1,..., z m )x a, where every polynomial g a is symmetric with respect to y i and with respect to z i. On the other hand, for every a we have g a (y 1,..., y n, z 1,..., z m ) = g a,d1,...,d m (y 1,..., y n )z d zdm m, d 1,...,d m 1 where every polynomial g a,d1,...,d m (y 1,..., y n ) is symmetric (with respect to y i ) and has integral coefficients. Lemma Let Ψ(y 1,..., y n ) Z[y 1,..., y n ] be a symmetric polynomial on y 1,..., y n, deg Ψ = N, and let α 1,..., α n C be the roots of a polynomial h(x) = a 0 +a 1 x+ +a n x n Z[x], a n 0. Then a N n Ψ(α 1,..., α n ) Z. Proof. By the Fundamental Theorem of Symmetric Polynomials, there exists a polynomial G(t 1,..., t n ) Z[t 1,..., t n ] such that Ψ(y 1,..., y n ) = G(σ 1,..., σ n ), deg G N. where σ i (y 1,..., y n ) are the elementary symmetric polynomials on y 1,..., y n. The Viet formulae imply σ k (α 1,..., α n ) = ( 1) k a n k, k = 1,..., n. Since deg G N, we obtain a N n Ψ(α 1,..., α n ) = a N n G The latter is an integer number. a n ( a n 1,..., ( 1) n a ) 0. a n a n

13 1.2. Diophantine approximations 13 Now we can use Lemma 1.15 for polynomials g a,d1,...,d m (y 1,..., y n ) to obtain g a,d1,...,d m (α 1,..., α n ) Z (in this case, a n = 1 since α i are algebraic integers). Hence, g a (α 1,..., α n, z 1,..., z m ) Z[z 1,..., z m ] are symmetric polynomials on z 1,..., z m, and by Lemma 1.15 we have Therefore, p(x) Z[x]. g a (α 1,..., α n, β 1,..., β m ) Z Diophantine approximations of algebraic numbers It is well-known from the course of Analysis that the set of rational numbers Q is a dense subset of the set of real numbers R, i. e., for every α R and for every ε > 0 there exists p Q such that q α p q < ε. (1.3) Given a natural number N, the set Q N = {p/q p Z, 0 < q N}, has the following obvious property: a b 1 N 2 for all a, b Q N. Hence, for every α / Q N (in particular, for an irrational one), we have min a α > 0. a Q N Therefore, in order to satisfy (1.3) for small ε, the number p should have a q an unboundedly large denominator q. This observation raises the following natural question: How to measure the accuracy of a rational approximation relative to the growing denominator values? 1. Diophantine approximation of degree ν. To estimate the accuracy of an approximation by rationals, we will compare the difference α p/q with a decreasing function q ν, ν > 0. Namely, let us consider the following quantity: { ε α,ν (q) = min q ν α p } p Z,p/q α q, ν > 0. (1.4) It turns out that the study of the behavior of ε α,ν (q) as q approaches infinity leads to a necessary condition for α to be algebraic.

14 1.2. Diophantine approximations 14 Definition 1.1. A real number α R possesses a Diophantine approximation of degree ν > 0 if lim ε α,ν (q) <. (1.5) q Let us state a useful criterion that allows to determine whether a given number possesses a Diophantine approximation of a given degree. Lemma A real number α possesses a Diophantine approximation of degree ν > 0 if and only if there exists a constant c > 0 such that the inequality α p q < c q ν (1.6) holds for infinitely many rationals p q Q. Proof. Denote by M = M(α, c, ν) the set of all pairs (p, q) Z N such that (1.6) holds. Suppose α possesses a Diophantine approximation of degree ν > 0. By Definition 1.1, there exists c > 0 such that ε α,ν (q) < c for infinitely many q N. Let us fix this c and note that for every such q N there exists p Z satisfying the inequality (1.6). Therefore, the set M is infinite, and there is no upper bound for the set {q (p, q) M for some p}. It remains to show that the set {p/q (p, q) M} is also infinite. Indeed, if it were finite then the left-hand side of (1.6) has a positive lower bound, but the right-hand side of (1.6) approaches zero since ν > 0 and q may be chosen to be as large as we need. Conversely, suppose there exists c such that (1.6) holds for infinitely many rationals p/q. Then the set M defined in the first part of the proof is infinite. Assume the set {q (p, q) M for some p} has an upper bound N, i.e., where {p/q (p, q) M} Q N, Q N = {p/q Q p Z, 0 < q N}. As we have already mentioned above, for every distinct a 1, a 2 A the inequality a 1 a 2 > 1/N 2 holds. Hence, any infinite subset S of Q N has infinite diameter, i.e., for any d > 0 one may find p 1 /q 1 and p 2 /q 2 in S such that p 1 /q 1 p 2 /q 2 > d.

15 1.2. Diophantine approximations 15 In particular, the set S = {p/q (p, q) M} Q N contains p 1 /q 1 and p 2 /q 2 such that p 1 /q 1 p 2 /q 2 > 2c. Since (1.6) holds for p i /q i, i = 1, 2, we have α p 1 < c, α p 2 < c q 1 q ν 1 which implies 2c < p 1 p ( 2 1 q 1 q 2 < c q1 ν + 1 ) q2 ν < 2c, a contradiction. Therefore, the set of denominators {q (p, q) M for some p} is infinite, so there exist infinitely many q such that ε α,ν (q) < c. Hence, the sequence {ε α,ν (q)} q N has a finite accumulation point, and (1.5) holds. Exercise Whether a rational number possesses a Diophantine approximation of degree 1? 2. Dirichlet approximation theorem. Theorem 1.18 (Dirichlet Approximation Theorem). For every α R and for every N N there exist p Z and q N such that α p q < 1 qn, q N. Proof. Consider the fractional parts of the numbers kα, k = 0,..., N: q 2 q ν 2 ξ k = {kα} = kα [kα] [0, 1). Divide the interval [0, 1) into N intervals of length 1/N as follows: [k/n, (k + 1)/N), k = 0,..., N 1. According to the combinatorial Dirichlet s Principle, when N + 1 numbers ξ 0,..., ξ N are set into N intervals, there exists at least one interval which contains at least two of these numbers, i.e., ξ k1 ξ k2 < 1/N for some k 1, k 2, 0 k 1 < k 2 N. Let p = [k 2 α] [k 1 α], q = k 2 k 1. Then α p q = 1 q α(k 2 k 1 ) [k 2 α] + [k 1 α] = 1 q ξ k 2 ξ k1 < 1 Nq.

16 1.2. Diophantine approximations 16 Corollary If α R\Q then α possesses a Diophantine approximation of degree ν = 2. Proof. Theorem 1.18 implies that for every natural number N there exist p N Z and q N N, q N N, such that α p N < 1. (1.7) Nq N q N Let us show that the sequence {q N } N 1 is not bounded. Assume the converse, i.e., suppose there exists a constant M such that q N M for all N. Then (1.7) implies min p/q Q M α p q < 1 1 Nq N N 0, N which is impossible. Therefore, (1.7) holds for infinitely many numbers q N. Since q N N, we have qn 2 α p N < 1 for infinitely many q N. Proposition Let α Q. Then for every function ϕ : N R + satisfying lim qϕ(q) = 0 there exist only a finite number of rationals p q q such that α p q < ϕ(q). (1.8) Proof. Denote by S the set of all rationals p/q, gcd(p, q) = 1, satisfying (1.8). Assume S is infinite. It is easy to see that the distance between any two numbers from S does not exceed 2 max ϕ(q) which is finite. Note q 1 that the set of denominators of all fractions from S has no upper bound: Otherwise, if S Q N, where Q N introduced in the proof of Lemma 1.16, q N except α itself the following in- then S has an infinite diameter. Let α = a b. Then for all rationals p q equality holds: a b p q = aq pb bq 1 bq. Since b is fixed, we have 1 bq > ϕ(q),

17 1.2. Diophantine approximations 17 which is impossible for sufficiently large q. proves that S is finite. The contradiction obtained Corollary A rational number α does not have a Diophantine approximation of degree ν > 1. The result obtained is in some sence paradoxical: All irrational numbers possess Diophantine approximations of degree 2, but neither of rationals has a Diophantine approximation of degree ν > 1. Therefore, the existence of a Diophantine approximation of degree ν 2 is a criterion of irrationality. Let us apply the criterion above to the base of the natural logarithm also called the Euler s number. e = ! + 1 2! n! +..., (1.9) Proposition The Euler s number is irrational. Proof. Consider the sum of the first n + 1 summands in (1.9) and reduce to the common denominator: Then e p n = n! ! + 1 2! n! = p n n!. ( 1 1 (n + 1)! n Define a function ϕ : N R as follows: ϕ(q) = If q ((n 1)!, n!] then 2 (n + 1)! so lim qϕ(q) = 0. However, q ) (n + 2)(n + 3) +... < 1 (n + 1)! ϕ(1) = 1, ( for (n 1)! < q n!, n 2. 2 qϕ(q) n! (n + 1)! = 2 n + 1, e p q < ϕ(q) ) = 2 (n + 1)!.

18 1.2. Diophantine approximations 18 for infinitely many p/q = p n /n!, n 2. Hence, e is irrational by Proposition Liouville theorem on Diophantine approximation of algebraic numbers. Theorem 1.23 (Liouville Theorem). Let α be an algebraic number of degree n 2. Then α has no Diophantine approximation of degree ν > n. Proof. First, let us find a constant M > 0 such that α p q > M q n (1.10) for all p Z and q N. Set h(x) to be a multiple of the minmal polynomial for α with integer coefficients, e.g., h(x) = q(h α )h α (x). Suppose α 1,..., α n C are the complex roots of this polynomial, and assume α 1 = α. Then n n h(x) = a n (x α k ) = a n (x α) (x α k ), k=1 where a n is the leading coefficient of h(x). Denote by M the following quantity: ( 1 n M = a n ( α + α k + 1)), k=2 and let us show that (1.10) holds. If p and q meet the inequality α p/q 1 then (1.10) is valid since M < 1 ( α > 0, n 2). Assume p and q satisfy the condition α p/q < 1. In this case, p q < 1 + α and thus h(p/q) = a n α p n q α k p q k=2 α p n ( ) q a n α k + p q < α p 1 q M. k=2 k=2

19 1.2. Diophantine approximations 19 Lemma 1.5 implies h(x) to be irreducible over Q, hence, it has no rational roots. Therefore, h(p/q) = a p 0 + a 1 q + + a p n n q n 1 q n, and (1.10) follows. Finally, apply (1.10) to show that α has no Diophantine approximation of degree ν > n. Assume the converse: Let there exist ν > n and c > 0 such that α p q < c q ν holds for infinitely many p/q Q. Then, as it was shown in the proof of Lemma 1.16, the last inequality holds for infinitely many denominators q N. Then for sufficiently large q we have in contradiction to (1.10). c q ν < M q n, The Liouville Theorem is a powerful tool that allows constructing explicit examples of transcendental numbers. Namely, we obtain the following sufficient condition of transcendentality. Corollary Let α be a real number. If for every N N it possesses a Diophantine approximation of degree ν N then α is transcendental. Example 1.2. The following number is transcendental: α = 10 n!. Proof. Given N N, consider Note that α p N = q N p N q N = n=n+1 N 10 n! = p N 10 N!. 10 n! < 2 10 (N+1)! = 2 q N+1 N.

20 1.3. Transcendentality of e and π 20 Therefore, for every ν > 0 there exist infinitely many rationals p N, N ν, q N satisfying α p N q N < 2 q N+1 < 2 q ν. N N Hence, α possesses a Diophantine approximation of any degree. Theorem 1.23 implies α is not algebraic Transcendentality of e and π The Liouville Theorem provides a sufficient condition for a real number to be transcendental. However, this condition is not necessary. There exist different methods to prove transcendentality of a series of important constants, e.g., e and π. In this section, we are going to study one of these methods known as the Hermite Method. Given an analytic function f(z) on the complex plane, denote by x x 0 f(z) dz the Riemann integral of f(z) along the straight segment starting at x 0 C and ending at x C (the Cauchy Integral Theorem implies that this integral does not depend on the choice of a path with the same endpoints x 0 and x, we choose the straight segment for convenience). 1. Hermite identity. Lemma 1.25 (Hermite Identity). Let α C, α 0, and let f(x) C[x], deg f 1. Then for every x C we have where x 0 f(t)e αt dt = F (0) F (x)e αx, (1.11) F (x) = f(x) α + f (x) α f (n) (x) α n+1. Proof. According to the Fundamental Theorem of Calculus (the Newton Leibniz Formula), d dx x 0 f(t)e αt dt = f(x)e αx.

21 1.3. Transcendentality of e and π 21 On the other hand, d dx F (x)e αx = (F (x) αf (x))e αx = f(x)e αx. Hence, the derivatives with respect to x of the both sides of (1.11) coincide. It remains to compare the values at x = 0 to obtain the desired equality. The main idea of the Hermite s method is to apply the Hermite identity (1.11) to a polynomial of the form H(h(x)) = 1 (p 1)! a(n 1)p n x p 1 h(x) p, (1.12) h(x) Z[x], n = deg h, p N. Let us establish the properties of H(h(x)). Lemma Let h(x) = a 0 + a 1 x + + a n x n Z[x], a 0, a n 0, n 1, and let β 1,..., β n C be the entire collection of roots of h(x) in which every root of multiplicity k appears k times. Then for every p N, p 2, the polynomial f(x) = H(h(x)) defined by (1.12) has the following properties: (1) f (j) (0) = 0, 0 j p 2; (2) f (j) (β i ) = 0, 0 j p 1, i = 1,..., n; (3) f (p 1) (0) = a (n 1)p n a p 0 ; (4) f (j) (0) pz, j p; n (5) f (j) (β i ) pz, j p. i=1 Proof. It is easy to see from the construction of f(x) that 0 is its root of multiplicity p 1 and every β i is a root of f(x) of multiplicity at least p. As we know from the Abstract Algebra course, a root of multiplicity k of a polynomial f(x) is also a root of f (x), f (x),..., f (k 1) (x). This implies (1) and (2). To prove (3), let us distribute all brackets in the definition of f(x) and find the term of lowest degree in x, namely, the term is 1 (p 1)! a(n 1)p n a p 0 xp 1. Its (p 1)th derivative is equal to a n (n 1)p a p 0. For all other terms in f(x), their (p 1)th derivatives contain x and thus turn into zero at x = 0. Before we proceed with the proof of the remaining statements, note the following general fact. Given a polynomial Φ(x) Z[x], its jth derivative

### Chapter 4, Arithmetic in F [x] Polynomial arithmetic and the division algorithm.

Chapter 4, Arithmetic in F [x] Polynomial arithmetic and the division algorithm. We begin by defining the ring of polynomials with coefficients in a ring R. After some preliminary results, we specialize

### CHAPTER SIX IRREDUCIBILITY AND FACTORIZATION 1. BASIC DIVISIBILITY THEORY

January 10, 2010 CHAPTER SIX IRREDUCIBILITY AND FACTORIZATION 1. BASIC DIVISIBILITY THEORY The set of polynomials over a field F is a ring, whose structure shares with the ring of integers many characteristics.

### Module MA3411: Abstract Algebra Galois Theory Appendix Michaelmas Term 2013

Module MA3411: Abstract Algebra Galois Theory Appendix Michaelmas Term 2013 D. R. Wilkins Copyright c David R. Wilkins 1997 2013 Contents A Cyclotomic Polynomials 79 A.1 Minimum Polynomials of Roots of

### U.C. Berkeley CS276: Cryptography Handout 0.1 Luca Trevisan January, 2009. Notes on Algebra

U.C. Berkeley CS276: Cryptography Handout 0.1 Luca Trevisan January, 2009 Notes on Algebra These notes contain as little theory as possible, and most results are stated without proof. Any introductory

### PROBLEM SET 6: POLYNOMIALS

PROBLEM SET 6: POLYNOMIALS 1. introduction In this problem set we will consider polynomials with coefficients in K, where K is the real numbers R, the complex numbers C, the rational numbers Q or any other

### The Dirichlet Unit Theorem

Chapter 6 The Dirichlet Unit Theorem As usual, we will be working in the ring B of algebraic integers of a number field L. Two factorizations of an element of B are regarded as essentially the same if

### MOP 2007 Black Group Integer Polynomials Yufei Zhao. Integer Polynomials. June 29, 2007 Yufei Zhao yufeiz@mit.edu

Integer Polynomials June 9, 007 Yufei Zhao yufeiz@mit.edu We will use Z[x] to denote the ring of polynomials with integer coefficients. We begin by summarizing some of the common approaches used in dealing

### Introduction to Finite Fields (cont.)

Chapter 6 Introduction to Finite Fields (cont.) 6.1 Recall Theorem. Z m is a field m is a prime number. Theorem (Subfield Isomorphic to Z p ). Every finite field has the order of a power of a prime number

### Mathematics Course 111: Algebra I Part IV: Vector Spaces

Mathematics Course 111: Algebra I Part IV: Vector Spaces D. R. Wilkins Academic Year 1996-7 9 Vector Spaces A vector space over some field K is an algebraic structure consisting of a set V on which are

### PUTNAM TRAINING POLYNOMIALS. Exercises 1. Find a polynomial with integral coefficients whose zeros include 2 + 5.

PUTNAM TRAINING POLYNOMIALS (Last updated: November 17, 2015) Remark. This is a list of exercises on polynomials. Miguel A. Lerma Exercises 1. Find a polynomial with integral coefficients whose zeros include

### Finite Fields and Error-Correcting Codes

Lecture Notes in Mathematics Finite Fields and Error-Correcting Codes Karl-Gustav Andersson (Lund University) (version 1.013-16 September 2015) Translated from Swedish by Sigmundur Gudmundsson Contents

### it is easy to see that α = a

21. Polynomial rings Let us now turn out attention to determining the prime elements of a polynomial ring, where the coefficient ring is a field. We already know that such a polynomial ring is a UF. Therefore

### FACTORING POLYNOMIALS IN THE RING OF FORMAL POWER SERIES OVER Z

FACTORING POLYNOMIALS IN THE RING OF FORMAL POWER SERIES OVER Z DANIEL BIRMAJER, JUAN B GIL, AND MICHAEL WEINER Abstract We consider polynomials with integer coefficients and discuss their factorization

### Unique Factorization

Unique Factorization Waffle Mathcamp 2010 Throughout these notes, all rings will be assumed to be commutative. 1 Factorization in domains: definitions and examples In this class, we will study the phenomenon

### 1 if 1 x 0 1 if 0 x 1

Chapter 3 Continuity In this chapter we begin by defining the fundamental notion of continuity for real valued functions of a single real variable. When trying to decide whether a given function is or

### r + s = i + j (q + t)n; 2 rs = ij (qj + ti)n + qtn.

Chapter 7 Introduction to finite fields This chapter provides an introduction to several kinds of abstract algebraic structures, particularly groups, fields, and polynomials. Our primary interest is in

### Continued Fractions and the Euclidean Algorithm

Continued Fractions and the Euclidean Algorithm Lecture notes prepared for MATH 326, Spring 997 Department of Mathematics and Statistics University at Albany William F Hammond Table of Contents Introduction

### A number field is a field of finite degree over Q. By the Primitive Element Theorem, any number

Number Fields Introduction A number field is a field of finite degree over Q. By the Primitive Element Theorem, any number field K = Q(α) for some α K. The minimal polynomial Let K be a number field and

### 7. Some irreducible polynomials

7. Some irreducible polynomials 7.1 Irreducibles over a finite field 7.2 Worked examples Linear factors x α of a polynomial P (x) with coefficients in a field k correspond precisely to roots α k [1] of

### Modern Algebra Lecture Notes: Rings and fields set 4 (Revision 2)

Modern Algebra Lecture Notes: Rings and fields set 4 (Revision 2) Kevin Broughan University of Waikato, Hamilton, New Zealand May 13, 2010 Remainder and Factor Theorem 15 Definition of factor If f (x)

### RESULTANT AND DISCRIMINANT OF POLYNOMIALS

RESULTANT AND DISCRIMINANT OF POLYNOMIALS SVANTE JANSON Abstract. This is a collection of classical results about resultants and discriminants for polynomials, compiled mainly for my own use. All results

### Factoring Polynomials

Factoring Polynomials Sue Geller June 19, 2006 Factoring polynomials over the rational numbers, real numbers, and complex numbers has long been a standard topic of high school algebra. With the advent

### The cyclotomic polynomials

The cyclotomic polynomials Notes by G.J.O. Jameson 1. The definition and general results We use the notation e(t) = e 2πit. Note that e(n) = 1 for integers n, e(s + t) = e(s)e(t) for all s, t. e( 1 ) =

### Basic Concepts of Point Set Topology Notes for OU course Math 4853 Spring 2011

Basic Concepts of Point Set Topology Notes for OU course Math 4853 Spring 2011 A. Miller 1. Introduction. The definitions of metric space and topological space were developed in the early 1900 s, largely

### Course 221: Analysis Academic year , First Semester

Course 221: Analysis Academic year 2007-08, First Semester David R. Wilkins Copyright c David R. Wilkins 1989 2007 Contents 1 Basic Theorems of Real Analysis 1 1.1 The Least Upper Bound Principle................

### 6. Fields I. 1. Adjoining things

6. Fields I 6.1 Adjoining things 6.2 Fields of fractions, fields of rational functions 6.3 Characteristics, finite fields 6.4 Algebraic field extensions 6.5 Algebraic closures 1. Adjoining things The general

### Complex Numbers and the Complex Exponential

Complex Numbers and the Complex Exponential Frank R. Kschischang The Edward S. Rogers Sr. Department of Electrical and Computer Engineering University of Toronto September 5, 2005 Numbers and Equations

### Mathematics Review for MS Finance Students

Mathematics Review for MS Finance Students Anthony M. Marino Department of Finance and Business Economics Marshall School of Business Lecture 1: Introductory Material Sets The Real Number System Functions,

### 9. POLYNOMIALS. Example 1: The expression a(x) = x 3 4x 2 + 7x 11 is a polynomial in x. The coefficients of a(x) are the numbers 1, 4, 7, 11.

9. POLYNOMIALS 9.1. Definition of a Polynomial A polynomial is an expression of the form: a(x) = a n x n + a n-1 x n-1 +... + a 1 x + a 0. The symbol x is called an indeterminate and simply plays the role

### Ideal Class Group and Units

Chapter 4 Ideal Class Group and Units We are now interested in understanding two aspects of ring of integers of number fields: how principal they are (that is, what is the proportion of principal ideals

### Primality - Factorization

Primality - Factorization Christophe Ritzenthaler November 9, 2009 1 Prime and factorization Definition 1.1. An integer p > 1 is called a prime number (nombre premier) if it has only 1 and p as divisors.

### Die ganzen zahlen hat Gott gemacht

Die ganzen zahlen hat Gott gemacht Polynomials with integer values B.Sury A quote attributed to the famous mathematician L.Kronecker is Die Ganzen Zahlen hat Gott gemacht, alles andere ist Menschenwerk.

### Factoring of Prime Ideals in Extensions

Chapter 4 Factoring of Prime Ideals in Extensions 4. Lifting of Prime Ideals Recall the basic AKLB setup: A is a Dedekind domain with fraction field K, L is a finite, separable extension of K of degree

### THE FUNDAMENTAL THEOREM OF ALGEBRA VIA PROPER MAPS

THE FUNDAMENTAL THEOREM OF ALGEBRA VIA PROPER MAPS KEITH CONRAD 1. Introduction The Fundamental Theorem of Algebra says every nonconstant polynomial with complex coefficients can be factored into linear

### MA651 Topology. Lecture 6. Separation Axioms.

MA651 Topology. Lecture 6. Separation Axioms. This text is based on the following books: Fundamental concepts of topology by Peter O Neil Elements of Mathematics: General Topology by Nicolas Bourbaki Counterexamples

### Solutions to Practice Problems

Solutions to Practice Problems March 205. Given n = pq and φ(n = (p (q, we find p and q as the roots of the quadratic equation (x p(x q = x 2 (n φ(n + x + n = 0. The roots are p, q = 2[ n φ(n+ ± (n φ(n+2

### Series. Chapter Convergence of series

Chapter 4 Series Divergent series are the devil, and it is a shame to base on them any demonstration whatsoever. Niels Henrik Abel, 826 This series is divergent, therefore we may be able to do something

### Irreducibility criteria for compositions and multiplicative convolutions of polynomials with integer coefficients

DOI: 10.2478/auom-2014-0007 An. Şt. Univ. Ovidius Constanţa Vol. 221),2014, 73 84 Irreducibility criteria for compositions and multiplicative convolutions of polynomials with integer coefficients Anca

### Quotient Rings and Field Extensions

Chapter 5 Quotient Rings and Field Extensions In this chapter we describe a method for producing field extension of a given field. If F is a field, then a field extension is a field K that contains F.

### Factorization Algorithms for Polynomials over Finite Fields

Degree Project Factorization Algorithms for Polynomials over Finite Fields Sajid Hanif, Muhammad Imran 2011-05-03 Subject: Mathematics Level: Master Course code: 4MA11E Abstract Integer factorization is

### Elementary Number Theory We begin with a bit of elementary number theory, which is concerned

CONSTRUCTION OF THE FINITE FIELDS Z p S. R. DOTY Elementary Number Theory We begin with a bit of elementary number theory, which is concerned solely with questions about the set of integers Z = {0, ±1,

### The Division Algorithm for Polynomials Handout Monday March 5, 2012

The Division Algorithm for Polynomials Handout Monday March 5, 0 Let F be a field (such as R, Q, C, or F p for some prime p. This will allow us to divide by any nonzero scalar. (For some of the following,

### 6. Metric spaces. In this section we review the basic facts about metric spaces. d : X X [0, )

6. Metric spaces In this section we review the basic facts about metric spaces. Definitions. A metric on a non-empty set X is a map with the following properties: d : X X [0, ) (i) If x, y X are points

### H/wk 13, Solutions to selected problems

H/wk 13, Solutions to selected problems Ch. 4.1, Problem 5 (a) Find the number of roots of x x in Z 4, Z Z, any integral domain, Z 6. (b) Find a commutative ring in which x x has infinitely many roots.

### The Fundamental Theorem of Arithmetic

Chapter 1 The Fundamental Theorem of Arithmetic 1.1 Prime numbers If a, b Z we say that a divides b (or is a divisor of b) and we write a b, if for some c Z. Thus 2 0 but 0 2. b = ac Definition 1.1 The

### Winter Camp 2011 Polynomials Alexander Remorov. Polynomials. Alexander Remorov alexanderrem@gmail.com

Polynomials Alexander Remorov alexanderrem@gmail.com Warm-up Problem 1: Let f(x) be a quadratic polynomial. Prove that there exist quadratic polynomials g(x) and h(x) such that f(x)f(x + 1) = g(h(x)).

### Course 421: Algebraic Topology Section 1: Topological Spaces

Course 421: Algebraic Topology Section 1: Topological Spaces David R. Wilkins Copyright c David R. Wilkins 1988 2008 Contents 1 Topological Spaces 1 1.1 Continuity and Topological Spaces...............

### Prime Numbers and Irreducible Polynomials

Prime Numbers and Irreducible Polynomials M. Ram Murty The similarity between prime numbers and irreducible polynomials has been a dominant theme in the development of number theory and algebraic geometry.

### Introduction to finite fields

Introduction to finite fields Topics in Finite Fields (Fall 2013) Rutgers University Swastik Kopparty Last modified: Monday 16 th September, 2013 Welcome to the course on finite fields! This is aimed at

### 3. INNER PRODUCT SPACES

. INNER PRODUCT SPACES.. Definition So far we have studied abstract vector spaces. These are a generalisation of the geometric spaces R and R. But these have more structure than just that of a vector space.

### 11 Ideals. 11.1 Revisiting Z

11 Ideals The presentation here is somewhat different than the text. In particular, the sections do not match up. We have seen issues with the failure of unique factorization already, e.g., Z[ 5] = O Q(

### The Mathematics of Origami

The Mathematics of Origami Sheri Yin June 3, 2009 1 Contents 1 Introduction 3 2 Some Basics in Abstract Algebra 4 2.1 Groups................................. 4 2.2 Ring..................................

### Continued fractions and good approximations.

Continued fractions and good approximations We will study how to find good approximations for important real life constants A good approximation must be both accurate and easy to use For instance, our

### Introduction to Algebraic Geometry. Bézout s Theorem and Inflection Points

Introduction to Algebraic Geometry Bézout s Theorem and Inflection Points 1. The resultant. Let K be a field. Then the polynomial ring K[x] is a unique factorisation domain (UFD). Another example of a

### Chapter 7. Induction and Recursion. Part 1. Mathematical Induction

Chapter 7. Induction and Recursion Part 1. Mathematical Induction The principle of mathematical induction is this: to establish an infinite sequence of propositions P 1, P 2, P 3,..., P n,... (or, simply

### Non-unique factorization of polynomials over residue class rings of the integers

Comm. Algebra 39(4) 2011, pp 1482 1490 Non-unique factorization of polynomials over residue class rings of the integers Christopher Frei and Sophie Frisch Abstract. We investigate non-unique factorization

### MATH 289 PROBLEM SET 1: INDUCTION. 1. The induction Principle The following property of the natural numbers is intuitively clear:

MATH 89 PROBLEM SET : INDUCTION The induction Principle The following property of the natural numbers is intuitively clear: Axiom Every nonempty subset of the set of nonnegative integers Z 0 = {0,,, 3,

### k=1 k2, and therefore f(m + 1) = f(m) + (m + 1) 2 =

Math 104: Introduction to Analysis SOLUTIONS Alexander Givental HOMEWORK 1 1.1. Prove that 1 2 +2 2 + +n 2 = 1 n(n+1)(2n+1) for all n N. 6 Put f(n) = n(n + 1)(2n + 1)/6. Then f(1) = 1, i.e the theorem

### Field Fundamentals. Chapter 3. 3.1 Field Extensions. 3.1.1 Definitions. 3.1.2 Lemma

Chapter 3 Field Fundamentals 3.1 Field Extensions If F is a field and F [X] is the set of all polynomials over F, that is, polynomials with coefficients in F, we know that F [X] is a Euclidean domain,

### Notes on Factoring. MA 206 Kurt Bryan

The General Approach Notes on Factoring MA 26 Kurt Bryan Suppose I hand you n, a 2 digit integer and tell you that n is composite, with smallest prime factor around 5 digits. Finding a nontrivial factor

### x if x 0, x if x < 0.

Chapter 3 Sequences In this chapter, we discuss sequences. We say what it means for a sequence to converge, and define the limit of a convergent sequence. We begin with some preliminary results about the

### Mathematical Methods of Engineering Analysis

Mathematical Methods of Engineering Analysis Erhan Çinlar Robert J. Vanderbei February 2, 2000 Contents Sets and Functions 1 1 Sets................................... 1 Subsets.............................

### FACTORING IN QUADRATIC FIELDS. 1. Introduction. This is called a quadratic field and it has degree 2 over Q. Similarly, set

FACTORING IN QUADRATIC FIELDS KEITH CONRAD For a squarefree integer d other than 1, let 1. Introduction K = Q[ d] = {x + y d : x, y Q}. This is called a quadratic field and it has degree 2 over Q. Similarly,

### 5 Indefinite integral

5 Indefinite integral The most of the mathematical operations have inverse operations: the inverse operation of addition is subtraction, the inverse operation of multiplication is division, the inverse

### Further linear algebra. Chapter I. Integers.

Further linear algebra. Chapter I. Integers. Andrei Yafaev Number theory is the theory of Z = {0, ±1, ±2,...}. 1 Euclid s algorithm, Bézout s identity and the greatest common divisor. We say that a Z divides

### 3 1. Note that all cubes solve it; therefore, there are no more

Math 13 Problem set 5 Artin 11.4.7 Factor the following polynomials into irreducible factors in Q[x]: (a) x 3 3x (b) x 3 3x + (c) x 9 6x 6 + 9x 3 3 Solution: The first two polynomials are cubics, so if

### Copy in your notebook: Add an example of each term with the symbols used in algebra 2 if there are any.

Algebra 2 - Chapter Prerequisites Vocabulary Copy in your notebook: Add an example of each term with the symbols used in algebra 2 if there are any. P1 p. 1 1. counting(natural) numbers - {1,2,3,4,...}

### Computer Algebra for Computer Engineers

p.1/14 Computer Algebra for Computer Engineers Preliminaries Priyank Kalla Department of Electrical and Computer Engineering University of Utah, Salt Lake City p.2/14 Notation R: Real Numbers Q: Fractions

### INTRODUCTORY SET THEORY

M.Sc. program in mathematics INTRODUCTORY SET THEORY Katalin Károlyi Department of Applied Analysis, Eötvös Loránd University H-1088 Budapest, Múzeum krt. 6-8. CONTENTS 1. SETS Set, equal sets, subset,

### some algebra prelim solutions

some algebra prelim solutions David Morawski August 19, 2012 Problem (Spring 2008, #5). Show that f(x) = x p x + a is irreducible over F p whenever a F p is not zero. Proof. First, note that f(x) has no

### PERIMETERS OF PRIMITIVE PYTHAGOREAN TRIANGLES. 1. Introduction

PERIMETERS OF PRIMITIVE PYTHAGOREAN TRIANGLES LINDSEY WITCOSKY ADVISOR: DR. RUSS GORDON Abstract. This paper examines two methods of determining whether a positive integer can correspond to the semiperimeter

### CHAPTER II THE LIMIT OF A SEQUENCE OF NUMBERS DEFINITION OF THE NUMBER e.

CHAPTER II THE LIMIT OF A SEQUENCE OF NUMBERS DEFINITION OF THE NUMBER e. This chapter contains the beginnings of the most important, and probably the most subtle, notion in mathematical analysis, i.e.,

### QUADRATIC RECIPROCITY IN CHARACTERISTIC 2

QUADRATIC RECIPROCITY IN CHARACTERISTIC 2 KEITH CONRAD 1. Introduction Let F be a finite field. When F has odd characteristic, the quadratic reciprocity law in F[T ] (see [4, Section 3.2.2] or [5]) lets

### The Ideal Class Group

Chapter 5 The Ideal Class Group We will use Minkowski theory, which belongs to the general area of geometry of numbers, to gain insight into the ideal class group of a number field. We have already mentioned

### Prime Numbers. Chapter Primes and Composites

Chapter 2 Prime Numbers The term factoring or factorization refers to the process of expressing an integer as the product of two or more integers in a nontrivial way, e.g., 42 = 6 7. Prime numbers are

### Review for Final Exam

Review for Final Exam Note: Warning, this is probably not exhaustive and probably does contain typos (which I d like to hear about), but represents a review of most of the material covered in Chapters

### Practice Problems for First Test

Mathematicians have tried in vain to this day to discover some order in the sequence of prime numbers, and we have reason to believe that it is a mystery into which the human mind will never penetrate.-

### Chapter 13: Basic ring theory

Chapter 3: Basic ring theory Matthew Macauley Department of Mathematical Sciences Clemson University http://www.math.clemson.edu/~macaule/ Math 42, Spring 24 M. Macauley (Clemson) Chapter 3: Basic ring

### 1 Homework 1. [p 0 q i+j +... + p i 1 q j+1 ] + [p i q j ] + [p i+1 q j 1 +... + p i+j q 0 ]

1 Homework 1 (1) Prove the ideal (3,x) is a maximal ideal in Z[x]. SOLUTION: Suppose we expand this ideal by including another generator polynomial, P / (3, x). Write P = n + x Q with n an integer not

### Some Notes on Taylor Polynomials and Taylor Series

Some Notes on Taylor Polynomials and Taylor Series Mark MacLean October 3, 27 UBC s courses MATH /8 and MATH introduce students to the ideas of Taylor polynomials and Taylor series in a fairly limited

### 1 Lecture: Integration of rational functions by decomposition

Lecture: Integration of rational functions by decomposition into partial fractions Recognize and integrate basic rational functions, except when the denominator is a power of an irreducible quadratic.

### Appendix A. Appendix. A.1 Algebra. Fields and Rings

Appendix A Appendix A.1 Algebra Algebra is the foundation of algebraic geometry; here we collect some of the basic algebra on which we rely. We develop some algebraic background that is needed in the text.

### MA257: INTRODUCTION TO NUMBER THEORY LECTURE NOTES

MA257: INTRODUCTION TO NUMBER THEORY LECTURE NOTES 2016 47 4. Diophantine Equations A Diophantine Equation is simply an equation in one or more variables for which integer (or sometimes rational) solutions

### 8.1 Examples, definitions, and basic properties

8 De Rham cohomology Last updated: May 21, 211. 8.1 Examples, definitions, and basic properties A k-form ω Ω k (M) is closed if dω =. It is exact if there is a (k 1)-form σ Ω k 1 (M) such that dσ = ω.

### Chapter 6. Number Theory. 6.1 The Division Algorithm

Chapter 6 Number Theory The material in this chapter offers a small glimpse of why a lot of facts that you ve probably nown and used for a long time are true. It also offers some exposure to generalization,

### LINEAR ALGEBRA W W L CHEN

LINEAR ALGEBRA W W L CHEN c W W L Chen, 1997, 2008 This chapter is available free to all individuals, on understanding that it is not to be used for financial gain, and may be downloaded and/or photocopied,

### MATH : HONORS CALCULUS-3 HOMEWORK 6: SOLUTIONS

MATH 16300-33: HONORS CALCULUS-3 HOMEWORK 6: SOLUTIONS 25-1 Find the absolute value and argument(s) of each of the following. (ii) (3 + 4i) 1 (iv) 7 3 + 4i (ii) Put z = 3 + 4i. From z 1 z = 1, we have

### APPLICATIONS OF THE ORDER FUNCTION

APPLICATIONS OF THE ORDER FUNCTION LECTURE NOTES: MATH 432, CSUSM, SPRING 2009. PROF. WAYNE AITKEN In this lecture we will explore several applications of order functions including formulas for GCDs and

### 1 Mathematical Induction

Extra Credit Homework Problems Note: these problems are of varying difficulty, so you might want to assign different point values for the different problems. I have suggested the point values each problem

### POLYNOMIAL RINGS AND UNIQUE FACTORIZATION DOMAINS

POLYNOMIAL RINGS AND UNIQUE FACTORIZATION DOMAINS RUSS WOODROOFE 1. Unique Factorization Domains Throughout the following, we think of R as sitting inside R[x] as the constant polynomials (of degree 0).

### Mathematics of Cryptography

CHAPTER 2 Mathematics of Cryptography Part I: Modular Arithmetic, Congruence, and Matrices Objectives This chapter is intended to prepare the reader for the next few chapters in cryptography. The chapter

### PYTHAGOREAN TRIPLES PETE L. CLARK

PYTHAGOREAN TRIPLES PETE L. CLARK 1. Parameterization of Pythagorean Triples 1.1. Introduction to Pythagorean triples. By a Pythagorean triple we mean an ordered triple (x, y, z) Z 3 such that x + y =

### Lectures on Number Theory. Lars-Åke Lindahl

Lectures on Number Theory Lars-Åke Lindahl 2002 Contents 1 Divisibility 1 2 Prime Numbers 7 3 The Linear Diophantine Equation ax+by=c 12 4 Congruences 15 5 Linear Congruences 19 6 The Chinese Remainder

### 4. Complex integration: Cauchy integral theorem and Cauchy integral formulas. Definite integral of a complex-valued function of a real variable

4. Complex integration: Cauchy integral theorem and Cauchy integral formulas Definite integral of a complex-valued function of a real variable Consider a complex valued function f(t) of a real variable

### Factorization in Polynomial Rings

Factorization in Polynomial Rings These notes are a summary of some of the important points on divisibility in polynomial rings from 17 and 18 of Gallian s Contemporary Abstract Algebra. Most of the important

### CONTINUED FRACTIONS, PELL S EQUATION, AND TRANSCENDENTAL NUMBERS

CONTINUED FRACTIONS, PELL S EQUATION, AND TRANSCENDENTAL NUMBERS JEREMY BOOHER Continued fractions usually get short-changed at PROMYS, but they are interesting in their own right and useful in other areas

### 61. REARRANGEMENTS 119

61. REARRANGEMENTS 119 61. Rearrangements Here the difference between conditionally and absolutely convergent series is further refined through the concept of rearrangement. Definition 15. (Rearrangement)

### 2 Complex Functions and the Cauchy-Riemann Equations

2 Complex Functions and the Cauchy-Riemann Equations 2.1 Complex functions In one-variable calculus, we study functions f(x) of a real variable x. Likewise, in complex analysis, we study functions f(z)