H/wk 13, Solutions to selected problems


 Melanie Marsh
 1 years ago
 Views:
Transcription
1 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. (a) By a direct check we verify that the only roots of x x = 0 in Z 4 are 0 and 1. Thus x x = 0 has roots in Z 4. For every element a of Z we have a = a and hence for every (a, b) Z Z we have (a, b) = (a, b ) = (a, b), so that (a, b) is a root of x x = 0. Thus x x = 0 has Z Z = 4 roots in Z Z. Suppose now that R is an integral domain. It is easy to see that x = 0 and x = 1 are roots of x x = 0 in R. We claim that there are no other roots. Indeed, suppose a R is a root, so that a a = 0 in R. Then 0 = a a = a(a 1). Since R is an integral domain, it follows that either a = 0 or a 1 = 0, that is, either a = 0 or a = 1. Thus x x = 0 has exactly roots in R. By a direct check we verify that x x = 0 has exactly 4 roots in Z 6, namely 0, 1, 3 and 4. (b) Consider the ring R = Z where R = {(a 1, a, a 3,... ) a i Z } and where (a 1, a, a 3,... ) + (b 1, b, b 3,... ) = (a 1 + b 1, a + b, a 3 + b 3,... ) (a 1, a, a 3,... ) (b 1, b, b 3,... ) = (a 1 b 1, a b, a 3 b 3,... ) for a i, b i Z. It is easy to see that R is a commutative ring. Moreover, since for every a Z we have a = a, it follows that for every x = (a 1, a, a 3,... ) R we have is also infinite and commutative, it satisfies all the required proper Since R = Z ties. x = (a 1, a, a 3,... ) = (a 1, a, a 3,... ) = x and hence x x = 0. Ch. 4.1, Problem 17 In each case factor f(x) into linear factors in F [x]. (a) f(x) = x 4 + 1, F = Z 13. We have 1 = 1 in Z 13. Hence in Z 13 [x] we have x = x 4 1 = x 4 1 = (x 1)(x + 1). Moreover, in Z 13 we have 1 = 5 = 5. Hence in Z 13 [x] we have x = (x 1)(x + 1) = (x 1 )(x 5 ) = (x 1)(x + 1)(x 5)(x + 5). (b) f(x) = x 3 + 1, F = Z 7. In Z 7 we have = 0, so that 1 is a root of f(x) = x in Z 7. Hence (x + 1) f(x) in Z 7 [x]. By performing division with the remainder, we get x = (x + 1)(x x + 1) in Z 7 [x]. We then look for roots of x x + 1 in Z 7. It is easy to see that 3 is such a root since = 7 0 mod 7. Dividing 1
2 x x + 1 with the remainder by x 3 in Z 7 [x] we get x x + 1 = (x 3)(x + ) in Z 7 [x]. Hence x = (x + 1)(x 3)(x + ) in Z 7 [x]. Ch. 4.1, Problem 3 In each case determine the multiplicity of a as a root of f(x). (b) f(x) = x 4 + x + x +, a = 1, R = Z 3. Dividing f(x) by x + 1 with the remainder in Z 3 [x], we get: f(x) = x 4 + x + x + = (x + 1)(x 3 x + ) in Z 3 [x]. Observe that a = 1 is a root of x 3 x + in Z 3 [x]. Dividing x 3 x + by x + 1 in Z 3 [x], we get: x 3 x + = (x + 1)(x x + ) We see that = 1 0 in Z 3, so that a = 1 is not a root of x x + in Z 3 [x]. Hence the multiplicity of a = 1 as a root of f(x) in Z 3 [x] is equal to. Ch. 4.1, Problem 4 If R is a commutative ring, a polynomial f(x) in R[x] is said to annihilate R if f(a) = 0 for every a R. (a) Show that x p x annihilates Z p for a prime p. By Fermat s theorem, for every a Z we have a p a mod p, that is a p = a, a p a = 0 in Z p. Thus indeed x p x annihilates Z p. (b) Show that x 5 x annihilates Z 10. Can be verified via a direct check and also follows from (c). (c) Show that if p is a prime then x p x annihilates Z p. Since p is an odd prime, we have gcd(, p) = 1. Hence by Corollary 1 to Theorem 8 in Ch 3.4 we have that Z p = Z Z p as rings. Thus it suffices to show that x p x annihilates Z Z p. By part (a) we already know that for every b Z p we have b p = b. A direct check shows that for every a Z we have a = a. Therefore for every a Z, b Z p we have (a, b) p = (a p, b p ) = (a, b) and hence (a, b) p (a, b) = (0, 0) so that x p x annihilates Z Z p as required. If p > 3 is a prime, show that x p x annihilates Z 3p. Since p > 3 is a prime, we have gcd(3, p) = 1. Hence by Corollary 1 to Theorem 8 in Ch 3.4 we have that Z 3p = Z3 Z p as rings. Thus it suffices to show that x p x annihilates Z 3 Z p.
3 3 By part (a) we know that x p x annihilates Z p. It is easy to see by a direct check that x p x annihilates Z 3. Indeed, in Z 3 we have 0 p = 0, 1 p = 1 and 1 p = 1, where the last equality holds since p > 3 is a prime and hence p is odd. Therefore for every a Z 3, b Z p we have (a, b) p = (a p, b p ) = (a, b) and hence (a, b) p (a, b) = (0, 0) so that x p x annihilates Z 3 Z p as required. (e) Does x 5 x or x 7 x annihilate Z 35? Since gcd(5, 7) = 1, it follows that Z 35 = Z5 Z 7 as rings. Thus x p x annihilates Z 35 if and only if it annihilates each of Z 5, Z 7. For Z 7 we have 5 = 3 = 4 in Z 7. Thus x 5 x does not annihilate Z 7 and hence it does not annihilate Z 35. Also, for Z 5 we have 7 = 18 = 3 in Z 5. Thus x 7 x does not annihilate Z 5 and hence it does not annihilate Z 35. (f) Show that there exists a polynomial of degree n in Z n [x] that annihilates Z n. Take f(x) = x(x 1)(x )... (x n 1) Z n [x]. Ch. 4., Problem 5 In each case determine whether the polynomial is irreducible over each of the fields Q, R, C, Z, Z 3, Z 5 and Z 7. (a) x 3. We claim that x 3 is irreducible over Q. Since deg(x 3) =, to show that x 3 is irreducible over Q it suffices to prove that x 3 has no rational roots. We have x 3 = (x 3)(x + 3) in R[x]. Since R is an integral domain, it follows that x 3 has exactly two roots in R, namely ± 3. Since both of these roots are irrational, it follows that x 3 has no rational roots and hence it is irreducible over Q. We have x 3 = (x 3)(x+ 3) in R[x] and in C[x]. Hence x 3 is reducible over R and over C. Also, the following holds in Z [x]: and hence x 3 is reducible over Z. Similar, in Z [x] we have: x 3 = x 1 = (x 1)(x + 1) x 3 = x 0 = x = x x and hence x 3 is reducible over Z 3. A direct check shows that x 3 has no roots in Z 5. Indeed, in Z 5 we have 0 3 = 3 0, 1 3 = 0, 3 = 1 0, 3 3 = 1 0 and 4 3 = 3 0. Hence x 3 is irreducible over Z 5. Similarly, a direct check shows that x 3 has no roots in Z 7 and hence it is irreducible over Z 7. (b) x + x + 1
4 4 Via applying the quadratic formula we find the complex roots of x +x+1: x 1, = 1± 3 = 1+± 3i, so that in C[x] we have x +x+1 = (x 1± 3 )(x+ 1± 3 ). Since C is an integral domain, it follows that x 1, = 1± 3 = 1+± 3i are the only roots of x + x + 1 in C. Since none of these roots belong to Q and none of them belong to R, it follows that x + x + 1 has no roots in Q and no roots in R. Since deg(x + x + 1) =, this implies that x + x + 1 is irreducible over Q and it is also irreducible over R. Since x + x + 1 = (x 1± 3 )(x + 1± 3 ) in C[x], it follows that x + x + 1 is reducible over C. A direct check shows that x +x+1 has no roots in Z. Indeed, = 1 0 and = 1 0 in Z. Hence x + x + 1 is irreducible over Z. On the other hand, x + x + 1 has a root in Z 3, namely 1. Hence x + x + 1 is reducible over Z 3. One can also see this directly by checking that in Z 3 [x] we have x + x + 1 = x x + 1 = (x 1). A direct check shows that x + x + 1 has no roots in Z 5. Indeed, in Z 5 we have = 1 0, = 3 0, = 0, = 3 0, = 1 0. Hence x + x + 1 is irreducible over Z 5. On the other hand, x + x + 1 has a root in Z 7, namely. Hence x + x + 1 is reducible over Z 7. (c) x 3 + x + 1. Note that lim x x 3 + x + 1 = and lim x x 3 + x + 1 =. Hence by the Intermediate Value Theorem from calculus there exists x 0 R such that x x = 0. Thus x 3 + x + 1 has a root in R and hence it is reducible over R. Since this real root x 0 also belongs to C, it follows that x 3 + x + 1 is reducible over C. We claim that x 3 + x + 1 is irreducible over Q. Indeed, suppose, on the contrary, that x 3 +x+1 is reducible over Q. Since deg(x 3 +x+1) = 3, it follows that x 3 +x+1 has a root r Q. Then Theorem 9 in Ch 9.1 implies that r = c d where c, d Z and c 1, d 1. Hence c {1, 1} and d {1, 1}. Therefore r = c din{1, 1}. However = 3 0 and ( 1) 3 + ( 1) + 1 = 1 0 in Q, yielding a contradiction. Thus indeed x 3 + x + 1 is irreducible over Q. Ch. 4., Problem 6 Let R be an integral domain and let f(x) R[x] be monic. If f(x) factors properly in R[x], show that it has a proper factorization f(x) = g(x)h(x) where g(x) and h(x) are both monic. Let n = deg(f). Let f(x) = g 1 (x)h 1 (x) be a proper factorization of f(x) in R[x]. Thus degg 1 = m, degh 1 = k where m + k = n and 0 < m, k < n. We have f(x) = x n + a n 1 x n a 0 (since f is monic), g 1 (x) = b m x m + + b 0, h 1 = c k x k + + c 0 where b m, c k R, b m 0, c k 0. Then the leading coefficient of g 1 h 1 is equal to b m c k. Since f is monic, it follows that b m c k = 1 in R. Since R is an integral domain (and thus is commutative) this means that b m and c k are units in R and b m = c 1 k.
5 5 Put g(x) = c k g 1 (x) = c k (b m x m + +b 0 ) = c k b m x m + +c k b 0 = x m + +c k b 0, so that g(x) is monic. Also put h(x) = b m h 1 (x) = b m (c k x k + + c 0 ) = b m c k x k + + b m c 0 = x k + + b m c 0, so that h(x) is monic. We also have f(x) = g 1 (x)h 1 (x) = c k b m g 1 (x)h 1 (x) = c k g 1 (x)b m h 1 (x) = g(x)h(x). Thus we have found a proper factorization of f(x) as a product of two monic polynomials in R[x].
The Mean Value Theorem
The Mean Value Theorem THEOREM (The Extreme Value Theorem): If f is continuous on a closed interval [a, b], then f attains an absolute maximum value f(c) and an absolute minimum value f(d) at some numbers
More informationz = i ± 9 2 2 so z = 2i or z = i are the solutions. (c) z 4 + 2z 2 + 4 = 0. By the quadratic formula,
91 Homework 8 solutions Exercises.: 18. Show that Z[i] is an integral domain, describe its field of fractions and find the units. There are two ways to show it is an integral domain. The first is to observe:
More informationit 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
More informationPROBLEM 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
More informationCHAPTER 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.
More informationIntroduction 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
More informationChapter 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
More informationModern 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)
More informationFactoring 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
More informationcalculating the result modulo 3, as follows: p(0) = 0 3 + 0 + 1 = 1 0,
Homework #02, due 1/27/10 = 9.4.1, 9.4.2, 9.4.5, 9.4.6, 9.4.7. Additional problems recommended for study: (9.4.3), 9.4.4, 9.4.9, 9.4.11, 9.4.13, (9.4.14), 9.4.17 9.4.1 Determine whether the following polynomials
More informationPUTNAM 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
More informationQuotient 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.
More information3 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
More information7. 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
More informationFactorization 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
More informationDefinition 2.1 The line x = a is a vertical asymptote of the function y = f(x) if y approaches ± as x approaches a from the right or left.
Vertical and Horizontal Asymptotes Definition 2.1 The line x = a is a vertical asymptote of the function y = f(x) if y approaches ± as x approaches a from the right or left. This graph has a vertical asymptote
More informationa 1 x + a 0 =0. (3) ax 2 + bx + c =0. (4)
ROOTS OF POLYNOMIAL EQUATIONS In this unit we discuss polynomial equations. A polynomial in x of degree n, where n 0 is an integer, is an expression of the form P n (x) =a n x n + a n 1 x n 1 + + a 1 x
More informationminimal polyonomial Example
Minimal Polynomials Definition Let α be an element in GF(p e ). We call the monic polynomial of smallest degree which has coefficients in GF(p) and α as a root, the minimal polyonomial of α. Example: We
More informationUnique 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
More information3 Factorisation into irreducibles
3 Factorisation into irreducibles Consider the factorisation of a nonzero, noninvertible integer n as a product of primes: n = p 1 p t. If you insist that primes should be positive then, since n could
More informationPOLYNOMIAL 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).
More informationFactorization Algorithms for Polynomials over Finite Fields
Degree Project Factorization Algorithms for Polynomials over Finite Fields Sajid Hanif, Muhammad Imran 20110503 Subject: Mathematics Level: Master Course code: 4MA11E Abstract Integer factorization is
More informationMath 34560 Abstract Algebra I Questions for Section 23: Factoring Polynomials over a Field
Math 34560 Abstract Algebra I Questions for Section 23: Factoring Polynomials over a Field 1. Throughout this section, F is a field and F [x] is the ring of polynomials with coefficients in F. We will
More informationMTH4100 Calculus I. Lecture notes for Week 8. Thomas Calculus, Sections 4.1 to 4.4. Rainer Klages
MTH4100 Calculus I Lecture notes for Week 8 Thomas Calculus, Sections 4.1 to 4.4 Rainer Klages School of Mathematical Sciences Queen Mary University of London Autumn 2009 Theorem 1 (First Derivative Theorem
More information2.5 ZEROS OF POLYNOMIAL FUNCTIONS. Copyright Cengage Learning. All rights reserved.
2.5 ZEROS OF POLYNOMIAL FUNCTIONS Copyright Cengage Learning. All rights reserved. What You Should Learn Use the Fundamental Theorem of Algebra to determine the number of zeros of polynomial functions.
More informationMOP 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
More informationThe 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,
More informationZeros of Polynomial Functions
Review: Synthetic Division Find (x 25x  5x 3 + x 4 ) (5 + x). Factor Theorem Solve 2x 35x 2 + x + 2 =0 given that 2 is a zero of f(x) = 2x 35x 2 + x + 2. Zeros of Polynomial Functions Introduction
More informationON GALOIS REALIZATIONS OF THE 2COVERABLE SYMMETRIC AND ALTERNATING GROUPS
ON GALOIS REALIZATIONS OF THE 2COVERABLE SYMMETRIC AND ALTERNATING GROUPS DANIEL RABAYEV AND JACK SONN Abstract. Let f(x) be a monic polynomial in Z[x] with no rational roots but with roots in Q p for
More informationFactoring Cubic Polynomials
Factoring Cubic Polynomials Robert G. Underwood 1. Introduction There are at least two ways in which using the famous Cardano formulas (1545) to factor cubic polynomials present more difficulties than
More information1 = (a 0 + b 0 α) 2 + + (a m 1 + b m 1 α) 2. for certain elements a 0,..., a m 1, b 0,..., b m 1 of F. Multiplying out, we obtain
Notes on realclosed fields These notes develop the algebraic background needed to understand the model theory of realclosed fields. To understand these notes, a standard graduate course in algebra is
More informationModule 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
More informationZero: If P is a polynomial and if c is a number such that P (c) = 0 then c is a zero of P.
MATH 11011 FINDING REAL ZEROS KSU OF A POLYNOMIAL Definitions: Polynomial: is a function of the form P (x) = a n x n + a n 1 x n 1 + + a x + a 1 x + a 0. The numbers a n, a n 1,..., a 1, a 0 are called
More informationFinite Fields and ErrorCorrecting Codes
Lecture Notes in Mathematics Finite Fields and ErrorCorrecting Codes KarlGustav Andersson (Lund University) (version 1.01316 September 2015) Translated from Swedish by Sigmundur Gudmundsson Contents
More information1 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
More informationInteger roots of quadratic and cubic polynomials with integer coefficients
Integer roots of quadratic and cubic polynomials with integer coefficients Konstantine Zelator Mathematics, Computer Science and Statistics 212 Ben Franklin Hall Bloomsburg University 400 East Second Street
More information9. 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 n1 x n1 +... + a 1 x + a 0. The symbol x is called an indeterminate and simply plays the role
More informationSOLVING POLYNOMIAL EQUATIONS
C SOLVING POLYNOMIAL EQUATIONS We will assume in this appendix that you know how to divide polynomials using long division and synthetic division. If you need to review those techniques, refer to an algebra
More informationJUST THE MATHS UNIT NUMBER 1.8. ALGEBRA 8 (Polynomials) A.J.Hobson
JUST THE MATHS UNIT NUMBER 1.8 ALGEBRA 8 (Polynomials) by A.J.Hobson 1.8.1 The factor theorem 1.8.2 Application to quadratic and cubic expressions 1.8.3 Cubic equations 1.8.4 Long division of polynomials
More informationContinuity. DEFINITION 1: A function f is continuous at a number a if. lim
Continuity DEFINITION : A function f is continuous at a number a if f(x) = f(a) REMARK: It follows from the definition that f is continuous at a if and only if. f(a) is defined. 2. f(x) and +f(x) exist.
More informationBasics of Polynomial Theory
3 Basics of Polynomial Theory 3.1 Polynomial Equations In geodesy and geoinformatics, most observations are related to unknowns parameters through equations of algebraic (polynomial) type. In cases where
More informationALGEBRA 2 CRA 2 REVIEW  Chapters 16 Answer Section
ALGEBRA 2 CRA 2 REVIEW  Chapters 16 Answer Section MULTIPLE CHOICE 1. ANS: C 2. ANS: A 3. ANS: A OBJ: 53.1 Using Vertex Form SHORT ANSWER 4. ANS: (x + 6)(x 2 6x + 36) OBJ: 64.2 Solving Equations by
More informationFACTORING 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
More informationFactoring 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
More informationCOMMUTATIVE RINGS. Definition: A domain is a commutative ring R that satisfies the cancellation law for multiplication:
COMMUTATIVE RINGS Definition: A commutative ring R is a set with two operations, addition and multiplication, such that: (i) R is an abelian group under addition; (ii) ab = ba for all a, b R (commutative
More information(a) Write each of p and q as a polynomial in x with coefficients in Z[y, z]. deg(p) = 7 deg(q) = 9
Homework #01, due 1/20/10 = 9.1.2, 9.1.4, 9.1.6, 9.1.8, 9.2.3 Additional problems for study: 9.1.1, 9.1.3, 9.1.5, 9.1.13, 9.2.1, 9.2.2, 9.2.4, 9.2.5, 9.2.6, 9.3.2, 9.3.3 9.1.1 (This problem was not assigned
More informationHomework # 3 Solutions
Homework # 3 Solutions February, 200 Solution (2.3.5). Noting that and ( + 3 x) x 8 = + 3 x) by Equation (2.3.) x 8 x 8 = + 3 8 by Equations (2.3.7) and (2.3.0) =3 x 8 6x2 + x 3 ) = 2 + 6x 2 + x 3 x 8
More information1 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.
More informationPolynomial Factoring. Ramesh Hariharan
Polynomial Factoring Ramesh Hariharan The Problem Factoring Polynomials overs Integers Factorization is unique (why?) (x^2 + 5x +6) (x+2)(x+3) Time: Polynomial in degree A Related Problem Factoring Integers
More informationsome 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
More informationProblem Set 7  Fall 2008 Due Tuesday, Oct. 28 at 1:00
18.781 Problem Set 7  Fall 2008 Due Tuesday, Oct. 28 at 1:00 Throughout this assignment, f(x) always denotes a polynomial with integer coefficients. 1. (a) Show that e 32 (3) = 8, and write down a list
More information317 1525 5 1510 25 32 5 0. 1b) since the remainder is 0 I need to factor the numerator. Synthetic division tells me this is true
Section 5.2 solutions #110: a) Perform the division using synthetic division. b) if the remainder is 0 use the result to completely factor the dividend (this is the numerator or the polynomial to the
More informationFIRST YEAR CALCULUS. Chapter 7 CONTINUITY. It is a parabola, and we can draw this parabola without lifting our pencil from the paper.
FIRST YEAR CALCULUS WWLCHENW L c WWWL W L Chen, 1982, 2008. 2006. This chapter originates from material used by the author at Imperial College, University of London, between 1981 and 1990. It It is is
More informationLagrange Interpolation is a method of fitting an equation to a set of points that functions well when there are few points given.
Polynomials (Ch.1) Study Guide by BS, JL, AZ, CC, SH, HL Lagrange Interpolation is a method of fitting an equation to a set of points that functions well when there are few points given. Sasha s method
More informationDie 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.
More informationSOLVING POLYNOMIAL EQUATIONS BY RADICALS
SOLVING POLYNOMIAL EQUATIONS BY RADICALS Lee Si Ying 1 and Zhang DeQi 2 1 Raffles Girls School (Secondary), 20 Anderson Road, Singapore 259978 2 Department of Mathematics, National University of Singapore,
More informationPrime 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.
More informationWinter Camp 2011 Polynomials Alexander Remorov. Polynomials. Alexander Remorov alexanderrem@gmail.com
Polynomials Alexander Remorov alexanderrem@gmail.com Warmup 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)).
More informationHOMEWORK 5 SOLUTIONS. n!f n (1) lim. ln x n! + xn x. 1 = G n 1 (x). (2) k + 1 n. (n 1)!
Math 7 Fall 205 HOMEWORK 5 SOLUTIONS Problem. 2008 B2 Let F 0 x = ln x. For n 0 and x > 0, let F n+ x = 0 F ntdt. Evaluate n!f n lim n ln n. By directly computing F n x for small n s, we obtain the following
More informationI. GROUPS: BASIC DEFINITIONS AND EXAMPLES
I GROUPS: BASIC DEFINITIONS AND EXAMPLES Definition 1: An operation on a set G is a function : G G G Definition 2: A group is a set G which is equipped with an operation and a special element e G, called
More informationGREATEST COMMON DIVISOR
DEFINITION: GREATEST COMMON DIVISOR The greatest common divisor (gcd) of a and b, denoted by (a, b), is the largest common divisor of integers a and b. THEOREM: If a and b are nonzero integers, then their
More informationCONTINUED FRACTIONS AND PELL S EQUATION. Contents 1. Continued Fractions 1 2. Solution to Pell s Equation 9 References 12
CONTINUED FRACTIONS AND PELL S EQUATION SEUNG HYUN YANG Abstract. In this REU paper, I will use some important characteristics of continued fractions to give the complete set of solutions to Pell s equation.
More informationReal Roots of Univariate Polynomials with Real Coefficients
Real Roots of Univariate Polynomials with Real Coefficients mostly written by Christina Hewitt March 22, 2012 1 Introduction Polynomial equations are used throughout mathematics. When solving polynomials
More information3.6 The Real Zeros of a Polynomial Function
SECTION 3.6 The Real Zeros of a Polynomial Function 219 3.6 The Real Zeros of a Polynomial Function PREPARING FOR THIS SECTION Before getting started, review the following: Classification of Numbers (Appendix,
More informationSection 3.7. Rolle s Theorem and the Mean Value Theorem. Difference Equations to Differential Equations
Difference Equations to Differential Equations Section.7 Rolle s Theorem and the Mean Value Theorem The two theorems which are at the heart of this section draw connections between the instantaneous rate
More informationRolle s Theorem. q( x) = 1
Lecture 1 :The Mean Value Theorem We know that constant functions have derivative zero. Is it possible for a more complicated function to have derivative zero? In this section we will answer this question
More informationPolynomials. Dr. philippe B. laval Kennesaw State University. April 3, 2005
Polynomials Dr. philippe B. laval Kennesaw State University April 3, 2005 Abstract Handout on polynomials. The following topics are covered: Polynomial Functions End behavior Extrema Polynomial Division
More informationRESULTANT 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
More informationAppendix 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.
More informationTHE 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
More informationPractice with Proofs
Practice with Proofs October 6, 2014 Recall the following Definition 0.1. A function f is increasing if for every x, y in the domain of f, x < y = f(x) < f(y) 1. Prove that h(x) = x 3 is increasing, using
More informationPYTHAGOREAN TRIPLES KEITH CONRAD
PYTHAGOREAN TRIPLES KEITH CONRAD 1. Introduction A Pythagorean triple is a triple of positive integers (a, b, c) where a + b = c. Examples include (3, 4, 5), (5, 1, 13), and (8, 15, 17). Below is an ancient
More informationThe Method of Partial Fractions Math 121 Calculus II Spring 2015
Rational functions. as The Method of Partial Fractions Math 11 Calculus II Spring 015 Recall that a rational function is a quotient of two polynomials such f(x) g(x) = 3x5 + x 3 + 16x x 60. The method
More informationIntroduction 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
More informationZeros of a Polynomial Function
Zeros of a Polynomial Function An important consequence of the Factor Theorem is that finding the zeros of a polynomial is really the same thing as factoring it into linear factors. In this section we
More informationMA107 Precalculus Algebra Exam 2 Review Solutions
MA107 Precalculus Algebra Exam 2 Review Solutions February 24, 2008 1. The following demand equation models the number of units sold, x, of a product as a function of price, p. x = 4p + 200 a. Please write
More informationProofs are short works of prose and need to be written in complete sentences, with mathematical symbols used where appropriate.
Advice for homework: Proofs are short works of prose and need to be written in complete sentences, with mathematical symbols used where appropriate. Even if a problem is a simple exercise that doesn t
More informationLecture 6: Finite Fields (PART 3) PART 3: Polynomial Arithmetic. Theoretical Underpinnings of Modern Cryptography
Lecture 6: Finite Fields (PART 3) PART 3: Polynomial Arithmetic Theoretical Underpinnings of Modern Cryptography Lecture Notes on Computer and Network Security by Avi Kak (kak@purdue.edu) January 29, 2015
More informationSince [L : K(α)] < [L : K] we know from the inductive assumption that [L : K(α)] s < [L : K(α)]. It follows now from Lemma 6.
Theorem 7.1. Let L K be a finite extension. Then a)[l : K] [L : K] s b) the extension L K is separable iff [L : K] = [L : K] s. Proof. Let M be a normal closure of L : K. Consider first the case when L
More informationGalois Theory III. 3.1. Splitting fields.
Galois Theory III. 3.1. Splitting fields. We know how to construct a field extension L of a given field K where a given irreducible polynomial P (X) K[X] has a root. We need a field extension of K where
More informationk, then n = p2α 1 1 pα k
Powers of Integers An integer n is a perfect square if n = m for some integer m. Taking into account the prime factorization, if m = p α 1 1 pα k k, then n = pα 1 1 p α k k. That is, n is a perfect square
More informationIrreducibility criteria for compositions and multiplicative convolutions of polynomials with integer coefficients
DOI: 10.2478/auom20140007 An. Şt. Univ. Ovidius Constanţa Vol. 221),2014, 73 84 Irreducibility criteria for compositions and multiplicative convolutions of polynomials with integer coefficients Anca
More information6 EXTENDING ALGEBRA. 6.0 Introduction. 6.1 The cubic equation. Objectives
6 EXTENDING ALGEBRA Chapter 6 Extending Algebra Objectives After studying this chapter you should understand techniques whereby equations of cubic degree and higher can be solved; be able to factorise
More information(x + a) n = x n + a Z n [x]. Proof. If n is prime then the map
22. A quick primality test Prime numbers are one of the most basic objects in mathematics and one of the most basic questions is to decide which numbers are prime (a clearly related problem is to find
More informationMA4001 Engineering Mathematics 1 Lecture 10 Limits and Continuity
MA4001 Engineering Mathematics 1 Lecture 10 Limits and Dr. Sarah Mitchell Autumn 2014 Infinite limits If f(x) grows arbitrarily large as x a we say that f(x) has an infinite limit. Example: f(x) = 1 x
More informationThe 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
More informationSome 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
More informationReview for Calculus Rational Functions, Logarithms & Exponentials
Definition and Domain of Rational Functions A rational function is defined as the quotient of two polynomial functions. F(x) = P(x) / Q(x) The domain of F is the set of all real numbers except those for
More informationZeros of Polynomial Functions
Zeros of Polynomial Functions The Rational Zero Theorem If f (x) = a n x n + a n1 x n1 + + a 1 x + a 0 has integer coefficients and p/q (where p/q is reduced) is a rational zero, then p is a factor of
More informationPrimality  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.
More informationSolutions to SelfTest for Chapter 4 c4sts  p1
Solutions to SelfTest for Chapter 4 c4sts  p1 1. Graph a polynomial function. Label all intercepts and describe the end behavior. a. P(x) = x 4 2x 3 15x 2. (1) Domain = R, of course (since this is a
More information1 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
More information1. PROOFS THAT THERE ARE INFINITELY MANY PRIMES
1. PROOFS THAT THERE ARE INFINITELY MANY PRIMES Introduction The fundamental theorem of arithmetic states that every positive integer may be factored into a product of primes in a unique way. Moreover
More informationIntroduction Proof by unique factorization in Z Proof with Gaussian integers Proof by geometry Applications. Pythagorean Triples
Pythagorean Triples Keith Conrad University of Connecticut August 4, 008 Introduction We seek positive integers a, b, and c such that a + b = c. Plimpton 3 Babylonian table of Pythagorean triples (1800
More informationConsequently, for the remainder of this discussion we will assume that a is a quadratic residue mod p.
Computing square roots mod p We now have very effective ways to determine whether the quadratic congruence x a (mod p), p an odd prime, is solvable. What we need to complete this discussion is an effective
More information4.3 Lagrange Approximation
206 CHAP. 4 INTERPOLATION AND POLYNOMIAL APPROXIMATION Lagrange Polynomial Approximation 4.3 Lagrange Approximation Interpolation means to estimate a missing function value by taking a weighted average
More information5.1 Commutative rings; Integral Domains
5.1 J.A.Beachy 1 5.1 Commutative rings; Integral Domains from A Study Guide for Beginner s by J.A.Beachy, a supplement to Abstract Algebra by Beachy / Blair 23. Let R be a commutative ring. Prove the following
More informationLecture 13  Basic Number Theory.
Lecture 13  Basic Number Theory. Boaz Barak March 22, 2010 Divisibility and primes Unless mentioned otherwise throughout this lecture all numbers are nonnegative integers. We say that A divides B, denoted
More informationGalois Fields and Hardware Design
Galois Fields and Hardware Design Construction of Galois Fields, Basic Properties, Uniqueness, Containment, Closure, Polynomial Functions over Galois Fields Priyank Kalla Associate Professor Electrical
More informationLimits and Continuity
Math 20C Multivariable Calculus Lecture Limits and Continuity Slide Review of Limit. Side limits and squeeze theorem. Continuous functions of 2,3 variables. Review: Limits Slide 2 Definition Given a function
More information