7. Some irreducible polynomials

Size: px
Start display at page:

Transcription

1 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 the equation P (x) = 0. This follows from unique factorization in the ring k[x]. Here we also look at some special higher-degree polynomials, over finite fields, where we useful structural interpretation of the polynomials. [2] Here we take for granted the existence of an algebraic closure k of a given field, as a fixed universe in which to consider roots of polynomial equations. 1. Irreducibles over a finite field [1.0.1] Proposition: Let (non-constant) M(x) be an irreducible in k[x], with field k. Let I be the ideal generated in k[x] by M(x). Let α be the image of x in the field K = k[x]/i. Then α is a root of the equation M(x) = 0. [3] Proof: The salient aspects of the ring structure in the quotient can be summarized by the point that the quotient map k[x] k[x]/i is a ring homomorphism, in fact, a k-algebra homomorphism. Thus, for any polynomial f, f(x) + I = f(x + I) In particular, M(x + I) = M(x) + I = I = 0 + I which shows that x + I is a root of the equations. /// [1] And this unique factorization follows from the Euclidean-ness of the polynomial ring. [2] All these are cyclotomic polynomials, that is, divisors of x n 1 for some n. A systematic investigation of these polynomials is best done with a little more preparation. But they do provide accessible examples immediately. [3] This is immediate, when one looks at the proof, but deserves complete explicitness. 107

2 108 Some irreducible polynomials [1.0.2] Proposition: [4] Let P (x) be a polynomial in k[x] for a field k. The equation P (x) = 0 has a root α generating [5] a degree d extension K of k if and only if P (x) has a degree d irreducible factor f(x) in k[x]. Proof: Let α be a root of P (x) = 0 generating a degree d extension [6] the minimal polynomial for α over k. Let k(α) = k[α] over k. Let M(x) be P = Q M + R in k[x] with deg R < deg M. Then, evaluating these polynomials at α, R(α) = 0, but the minimality of the degree of M with this property assures that R = 0. That is, M divides P. On the other hand, for an irreducible (monic, without loss of generality) M(x) dividing P (x), the quotient K = k[x]/ M(x) is a field containing (a canonical copy of) k, and the image α of x in that extension is a root of M(x) = 0. Letting P = Q M, P (α) = Q(α) M(α) = Q(α) 0 = 0 showing that P (x) = 0 has root α. /// The first two examples use only the correspondence between linear factors and roots in the ground field. [1.0.3] Example: x is irreducible over k = Z/p for any prime p = 3 mod 4. Indeed, if x had a linear factor then the equation x = 0 would have a root α in k. This alleged root would have the property that α 2 = 1. Thus, α 1, α 1, but α 4 = 1. That is, the order of α in k is 4. But the order of (Z/p) is p 1. The hypothesis p = 3 mod 4 was exactly designed to deny the existence of an element of order 4 in (Z/p). Thus, x is irreducible in such k[x]. [1.0.4] Example: x 2 + x + 1 is irreducible over k = Z/p for any prime p = 2 mod 3. If x 2 + x + 1 had a linear factor then x 2 + x + 1 = 0 would have a root α in k, and, since x 3 1 = (x 1)(x 2 + x + 1) α 3 = 1 but α 1 since in k. That is, the order of α in k is 3. But the order of (Z/p) is p 1, and the hypothesis p = 2 mod 3 exactly precludes any element of order 3 in (Z/p). Thus, x 2 + x + 1 is irreducible in such k[x]. [1.0.5] Example: P (x) = x 4 + x 3 + x 2 + x + 1 is irreducible over k = Z/p for prime p ±1 mod 5 and p 5. Note that x 5 1 = (x 1)(x 4 + x 3 + x 2 + x + 1) Thus, any root of P (x) = 0 has order [7] 5 or 1 (in whatever field it lies). The only element of order 1 is the identity element 1. If P (x) had a linear factor in k[x], then P (x) = 0 would have a root in k. Since [4] This assertion should not be surprising, when one looks at the technique of the proof, which is nearly identical to the proof that linear factors correspond to roots in the base field. [5] As earlier, the field extension k(α) generated by α makes sense only inside a fixed larger field. Throughout the present discussion we fix an algebraic closure of any ground field k and consider extensions inside that algebraic closure. [6] Since the degree of the extension is finite, it is equal to polynomials in α over k, as we saw earlier. [7] By Lagrange.

4 110 Some irreducible polynomials Again, any root of P (x) = 0 has order 11 or 1 (in whatever field it lies). The only element of order 1 is the identity element 1. If P (x) had a linear factor in k[x], then P (x) = 0 would have a root in k. Since 11 0 in k, 1 is not a root, so any possible root must have order 11. But the order of k = (Z/p) is p 1, which is not divisible by 11, so there is no root in the base field k. If P (x) had an irreducible degree d factor q(x) in k[x], then P (x) = 0 would have a root in a degree d extension K of k. The field K has p d elements, so K = p d 1 By Lagrange, the order of any element of K is a divisor of p d 1, but 11 divides none of p 1, p 2 1, p 3 1, p 4 1,..., p 9 1, by design. 2. Worked examples [7.1] (Lagrange interpolation) Let α 1,..., α n be distinct elements in a field k, and let β 1,..., β n be any elements of k. Prove that there is a unique polynomial P (x) of degree < n in k[x] such that, for all indices i, P (α i ) = β i Indeed, letting show that Since the α i are distinct, P (x) = Q(x) = n i=1 n (x α i ) i=1 Q(x) (x α i ) Q (α i ) β i Q (α i ) = j i(α i α j ) 0 (One could say more about purely algebraic notions of derivative, but maybe not just now.) Evaluating P (x) at x α i, { Q(x) 1 (for j = i) (x α j ) evaluated at x α i = 0 (for j = i) Thus, all terms but the i th vanish in the sum, and the i th one, by design, gives β i. For uniqueness, suppose R(x) were another polynomial of degree < n taking the same values at n distinct points α i as does Q(x). Then Q R is of degree < n and vanishes at n points. A non-zero degree l polynomial has at most l zeros, so it must be that Q R is the 0 polynomial. [7.2] (Simple case of partial fractions) Let α 1,..., α n be distinct elements in a field k. Let R(x) be any polynomial in k[x] of degree < n. Show that there exist unique constants c i k such that in the field of rational functions k(x) R(x) (x α 1 )... (x α n ) = c c n x α 1 x α n In particular, let and show that Q(x) = n (x α i ) i=1 c i = R(α i) Q (α i )

5 Garrett: Abstract Algebra 111 We might emphasize that the field of rational functions k(x) is most precisely the field of fractions of the polynomial ring k[x]. Thus, in particular, equality r/s = r /s is exactly equivalent to the equality rs = r s (as in elementary school). Thus, to test whether or not the indicated expression performs as claimed, we test whether or not R(x) = ( ) R(αi ) Q (α i i ) Q(x) x α i One might notice that this is the previous problem, in case β i = R(α i ), so its correctness is just a special case of that, as is the uniqueness (since deg R < n). [7.3] Show that the ideal I generated in Z[x] by x and 5 is not maximal. We will show that the quotient is not a field, which implies (by the standard result proven above) that the ideal is not maximal (proper). First, let us make absolutely clear that the quotient of a ring R by an ideal I = Rx + Ry generated by two elements can be expressed as a two-step quotient, namely (R/ x )/ ȳ R/(Rx + Ry) where the ȳ is the principal ideal generated by the image ȳ of y in the quotient R/ x. The principal ideal generated by y in the quotient R/ x is the set of cosets ȳ = {(r + Rx) (y + Rx) : r R} = {ry + Rx : r R} noting that the multiplication of cosets in the quotient ring is not just the element-wise multiplication of the cosets. With this explication, the natural map is r + x = r + x r + x + y = r + (Rx + Rx) which is visibly the same as taking the quotient in a single step. Thus, first Z[x]/ 5 (Z/5)[x] by the map which reduces the coefficients of a polynomial modulo 5. In (Z/5)[x], the polynomial x does factor, as x = (x 2)(x + 2) (where these 2s are in Z/5, not in Z). Thus, the quotient (Z/5)[x]/ x has proper zero divisors x 2 and x + 2, where x is the image of x in the quotient. Thus, it s not even an integral domain, much less a field. [7.4] Show that the ideal I generated in Z[x] by x 2 + x + 1 and 7 is not maximal. As in the previous problem, we compute the quotient in two steps. First, Z[x]/ 7 (Z/7)[x] by the map which reduces the coefficients of a polynomial modulo 7. In (Z/7)[x], the polynomial x 2 + x + 1 does factor, as x 2 + x + 1 = (x 2)(x 4) (where 2 and 4 are in Z/7). Thus, the quotient (Z/7)[x]/ x 2 + x + 1 has proper zero divisors x 2 and x 4, where x is the image of x in the quotient. Thus, it s not even an integral domain, so certainly not a field.

6 112 Some irreducible polynomials Exercises 7.[2.0.1] Show that x 2 + x + 1 is irreducible in F 5 [x], and in F 29 [x]. 7.[2.0.2] Show that x 3 a is irreducible in F 7 [x] unless a = 0 or ±1. 7.[2.0.3] Determine how x factors into irreducibles in F 2 [x]. 7.[2.0.4] Exhibit an irreducible quintic in F 11 [x]. 7.[2.0.5] Show that the ideal generated by x 2 x + 1 and 13 in Z[x] is not maximal. 7.[2.0.6] Show that the ideal generated by x 2 x + 1 and 17 in Z[x] is maximal. 7.[2.0.7] Let α 1,..., α n be distinct elements of a field k not of characteristic 2. Show that there are elements a 1, b 1,..., a n, b n in k such that 1 (x α 1 ) 2... (x α n ) 2 = a 1 x α 1 + b 1 (x α 1 ) a n x α n + b n (x α n ) 2

minimal 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

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

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)

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.

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

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.

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

calculating 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

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

How To Prove 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

a 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

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

(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

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.

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

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

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

FACTORING AFTER DEDEKIND

FACTORING AFTER DEDEKIND KEITH CONRAD Let K be a number field and p be a prime number. When we factor (p) = po K into prime ideals, say (p) = p e 1 1 peg g, we refer to the data of the e i s, the exponents

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

SOLVING 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

Cyclotomic Extensions

Chapter 7 Cyclotomic Extensions A cyclotomic extension Q(ζ n ) of the rationals is formed by adjoining a primitive n th root of unity ζ n. In this chapter, we will find an integral basis and calculate

15. Symmetric polynomials

15. Symmetric polynomials 15.1 The theorem 15.2 First examples 15.3 A variant: discriminants 1. The theorem Let S n be the group of permutations of {1,, n}, also called the symmetric group on n things.

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).

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

3 Factorisation into irreducibles

3 Factorisation into irreducibles Consider the factorisation of a non-zero, non-invertible integer n as a product of primes: n = p 1 p t. If you insist that primes should be positive then, since n could

Math 345-60 Abstract Algebra I Questions for Section 23: Factoring Polynomials over a Field

Math 345-60 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

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)).

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

JUST 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

Galois 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

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 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,

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

Basics 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

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

SOLVING POLYNOMIAL EQUATIONS BY RADICALS Lee Si Ying 1 and Zhang De-Qi 2 1 Raffles Girls School (Secondary), 20 Anderson Road, Singapore 259978 2 Department of Mathematics, National University of Singapore,

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

ON GALOIS REALIZATIONS OF THE 2-COVERABLE SYMMETRIC AND ALTERNATING GROUPS

ON GALOIS REALIZATIONS OF THE 2-COVERABLE 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

EXERCISES FOR THE COURSE MATH 570, FALL 2010

EXERCISES FOR THE COURSE MATH 570, FALL 2010 EYAL Z. GOREN (1) Let G be a group and H Z(G) a subgroup such that G/H is cyclic. Prove that G is abelian. Conclude that every group of order p 2 (p a prime

Lagrange 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

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.

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

6 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

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 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.

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

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.

Generic Polynomials of Degree Three

Generic Polynomials of Degree Three Benjamin C. Wallace April 2012 1 Introduction In the nineteenth century, the mathematician Évariste Galois discovered an elegant solution to the fundamental problem

20. Cyclotomic III. 1. Prime-power cyclotomic polynomials over Q. The proof of the following is just a slight generalization of the prime-order case.

0. Cyclotomic III 0.1 Prime-power cyclotomic polynomials over Q 0. Irreducibility of cyclotomic polynomials over Q 0.3 Factoring Φ n (x) in F p [x] with p n 0.4 Worked examples The main goal is to prove

Math 4310 Handout - Quotient Vector Spaces

Math 4310 Handout - Quotient Vector Spaces Dan Collins The textbook defines a subspace of a vector space in Chapter 4, but it avoids ever discussing the notion of a quotient space. This is understandable

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,...}

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 = (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 real-closed fields These notes develop the algebraic background needed to understand the model theory of real-closed fields. To understand these notes, a standard graduate course in algebra is

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

Application. Outline. 3-1 Polynomial Functions 3-2 Finding Rational Zeros of. Polynomial. 3-3 Approximating Real Zeros of.

Polynomial and Rational Functions Outline 3-1 Polynomial Functions 3-2 Finding Rational Zeros of Polynomials 3-3 Approximating Real Zeros of Polynomials 3-4 Rational Functions Chapter 3 Group Activity:

4.3 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

The Factor Theorem and a corollary of the Fundamental Theorem of Algebra

Math 421 Fall 2010 The Factor Theorem and a corollary of the Fundamental Theorem of Algebra 27 August 2010 Copyright 2006 2010 by Murray Eisenberg. All rights reserved. Prerequisites Mathematica Aside

3.6. The factor theorem

3.6. The factor theorem Example 1. At the right we have drawn the graph of the polynomial y = x 4 9x 3 + 8x 36x + 16. Your problem is to write the polynomial in factored form. Does the geometry of the

ALGEBRA HW 5 CLAY SHONKWILER

ALGEBRA HW 5 CLAY SHONKWILER 510.5 Let F = Q(i). Prove that x 3 and x 3 3 are irreducible over F. Proof. If x 3 is reducible over F then, since it is a polynomial of degree 3, it must reduce into a product

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

Revised Version of Chapter 23. We learned long ago how to solve linear congruences. ax c (mod m)

Chapter 23 Squares Modulo p Revised Version of Chapter 23 We learned long ago how to solve linear congruences ax c (mod m) (see Chapter 8). It s now time to take the plunge and move on to quadratic equations.

MA3D5 Galois theory. Miles Reid. Jan Mar 2004 printed Jan 2014

MA3D5 Galois theory Miles Reid Jan Mar 2004 printed Jan 2014 Contents 1 The theory of equations 3 1.1 Primitive question........................ 3 1.2 Quadratic equations....................... 3 1.3 The

SOLUTIONS TO PROBLEM SET 3

SOLUTIONS TO PROBLEM SET 3 MATTI ÅSTRAND The General Cubic Extension Denote L = k(α 1, α 2, α 3 ), F = k(a 1, a 2, a 3 ) and K = F (α 1 ). The polynomial f(x) = x 3 a 1 x 2 + a 2 x a 3 = (x α 1 )(x α 2

Algebra 3: algorithms in algebra

Algebra 3: algorithms in algebra Hans Sterk 2003-2004 ii Contents 1 Polynomials, Gröbner bases and Buchberger s algorithm 1 1.1 Introduction............................ 1 1.2 Polynomial rings and systems

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

3.2 The Factor Theorem and The Remainder Theorem

3. The Factor Theorem and The Remainder Theorem 57 3. The Factor Theorem and The Remainder Theorem Suppose we wish to find the zeros of f(x) = x 3 + 4x 5x 4. Setting f(x) = 0 results in the polynomial

10 Splitting Fields. 2. The splitting field for x 3 2 over Q is Q( 3 2,ω), where ω is a primitive third root of 1 in C. Thus, since ω = 1+ 3

10 Splitting Fields We have seen how to construct a field K F such that K contains a root α of a given (irreducible) polynomial p(x) F [x], namely K = F [x]/(p(x)). We can extendthe procedure to build

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,

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

Integer 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

Zeros of Polynomial Functions

Review: Synthetic Division Find (x 2-5x - 5x 3 + x 4 ) (5 + x). Factor Theorem Solve 2x 3-5x 2 + x + 2 =0 given that 2 is a zero of f(x) = 2x 3-5x 2 + x + 2. Zeros of Polynomial Functions Introduction

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

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

Factoring 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

The finite field with 2 elements The simplest finite field is

The finite field with 2 elements The simplest finite field is GF (2) = F 2 = {0, 1} = Z/2 It has addition and multiplication + and defined to be 0 + 0 = 0 0 + 1 = 1 1 + 0 = 1 1 + 1 = 0 0 0 = 0 0 1 = 0

ALGEBRA 2 CRA 2 REVIEW - Chapters 1-6 Answer Section

ALGEBRA 2 CRA 2 REVIEW - Chapters 1-6 Answer Section MULTIPLE CHOICE 1. ANS: C 2. ANS: A 3. ANS: A OBJ: 5-3.1 Using Vertex Form SHORT ANSWER 4. ANS: (x + 6)(x 2 6x + 36) OBJ: 6-4.2 Solving Equations by

The van Hoeij Algorithm for Factoring Polynomials

The van Hoeij Algorithm for Factoring Polynomials Jürgen Klüners Abstract In this survey we report about a new algorithm for factoring polynomials due to Mark van Hoeij. The main idea is that the combinatorial

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,

3.6 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,

Short Programs for functions on Curves

Short Programs for functions on Curves Victor S. Miller Exploratory Computer Science IBM, Thomas J. Watson Research Center Yorktown Heights, NY 10598 May 6, 1986 Abstract The problem of deducing a function

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

Partial Fractions. Combining fractions over a common denominator is a familiar operation from algebra:

Partial Fractions Combining fractions over a common denominator is a familiar operation from algebra: From the standpoint of integration, the left side of Equation 1 would be much easier to work with than

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 ) =

Zeros 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

SECTION 0.6: POLYNOMIAL, RATIONAL, AND ALGEBRAIC EXPRESSIONS

(Section 0.6: Polynomial, Rational, and Algebraic Expressions) 0.6.1 SECTION 0.6: POLYNOMIAL, RATIONAL, AND ALGEBRAIC EXPRESSIONS LEARNING OBJECTIVES Be able to identify polynomial, rational, and algebraic

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

Polynomial and Synthetic Division. Long Division of Polynomials. Example 1. 6x 2 7x 2 x 2) 19x 2 16x 4 6x3 12x 2 7x 2 16x 7x 2 14x. 2x 4.

_.qd /7/5 9: AM Page 5 Section.. Polynomial and Synthetic Division 5 Polynomial and Synthetic Division What you should learn Use long division to divide polynomials by other polynomials. Use synthetic

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

Some Polynomial Theorems. John Kennedy Mathematics Department Santa Monica College 1900 Pico Blvd. Santa Monica, CA 90405 rkennedy@ix.netcom.

Some Polynomial Theorems by John Kennedy Mathematics Department Santa Monica College 1900 Pico Blvd. Santa Monica, CA 90405 rkennedy@ix.netcom.com This paper contains a collection of 31 theorems, lemmas,

Lecture 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

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

Row Ideals and Fibers of Morphisms

Michigan Math. J. 57 (2008) Row Ideals and Fibers of Morphisms David Eisenbud & Bernd Ulrich Affectionately dedicated to Mel Hochster, who has been an inspiration to us for many years, on the occasion

FOUNDATIONS OF ALGEBRAIC GEOMETRY CLASS 22

FOUNDATIONS OF ALGEBRAIC GEOMETRY CLASS 22 RAVI VAKIL CONTENTS 1. Discrete valuation rings: Dimension 1 Noetherian regular local rings 1 Last day, we discussed the Zariski tangent space, and saw that it

FACTORING SPARSE POLYNOMIALS

FACTORING SPARSE POLYNOMIALS Theorem 1 (Schinzel): Let r be a positive integer, and fix non-zero integers a 0,..., a r. Let F (x 1,..., x r ) = a r x r + + a 1 x 1 + a 0. Then there exist finite sets S

3.3. Solving Polynomial Equations. Introduction. Prerequisites. Learning Outcomes

Solving Polynomial Equations 3.3 Introduction Linear and quadratic equations, dealt within Sections 3.1 and 3.2, are members of a class of equations, called polynomial equations. These have the general

The 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

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.

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