# Quasi-reducible Polynomials

Size: px
Start display at page:

Transcription

1 Quasi-reducible Polynomials Jacques Willekens 06-Dec-2008 Abstract In this article, we investigate polynomials that are irreducible over Q, but are reducible modulo any prime number. 1 Introduction Let f(x) be a polynomial with integer coefficients. As Gauss has shown, if f can be factored over Q, it can also be factored over Z; in other words, the factors can be expressed using integer coefficients. For a given prime p, let f p (x) be the polynomial over GF(p) (the finite field of p elements) obtained by reducing the coefficients modulo p. If G is the Galois group of f over Q, we will write G p for the Galois group of f p over GF(p). As this defines a homomorphism from Z[x] to GF(p)[x], any factorization of f yields a corresponding factorization of f p, obtained by reducing each factor separately. This shows that, if f is reducible (over Z), then f p is also reducible (over GF(p)) for any prime p. In many cases, this allows to prove that f is irreducible: if there is any prime p such that f p is irreducible, then f is irreducible. For example, this can be used to prove that a quadratic equation with odd integer coefficients has no rational root. Indeed, in this case, we have: f 2 (x) = x 2 + x + 1 and this polynomial is irreducible modulo 2; therefore, the original polynomial is irreducible, and, in the case of a second degree polynomial, this means that it has no rational root. However, the converse of that proposition is not true: there are polynomials that are irreducible over Q, but become reducible modulo any prime number. For the lack of a better word, we call these polynomials quasi-reducible. As we will show, there are infinitely many such polynomials; in fact, they are more frequent than one may expect. 2 General approach We first recall some well-known results of Galois theory: If G is the Galois group of f, G p is the Galois group of f p, then G p is a subgroup of G (as an abstract group). The idea behind this is that f p may be reducible, and any permutation of the roots must act separately on the roots of the irreducible factors of f p. 1

2 Furthermore, if f p has no repeated factors, there is a direct correspondence between the degrees of the irreducible factors of f p and the cycle structure of an element of G, as a permutation group on the roots of f. Over any finite field GF(q), the splitting field of a polynomial is GF(q k ) for some k, and the corresponding Galois group is cyclic and generated by the automorphims x x q. The degree of the splitting field is the LCM of the degrees of the irreducible factors of the polynomial. We will consider normal polynomials, i.e., minimal polynomials of a primitive element of a finite extension of Q. Such a polynomial splits completely in the extension generated by a single root; therefore, the order of the Galois group is equal to the degree of the polynomial. We can find a monic integral normal polynomial for any finite extension, by choosing an algebraic integer as primitive element; we will only use such polynomials, to ensure that reduction modulo p preserves the degree of the polynomial. Let f(x) be a normal (monic) polynomial of degree d. If the Galois group G is not cyclic, G cannot be the Galois group of a polynomial over GF(p). Therefore, G p is a proper subgroup of G, and has order less than d, which shows that f p is reducible. As this holds for any prime p, f is quasi-reducible. If e is the exponent of the Galois group G of f(x), and e p the exponent of G p, then e p e, as G p is a subgroup of G. The degree of any irreducible factor of f p divides e p ; this means that, if we know e, this gives an upper bound on the degree of the irreducible factors of f p. 3 Cyclotomic polynomials Let us denote by Φ n (x) the cyclotomic polynomial of order n, i.e., the monic polynomial whose roots are the primitive n th roots of unity. It is well known that Φ n (x) has rational integer coefficients and is irreducible over Q (see [1, 3.7]). Φ n (x) has degree φ(n) (Euler s totient function), and its Galois group is U(n), the multiplicative group of units of Z n. U(n) is the direct product of the groups U(p k ), where the p k are the prime powers occurring in the factorization of n. If p k is such an odd factor, U(p k ) has even order, and therefore contains a subgroup isomorphic to C 2. Therefore, if n is divisible by two distinct odd primes, U(n) contains a subgroup isomorphic to C 2 C 2, and is not cyclic, which shows that Φ n (x) is quasi-reducible. Furthermore, as U(4) = C 2 and U(8) = C 2 C 2, we can get the more general result: Proposition 3.1. If n is divisible by two distinct odd primes, or n is a proper multiple of 4, then Φ n (x) is quasi-reducible. In this case, we can find explicitly the degrees of the irreducible factors of Φ n modulo p for a given prime p. Assume first that p does not divide n. As the Galois group of an irreducible factor is generated by the automorphism α α p, and α n = 1, we see that the order of the group is the multiplicative order of p modulo n. Since x n 1 has no repeated roots modulo p in this case, all the irreducible factors of Φ n are distinct, and have the same degree. If, on the other hand, p n, x n 1 has repeated roots modulo p, since its derivative is 0. 2

3 In this case, we make use of the identities: { Φ m (x p Φ mp (x)φ m (x) if p m ) = Φ mp (x) if p m Since, in GF(p), we have f(x p ) = f(x) p for any polynomial f, we get: { Φ m (x) p 1 if p m Φ mp (x) = Φ m (x) p if p m and we can repeat the process if p m. If n = p k r, where p r, we obtain finally: Φ n (x) = Φ r (x) φ(pk ) where Φ r (x) may be further reducible. Consider, for example, the polynomial: f(x) = Φ 15 (x) = (x15 1)(x 1) (x 3 1)(x 5 1) = x8 x 7 + x 5 x 4 + x 3 x + 1 As U(15) = U(3) U(5) = C 2 C 4, the exponent of G is 4, and any irreducible factor of f p has degree at most 4. If p 15, the degree of these factors is the multiplicative order of p modulo 15. For example, we have: f(x) = (x 4 + 2x 3 + 4x 2 + x + 2)(x 4 + 4x 3 + 2x 2 + x + 4) (mod 7) = (x 2 + 3x + 9)(x 2 + 4x + 5)(x 2 + 5x + 3)(x 2 + 9x + 4) (mod 11) and, modulo 31, f(x) splits into 8 linear factors. For p = 3 and p = 5, we have: f(x) = (x 4 + x 3 + x 2 + x + 1) 2 = Φ 5 (x) 2 (mod 3) = (x 2 + x + 1) 4 = Φ 3 (x) 4 (mod 5) 4 Bi-quadratic equations We investigate now the polynomials of the form: f(x) = x 4 + ux 2 + v where v is a square. These polynomials have Galois group C 2 C 2, their roots can be expressed as (± a ± b), where a and b need not be positive. By using different groupings of the factors, we obtain the following factorizations over C: f(x) = (x 2 a 2 b ab)(x 2 a 2 b 2 2 ab) = (x a + a b)(x 2 2 a + a b) = (x b a + b)(x 2 2 b a + b) 3

4 These factorizations can be used over GF(p), provided that the required square roots exist, i.e., if the expressions under the radicals are quadratic residues. Now, if (k/p) denotes the Legendre symbol, we have: (a/p)(b/p)(ab/p) = (a 2 b 2 /p) = +1 or 0 which shows that at least one of a, b, or ab is a quadratic residue, and we can use the corresponding factorization. Furthermore, if both a and b are quadratic residues, the roots are in GF(p) and f splits into linear factors. Consider, for example, the polynomial: f(x) = x 4 10x whose roots are (± 2 ± 3). For p = 2 and p = 3, we have the factorizations: f(x) = (x + 1) 4 (mod 2) = (x 2 + 1) 2 (mod 3) For the other primes, we use the theory of quadratic reciprocity to find the values of the Legendre symbols (2/p, (3/p) and (6/p) depending on the congruence class of p modulo 24. The results are shown in table 4.1. Using these values, we can exhibit an example of each kind of factorization: p mod (2/p) (3/p) (6/p) Table 4.1: Legendre symbols (2/p), (3/p) and (6/p) f(x) = (x 2 + 2)(x 2 + 3) (mod 5) = (x 2 + x + 6)(x 2 + 6x + 6) (mod 7) = (x 2 + x + 1)(x x + 1) (mod 11) = (x + 2)(x + 11)(x + 12)(x + 21) (mod 23) We may also mention that the cyclotomic polynomial Φ 8 (x) = x can also be analyzed using this technique, with a = 1/2 and b = 1/2. 5 Non-abelian groups A random polynomial has usually a non-abelian Galois group. Therefore, in this case, any normal polynomial generating the splitting field will be quasi-reducible. For example, let us consider the polynomial: g(x) = x 3 + 2x 2 x + 3 (5.1) 4

5 The discriminant of this polynomial is 439, and the Galois group is therefore S 3. To find a normal polynomial, we compute the minimal polynomial of ζ = (α β), where α and β are roots of g. This gives the polynomial: f(x) = x 6 14x x (5.2) with discriminant equal to As f has degree 6, it is a normal polynomial, and is therefore quasi-reducible, since S 3 is not abelian (and therefore not cyclic). We note that, in this case, we have e = G = 6. Factorization modulo some small primes produces various patterns; for example: f(x) = (x 3 + x + 1) 2 (mod 2) = (x 2 + 1)(x 2 + x + 2)(x 2 + 2x + 2) (mod 3) = (x + 1)(x + 2) 2 (x + 3) 2 (x + 4) (mod 5) = (x 3 + 4x + 1)(x 3 + 4x + 10) (mod 11) = (x + 1)(x + 2)(x + 3)(x + 16)(x + 17)(x + 18) (mod 19) = (x 2 + 6)(x ) 2 (mod 23) = x 2 (x + 165) 2 (x + 274) 2 (mod 439) We can make a few preliminary observations: The degrees of the factors are divisors of 6, as expected, since G p is a subgroup of S 3. We never have a factor of degree 2 and a factor of degree 3 in the same factorization, since the resulting Galois group would be C 6, which is not isomorphic to a subgroup of S 3. There are repeated factors for the primes that divide the discriminant of f; this corresponds to the definition of the discriminant. Let us now analyze in more detail the structure of the possible factorizations. In what follows, we will use the following notations and conventions: f(x) = x 3 + ux 2 + vx + w is a separable polynomial (possible reducible) over a field K, not of characteristic 2. The roots of f are α, β, and γ; the discriminant is. The splitting field of f is L. By a common abuse of language, we call an element of L rational if it belongs to K. Proposition 5.1. With the above notations, L = K(α, ), where α is any root (possibly rational) of f. Proof. It is obvious that L contains α and = (α β)(β γ)(α γ), and therefore K(α, ) L. To show the reverse inclusion, we must prove that β K(α, ), since L = K(α, β). We have: = (α β)(β γ)(α γ) = (β γ)(α 2 α(β + γ) + βγ) = (β γ)θ θ = (α 2 α(β + γ) + βγ) 5

6 We note that θ K(α), since β+γ = u α and βγ = v α(β+γ) = v+α(u+α). If θ = 0, either α = β or α = γ and β = u 2α. In either case, we have β K(α) K(α, ). If θ 0, we have: β γ = θ K(α, ) and, since β + γ K(α), and chark 2, we conclude that β K(α, ). We now return to the investigation of the degrees of the irreducible factors of f p over K = GF(p) (for odd p) and their relation with the factors of g p. As remarked before, the splitting field L cannot have degree 6, since the Galois group would have to be C 6, which is not a subgroup of S 3. If is not a quadratic residue modulo p, K( ) is an extension of degree 2. By the remark above, we have L = K( ), which means that there is at least one factor of degree 2 and no factors of degree 3. If is a quadratic residue, L = K(α). If g p is irreducible, L has degree 3, and the possible degrees of the factors of f p are (3, 3) or (3, 1, 1, 1). If, on the other hand, g p is reducible, it has at least one linear factor. By choosing the rational root as α, we find that L = K, which means that f p splits into 6 linear factors. This also implies that g p splits into three linear factors. In the above examples, we notice that the degrees of the irreducible factors of f p are the same, which suggests the following conjecture: Conjecture 5.2. If f(x) is a quasi-reducible polynomial, the irreducible factors of f p have the same degree. We know that the conjecture is true for cyclotomic and bi-quadratic polynomials with Galois group C 2 C 2. As we will see, the conjecture is false in general, but counter-examples are somewhat hard to find. In particular, the conjecture is true for the specific polynomial given by (5.2); this is related to the choice of ζ = α β as primitive element. In the case of a cubic polynomial over Q, we can choose as primitive element any element of the splitting field that does not belong to Q(α), for any root α. Such an element can be expressed as a polynomial P (α, β), and we can independently choose the roots α and β, as the Galois group is S 3. The 6 roots of f correspond to the 6 possible choices, and all these roots have the same minimal polynomial (which is, of course, f itself). However, over a finite field GF(p), this is no longer true, because the Galois group is generated by the Frobenius automorphism σ : α α p. Once a root α has been chosen, the other roots of the corresponding irreducible factor of g can be identified indivually as α p, α p2.... We may therefore expect that different choices of β could produce different results. Before showing how this idea can be used to produce counter-examples, let us first first show that the conjecture holds for the particular polynomial f(x) given by (5.2), at least for odd primes. We have already seen that f p splits into linear factors if and only if g p does. There are two remaining cases: 1. g p is irreducible. In this case, the splitting field is GF(p 3 ), and f p (x) has an irreducible factor q p (x) of degree 3. We notice that f p is an even polynomial, because, if ζ = α β 6

7 is a root of f, ζ is also a root. As q p is an irreducible cubic, it contains a term x 3 and a constant term, and is therefore neither odd nor even. This means that we must have f p (x) = q p (x)q p ( x), and, if q p (x) is irreducible, so is q p ( x). 2. g p factors into a linear factor and an irreducible quadratic. In this case, the splitting field is GF(p 2 ). If α and β are the roots of the quadratic, α β is irrational (because α + β is rational). If one of α or β is the rational root, α β is also irrational. We may therefore conclude that all the roots of f p are irrational, and the corresponding minimal polynomials are of degree 2, since f p splits in GF(p 2 ). Let us now find counter-examples to conjecture 5.2. We first notice that the proof above fails for p = 2. This is related to the fact that, over GF(2), the polynomial x 2 +x+1 is irreducible, although its discriminant is equal to 1. We start with the polynomial: g(x) = x(x 2 + x + 1) + 2 = x 3 + x 2 + x + 2 and we compute, as before, the minimal polynomial of α β: f(x) = x 6 + 4x 4 + 4x Over GF(2), we have the factorization: f 2 (x) = (x + 1) 2 (x 2 + x + 1) 2 which provides a first counter-example to the conjecture. As the case p = 2 is rather special, we now look for counter-examples modulo other primes. We take p = 11, and g(x) as in (5.1); g 11 is irreducible. If α is a root of g 11 in GF(11 3 ), the other two roots are α 11 and α 121. As g 11 (α) = 0, we obtain, by division: α 11 = 4α 2 + 4α + 5 (5.3) We define the primitive element as: ζ = 4α 2 + 4α + 5 β (5.4) Now, (5.4) can be taken as a definition over Q, and we obtain the minimal polynomial: h(x) = x 6 66x x x x x which is quasi-reducible. When p = 11, if we choose β = α 11, we have ζ = 0, and the minimial polynomial of that particular root of f 11 is x. Indeed, over GF(11), we have the factorization: h 11 (x) = x 3 (x 3 + 4x + 10) which is a counter-example to conjecture 5.2, since the degrees of the irreducible factors are (3, 1, 1, 1). To obtain an example with a quadratic factor, we take p = 3, in which case g 3 splits as: g 3 (x) = x(x 2 + 2x + 2) 7

8 We take the primitive element: ζ = α 3 β = 2α + 1 β and obtain the quasi-reducible polynomial: k(x) = x 6 2x 5 31x x x x which factors over GF(3) as: k 3 (x) = x 2 (x 2 + 2x + 2) 2 where the degrees of the irreducible factors are (2, 2, 1, 1). In each of the counter-examples above, we find that we get a repeated factor. This is due to the fact that the number of possible inequivalent choices for the pair (α, β) gives an upper bound on the number of distinct irreducible factors. For example, if g is irreducible, the choice of α is arbitrary, and there are two possible choices for β, and therefore there can be at most two distinct irreducible factors, corresponding to the minimal polynomials of the corresponding ζ. If g factors as (2, 1), there are three possible choices, depending on which root (α,β, or γ) is taken as the rational root. As the factorization of (5.2) modulo 5 shows, this is only an upper bound. References [1] Paul M. Cohn. Algebra, volume 2. Wiley, second edition, ISBN

### Galois theory (Part II)( ) Example Sheet 1

Galois theory (Part II)(2015 2016) Example Sheet 1 c.birkar@dpmms.cam.ac.uk (1) Find the minimal polynomial of 2 + 3 over Q. (2) Let K L be a finite field extension such that [L : K] is prime. Show that

### Public-key Cryptography: Theory and Practice

Public-key Cryptography Theory and Practice Department of Computer Science and Engineering Indian Institute of Technology Kharagpur Chapter 2: Mathematical Concepts Divisibility Congruence Quadratic Residues

### Galois Theory TCU Graduate Student Seminar George Gilbert October 2015

Galois Theory TCU Graduate Student Seminar George Gilbert October 201 The coefficients of a polynomial are symmetric functions of the roots {α i }: fx) = x n s 1 x n 1 + s 2 x n 2 + + 1) n s n, where s

### 1 The Galois Group of a Quadratic

Algebra Prelim Notes The Galois Group of a Polynomial Jason B. Hill University of Colorado at Boulder Throughout this set of notes, K will be the desired base field (usually Q or a finite field) and F

### Section VI.33. Finite Fields

VI.33 Finite Fields 1 Section VI.33. Finite Fields Note. In this section, finite fields are completely classified. For every prime p and n N, there is exactly one (up to isomorphism) field of order p n,

### Galois fields/1. (M3) There is an element 1 (not equal to 0) such that a 1 = a for all a.

Galois fields 1 Fields A field is an algebraic structure in which the operations of addition, subtraction, multiplication, and division (except by zero) can be performed, and satisfy the usual rules. More

### The Galois group of a polynomial f(x) K[x] is the Galois group of E over K where E is a splitting field for f(x) over K.

The third exam will be on Monday, April 9, 013. The syllabus for Exam III is sections 1 3 of Chapter 10. Some of the main examples and facts from this material are listed below. If F is an extension field

### Section X.55. Cyclotomic Extensions

X.55 Cyclotomic Extensions 1 Section X.55. Cyclotomic Extensions Note. In this section we return to a consideration of roots of unity and consider again the cyclic group of roots of unity as encountered

### THE JOHNS HOPKINS UNIVERSITY Faculty of Arts and Sciences FINAL EXAM - SPRING SESSION ADVANCED ALGEBRA II.

THE JOHNS HOPKINS UNIVERSITY Faculty of Arts and Sciences FINAL EXAM - SPRING SESSION 2006 110.402 - ADVANCED ALGEBRA II. Examiner: Professor C. Consani Duration: 3 HOURS (9am-12:00pm), May 15, 2006. No

### Classification of Finite Fields

Classification of Finite Fields In these notes we use the properties of the polynomial x pd x to classify finite fields. The importance of this polynomial is explained by the following basic proposition.

### Math 201C Homework. Edward Burkard. g 1 (u) v + f 2(u) g 2 (u) v2 + + f n(u) a 2,k u k v a 1,k u k v + k=0. k=0 d

Math 201C Homework Edward Burkard 5.1. Field Extensions. 5. Fields and Galois Theory Exercise 5.1.7. If v is algebraic over K(u) for some u F and v is transcendental over K, then u is algebraic over K(v).

### but no smaller power is equal to one. polynomial is defined to be

13. Radical and Cyclic Extensions The main purpose of this section is to look at the Galois groups of x n a. The first case to consider is a = 1. Definition 13.1. Let K be a field. An element ω K is said

### School of Mathematics and Statistics. MT5836 Galois Theory. Handout 0: Course Information

MRQ 2017 School of Mathematics and Statistics MT5836 Galois Theory Handout 0: Course Information Lecturer: Martyn Quick, Room 326. Prerequisite: MT3505 (or MT4517) Rings & Fields Lectures: Tutorials: Mon

### Page Points Possible Points. Total 200

Instructions: 1. The point value of each exercise occurs adjacent to the problem. 2. No books or notes or calculators are allowed. Page Points Possible Points 2 20 3 20 4 18 5 18 6 24 7 18 8 24 9 20 10

### Part II. Number Theory. Year

Part II Year 2017 2016 2015 2014 2013 2012 2011 2010 2009 2008 2007 2006 2005 2017 Paper 3, Section I 1G 70 Explain what is meant by an Euler pseudoprime and a strong pseudoprime. Show that 65 is an Euler

### Solutions of exercise sheet 6

D-MATH Algebra I HS 14 Prof. Emmanuel Kowalski Solutions of exercise sheet 6 1. (Irreducibility of the cyclotomic polynomial) Let n be a positive integer, and P Z[X] a monic irreducible factor of X n 1

### arxiv: v1 [math.nt] 2 Jul 2009

About certain prime numbers Diana Savin Ovidius University, Constanţa, Romania arxiv:0907.0315v1 [math.nt] 2 Jul 2009 ABSTRACT We give a necessary condition for the existence of solutions of the Diophantine

### Field Theory Qual Review

Field Theory Qual Review Robert Won Prof. Rogalski 1 (Some) qual problems ˆ (Fall 2007, 5) Let F be a field of characteristic p and f F [x] a polynomial f(x) = i f ix i. Give necessary and sufficient conditions

### TC10 / 3. Finite fields S. Xambó

TC10 / 3. Finite fields S. Xambó The ring Construction of finite fields The Frobenius automorphism Splitting field of a polynomial Structure of the multiplicative group of a finite field Structure of the

### ALGEBRA PH.D. QUALIFYING EXAM September 27, 2008

ALGEBRA PH.D. QUALIFYING EXAM September 27, 2008 A passing paper consists of four problems solved completely plus significant progress on two other problems; moreover, the set of problems solved completely

### (January 14, 2009) q n 1 q d 1. D = q n = q + d

(January 14, 2009) [10.1] Prove that a finite division ring D (a not-necessarily commutative ring with 1 in which any non-zero element has a multiplicative inverse) is commutative. (This is due to Wedderburn.)

### Finite Fields. [Parts from Chapter 16. Also applications of FTGT]

Finite Fields [Parts from Chapter 16. Also applications of FTGT] Lemma [Ch 16, 4.6] Assume F is a finite field. Then the multiplicative group F := F \ {0} is cyclic. Proof Recall from basic group theory

### FIELD THEORY. Contents

FIELD THEORY MATH 552 Contents 1. Algebraic Extensions 1 1.1. Finite and Algebraic Extensions 1 1.2. Algebraic Closure 5 1.3. Splitting Fields 7 1.4. Separable Extensions 8 1.5. Inseparable Extensions

### A BRIEF INTRODUCTION TO LOCAL FIELDS

A BRIEF INTRODUCTION TO LOCAL FIELDS TOM WESTON The purpose of these notes is to give a survey of the basic Galois theory of local fields and number fields. We cover much of the same material as [2, Chapters

### arxiv: v1 [math.gr] 3 Feb 2019

Galois groups of symmetric sextic trinomials arxiv:1902.00965v1 [math.gr] Feb 2019 Alberto Cavallo Max Planck Institute for Mathematics, Bonn 5111, Germany cavallo@mpim-bonn.mpg.de Abstract We compute

### NAP Module 4. Homework 2 Friday, July 14, 2017.

NAP 2017. Module 4. Homework 2 Friday, July 14, 2017. These excercises are due July 21, 2017, at 10 pm. Nepal time. Please, send them to nap@rnta.eu, to laurageatti@gmail.com and schoo.rene@gmail.com.

### Name: Solutions Final Exam

Instructions. Answer each of the questions on your own paper, and be sure to show your work so that partial credit can be adequately assessed. Put your name on each page of your paper. 1. [10 Points] For

### Real Analysis Prelim Questions Day 1 August 27, 2013

Real Analysis Prelim Questions Day 1 August 27, 2013 are 5 questions. TIME LIMIT: 3 hours Instructions: Measure and measurable refer to Lebesgue measure µ n on R n, and M(R n ) is the collection of measurable

Graduate Preliminary Examination Algebra II 18.2.2005: 3 hours Problem 1. Prove or give a counter-example to the following statement: If M/L and L/K are algebraic extensions of fields, then M/K is algebraic.

### MATH 101A: ALGEBRA I, PART D: GALOIS THEORY 11

MATH 101A: ALGEBRA I, PART D: GALOIS THEORY 11 3. Examples I did some examples and explained the theory at the same time. 3.1. roots of unity. Let L = Q(ζ) where ζ = e 2πi/5 is a primitive 5th root of

### 1 Rings 1 RINGS 1. Theorem 1.1 (Substitution Principle). Let ϕ : R R be a ring homomorphism

1 RINGS 1 1 Rings Theorem 1.1 (Substitution Principle). Let ϕ : R R be a ring homomorphism (a) Given an element α R there is a unique homomorphism Φ : R[x] R which agrees with the map ϕ on constant polynomials

### 1. Group Theory Permutations.

1.1. Permutations. 1. Group Theory Problem 1.1. Let G be a subgroup of S n of index 2. Show that G = A n. Problem 1.2. Find two elements of S 7 that have the same order but are not conjugate. Let π S 7

### Information Theory. Lecture 7

Information Theory Lecture 7 Finite fields continued: R3 and R7 the field GF(p m ),... Cyclic Codes Intro. to cyclic codes: R8.1 3 Mikael Skoglund, Information Theory 1/17 The Field GF(p m ) π(x) irreducible

### A Generalization of Wilson s Theorem

A Generalization of Wilson s Theorem R. Andrew Ohana June 3, 2009 Contents 1 Introduction 2 2 Background Algebra 2 2.1 Groups................................. 2 2.2 Rings.................................

### Some algebraic number theory and the reciprocity map

Some algebraic number theory and the reciprocity map Ervin Thiagalingam September 28, 2015 Motivation In Weinstein s paper, the main problem is to find a rule (reciprocity law) for when an irreducible

### Section V.8. Cyclotomic Extensions

V.8. Cyclotomic Extensions 1 Section V.8. Cyclotomic Extensions Note. In this section we explore splitting fields of x n 1. The splitting fields turn out to be abelian extensions (that is, algebraic Galois

### Primes, Polynomials, Progressions. B.Sury Indian Statistical Institute Bangalore NISER Bhubaneshwar February 15, 2016

Indian Statistical Institute Bangalore NISER Bhubaneshwar February 15, 2016 Primes and polynomials It is unknown if there exist infinitely many prime numbers of the form n 2 + 1. Primes and polynomials

### Irreducible Polynomials. Finite Fields of Order p m (1) Primitive Polynomials. Finite Fields of Order p m (2)

S-72.3410 Finite Fields (2) 1 S-72.3410 Finite Fields (2) 3 Irreducible Polynomials Finite Fields of Order p m (1) The following results were discussed in the previous lecture: The order of a finite field

### Coding Theory and Applications. Solved Exercises and Problems of Cyclic Codes. Enes Pasalic University of Primorska Koper, 2013

Coding Theory and Applications Solved Exercises and Problems of Cyclic Codes Enes Pasalic University of Primorska Koper, 2013 Contents 1 Preface 3 2 Problems 4 2 1 Preface This is a collection of solved

### Abstract Algebra, Second Edition, by John A. Beachy and William D. Blair. Corrections and clarifications

1 Abstract Algebra, Second Edition, by John A. Beachy and William D. Blair Corrections and clarifications Note: Some corrections were made after the first printing of the text. page 9, line 8 For of the

### Part II Galois Theory

Part II Galois Theory Definitions Based on lectures by C. Birkar Notes taken by Dexter Chua Michaelmas 2015 These notes are not endorsed by the lecturers, and I have modified them (often significantly)

### Bulletin of the Transilvania University of Braşov Vol 7(56), No Series III: Mathematics, Informatics, Physics, 59-64

Bulletin of the Transilvania University of Braşov Vol 756), No. 2-2014 Series III: Mathematics, Informatics, Physics, 59-64 ABOUT QUATERNION ALGEBRAS AND SYMBOL ALGEBRAS Cristina FLAUT 1 and Diana SAVIN

### Chapter 4. Fields and Galois Theory

Chapter 4 Fields and Galois Theory 63 64 CHAPTER 4. FIELDS AND GALOIS THEORY 4.1 Field Extensions 4.1.1 K[u] and K(u) Def. A field F is an extension field of a field K if F K. Obviously, F K = 1 F = 1

### 18. Cyclotomic polynomials II

18. Cyclotomic polynomials II 18.1 Cyclotomic polynomials over Z 18.2 Worked examples Now that we have Gauss lemma in hand we can look at cyclotomic polynomials again, not as polynomials with coefficients

### ϕ : Z F : ϕ(t) = t 1 =

1. Finite Fields The first examples of finite fields are quotient fields of the ring of integers Z: let t > 1 and define Z /t = Z/(tZ) to be the ring of congruence classes of integers modulo t: in practical

### CSIR - Algebra Problems

CSIR - Algebra Problems N. Annamalai DST - INSPIRE Fellow (SRF) Department of Mathematics Bharathidasan University Tiruchirappalli -620024 E-mail: algebra.annamalai@gmail.com Website: https://annamalaimaths.wordpress.com

### Galois Theory, summary

Galois Theory, summary Chapter 11 11.1. UFD, definition. Any two elements have gcd 11.2 PID. Every PID is a UFD. There are UFD s which are not PID s (example F [x, y]). 11.3 ED. Every ED is a PID (and

### Explicit Methods in Algebraic Number Theory

Explicit Methods in Algebraic Number Theory Amalia Pizarro Madariaga Instituto de Matemáticas Universidad de Valparaíso, Chile amaliapizarro@uvcl 1 Lecture 1 11 Number fields and ring of integers Algebraic

### Part II Galois Theory

Part II Galois Theory Theorems Based on lectures by C. Birkar Notes taken by Dexter Chua Michaelmas 2015 These notes are not endorsed by the lecturers, and I have modified them (often significantly) after

### MT5836 Galois Theory MRQ

MT5836 Galois Theory MRQ May 3, 2017 Contents Introduction 3 Structure of the lecture course............................... 4 Recommended texts..................................... 4 1 Rings, Fields and

### Math 603, Spring 2003, HW 6, due 4/21/2003

Math 603, Spring 2003, HW 6, due 4/21/2003 Part A AI) If k is a field and f k[t ], suppose f has degree n and has n distinct roots α 1,..., α n in some extension of k. Write Ω = k(α 1,..., α n ) for the

### GALOIS THEORY. Contents

GALOIS THEORY MARIUS VAN DER PUT & JAAP TOP Contents 1. Basic definitions 1 1.1. Exercises 2 2. Solving polynomial equations 2 2.1. Exercises 4 3. Galois extensions and examples 4 3.1. Exercises. 6 4.

### Algebra Review. Instructor: Laszlo Babai Notes by Vincent Lucarelli and the instructor. June 15, 2001

Algebra Review Instructor: Laszlo Babai Notes by Vincent Lucarelli and the instructor June 15, 2001 1 Groups Definition 1.1 A semigroup (G, ) is a set G with a binary operation such that: Axiom 1 ( a,

### Section 33 Finite fields

Section 33 Finite fields Instructor: Yifan Yang Spring 2007 Review Corollary (23.6) Let G be a finite subgroup of the multiplicative group of nonzero elements in a field F, then G is cyclic. Theorem (27.19)

### Maximal Class Numbers of CM Number Fields

Maximal Class Numbers of CM Number Fields R. C. Daileda R. Krishnamoorthy A. Malyshev Abstract Fix a totally real number field F of degree at least 2. Under the assumptions of the generalized Riemann hypothesis

### A connection between number theory and linear algebra

A connection between number theory and linear algebra Mark Steinberger Contents 1. Some basics 1 2. Rational canonical form 2 3. Prime factorization in F[x] 4 4. Units and order 5 5. Finite fields 7 6.

### M3P11/M4P11/M5P11. Galois Theory

BSc and MSci EXAMINATIONS (MATHEMATICS) May-June 2014 This paper is also taken for the relevant examination for the Associateship of the Royal College of Science. M3P11/M4P11/M5P11 Galois Theory Date:

### CYCLOTOMIC POLYNOMIALS

CYCLOTOMIC POLYNOMIALS 1. The Derivative and Repeated Factors The usual definition of derivative in calculus involves the nonalgebraic notion of limit that requires a field such as R or C (or others) where

### CDM. Finite Fields. Klaus Sutner Carnegie Mellon University. Fall 2018

CDM Finite Fields Klaus Sutner Carnegie Mellon University Fall 2018 1 Ideals The Structure theorem Where Are We? 3 We know that every finite field carries two apparently separate structures: additive and

### Galois Theory and Some Applications

Galois Theory and Some Applications Aparna Ramesh July 19, 2015 Introduction In this project, we study Galois theory and discuss some applications. The theory of equations and the ancient Greek problems

### Galois groups of polynomials and the construction of finite fields

Pure and Applied Mathematics Journal 01; 1(1) : 10-16 Published online December 0, 01 (http://wwwsciencepublishinggroupcom/j/pamj) doi: 1011648/jpamj0101011 Galois groups of polynomials and the construction

### 1 Spring 2002 Galois Theory

1 Spring 2002 Galois Theory Problem 1.1. Let F 7 be the field with 7 elements and let L be the splitting field of the polynomial X 171 1 = 0 over F 7. Determine the degree of L over F 7, explaining carefully

### ALGEBRA 11: Galois theory

Galois extensions Exercise 11.1 (!). Consider a polynomial P (t) K[t] of degree n with coefficients in a field K that has n distinct roots in K. Prove that the ring K[t]/P of residues modulo P is isomorphic

### Algebraic proofs of Dirichlet s theorem on arithmetic progressions

Charles University in Prague Faculty of Mathematics and Physics BACHELOR THESIS Martin Čech Algebraic proofs of Dirichlet s theorem on arithmetic progressions Department of Algebra Supervisor of the bachelor

### CYCLOTOMIC POLYNOMIALS

CYCLOTOMIC POLYNOMIALS 1. The Derivative and Repeated Factors The usual definition of derivative in calculus involves the nonalgebraic notion of limit that requires a field such as R or C (or others) where

### MATH 361: NUMBER THEORY TENTH LECTURE

MATH 361: NUMBER THEORY TENTH LECTURE The subject of this lecture is finite fields. 1. Root Fields Let k be any field, and let f(x) k[x] be irreducible and have positive degree. We want to construct a

### Polynomials with nontrivial relations between their roots

ACTA ARITHMETICA LXXXII.3 (1997) Polynomials with nontrivial relations between their roots by John D. Dixon (Ottawa, Ont.) 1. Introduction. Consider an irreducible polynomial f(x) over a field K. We are

### Course 311: Abstract Algebra Academic year

Course 311: Abstract Algebra Academic year 2007-08 D. R. Wilkins Copyright c David R. Wilkins 1997 2007 Contents 3 Introduction to Galois Theory 41 3.1 Field Extensions and the Tower Law..............

### Algebra Exam Fall Alexander J. Wertheim Last Updated: October 26, Groups Problem Problem Problem 3...

Algebra Exam Fall 2006 Alexander J. Wertheim Last Updated: October 26, 2017 Contents 1 Groups 2 1.1 Problem 1..................................... 2 1.2 Problem 2..................................... 2

### Fundamental Theorem of Algebra

EE 387, Notes 13, Handout #20 Fundamental Theorem of Algebra Lemma: If f(x) is a polynomial over GF(q) GF(Q), then β is a zero of f(x) if and only if x β is a divisor of f(x). Proof: By the division algorithm,

### ALGEBRA PH.D. QUALIFYING EXAM SOLUTIONS October 20, 2011

ALGEBRA PH.D. QUALIFYING EXAM SOLUTIONS October 20, 2011 A passing paper consists of four problems solved completely plus significant progress on two other problems; moreover, the set of problems solved

### Lemma 1.1. The field K embeds as a subfield of Q(ζ D ).

Math 248A. Quadratic characters associated to quadratic fields The aim of this handout is to describe the quadratic Dirichlet character naturally associated to a quadratic field, and to express it in terms

### Section V.7. Cyclic Extensions

V.7. Cyclic Extensions 1 Section V.7. Cyclic Extensions Note. In the last three sections of this chapter we consider specific types of Galois groups of Galois extensions and then study the properties of

### GALOIS THEORY: LECTURE 24

GALOIS THEORY: LECTURE 4 LEO GOLDMAKHER 1. CONSTRUCTING FINITE FIELDS Although most of the semester we stated and proved theorems about general field extensions L/K, in practice we ve barely touched the

### NOTES FOR DRAGOS: MATH 210 CLASS 12, THURS. FEB. 22

NOTES FOR DRAGOS: MATH 210 CLASS 12, THURS. FEB. 22 RAVI VAKIL Hi Dragos The class is in 381-T, 1:15 2:30. This is the very end of Galois theory; you ll also start commutative ring theory. Tell them: midterm

### 17 Galois Fields Introduction Primitive Elements Roots of Polynomials... 8

Contents 17 Galois Fields 2 17.1 Introduction............................... 2 17.2 Irreducible Polynomials, Construction of GF(q m )... 3 17.3 Primitive Elements... 6 17.4 Roots of Polynomials..........................

### 22M: 121 Final Exam. Answer any three in this section. Each question is worth 10 points.

22M: 121 Final Exam This is 2 hour exam. Begin each question on a new sheet of paper. All notations are standard and the ones used in class. Please write clearly and provide all details of your work. Good

### Solving the general quadratic congruence. y 2 Δ (mod p),

Quadratic Congruences Solving the general quadratic congruence ax 2 +bx + c 0 (mod p) for an odd prime p (with (a, p) = 1) is equivalent to solving the simpler congruence y 2 Δ (mod p), where Δ = b 2 4ac

### Chapter 5. Modular arithmetic. 5.1 The modular ring

Chapter 5 Modular arithmetic 5.1 The modular ring Definition 5.1. Suppose n N and x, y Z. Then we say that x, y are equivalent modulo n, and we write x y mod n if n x y. It is evident that equivalence

### ALGEBRA QUALIFYING EXAM SPRING 2012

ALGEBRA QUALIFYING EXAM SPRING 2012 Work all of the problems. Justify the statements in your solutions by reference to specific results, as appropriate. Partial credit is awarded for partial solutions.

### Algebraic Structures Exam File Fall 2013 Exam #1

Algebraic Structures Exam File Fall 2013 Exam #1 1.) Find all four solutions to the equation x 4 + 16 = 0. Give your answers as complex numbers in standard form, a + bi. 2.) Do the following. a.) Write

### Homework 10 M 373K by Mark Lindberg (mal4549)

Homework 10 M 373K by Mark Lindberg (mal4549) 1. Artin, Chapter 11, Exercise 1.1. Prove that 7 + 3 2 and 3 + 5 are algebraic numbers. To do this, we must provide a polynomial with integer coefficients

### EE 229B ERROR CONTROL CODING Spring 2005

EE 9B ERROR CONTROL CODING Spring 005 Solutions for Homework 1. (Weights of codewords in a cyclic code) Let g(x) be the generator polynomial of a binary cyclic code of length n. (a) Show that if g(x) has

### Field Theory Problems

Field Theory Problems I. Degrees, etc. 1. Answer the following: (a Find u R such that Q(u = Q( 2, 3 5. (b Describe how you would find all w Q( 2, 3 5 such that Q(w = Q( 2, 3 5. 2. If a, b K are algebraic

### Factorization in Polynomial Rings

Factorization in Polynomial Rings Throughout these notes, F denotes a field. 1 Long division with remainder We begin with some basic definitions. Definition 1.1. Let f, g F [x]. We say that f divides g,

### Determining the Galois group of a rational polynomial

JAH 1 Determining the Galois group of a rational polynomial Alexander Hulpke Department of Mathematics Colorado State University Fort Collins, CO, 80523 hulpke@math.colostate.edu http://www.math.colostate.edu/

### IUPUI Qualifying Exam Abstract Algebra

IUPUI Qualifying Exam Abstract Algebra January 2017 Daniel Ramras (1) a) Prove that if G is a group of order 2 2 5 2 11, then G contains either a normal subgroup of order 11, or a normal subgroup of order

### Algebra Ph.D. Preliminary Exam

RETURN THIS COVER SHEET WITH YOUR EXAM AND SOLUTIONS! Algebra Ph.D. Preliminary Exam August 18, 2008 INSTRUCTIONS: 1. Answer each question on a separate page. Turn in a page for each problem even if you

### Finite Fields. Saravanan Vijayakumaran Department of Electrical Engineering Indian Institute of Technology Bombay

1 / 25 Finite Fields Saravanan Vijayakumaran sarva@ee.iitb.ac.in Department of Electrical Engineering Indian Institute of Technology Bombay September 25, 2014 2 / 25 Fields Definition A set F together

### Department of Mathematics, University of California, Berkeley

ALGORITHMIC GALOIS THEORY Hendrik W. Lenstra jr. Mathematisch Instituut, Universiteit Leiden Department of Mathematics, University of California, Berkeley K = field of characteristic zero, Ω = algebraically

### RUDIMENTARY GALOIS THEORY

RUDIMENTARY GALOIS THEORY JACK LIANG Abstract. This paper introduces basic Galois Theory, primarily over fields with characteristic 0, beginning with polynomials and fields and ultimately relating the

### August 2015 Qualifying Examination Solutions

August 2015 Qualifying Examination Solutions If you have any difficulty with the wording of the following problems please contact the supervisor immediately. All persons responsible for these problems,

### Solutions for Problem Set 6

Solutions for Problem Set 6 A: Find all subfields of Q(ζ 8 ). SOLUTION. All subfields of K must automatically contain Q. Thus, this problem concerns the intermediate fields for the extension K/Q. In a

### Polynomials. Chapter 4

Chapter 4 Polynomials In this Chapter we shall see that everything we did with integers in the last Chapter we can also do with polynomials. Fix a field F (e.g. F = Q, R, C or Z/(p) for a prime p). Notation

### Name: Solutions Final Exam

Instructions. Answer each of the questions on your own paper. Be sure to show your work so that partial credit can be adequately assessed. Put your name on each page of your paper. 1. [10 Points] All of

### Continuing the pre/review of the simple (!?) case...

Continuing the pre/review of the simple (!?) case... Garrett 09-16-011 1 So far, we have sketched the connection between prime numbers, and zeros of the zeta function, given by Riemann s formula p m

### Factorization in Integral Domains II

Factorization in Integral Domains II 1 Statement of the main theorem Throughout these notes, unless otherwise specified, R is a UFD with field of quotients F. The main examples will be R = Z, F = Q, and

### be any ring homomorphism and let s S be any element of S. Then there is a unique ring homomorphism

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 UFD. Therefore