# Homework problems from Chapters IV-VI: answers and solutions

Size: px
Start display at page:

Transcription

1 Homework problems from Chapters IV-VI: answers and solutions IV In this problem we have to describe the field F of quotients of the domain D. Note that by definition, F is the set of equivalence classes [(n 1 + m 1 i, n 2 + m 2 i)] of D D with equivalence relation if and only if (n 1 + m 1 i, n 2 + m 2 i) (k 1 + l 1 i, k 2 + l 2 i) (n 1 + m 1 i)(k 2 + l 2 i) = (n 2 + m 2 i)(k 1 + l 1 i). (1) We will show that F is isomorphic to the following field: Q[i] := {a + bi a, b Q}. The isomorphism ϕ : F Q[i] is given by the formula ϕ([(n 1 + m 1 i, n 2 + m 2 i)]) := n 1 + m 1 i n 2 + m 2 i or in short ϕ([z, w]) := z w for (z, w) D D. In what follows we show that ϕ is an isomorphism. The map ϕ is well-defined because of (1). The map ϕ is a homomorphism of rings because: ϕ([(z 1, w 1 )]+[(z 2, w 2 )]) = ϕ([z 1 w 2 +w 1 z 2, w 1 w 2 ]) = z 1w 2 + w 1 z 2 w 1 w 2 = z 1 w 1 + z 1 w 1 = ϕ([(z 1, w 1 )])+ϕ([(z 2, w 2 )]) Similarly ϕ([(z 1, w 1 )] [(z 2, w 2 )]) = ϕ([(z 1, w 1 )]) ϕ([(z 2, w 2 )]). The map ϕ is onto because for z w Q[i] we have ϕ([(z, w)]) = z w. The map ϕ is one-to-one because if ϕ([(z, w)]) = 0 then z = 0, i.e. [(z, w)] = [(0, 1)]. IV Solved in class. IV The elements of Q(Z 4, {1, 3})are [(0, 1)] = [(0, 3)], [(1, 1)] = [(3, 3)], [(1, 3)], [(2, 1)], [(2, 3)], [(3, 1)], i.e. total 6 distinct elements. IV f(x)g(x) = x IV We have that 106 = , 99 = , 53 = , and thus φ 4 (3x x x 53 ) = (4 6 ) (4 6 ) (4 6 ) 8 = =...( mod 7), since 4 6 = 1( mod 7) by Fermat s Little Theorem.

2 IV Solved in class. IV (a) The units of D[x] are the constant polynomials f(x) = u, where u is a unit of D. (b) The units of Z[x] are the constant polynomials f(x) = ±1. (b) The units of Z 7 [x] are the constant polynomials f(x) = u, for u = 1, 2, 3, 4, 5, 6. IV (a) You must show that D is a homomorphism without using any properties of derivatives (done in class). (b) The kernel is the set of all constant functions. (c) The image F [x] is F [x] itself, i.e. D is onto (this needs a proof). IV The factorization is x = (x + 1)(x + 2)(x + 3)(x + 4) (you simply verify that any nonzero element of Z 4 is a root of x = 0). IV Verify that there f(x) has no rational root.the possible rational roots are ±1. ± 2, ±4, ±8. Note that one can not apply Eisenstein s Criterion for any p in this case. IV Eisenstein s Criterion does not apply for any p. IV Eisenstein s Criterion applies for p = 5. IV The polynomial has always a linear factor x + a, since a is a root (due to Fermat s Little Theorem). IV Done in class. V The sets are isomorphic as abelian groups (identical addition tables), but not as rings (as the multiplication tables are different). One has to write down the tables. V One must show that (a + b) p = a p + b p and (ab) p = a p b p. The first one is done in class using the binomial formula and the second one follows by the fact that ab = ba. V Let N(R) := {a R a n = 0 for some n Z + }. be the nilradical of R. One must show that:

3 (1) N(R) is an abelian subgroup of R, i.e. a + b N(R), 0 R N(R), a N(R), whenever a, b N(R); (2) If a N(R) and b R then ab N(R). For the addition part of (1) one uses the binomial formula to show that, if a n = 0 and b m = 0, then (a + b) n+m = 0 (done in class). Not that in order to prove ( a) n = 0 one must use the definition of a and not the formula ( a) n = ( 1) n a n as R might have not identity. The second axiom follows from the identity (ab) n = a n b n. V The reasoning is similar to the on in V V Take R = Q[x]. (a) One example is N := x 2, i.e. the principal ideal of R generated by the polynomial f(x) = x 2. Then x N, but x / N. (b) One example is N := x, i.e. the principal ideal of R generated by the polynomial g(x) = x. Then N = N. V The prime and maximal ideals of Z 12 (all primes are maximals in this case) are those that are isomorphic to Z 4 and Z 6, since M is prime iff Z 12 /M is an integral domain iff Z 12 /M = Z 2, Z 3. Thus the answer is: I 1 = 3 = 9 = Z 4 and I 2 = 2 = 10 = Z 6. V All nonzero proper ideals are both maximal and prime. V Solved in class. The shortest way to find the prime and maximal ideals is again to apply the theorems: M is a maximal (respectively, prime) ideal of a commutative ring R iff R/M is a field (resp., integral domain). So we must find all M for which Z 2 Z 4 /M is an integral domain. But if M is proper and nontrivial then Z 2 Z 4 /M, as an abelian group, is isomorphic to one of the following: Z 2, Z 4, Z 2 Z 2. In this list there is only one integral domain: Z 2, so we need to list all ideals M of Z 2 Z 4 which are isomorphic as abelian groups to Z 2 Z 2 or Z 4. These will be both the maximal and prime ideals. The answer is Z 2 {0, 2}, {0} Z 4. V Find all c for which x 2 + x + c is irreducible, i.e. does not have a root. The answer is c = 1, 2 V Same reasoning as in V V Solved in class. Relates to the Theorem on page 181. V Solved in class. Must show that if a principal ideal f(x) is prime then f(x) is irreducible (here we use theorems and 27.25). The proof is by contradiction.

4 V The second statement follows by using a proof by contradiction. Namely, one assumes that f and g are irreducible and shows that N = F [x]. The proof relies on the polynomial version of Euclidean Algorithm whose general statement and proof can be found on p age 404. VI f(x) = (x 2 5) 2 24 = x 4 10x VI f(x) = (x 1) VI f(x) = (x 2 1) 3 2. VI irr(α, Q) = (x 1) = x 2 2x + 9 and deg(α, Q) = 2. One shows that x 2 2x + 9 is irreducible by verifying that it does not have a rational root. VI irr(α, R) = (x 1) = x 2 2x + 2 and deg(α, R) = 2. VI irr(α, R) = x π and deg(α, R) = 1. VI π 2 is transcendental over Q. Proof by contradiction: must show that if π 2 is algebraic, then π is also algebraic over Q. VI irr(α, Q(π)) = x π 2 and deg(α, Q(π)) = 1. In particular, F (α) = F. VI Solved in class. irr(α, Q(π 3 )) = x 3 π 6 and deg(α, Q(π 3 )) = 3. Must show that x 3 π 6 is irreducible in Q(π 3 )[x] (done in class). VI Solved in class. VI Since α is algebraic over F (β) one has that a 0 + a 1 α a n α n = 0 for some a i F (β). Every element u of F (β) has the form u = f(β) g(β), for some polynomials f, g F [x]. Thus there are polynomials p i, q i, i = 0,..., n for which a i = p i(β) q. But then i(β) p0(β) q 0 (β) + p 1(β) q 1 (β) α p n(β) q n (β) αn = 0 Multiplying by the least common denominator q 0 (β)...q n (β) we find that P (α, β) = 0 for some polynomial P (x, y) of two variables in F [x, y]. Then if Q(x) := P (α, x) F [α][x] F (α)[x] we have that Q(β) = 0 and we complete the proof. VI (a) The shortest way to solve this problem is first to find all irreducible (and reducible) monic polynomials in Z 3 of degree 2. We have that all reducible ones are x 2, x(x+1), x(x+2), (x+1) 2, (x+1)(x+2), (x+2) 2. Thus we

5 have 3 irreducible ones. Then we easily count all reducible monic polynomials of degree 3: we have 9 that are product of irreducibles of degree one and degree two, and 10 more that are product of three (not necessarily different!) of degree one. So we have total 19 reducible monic polynomials and thus 8 irreducible ones. VI [Q( 2, 3, 18) : Q] = 4, and a basis is {1, 2, 3, 6}. Here we prove Q( 2, 3, 18) = Q( 2, 3) and prove that f(x) = (x 2 5) 2 24 = x 4 10x is irreducible. (Must show more work!) VI [Q( 3 2, 3) : Q] = 6, and basis is {1, 3, 3 2, 3 2 3, 3 4, 3 4 3}. (Must show more work!) VI [Q( 2 + 3) : Q( 3)] = 2, and basis is {1, 2}. Here we use Q( 2 + 3) = Q( 2, 3). VI They are the same fields. (Must show more work!) VI Use the hint I provided in class. VI Solved in class. VI In class we showed Q( 2+ 3) = Q( 2, 3). Use similar reasoning. VI Solved during the review session. We look at [F (α) : F (α 2 )]. This degree is either 1 or 2 since f(α 2 ) = 0 for f(x) = x α 2 F (α 2 )[x]. We show that it can not be 2 using the assumption in the problem. Then we use the fact that [F (α) : F ] = [F (α) : F (α 2 )][F (α 2 ) : F ].

### MATH 431 PART 2: POLYNOMIAL RINGS AND FACTORIZATION

MATH 431 PART 2: POLYNOMIAL RINGS AND FACTORIZATION 1. Polynomial rings (review) Definition 1. A polynomial f(x) with coefficients in a ring R is n f(x) = a i x i = a 0 + a 1 x + a 2 x 2 + + a n x n i=0

### 6]. (10) (i) Determine the units in the rings Z[i] and Z[ 10]. If n is a squarefree

Quadratic extensions Definition: Let R, S be commutative rings, R S. An extension of rings R S is said to be quadratic there is α S \R and monic polynomial f(x) R[x] of degree such that f(α) = 0 and S

### Modern Algebra I. Circle the correct answer; no explanation is required. Each problem in this section counts 5 points.

1 2 3 style total Math 415 Please print your name: Answer Key 1 True/false Circle the correct answer; no explanation is required. Each problem in this section counts 5 points. 1. Every group of order 6

### 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

### Math 121 Homework 2 Solutions

Math 121 Homework 2 Solutions Problem 13.2 #16. Let K/F be an algebraic extension and let R be a ring contained in K that contains F. Prove that R is a subfield of K containing F. We will give two proofs.

### Algebra Homework, Edition 2 9 September 2010

Algebra Homework, Edition 2 9 September 2010 Problem 6. (1) Let I and J be ideals of a commutative ring R with I + J = R. Prove that IJ = I J. (2) Let I, J, and K be ideals of a principal ideal domain.

### Homework 8 Solutions to Selected Problems

Homework 8 Solutions to Selected Problems June 7, 01 1 Chapter 17, Problem Let f(x D[x] and suppose f(x is reducible in D[x]. That is, there exist polynomials g(x and h(x in D[x] such that g(x and h(x

### Math Introduction to Modern Algebra

Math 343 - Introduction to Modern Algebra Notes Field Theory Basics Let R be a ring. M is called a maximal ideal of R if M is a proper ideal of R and there is no proper ideal of R that properly contains

### RINGS: SUMMARY OF MATERIAL

RINGS: SUMMARY OF MATERIAL BRIAN OSSERMAN This is a summary of terms used and main results proved in the subject of rings, from Chapters 11-13 of Artin. Definitions not included here may be considered

### QUALIFYING EXAM IN ALGEBRA August 2011

QUALIFYING EXAM IN ALGEBRA August 2011 1. There are 18 problems on the exam. Work and turn in 10 problems, in the following categories. I. Linear Algebra 1 problem II. Group Theory 3 problems III. Ring

### MATH 113 FINAL EXAM December 14, 2012

p.1 MATH 113 FINAL EXAM December 14, 2012 This exam has 9 problems on 18 pages, including this cover sheet. The only thing you may have out during the exam is one or more writing utensils. You have 180

### φ(xy) = (xy) n = x n y n = φ(x)φ(y)

Groups 1. (Algebra Comp S03) Let A, B and C be normal subgroups of a group G with A B. If A C = B C and AC = BC then prove that A = B. Let b B. Since b = b1 BC = AC, there are a A and c C such that b =

### Math 581 Problem Set 3 Solutions

Math 581 Problem Set 3 Solutions 1. Prove that complex conjugation is a isomorphism from C to C. Proof: First we prove that it is a homomorphism. Define : C C by (z) = z. Note that (1) = 1. The other properties

### 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

### MATH 3030, Abstract Algebra Winter 2012 Toby Kenney Sample Midterm Examination Model Solutions

MATH 3030, Abstract Algebra Winter 2012 Toby Kenney Sample Midterm Examination Model Solutions Basic Questions 1. Give an example of a prime ideal which is not maximal. In the ring Z Z, the ideal {(0,

### Polynomial Rings. i=0. i=0. n+m. i=0. k=0

Polynomial Rings 1. Definitions and Basic Properties For convenience, the ring will always be a commutative ring with identity. Basic Properties The polynomial ring R[x] in the indeterminate x with coefficients

### 2a 2 4ac), provided there is an element r in our

MTH 310002 Test II Review Spring 2012 Absractions versus examples The purpose of abstraction is to reduce ideas to their essentials, uncluttered by the details of a specific situation Our lectures built

### Computations/Applications

Computations/Applications 1. Find the inverse of x + 1 in the ring F 5 [x]/(x 3 1). Solution: We use the Euclidean Algorithm: x 3 1 (x + 1)(x + 4x + 1) + 3 (x + 1) 3(x + ) + 0. Thus 3 (x 3 1) + (x + 1)(4x

### Rings. Chapter 1. Definition 1.2. A commutative ring R is a ring in which multiplication is commutative. That is, ab = ba for all a, b R.

Chapter 1 Rings We have spent the term studying groups. A group is a set with a binary operation that satisfies certain properties. But many algebraic structures such as R, Z, and Z n come with two binary

### Part VI. Extension Fields

VI.29 Introduction to Extension Fields 1 Part VI. Extension Fields Section VI.29. Introduction to Extension Fields Note. In this section, we attain our basic goal and show that for any polynomial over

### Fields and Galois Theory. Below are some results dealing with fields, up to and including the fundamental theorem of Galois theory.

Fields and Galois Theory Below are some results dealing with fields, up to and including the fundamental theorem of Galois theory. This should be a reasonably logical ordering, so that a result here should

### Math 547, Exam 2 Information.

Math 547, Exam 2 Information. 3/19/10, LC 303B, 10:10-11:00. Exam 2 will be based on: Homework and textbook sections covered by lectures 2/3-3/5. (see http://www.math.sc.edu/ boylan/sccourses/547sp10/547.html)

### 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

### 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.)

### Homework 9 Solutions to Selected Problems

Homework 9 Solutions to Selected Problems June 11, 2012 1 Chapter 17, Problem 12 Since x 2 + x + 4 has degree 2 and Z 11 is a eld, we may use Theorem 17.1 and show that f(x) is irreducible because it has

### 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).

### 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

### Algebra Ph.D. Entrance Exam Fall 2009 September 3, 2009

Algebra Ph.D. Entrance Exam Fall 2009 September 3, 2009 Directions: Solve 10 of the following problems. Mark which of the problems are to be graded. Without clear indication which problems are to be graded

### 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

### CHAPTER I. Rings. Definition A ring R is a set with two binary operations, addition + and

CHAPTER I Rings 1.1 Definitions and Examples Definition 1.1.1. A ring R is a set with two binary operations, addition + and multiplication satisfying the following conditions for all a, b, c in R : (i)

### Abstract Algebra: Chapters 16 and 17

Study polynomials, their factorization, and the construction of fields. Chapter 16 Polynomial Rings Notation Let R be a commutative ring. The ring of polynomials over R in the indeterminate x is the set

### Ph.D. Qualifying Examination in Algebra Department of Mathematics University of Louisville January 2018

Ph.D. Qualifying Examination in Algebra Department of Mathematics University of Louisville January 2018 Do 6 problems with at least 2 in each section. Group theory problems: (1) Suppose G is a group. The

### Direction: You are required to complete this test by Monday (April 24, 2006). In order to

Test 4 April 20, 2006 Name Math 522 Student Number Direction: You are required to complete this test by Monday (April 24, 2006). In order to receive full credit, answer each problem completely and must

### 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

### University of Ottawa

University of Ottawa Department of Mathematics and Statistics MAT3143: Ring Theory Professor: Hadi Salmasian Final Exam April 21, 2015 Surname First Name Instructions: (a) You have 3 hours to complete

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.

### Lecture 7.3: Ring homomorphisms

Lecture 7.3: Ring homomorphisms Matthew Macauley Department of Mathematical Sciences Clemson University http://www.math.clemson.edu/~macaule/ Math 4120, Modern Algebra M. Macauley (Clemson) Lecture 7.3:

### Selected Math 553 Homework Solutions

Selected Math 553 Homework Solutions HW6, 1. Let α and β be rational numbers, with α 1/2, and let m > 0 be an integer such that α 2 mβ 2 = 1 δ where 0 δ < 1. Set ǫ:= 1 if α 0 and 1 if α < 0. Show that

### M2P4. Rings and Fields. Mathematics Imperial College London

M2P4 Rings and Fields Mathematics Imperial College London ii As lectured by Professor Alexei Skorobogatov and humbly typed by as1005@ic.ac.uk. CONTENTS iii Contents 1 Basic Properties Of Rings 1 2 Factorizing

### 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

### Extension fields II. Sergei Silvestrov. Spring term 2011, Lecture 13

Extension fields II Sergei Silvestrov Spring term 2011, Lecture 13 Abstract Contents of the lecture. Algebraic extensions. Finite fields. Automorphisms of fields. The isomorphism extension theorem. Splitting

### May 6, Be sure to write your name on your bluebook. Use a separate page (or pages) for each problem. Show all of your work.

Math 236H May 6, 2008 Be sure to write your name on your bluebook. Use a separate page (or pages) for each problem. Show all of your work. 1. (15 points) Prove that the symmetric group S 4 is generated

### 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

### g(x) = 1 1 x = 1 + x + x2 + x 3 + is not a polynomial, since it doesn t have finite degree. g(x) is an example of a power series.

6 Polynomial Rings We introduce a class of rings called the polynomial rings, describing computation, factorization and divisibility in such rings For the case where the coefficients come from an integral

### Math 210B:Algebra, Homework 2

Math 210B:Algebra, Homework 2 Ian Coley January 21, 2014 Problem 1. Is f = 2X 5 6X + 6 irreducible in Z[X], (S 1 Z)[X], for S = {2 n, n 0}, Q[X], R[X], C[X]? To begin, note that 2 divides all coefficients

### Math 120 HW 9 Solutions

Math 120 HW 9 Solutions June 8, 2018 Question 1 Write down a ring homomorphism (no proof required) f from R = Z[ 11] = {a + b 11 a, b Z} to S = Z/35Z. The main difficulty is to find an element x Z/35Z

### Honors Algebra 4, MATH 371 Winter 2010 Assignment 4 Due Wednesday, February 17 at 08:35

Honors Algebra 4, MATH 371 Winter 2010 Assignment 4 Due Wednesday, February 17 at 08:35 1. Let R be a commutative ring with 1 0. (a) Prove that the nilradical of R is equal to the intersection of the prime

### 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

### Chapter 4. Remember: F will always stand for a field.

Chapter 4 Remember: F will always stand for a field. 4.1 10. Take f(x) = x F [x]. Could there be a polynomial g(x) F [x] such that f(x)g(x) = 1 F? Could f(x) be a unit? 19. Compare with Problem #21(c).

### Algebraic structures I

MTH5100 Assignment 1-10 Algebraic structures I For handing in on various dates January March 2011 1 FUNCTIONS. Say which of the following rules successfully define functions, giving reasons. For each one

### Honors Algebra 4, MATH 371 Winter 2010 Assignment 3 Due Friday, February 5 at 08:35

Honors Algebra 4, MATH 371 Winter 2010 Assignment 3 Due Friday, February 5 at 08:35 1. Let R 0 be a commutative ring with 1 and let S R be the subset of nonzero elements which are not zero divisors. (a)

### NOTES ON FINITE FIELDS

NOTES ON FINITE FIELDS AARON LANDESMAN CONTENTS 1. Introduction to finite fields 2 2. Definition and constructions of fields 3 2.1. The definition of a field 3 2.2. Constructing field extensions by adjoining

### 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

### PRACTICE FINAL MATH , MIT, SPRING 13. You have three hours. This test is closed book, closed notes, no calculators.

PRACTICE FINAL MATH 18.703, MIT, SPRING 13 You have three hours. This test is closed book, closed notes, no calculators. There are 11 problems, and the total number of points is 180. Show all your work.

### MATH 3030, Abstract Algebra FALL 2012 Toby Kenney Midyear Examination Friday 7th December: 7:00-10:00 PM

MATH 3030, Abstract Algebra FALL 2012 Toby Kenney Midyear Examination Friday 7th December: 7:00-10:00 PM Basic Questions 1. Compute the factor group Z 3 Z 9 / (1, 6). The subgroup generated by (1, 6) is

### 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

### Abstract Algebra FINAL EXAM May 23, Name: R. Hammack Score:

Abstract Algebra FINAL EXAM May 23, 2003 Name: R. Hammack Score: Directions: Please answer the questions in the space provided. To get full credit you must show all of your work. Use of calculators and

### 2 (17) Find non-trivial left and right ideals of the ring of 22 matrices over R. Show that there are no nontrivial two sided ideals. (18) State and pr

MATHEMATICS Introduction to Modern Algebra II Review. (1) Give an example of a non-commutative ring; a ring without unit; a division ring which is not a eld and a ring which is not a domain. (2) Show that

### Homework 6 Solution. Math 113 Summer 2016.

Homework 6 Solution. Math 113 Summer 2016. 1. For each of the following ideals, say whether they are prime, maximal (hence also prime), or neither (a) (x 4 + 2x 2 + 1) C[x] (b) (x 5 + 24x 3 54x 2 + 6x

### MTH310 EXAM 2 REVIEW

MTH310 EXAM 2 REVIEW SA LI 4.1 Polynomial Arithmetic and the Division Algorithm A. Polynomial Arithmetic *Polynomial Rings If R is a ring, then there exists a ring T containing an element x that is not

### ALGEBRA AND NUMBER THEORY II: Solutions 3 (Michaelmas term 2008)

ALGEBRA AND NUMBER THEORY II: Solutions 3 Michaelmas term 28 A A C B B D 61 i If ϕ : R R is the indicated map, then ϕf + g = f + ga = fa + ga = ϕf + ϕg, and ϕfg = f ga = faga = ϕfϕg. ii Clearly g lies

### Math 121 Homework 5: Notes on Selected Problems

Math 121 Homework 5: Notes on Selected Problems 12.1.2. Let M be a module over the integral domain R. (a) Assume that M has rank n and that x 1,..., x n is any maximal set of linearly independent elements

### Eighth Homework Solutions

Math 4124 Wednesday, April 20 Eighth Homework Solutions 1. Exercise 5.2.1(e). Determine the number of nonisomorphic abelian groups of order 2704. First we write 2704 as a product of prime powers, namely

Mathematical Olympiad Training Polynomials Definition A polynomial over a ring R(Z, Q, R, C) in x is an expression of the form p(x) = a n x n + a n 1 x n 1 + + a 1 x + a 0, a i R, for 0 i n. If a n 0,

### 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,

### Solutions of exercise sheet 11

D-MATH Algebra I HS 14 Prof Emmanuel Kowalski Solutions of exercise sheet 11 The content of the marked exercises (*) should be known for the exam 1 For the following values of α C, find the minimal polynomial

### Math 611 Homework 6. Paul Hacking. November 19, All rings are assumed to be commutative with 1.

Math 611 Homework 6 Paul Hacking November 19, 2015 All rings are assumed to be commutative with 1. (1) Let R be a integral domain. We say an element 0 a R is irreducible if a is not a unit and there does

### Math 4400, Spring 08, Sample problems Final Exam.

Math 4400, Spring 08, Sample problems Final Exam. 1. Groups (1) (a) Let a be an element of a group G. Define the notions of exponent of a and period of a. (b) Suppose a has a finite period. Prove that

### ALGEBRA QUALIFYING EXAM, WINTER SOLUTIONS

ALGEBRA QUALIFYING EXAM, WINTER 2017. SOLUTIONS Your Name: Conventions: all rings and algebras are assumed to be unital. Part I. True or false? If true provide a brief explanation, if false provide a counterexample

### 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,

### GEOMETRIC CONSTRUCTIONS AND ALGEBRAIC FIELD EXTENSIONS

GEOMETRIC CONSTRUCTIONS AND ALGEBRAIC FIELD EXTENSIONS JENNY WANG Abstract. In this paper, we study field extensions obtained by polynomial rings and maximal ideals in order to determine whether solutions

### 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

### Contradiction. Theorem 1.9. (Artin) Let G be a finite group of automorphisms of E and F = E G the fixed field of G. Then [E : F ] G.

1. Galois Theory 1.1. A homomorphism of fields F F is simply a homomorphism of rings. Such a homomorphism is always injective, because its kernel is a proper ideal (it doesnt contain 1), which must therefore

### Solutions of exercise sheet 8

D-MATH Algebra I HS 14 Prof. Emmanuel Kowalski Solutions of exercise sheet 8 1. In this exercise, we will give a characterization for solvable groups using commutator subgroups. See last semester s (Algebra

### 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,

### 4.4 Noetherian Rings

4.4 Noetherian Rings Recall that a ring A is Noetherian if it satisfies the following three equivalent conditions: (1) Every nonempty set of ideals of A has a maximal element (the maximal condition); (2)

### Supplement. Dr. Bob s Modern Algebra Glossary Based on Fraleigh s A First Course on Abstract Algebra, 7th Edition, Sections 0 through IV.

Glossary 1 Supplement. Dr. Bob s Modern Algebra Glossary Based on Fraleigh s A First Course on Abstract Algebra, 7th Edition, Sections 0 through IV.23 Abelian Group. A group G, (or just G for short) is

### 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

### Chapter 3. Rings. The basic commutative rings in mathematics are the integers Z, the. Examples

Chapter 3 Rings Rings are additive abelian groups with a second operation called multiplication. The connection between the two operations is provided by the distributive law. Assuming the results of Chapter

### 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

### Some practice problems for midterm 2

Some practice problems for midterm 2 Kiumars Kaveh November 14, 2011 Problem: Let Z = {a G ax = xa, x G} be the center of a group G. Prove that Z is a normal subgroup of G. Solution: First we prove Z is

### Section 15 Factor-group computation and simple groups

Section 15 Factor-group computation and simple groups Instructor: Yifan Yang Fall 2006 Outline Factor-group computation Simple groups The problem Problem Given a factor group G/H, find an isomorphic group

### Total 100

Math 542 Midterm Exam, Spring 2016 Prof: Paul Terwilliger Your Name (please print) SOLUTIONS NO CALCULATORS/ELECTRONIC DEVICES ALLOWED. MAKE SURE YOUR CELL PHONE IS OFF. Problem Value 1 10 2 10 3 10 4

### 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

### 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

### 1 First Theme: Sums of Squares

I will try to organize the work of this semester around several classical questions. The first is, When is a prime p the sum of two squares? The question was raised by Fermat who gave the correct answer

### 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.

### 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

### 38 Irreducibility criteria in rings of polynomials

38 Irreducibility criteria in rings of polynomials 38.1 Theorem. Let p(x), q(x) R[x] be polynomials such that p(x) = a 0 + a 1 x +... + a n x n, q(x) = b 0 + b 1 x +... + b m x m and a n, b m 0. If b m

### Extension theorems for homomorphisms

Algebraic Geometry Fall 2009 Extension theorems for homomorphisms In this note, we prove some extension theorems for homomorphisms from rings to algebraically closed fields. The prototype is the following

### Section III.6. Factorization in Polynomial Rings

III.6. Factorization in Polynomial Rings 1 Section III.6. Factorization in Polynomial Rings Note. We push several of the results in Section III.3 (such as divisibility, irreducibility, and unique factorization)

### 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

### Math 547, Exam 1 Information.

Math 547, Exam 1 Information. 2/10/10, LC 303B, 10:10-11:00. Exam 1 will be based on: Sections 5.1, 5.2, 5.3, 9.1; The corresponding assigned homework problems (see http://www.math.sc.edu/ boylan/sccourses/547sp10/547.html)

### Algebra Qualifying Exam Solutions. Thomas Goller

Algebra Qualifying Exam Solutions Thomas Goller September 4, 2 Contents Spring 2 2 2 Fall 2 8 3 Spring 2 3 4 Fall 29 7 5 Spring 29 2 6 Fall 28 25 Chapter Spring 2. The claim as stated is false. The identity

### Algebra Prelim Notes

Algebra Prelim Notes Eric Staron Summer 2007 1 Groups Define C G (A) = {g G gag 1 = a for all a A} to be the centralizer of A in G. In particular, this is the subset of G which commuted with every element

### 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

### 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