# COMMUTATIVE RINGS. Definition 3: A domain is a commutative ring R that satisfies the cancellation law for multiplication:

Size: px
Start display at page:

Download "COMMUTATIVE RINGS. Definition 3: A domain is a commutative ring R that satisfies the cancellation law for multiplication:"

Transcription

1 COMMUTATIVE RINGS Definition 1: A commutative ring R is a set with two operations, addition and multiplication, such that: (i) R is an abelian group under addition; (ii) ab = ba for all a, b R (commutative law); (iii) a(bc) = (ab)c for any a, b, c R (associative law); (iv) there is an element 1 R with 1 0 and with 1 a = a 1 = a for any a R; (v) a(b + c) = ab + ac for any a, b, c R (distributive law). Example: 1. Z, Q, R, C are commutative rings. 2. Z[i] = {a + bi : a, b Z} is a commutative ring. 3. {a + b 2 : a, b Z} is a commutative ring. 4. {a + b 3 2 : a, b Z} is not a ring. Moreover, {a + b 3 2 : a, b Q} is not a ring. 5. The set of all 2 2 matrices is a noncommutative ring. 6. Z m is a commutative ring. Theorem 1: Let R be a commutative ring. Then: (i) 0 a = 0 for any a R. (ii) If a is that number which, when added to a, gives 0, then ( 1)( a) = a for any a R. (iii) ( 1)a = a for any a R. Definition 2: A subset S of a commutative ring R is a subring of R if: (i) 1 S; (ii) if a, b S, then a b S; (iii) if a, b S, then ab S. Example: 1. Z is a subring of Q; Q is a subring of R; R is a subring of C; 2. Z[i] is a subring of C. Definition 3: A domain is a commutative ring R that satisfies the cancellation law for multiplication: if ca = cb and c 0, then a = b. 1

2 Example: 1. Z, Q, R, C are domains. 2. Z 4 is not a domain, since from [2][2] = [2][0] does not follow [2] = [0]. Theorem 2: A commutative ring R is a domain if and only if the product of any two nonzero elements of R is nonzero. Corollary: Z m is a domain if and only if m is a prime. Definition 4: FIELDS A field F is a commutative ring such that for any nonzero a F there exists a 1 F. Example: Q, R, C are fields. Z is not a field. Definition 5: An element u in a commutative ring R is called a unit if there exists u 1 R. Example: All nonzero elements of Q, R, and C are units. Definition 4 : A field F is a commutative ring in which every nonzero element of F is a unit. Theorem 3: Every field F is a domain. Remark: The converse of Theorem 3 is false. For example, Z is a domain, but not a field. Theorem 4: The commutative ring Z m is a field if and only if m is a prime. 2

3 Definition 6: If R is a commutative ring, then is called the ring of polynomials over R. If POLYNOMIALS R[x] = {s n x n s 1 x + s 0, s i R} f(x) = s n x n s 1 x + s 0 is a polynomial over R, then s n is called the leading coefficient and n is called the degree of f(x), denoted by deg(f). If s n = 1, then f(x) is called a monic polynomial. Theorem 5 (Division Algorithm): Assume that k is a field and that f(x), g(x) k[x] with f(x) 0. Then there are unique polynomials q(x), r(x) k[x] such that and either r(x) = 0 or deg(r) < deg(f). Example: g(x) = f(x)q(x) + r(x) Let g(x) = x 2 2x + 1, f(x) = x 3, then x 2 2x + 1 = (x 3)(x + 1) + 4. Definition 7: ROOTS OF POLYNOMIALS If f(x) k[x], where k is a field, then a root of f(x) in k is an element a k with f(a) = 0. Example: The polynomial f(x) = x R[x] has no roots in R. However, since f(x) is also from C[x], it has roots x 1 = i, x 2 = i in C. Lemma 1: Let k be a field and let f(x) k[x]. Then for any a k there exists q(x) k[x] such that Proof: By the Division Algorithm we have f(x) = (x a)q(x) + f(a). f(x) = (x a)q(x) + r(x), where deg(r) < deg(x a) = 1, and therefore deg(r) = 0, i.e. r(x) = r is a constant. So, f(x) = (x a)q(x) + r = f(a) = (a a)q(a) + r = r = r(x). 3

4 Lemma 2: Let k be a field and let f(x) k[x]. Then a k is a root of f(x) in k if and only if x a divides f(x) in k[x], i.e. there exists q(x) k[x] such that Proof: f(x) = (x a)q(x). = ) If a is a root of f(x) in k, then f(a) = 0, therefore by Lemma 1 we get =) If f(x) = (x a)q(x), then which means that a is a root of f(x). Theorem 6: f(x) = (x a)q(x) + f(a) = (x a)q(x). f(a) = (a a)q(a) = 0, Let k be a field and let f(x) k[x]. Let also f(x) has degree n. Then f(x) has at most n roots in k. Proof: We will use induction by n. If n = 0, then f(x) is a constant, and we are done. Suppose the theorem is true for some n = m 0. We prove it for n = m + 1. In fact, if f(x) has no roots in k, then we are done, since 0 < n. Otherwise, we may assume that there exists a k such that f(a) = 0. Then by Lemma 2 we have If there is a root b k with b a, then f(x) = q(x)(x a) and deg(q) = m. 0 = f(b) = q(b)(b a). Since b a 0, we have q(b) = 0, so that b is a root of q(x). Now deg(q) = m, so that the inductive hypothesis says that q(x) has at most m roots in k. Therefore, f(x) has at most m+1 roots in k. Corollary 1: Let k be a field and let f(x), g(x) k[x]. If deg(f) deg(g) = n and if f(a) = g(a) for n + 1 values a k, then f(x) = g(x). Proof: Suppose to the contrary that f(x) g(x). Then h(x) = f(x) g(x) 0 4

5 and deg(h) max(deg(f), deg(g)) = n. By hypothesis, there are n + 1 elements a k such that which contradicts Theorem 6.. Corollary 2: h(a) = f(a) g(a) = 0, Let k be a field and let f(x) k[x]. Let also f(x) has degree n and α 1,..., α n are distinct roots of f(x) in k. Then there exists c k such that Definition 8: f(x) = c(x α 1 )... (x α n ). GREATEST COMMON DIVISOR Let k be a field and let f(x), g(x) k[x]. A common divisor of f(x) and g(x) is a polynomial c(x) k[x] such that c(x) f(x) and c(x) g(x). The greatest common divisor (gcd) is a monic common divisor of the highest degree. Theorem 7: Let k be a field and let f(x), g(x) k[x]. Then their gcd is a linear combination of f(x) and g(x). Proof: Part I. Consider the following set: I = {s(x)f(x) + t(x)g(x) : s(x), t(x) k[x]}. Pick a polynomial d(x) I of the smallest degree. We have By the Division Algorithm we obtain d(x) = s(x)f(x) + t(x)g(x). f(x) = d(x)q(x) + r(x), where r(x) = 0 or deg(r) < deg(d). ( ) From ( ) and ( ) it follows that r(x) = f(x) d(x)q(x) = f(x) [s(x)f(x) + t(x)g(x)]q(x) = [1 s(x)q(x)]f(x) t(x)q(x)g(x) I. So, r(x) is a linear combination of f(x) and g(x) and deg(r) < deg(d). This is a contradiction. Therefore r(x) = 0, which means d(x) f(x). A similar argument shows that d(x) g(x). Part II. We now prove that d(x) is the greatest common divisor of f(x) and g(x). In fact, assume to the contrary that c(x) is the gcd of of f(x) and g(x). Then c(x) f(x) and c(x) g(x). Hence c(x) d(x) = s(x)f(x) + t(x)g(x). Since c(x) and d(x) are monic polynomials, it follows that deg(c) < deg(d), which is a contradiction. 5 ( )

6 EUCLIDEAN ALGORITHM Theorem 8 (Euclidean Algorithm ): Let k be a field and let f(x), g(x) k[x]. Then there is an algorithm for computing the gcd (f(x), g(x)). Proof: To find the greatest common divisor of two polynomials f(x) and g(x) we apply the algorithm, similar to the Euclidean Algorithm in Z : therefore the gcd (f(x), g(x)) = r n (x). Example: g(x) = f(x)q 1 (x) + r 1 (x) f(x) = r 1 (x)q 2 (x) + r 2 (x) r 1 (x) = r 2 (x)q 3 (x) + r 3 (x)... r n 2 (x) = r n 1 (x)q n (x) + r n (x) r n 1 (x) = r n (x)q n+1 (x), Find the gcd (x 3 + 1, x 5 + 1) and express it as a linear combination of x and x Solution: We have x = (x 3 + 1)x 2 + ( x 2 + 1) x = ( x 2 + 1)( x) + (x + 1) x = (x + 1)( x + 1) therefore the gcd (x 3 + 1, x 5 + 1) = x + 1. Finally, it follows that x+1 = (x 3 +1) ( x 2 +1)( x) = (x 3 +1) [x 5 +1 (x 3 +1)x 2 ]( x) = (x 3 +1)( x 3 +1)+(x 5 +1)x. 6

7 Definition 1: A commutative ring R is a set with two operations, addition and multiplication, such that: (i) R is an abelian group under addition; (ii) ab = ba for all a, b R (commutative law); (iii) a(bc) = (ab)c for any a, b, c R (associative law); (iv) there is an element 1 R with 1 0 and with 1 a = a 1 = a for any a R; (v) a(b+c) = ab+ac for any a, b, c R (distributive law).

8 Example: 1. Z, Q, R, C are commutative rings. 2. Z[i] = {a + bi : a, b Z} is a commutative ring. 3. {a+b 2 : a, b Z} is a commutative ring. 4. {a + b 3 2 : a, b Z} is not a ring. Moreover, {a + b 3 2 : a, b Q} is not a ring. 5. The set of all 2 2 matrices is a noncommutative ring. 6. Z m is a commutative ring.

9 Theorem 1: Let R be a commutative ring. Then: (i) 0 a = 0 for any a R. (ii) If a is that number which, when added to a, gives 0, then ( 1)( a) = a for any a R. (iii) ( 1)a = a for any a R.

10 Definition 2: A subset S of a commutative ring R is a subring of R if: (i) 1 S; (ii) if a, b S, then a b S; (iii) if a, b S, then ab S. Example: 1. Z is a subring of Q; Q is a subring of R; R is a subring of C; 2. Z[i] is a subring of C.

11 Definition 3: A domain is a commutative ring R that satisfies the cancellation law for multiplication: if ca = cb and c 0, then a = b. Example: 1. Z, Q, R, C are domains. 2. Z 4 is not a domain, since from [2][2] = [2][0] does not follow [2] = [0].

12 Theorem 2: A commutative ring R is a domain if and only if the product of any two nonzero elements of R is nonzero. Corollary: Z m is a domain if and only if m is a prime.

13 Definition 4: A field F is a commutative ring such that for any nonzero a F there exists a 1 F. Example: Q, R, C are fields. Z is not a field.

14 Definition 5: An element u in a commutative ring R is called a unit if there exists u 1 R. Example: All nonzero elements of Q, R, and C are units. Definition 4 : A field F is a commutative ring in which every nonzero element of F is a unit.

15 Theorem 3: Every field F is a domain. Remark: The converse of Theorem 3 is false. For example, Z is a domain, but not a field. Theorem 4: The commutative ring Z m is a field if and only if m is a prime.

16 Definition: If R is a commutative ring, then R[x] = {s n x n s 1 x + s 0, s i R} is called the ring of polynomials over R. If f(x) = s n x n s 1 x + s 0 is a polynomial over R, then s n is called the leading coefficient and n is called the degree of f(x), denoted by deg(f). If s n = 1, then f(x) is called a monic polynomial.

17 Theorem (Division Algorithm): Assume that k is a field and that f(x), g(x) k[x] with f(x) 0. Then there are unique polynomials q(x), r(x) k[x] with g(x) = f(x)q(x) + r(x) and either r(x) = 0 or deg(r) < deg(f). Example: Let g(x) = x 2 2x + 1, f(x) = x 3, then x 2 2x + 1 = (x 3)(x + 1) + 4.

18 Definition: If f(x) k[x], where k is a field, then a root of f(x) in k is an element a k with f(a) = 0. Example: The polynomial f(x) = x R[x] has no roots in R. However, since f(x) is also from C[x], it has roots x 1 = i, x 2 = i in C.

19 Lemma 1: Let k be a field and let f(x) k[x]. Then for any a k there exists q(x) k[x] such that f(x) = (x a)q(x) + f(a). Proof: By the Division Algorithm we have f(x) = (x a)q(x) + r(x), where deg(r) < deg(x a) = 1, and therefore deg(r) = 0, i.e. r(x) = r is a constant. So, f(x) = (x a)q(x)+r = f(a) = (a a)q

20 Lemma 2: Let k be a field and let f(x) k[x]. Then a k is a root of f(x) in k if and only if x a divides f(x) in k[x], i.e. there exists q(x) k[x] such that f(x) = (x a)q(x). Proof: = ) If a is a root of f(x) in k, then f(a) = 0, therefore by Lemma 1 we get f(x) = (x a)q(x)+f(a) = (x a)q(x). =) If f(x) = (x a)q(x), then f(a) = (a a)q(a) = 0, which means that a is a root of f(x).

21 Theorem 6: Let k be a field and let f(x) k[x]. Let also f(x) has degree n. Then f(x) has at most n roots in k. Proof: We will use induction by n. If n = 0, then f(x) is a constant, and we are done. Suppose the theorem is true for some n = m 0. We prove it for n = m + 1. In fact, if f(x) has no roots in k, then we are done, since 0 < n. Otherwise, we may assume that there exists a k such that f(a) = 0. Then by Lemma 2 we have f(x) = q(x)(x a) and deg(q) = m. If there is a root b k with b a, then 0 = f(b) = q(b)(b a). Since b a 0, we have q(b) = 0, so that b is a root of q(x). Now deg(q) =

22 m, so that the inductive hypothesis says that q(x) has at most m roots in k. Therefore, f(x) has at most m + 1 roots in k. Corollary 1: Let k be a field and let f(x), g(x) k[x]. If deg(f) deg(g) = n and if f(a) = g(a) for n + 1 values a k, then f(x) = g(x). Proof: Suppose to the contrary that f(x) g(x). Then and h(x) = f(x) g(x) 0 deg(h) max(deg(f), deg(g)) = n. By hypothesis, there are n + 1 elements a k such that h(a) = f(a) g(a) = 0,

23 which contradicts Theorem 6.. Corollary 2: Let k be a field and let f(x) k[x]. Let also f(x) has degree n and α 1,..., α n are distinct roots of f(x) in k. Then there exists c k such that f(x) = c(x α 1 )... (x α n ). GREATEST COMMON DIVISOR Definition 8: Let k be a field and let f(x), g(x) k[x]. A common divisor of f(x) and g(x) is a polynomial c(x) k[x] such that c(x) f(x) and c(x) g(x). The greatest comm (gcd) is a monic common divisor of the highest degree. Theorem 7: Let k be a field and let f(x), g(x) k[x]. Then their gcd is a linear combination

24 of f(x) and g(x). Proof: Part I. Consider the following set: I = {s(x)f(x)+t(x)g(x) : s(x), t(x) k[x]}. Pick a polynomial d(x) I of the smallest degree. We have d(x) = s(x)f(x) + t(x)g(x). ( ) By the Division Algorithm we obtain f(x) = d(x)q(x)+r(x), where r(x) = 0 o ( ) From ( ) and ( ) it follows that r(x) = f(x) d(x)q(x) = f(x) [s(x)f(x) + = [1 s(x)q(x)]f(x) So, r(x) is a linear combination of f(x) and g(x) and deg(r) < deg(d). This is a contradiction. Therefore r(x) = 0, which means d(x) f(x). A similar argument shows that d(x) g(x). Part II. We now prove that d(x) is the greatest common divisor of f(x) and g(x).

25 In fact, assume to the contrary that c(x) is the gcd of of f(x) and g(x). Then c(x) f(x) and c(x) g(x). Hence c(x) d(x) = s(x)f(x) + t(x)g(x). Since c(x) and d(x) are monic polynomials, it follows that deg(c) < deg(d), which is a contradiction. EUCLIDEAN ALGORITHM Theorem 8 (Euclidean Algorithm ): Let k be a field and let f(x), g(x) k[x]. Then there is an algorithm for computing the gcd (f(x), g(x)). Proof: To find the greatest common divisor of two polynomials f(x) and g(x) we apply the algorithm, similar to the Euclidean

26 Algorithm in Z : g(x) = f(x)q 1 (x) + r 1 (x) f(x) = r 1 (x)q 2 (x) + r 2 (x) r 1 (x) = r 2 (x)q 3 (x) + r 3 (x)... r n 2 (x) = r n 1 (x)q n (x) + r n (x) r n 1 (x) = r n (x)q n+1 (x), therefore the gcd (f(x), g(x)) = r n (x). Example: Find the gcd (x 3 +1, x 5 +1) and express it as a linear combination of x and x Solution: We have x = (x 3 + 1)x 2 + ( x 2 + 1) x = ( x 2 + 1)( x) + (x + 1) x = (x + 1)( x + 1) therefore the gcd (x 3 +1, x 5 +1) = x+1.

27 Finally, it follows that x+1 = (x 3 +1) ( x 2 +1)( x) = (x 3 +1) [x 5

### Further linear algebra. Chapter II. Polynomials.

Further linear algebra. Chapter II. Polynomials. Andrei Yafaev 1 Definitions. In this chapter we consider a field k. Recall that examples of felds include Q, R, C, F p where p is prime. A polynomial is

### 1. Algebra 1.5. Polynomial Rings

1. ALGEBRA 19 1. Algebra 1.5. Polynomial Rings Lemma 1.5.1 Let R and S be rings with identity element. If R > 1 and S > 1, then R S contains zero divisors. Proof. The two elements (1, 0) and (0, 1) are

### Notes 6: Polynomials in One Variable

Notes 6: Polynomials in One Variable Definition. Let f(x) = b 0 x n + b x n + + b n be a polynomial of degree n, so b 0 0. The leading term of f is LT (f) = b 0 x n. We begin by analyzing the long division

### 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)

### 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)

### 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

### Polynomial Rings : Linear Algebra Notes

Polynomial Rings : Linear Algebra Notes Satya Mandal September 27, 2005 1 Section 1: Basics Definition 1.1 A nonempty set R is said to be a ring if the following are satisfied: 1. R has two binary operations,

### Polynomial Rings. i=0

Polynomial Rings 4-15-2018 If R is a ring, the ring of polynomials in x with coefficients in R is denoted R[x]. It consists of all formal sums a i x i. Here a i = 0 for all but finitely many values of

### CHAPTER 10: POLYNOMIALS (DRAFT)

CHAPTER 10: POLYNOMIALS (DRAFT) LECTURE NOTES FOR MATH 378 (CSUSM, SPRING 2009). WAYNE AITKEN The material in this chapter is fairly informal. Unlike earlier chapters, no attempt is made to rigorously

### (Rgs) Rings Math 683L (Summer 2003)

(Rgs) Rings Math 683L (Summer 2003) We will first summarise the general results that we will need from the theory of rings. A unital ring, R, is a set equipped with two binary operations + and such that

### Polynomial Rings. (Last Updated: December 8, 2017)

Polynomial Rings (Last Updated: December 8, 2017) These notes are derived primarily from Abstract Algebra, Theory and Applications by Thomas Judson (16ed). Most of this material is drawn from Chapters

### 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)

### U + V = (U V ) (V U), UV = U V.

Solution of Some Homework Problems (3.1) Prove that a commutative ring R has a unique 1. Proof: Let 1 R and 1 R be two multiplicative identities of R. Then since 1 R is an identity, 1 R = 1 R 1 R. Since

### 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

### (a + b)c = ac + bc and a(b + c) = ab + ac.

2. R I N G S A N D P O LY N O M I A L S The study of vector spaces and linear maps between them naturally leads us to the study of rings, in particular the ring of polynomials F[x] and the ring of (n n)-matrices

### Algebra Review 2. 1 Fields. A field is an extension of the concept of a group.

Algebra Review 2 1 Fields A field is an extension of the concept of a group. Definition 1. A field (F, +,, 0 F, 1 F ) is a set F together with two binary operations (+, ) on F such that the following conditions

### 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

### 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

### 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,

### Class Notes; Week 7, 2/26/2016

Class Notes; Week 7, 2/26/2016 Day 18 This Time Section 3.3 Isomorphism and Homomorphism [0], [2], [4] in Z 6 + 0 4 2 0 0 4 2 4 4 2 0 2 2 0 4 * 0 4 2 0 0 0 0 4 0 4 2 2 0 2 4 So {[0], [2], [4]} is a subring.

### Moreover this binary operation satisfies the following properties

Contents 1 Algebraic structures 1 1.1 Group........................................... 1 1.1.1 Definitions and examples............................. 1 1.1.2 Subgroup.....................................

### Section 19 Integral domains

Section 19 Integral domains Instructor: Yifan Yang Spring 2007 Observation and motivation There are rings in which ab = 0 implies a = 0 or b = 0 For examples, Z, Q, R, C, and Z[x] are all such rings There

### Rings. Chapter Definitions and Examples

Chapter 5 Rings Nothing proves more clearly that the mind seeks truth, and nothing reflects more glory upon it, than the delight it takes, sometimes in spite of itself, in the driest and thorniest researches

### Introduction to the Galois Correspondence Maureen Fenrick

Introduction to the Galois Correspondence Maureen Fenrick A Primer on the Integers 1. Division algorithm: given n and d 0 one can find q and r such that 0 r < d and n = qd + r. 2. d n when r = 0. 3. Greatest

### 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

### Math 581: Skeleton Notes

Math 581: Skeleton Notes Bart Snapp June 7, 2010 Chapter 1 Rings Definition 1 A ring is a set R with two operations: + called addition and called multiplication such that: (i) (R,+) is an abelian group

### Coding Theory ( Mathematical Background I)

N.L.Manev, Lectures on Coding Theory (Maths I) p. 1/18 Coding Theory ( Mathematical Background I) Lector: Nikolai L. Manev Institute of Mathematics and Informatics, Sofia, Bulgaria N.L.Manev, Lectures

### Part IX ( 45-47) Factorization

Part IX ( 45-47) Factorization Satya Mandal University of Kansas, Lawrence KS 66045 USA January 22 45 Unique Factorization Domain (UFD) Abstract We prove evey PID is an UFD. We also prove if D is a UFD,

### CHAPTER 14. Ideals and Factor Rings

CHAPTER 14 Ideals and Factor Rings Ideals Definition (Ideal). A subring A of a ring R is called a (two-sided) ideal of R if for every r 2 R and every a 2 A, ra 2 A and ar 2 A. Note. (1) A absorbs elements

### Congruences and Residue Class Rings

Congruences and Residue Class Rings (Chapter 2 of J. A. Buchmann, Introduction to Cryptography, 2nd Ed., 2004) Shoichi Hirose Faculty of Engineering, University of Fukui S. Hirose (U. Fukui) Congruences

### a + bi by sending α = a + bi to a 2 + b 2. To see properties (1) and (2), it helps to think of complex numbers in polar coordinates:

5. Types of domains It turns out that in number theory the fact that certain rings have unique factorisation has very strong arithmetic consequences. We first write down some definitions. Definition 5.1.

### 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. PIDs Definition 1 A principal ideal domain (PID) is an integral

Algebra Qualifying Exam August 2001 Do all 5 problems. 1. Let G be afinite group of order 504 = 23 32 7. a. Show that G cannot be isomorphic to a subgroup of the alternating group Alt 7. (5 points) b.

### 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).

### 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

### D-MATH Algebra I HS18 Prof. Rahul Pandharipande. Solution 6. Unique Factorization Domains

D-MATH Algebra I HS18 Prof. Rahul Pandharipande Solution 6 Unique Factorization Domains 1. Let R be a UFD. Let that a, b R be coprime elements (that is, gcd(a, b) R ) and c R. Suppose that a c and b c.

### The Cayley-Hamilton Theorem and Minimal Polynomials

Math 5327 Spring 208 The Cayley-Hamilton Theorem and Minimal Polynomials Here are some notes on the Cayley-Hamilton Theorem, with a few extras thrown in. I will start with a proof of the Cayley-Hamilton

### 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

### Rings. Chapter Homomorphisms and ideals

Chapter 2 Rings This chapter should be at least in part a review of stuff you ve seen before. Roughly it is covered in Rotman chapter 3 and sections 6.1 and 6.2. You should *know* well all the material

### 36 Rings of fractions

36 Rings of fractions Recall. If R is a PID then R is a UFD. In particular Z is a UFD if F is a field then F[x] is a UFD. Goal. If R is a UFD then so is R[x]. Idea of proof. 1) Find an embedding R F where

### 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

### Finite Fields. Sophie Huczynska. Semester 2, Academic Year

Finite Fields Sophie Huczynska Semester 2, Academic Year 2005-06 2 Chapter 1. Introduction Finite fields is a branch of mathematics which has come to the fore in the last 50 years due to its numerous applications,

### Linear Cyclic Codes. Polynomial Word 1 + x + x x 4 + x 5 + x x + x f(x) = q(x)h(x) + r(x),

Coding Theory Massoud Malek Linear Cyclic Codes Polynomial and Words A polynomial of degree n over IK is a polynomial p(x) = a 0 + a 1 + + a n 1 x n 1 + a n x n, where the coefficients a 1, a 2,, a n are

### Math 4310 Solutions to homework 7 Due 10/27/16

Math 4310 Solutions to homework 7 Due 10/27/16 1. Find the gcd of x 3 + x 2 + x + 1 and x 5 + 2x 3 + x 2 + x + 1 in Rx. Use the Euclidean algorithm: x 5 + 2x 3 + x 2 + x + 1 = (x 3 + x 2 + x + 1)(x 2 x

### Chapter 5: The Integers

c Dr Oksana Shatalov, Fall 2014 1 Chapter 5: The Integers 5.1: Axioms and Basic Properties Operations on the set of integers, Z: addition and multiplication with the following properties: A1. Addition

### 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

### 8 Appendix: Polynomial Rings

8 Appendix: Polynomial Rings Throughout we suppose, unless otherwise specified, that R is a commutative ring. 8.1 (Largely) a reminder about polynomials A polynomial in the indeterminate X with coefficients

### 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

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,

### Introduction to finite fields

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

### Chapter 14: Divisibility and factorization

Chapter 14: Divisibility and factorization Matthew Macauley Department of Mathematical Sciences Clemson University http://www.math.clemson.edu/~macaule/ Math 4120, Summer I 2014 M. Macauley (Clemson) Chapter

### NOTES ON SIMPLE NUMBER THEORY

NOTES ON SIMPLE NUMBER THEORY DAMIEN PITMAN 1. Definitions & Theorems Definition: We say d divides m iff d is positive integer and m is an integer and there is an integer q such that m = dq. In this case,

### Rings. EE 387, Notes 7, Handout #10

Rings EE 387, Notes 7, Handout #10 Definition: A ring is a set R with binary operations, + and, that satisfy the following axioms: 1. (R, +) is a commutative group (five axioms) 2. Associative law for

### 1/30: Polynomials over Z/n.

1/30: Polynomials over Z/n. Last time to establish the existence of primitive roots we rely on the following key lemma: Lemma 6.1. Let s > 0 be an integer with s p 1, then we have #{α Z/pZ α s = 1} = s.

### NOTES ON INTEGERS. 1. Integers

NOTES ON INTEGERS STEVEN DALE CUTKOSKY The integers 1. Integers Z = {, 3, 2, 1, 0, 1, 2, 3, } have addition and multiplication which satisfy familar rules. They are ordered (m < n if m is less than n).

### 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

### Section IV.23. Factorizations of Polynomials over a Field

IV.23 Factorizations of Polynomials 1 Section IV.23. Factorizations of Polynomials over a Field Note. Our experience with classical algebra tells us that finding the zeros of a polynomial is equivalent

### D-MATH Algebra I HS 2013 Prof. Brent Doran. Exercise 11. Rings: definitions, units, zero divisors, polynomial rings

D-MATH Algebra I HS 2013 Prof. Brent Doran Exercise 11 Rings: definitions, units, zero divisors, polynomial rings 1. Show that the matrices M(n n, C) form a noncommutative ring. What are the units of M(n

### Notes for Math 345. Dan Singer Minnesota State University, Mankato. August 23, 2006

Notes for Math 345 Dan Singer Minnesota State University, Mankato August 23, 2006 Preliminaries 1. Read the To The Student section pp. xvi-xvii and the Thematic Table of Contents. 2. Read Appendix A: Logic

### Finite Fields: An introduction through exercises Jonathan Buss Spring 2014

Finite Fields: An introduction through exercises Jonathan Buss Spring 2014 A typical course in abstract algebra starts with groups, and then moves on to rings, vector spaces, fields, etc. This sequence

### Algebra I: Chapter 7 A Brief Introduction to Theory of Rings 7.1 Rings, Homomorphisms and Ideals.

Notes: c F.P. Greenleaf 2003-2014 v43-s14rings.tex, version 4/5/2014 Algebra I: Chapter 7 A Brief Introduction to Theory of Rings 7.1 Rings, Homomorphisms and Ideals. A ring is a set R equipped with two

### 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

### Lecture 7: Polynomial rings

Lecture 7: Polynomial rings Rajat Mittal IIT Kanpur You have seen polynomials many a times till now. The purpose of this lecture is to give a formal treatment to constructing polynomials and the rules

### AN INTRODUCTION TO GALOIS THEORY

AN INTRODUCTION TO GALOIS THEORY STEVEN DALE CUTKOSKY In these notes we consider the problem of constructing the roots of a polynomial. Suppose that F is a subfield of the complex numbers, and f(x) is

### Math 2070BC Term 2 Weeks 1 13 Lecture Notes

Math 2070BC 2017 18 Term 2 Weeks 1 13 Lecture Notes Keywords: group operation multiplication associative identity element inverse commutative abelian group Special Linear Group order infinite order cyclic

### 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.

### A few exercises. 1. Show that f(x) = x 4 x 2 +1 is irreducible in Q[x]. Find its irreducible factorization in

A few exercises 1. Show that f(x) = x 4 x 2 +1 is irreducible in Q[x]. Find its irreducible factorization in F 2 [x]. solution. Since f(x) is a primitive polynomial in Z[x], by Gauss lemma it is enough

### 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

### Lecture 2: Gröbner Basis and SAGBI Basis

Lecture 2: Gröbner Basis and SAGBI Basis Mohammed Tessema Suppose we have a graph. Suppose we color the graph s vertices with 3 colors so that if the vertices are adjacent they are not the same colors.

### Rings If R is a commutative ring, a zero divisor is a nonzero element x such that xy = 0 for some nonzero element y R.

Rings 10-26-2008 A ring is an abelian group R with binary operation + ( addition ), together with a second binary operation ( multiplication ). Multiplication must be associative, and must distribute over

### Maximal Subrings of small C-algebras

Maximal Subrings of small C-algebras Student: Klevis Ymeri School of Engineering and Sciences Jacobs University Bremen Advisor: Dr. Stefan Maubach December 22, 2013 Abstract If k is a field, C is a k-algebra,

### Homework #2 solutions Due: June 15, 2012

All of the following exercises are based on the material in the handout on integers found on the class website. 1. Find d = gcd(475, 385) and express it as a linear combination of 475 and 385. That is

### THROUGH THE FIELDS AND FAR AWAY

THROUGH THE FIELDS AND FAR AWAY JONATHAN TAYLOR I d like to thank Prof. Stephen Donkin for helping me come up with the topic of my project and also guiding me through its various complications. Contents

### a b (mod m) : m b a with a,b,c,d real and ad bc 0 forms a group, again under the composition as operation.

Homework for UTK M351 Algebra I Fall 2013, Jochen Denzler, MWF 10:10 11:00 Each part separately graded on a [0/1/2] scale. Problem 1: Recalling the field axioms from class, prove for any field F (i.e.,

### Skew Cyclic Codes Of Arbitrary Length

Skew Cyclic Codes Of Arbitrary Length Irfan Siap Department of Mathematics, Adıyaman University, Adıyaman, TURKEY, isiap@adiyaman.edu.tr Taher Abualrub Department of Mathematics and Statistics, American

### Contents. 4 Arithmetic and Unique Factorization in Integral Domains. 4.1 Euclidean Domains and Principal Ideal Domains

Ring Theory (part 4): Arithmetic and Unique Factorization in Integral Domains (by Evan Dummit, 018, v. 1.00) Contents 4 Arithmetic and Unique Factorization in Integral Domains 1 4.1 Euclidean Domains and

### Prime Rational Functions and Integral Polynomials. Jesse Larone, Bachelor of Science. Mathematics and Statistics

Prime Rational Functions and Integral Polynomials Jesse Larone, Bachelor of Science Mathematics and Statistics Submitted in partial fulfillment of the requirements for the degree of Master of Science Faculty

### Linear Cyclic Codes. Polynomial Word 1 + x + x x 4 + x 5 + x x + x

Coding Theory Massoud Malek Linear Cyclic Codes Polynomial and Words A polynomial of degree n over IK is a polynomial p(x) = a 0 + a 1 x + + a n 1 x n 1 + a n x n, where the coefficients a 0, a 1, a 2,,

### Lecture 7.5: Euclidean domains and algebraic integers

Lecture 7.5: Euclidean domains and algebraic integers Matthew Macauley Department of Mathematical Sciences Clemson University http://www.math.clemson.edu/~macaule/ Math 4120, Modern Algebra M. Macauley

### 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,

### where c R and the content of f is one. 1

9. Gauss Lemma Obviously it would be nice to have some more general methods of proving that a given polynomial is irreducible. The first is rather beautiful and due to Gauss. The basic idea is as follows.

### Lecture Notes Math 371: Algebra (Fall 2006) by Nathanael Leedom Ackerman

Lecture Notes Math 371: Algebra (Fall 2006) by Nathanael Leedom Ackerman October 31, 2006 TALK SLOWLY AND WRITE NEATLY!! 1 0.1 Symbolic Adjunction of Roots When dealing with subfields of C it is easy to

### ANALYSIS OF SMALL GROUPS

ANALYSIS OF SMALL GROUPS 1. Big Enough Subgroups are Normal Proposition 1.1. Let G be a finite group, and let q be the smallest prime divisor of G. Let N G be a subgroup of index q. Then N is a normal

### CANONICAL FORMS FOR LINEAR TRANSFORMATIONS AND MATRICES. D. Katz

CANONICAL FORMS FOR LINEAR TRANSFORMATIONS AND MATRICES D. Katz The purpose of this note is to present the rational canonical form and Jordan canonical form theorems for my M790 class. Throughout, we fix

### 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

### Finite Fields and Error-Correcting Codes

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

### (Inv) Computing Invariant Factors Math 683L (Summer 2003)

(Inv) Computing Invariant Factors Math 683L (Summer 23) We have two big results (stated in (Can2) and (Can3)) concerning the behaviour of a single linear transformation T of a vector space V In particular,

### Lecture Notes Math 371: Algebra (Fall 2006) by Nathanael Leedom Ackerman

Lecture Notes Math 371: Algebra (Fall 2006) by Nathanael Leedom Ackerman October 17, 2006 TALK SLOWLY AND WRITE NEATLY!! 1 0.1 Factorization 0.1.1 Factorization of Integers and Polynomials Now we are going

### Discrete valuation rings. Suppose F is a field. A discrete valuation on F is a function v : F {0} Z such that:

Discrete valuation rings Suppose F is a field. A discrete valuation on F is a function v : F {0} Z such that: 1. v is surjective. 2. v(ab) = v(a) + v(b). 3. v(a + b) min(v(a), v(b)) if a + b 0. Proposition:

### 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

### Basic Algebra. Final Version, August, 2006 For Publication by Birkhäuser Boston Along with a Companion Volume Advanced Algebra In the Series

Basic Algebra Final Version, August, 2006 For Publication by Birkhäuser Boston Along with a Companion Volume Advanced Algebra In the Series Cornerstones Selected Pages from Chapter I: pp. 1 15 Anthony

### Elementary Algebra Chinese Remainder Theorem Euclidean Algorithm

Elementary Algebra Chinese Remainder Theorem Euclidean Algorithm April 11, 2010 1 Algebra We start by discussing algebraic structures and their properties. This is presented in more depth than what we

### MTH 401: Fields and Galois Theory

MTH 401: Fields and Galois Theory Semester 1, 2016-2017 Dr. Prahlad Vaidyanathan Contents Classical Algebra 3 I. Polynomials 6 1. Ring Theory.................................. 6 2. Polynomial Rings...............................

### 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

### LECTURE NOTES IN CRYPTOGRAPHY

1 LECTURE NOTES IN CRYPTOGRAPHY Thomas Johansson 2005/2006 c Thomas Johansson 2006 2 Chapter 1 Abstract algebra and Number theory Before we start the treatment of cryptography we need to review some basic

### 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

### Definitions. Notations. Injective, Surjective and Bijective. Divides. Cartesian Product. Relations. Equivalence Relations

Page 1 Definitions Tuesday, May 8, 2018 12:23 AM Notations " " means "equals, by definition" the set of all real numbers the set of integers Denote a function from a set to a set by Denote the image of

### 50 Algebraic Extensions

50 Algebraic Extensions Let E/K be a field extension and let a E be algebraic over K. Then there is a nonzero polynomial f in K[x] such that f(a) = 0. Hence the subset A = {f K[x]: f(a) = 0} of K[x] does