# Integral Domains. As always in this course, a ring R is understood to be a commutative ring with unity.

Save this PDF as:

Size: px
Start display at page:

Download "Integral Domains. As always in this course, a ring R is understood to be a commutative ring with unity."

## Transcription

1 Integral Domains As always in this course, a ring R is understood to be a commutative ring with unity. 1 First definitions and properties Definition 1.1. Let R be a ring. A divisor of zero or zero divisor in R is an element r 0, such that there exists an s R with s 0 and rs = 0. Example: in Z/6Z, 0 = 2 3, hence both 2 and 3 are divisors of zero. One way to find divisors of zero is as follows: Definition 1.2. Let R be a ring. A nilpotent element of R is an element r, such that there exists an n N such that r n = 0. Note that 0 is allowed to be nilpotent. Lemma 1.3. Let R be a ring and let r R be nilpotent. If r 0, then r is a zero divisor. Proof. The set of n N such that r n = 0 is nonempty, so let m be the smallest such natural number. By assumption, r 0, hence m > 1. Then 0 = r r m 1, where m 1 1 and hence m 1 N. Since m 1 < m, r m 1 0. Hence r r m 1 = 0, with neither factor equal to 0, so that r is a divisor of zero. Example: in Z/16Z, 0 = 2 4 = 2 2 3, hence 2 is a divisor of zero. But in Z/6Z, neither 2 nor 3 is nilpotent, so there are examples of divisors of zero which are not nilpotent. Definition 1.4. A ring R is an integral domain if R {0}, or equivalently 1 0, and there do not exist zero divisors in R. Equivalently, a nonzero ring R is an integral domain if, for all r, s R with r 0, s 0, the product rs 0. 1

2 Definition 1.5. Let R be a ring. The cancellation law holds in R if, for all r, s, t R such that t 0, if tr = ts, then r = s. Lemma 1.6. A ring R {0} is an integral domain the cancellation law holds in R. Proof. = : if tr = ts and t 0, then tr ts = t(r s) = 0. Since t 0 and R is an integral domain, r s = 0 so that r = s. = : Suppose that rs = 0. We must show that either r or s is 0. If r 0, then apply cancellation to rs = 0 = r0 to conclude that s = 0. The following are examples of integral domains: 1. A field is an integral domain. In fact, if F is a field, r, s F with r 0 and rs = 0, then 0 = r 1 0 = r 1 (rs) = (r 1 r)s = 1s = s. Hence s = 0. (Recall that 1 0 in a field, so the condition that F 0 is automatic.) This argument also shows that, in any ring R 0, a unit is not a zero divisor. 2. If S is an integral domain and R S, then R is an integral domain. In particular, a subring of a field is an integral domain. (Note that, if R S and 1 0 in S, then 1 0 in R.) Examples: any subring of R or C is an integral domain. Thus for example Z[ 2], Q( 2) are integral domains. 3. For n N, the ring Z/nZ is an integral domain n is prime. In fact, we have already seen that Z/pZ = F p is a field, hence an integral domain. Conversely, if n is not prime, say n = ab with a, b N, then, as elements of Z/nZ, a 0, b 0, but ab = n = 0. Hence Z/nZ is not an integral domain. 4. If R is an integral domain, then, as we shall see in a minute, R[x] is an integral domain. Hence, by induction, if F is a field, F [x 1,..., x n ] is an integral domain, as is Z[x 1,..., x n ]. To prove the last statement (4) above, we show in fact: Lemma 1.7. Let R be an integral domain. Then, if f, g R[x] are both nonzero, then fg 0 and deg(fg) = deg f + deg g. Proof. Let d = deg f and e = deg g. Then f = d i=0 a ix i and g = e j=0 b jx j with a d, b e 0. Since a d b e 0, the leading term of fg is a d b e x d+e. Hence fg 0 and deg(fg) = d + e = deg f + deg g. 2

3 Corollary 1.8. Let R be an integral domain. Then the group of units (R[x]) in the polynomial ring R[x] is just the group of units R in R (viewed as constant polynomials). Proof. Clearly, if u is a unit in R, then it is a unit in R[x], so that R (R[x]). Conversely, if f (R[x]), then there exists a g R[x] such that fg = 1. Clearly, neither f nor g is the zero polynomial, and hence 0 = deg 1 = deg(fg) = deg f + deg g. Thus, deg f = deg g = 0, so that f, g are elements of R and clearly they are units in R. Hence f R, so that (R[x]) R. It follows that (R[x]) = R. The corollary fails if the ring R has nonzero nilpotent elements. example, in (Z/4Z)[x], For (1 + 2x)(1 + 2x) = (1 + 2x) 2 = 1 + 4x + 4x 2 = 1, so that 1 + 2x is a unit in (Z/4Z)[x]. Finally, we note the following: Proposition 1.9. A finite integral domain R is a field. Proof. Suppose r R with r 0. The elements 1 = r 0, r, r 2,... cannot all be different, since otherwise R would be infinite. Hence there exist 0 n < m with r n = r m. Writing m = n + k with k 1, we see that r n = r m = r n+k = r n r k. By induction, since R is an integral domain and r 0, r n 0 for all n 0. Applying cancellation to r n = r n 1 = r n r k gives r k = 1. Finally since r k = r r k 1, we see that r is invertible, with r 1 = r k 1. 2 The characteristic of an integral domain Let R be an integral domain. As we have seen in the homework, the function f : Z R defined by f(n) = n 1 is a ring homomorphism and its image is 1, the cyclic subgroup of (R, +) generated by 1. There are two possibilities: (1) 1 has finite order n, in which case 1 = Z/nZ, or (2) 1 has infinite order, in which case 1 = Z. Proposition 2.1. With notation as above, (i) If 1 has finite order n, then n = p is a prime number, and every nonzero element of R has order p. 3

4 (ii) If 1 has infinite order, then every nonzero element of R has infinite order. Proof. (i) By definition, n is the smallest positive integer such that n 1 = 0. If n = ab, where a, b N, then (using homework) 0 = n 1 = (a 1)(b 1). Since R is an integral domain, one of a 1, b 1 is 0. Say a 1 = 0. Then a n, but since a divides n, we must have a = n. Hence in every factorization of n, one of the factors is n, so by definition n is a prime p. Moreover, for every r R, p r = (p 1)r = 0, so that the order of r divides p. If r 0, then its order is greater than 1, hence must equal p. (ii) Let r R, and suppose that r has (finite) order n N, so that n r = 0. As in the proof of (i), write n r = (n 1)r. Since 1 has infinite order, n 1 0, and hence r = 0. Thus, if r 0, then n r 0 for every n N. Thus r has infinite order. Definition 2.2. Let R be an integral domain. If 1 R has infinite order, we say that the characteristic of R is zero. If 1 R has finite order, necessarily a prime p, we say that the characteristic of R is p. In either case we write char R for the characteristic of R, so that char R is either 0 or a prime number. Examples: Clearly, the characteristic of Z is 0. Also, if R and S are integral domains with R S, then clearly char R = char S. Thus char Q, char R, char C, char Q( 2), etc. are all 0. On the other hand, the characteristic of F p = Z/pZ is p. Thus, the characteristic of F p [x] is also p, so that F p [x] is an example of an infinite integral domain with characteristic p 0, and F p [x] is not a field. (Note however that a finite integral domain, which automatically has positive characteristic, is always a field.) 3 The field of quotients of an integral domain We first begin with some general remarks about fields. If F is a field and r, s F with s 0, we write (as usual) rs 1 = r/s. Note that r/s = t/w rw = st, since rw = (sw)r/s and st = (sw)t/w, and by cancellation. Then the laws for adding and multiplying fractions are forced by associativity and distributivity in F : for example, r/s + t/w = rs 1 + tw 1 = (rw)(sw) 1 + (ts)(sw) 1 = (rw + ts)(sw) 1 = (rw + ts)/(sw). 4

5 Now suppose that R is an integral domain. We would like to enlarge R to a field, in much the same way that we enlarge Z to Q. To this end, we construct a set whose elements are fractions r/s with r, s R and s 0. Two fractions r/s and t/w are identified if, as in the discussion above for fields, rw = st. The correct way to say this is via equivalence classes: on the set R (R {0}), define the relation on pairs (r, s) by: (r, s) (t, w) rw = st. Lemma 3.1. is an equivalence relation. Proof. We must show is reflexive, symmetric, and transitive. Reflexive: (r, s) (r, s) rs = sr, which holds since R is commutative. Symmetric: (r, s) (t, w) rw = st, in which case ts = wr, hence (t, w) (r, s). Transitive (it is here that we use the fact that R is an integral domain): suppose that (r, s) (t, w) and that (t, w) (u, v), with s, w, v 0. By definiton rw = st and tv = wu. Then rwv = stv = swu, hence w(rv) = w(su). Since w 0 and R is an integral domain, rv = su, hence (r, s) (u, v). Thus is transitive. Define Q(R), the field of quotients of R, to be the set of equivalence classes (R (R {0}))/. Next we need operations of addition and multiplication on Q(R). As is usually the case with equivalence relations, we define these operations by defining them on representative of equivalence classes, and then check that the operations are in fact well-defined. Define [(r, s)] + [(t, w)] = [(rw + st, sw)]; [(r, s)] [(t, w)] = [(rt, sw)]. Lemma 3.2. Let and Q(R) be as above. (i) The operations of addition and multiplication are well-defined. (ii) (Q(R), +, ) is a field. (iii) The function ρ: R Q(R) defined by ρ(r) = [(r, 1)] is an injective homomorphism. Proof. These are all straightforward if sometimes tedious calculations. For example, to see (i), suppose that (r, s) (r, s ). We shall show that (rw + st, sw) (r w + s t, s w) and that (rt, sw) (r t, s w). By definition, rs = sr. Then (rw + st)(s w) = rws w + sts w = (rs )(w 2 ) + (ss )(tw) = (r s)(w 2 ) + (ss )(tw) = (r w + s t)(sw). 5

6 Hence (rw + st, sw) (r w + s t, s w). Moreover, (rt)(s w) = (rs )(tw) = (r s)(tw) = (r t)(sw). Hence (rt, sw) (r t, s w). Similarly, if (t, w) (t, w ), then (rw+st, sw) (rw + st, sw ) and that (rt, sw) (rt, sw ). To see (ii), we must show first that (Q(R), +) is an abelian group and that multiplication is associative, commutative, and distributes over addition. These are all completely straightforward if long computations. Note that [(0, 1)] = [(0, r)] is the additive identity, that [(r, s)] [(0, 1)] r = 0, and that [(1, 1)] = [(r, r)] is a multiplicative identity. Finally, if [(r, s)] [(0, 1)], so that r 0, then [(s, r)] Q(R) and [(r, s)][(s, r)] = [(rs, rs)] = [(1, 1)]. Thus Q(R) is a field. To see (iii), defining ρ(r) = [(r, 1)], we see that ρ(r + s) = [(r + s, 1)] = [(r, 1)] + [(s, 1)] = ρ(r) + ρ(s); ρ(rs) = [(rs, 1)] = [(r, 1)][(s, 1)] = ρ(r)ρ(s). Thus ρ is a homomorphism. It is injective since ρ(r) = ρ(s) (r, 1) (s, 1) r = s. From now on we write [(r, s)] as r/s or as rs 1 and identify r R with its image r/1 Q(R). In this way we view R as a subring of Q(R). Example: 1) let F be a field and F [x] the polynomial ring with coefficients in F. Then we denote Q(F [x]) by F (x). By definition, the elements of F (x) are quotients f/g, where f, g are polynomials with coefficients in F. We call F (x) the field of rational functions with coefficients in F. In particular, taking F = F p, the field of rational functions F p (x) is an example of an infinite field (since it contains a subring isomorphic to the polynomial ring F p [x], which is infinite), whose characteristic is p > 0. 2) If R = F is already a field, then (r, s) (rs 1, 1). Thus the injective homomorphism ρ is also surjective, hence an isomorphism, so that Q(R) = R. Remark: In the field of quotients Q = Q(Z) of Z, we can always put a fraction n/m in lowest terms, i.e. we can assume that gcd(n, m) = 1. This says that the equivalence class [(n, m)] has a best representative, if we require in addition, say, that m > 0. Such a choice depends on results about factorization in Z, and is not possible in a general integral domain. Finally, we show that Q(R) has a very general property with respect to injective homomorphisms from R to a field: 6

7 Proposition 3.3. Let R be an integral domain, F a field, and φ: R F be an injective homomorphism. Then there exists a unique injective homomorphism φ: Q(R) F such that φ(r/1) = φ(r). Finally, if every element of F is of the form φ(r)/φ(s) for some r, s R with s 0, then φ: Q(R) F is an isomorphism, and in particular Q(R) = F. Proof. Clearly, if φ exists, then we must have φ(r/s) = φ(rs 1 ) = φ(r) φ(s 1 ) = φ(r) φ(s) 1 = φ(r)/ φ(s) = φ(r)/φ(s). This proves that φ is unique, if it exists. Conversely, we try to define φ by the formula φ(r/s) = φ(r)/φ(s). Here r/s is shorthand for the equivalence class [(r, s)] Q(R), and the fraction φ(r)/φ(s) = φ(r)/φ(s) 1 is well-defined in F since, as φ is injective and s 0, φ(s) 0. We must first show that φ is well-defined, i.e. independent of the choice of representative (r, s) [(r, s)]. Choosing another representative (r, s ) [(r, s)], we have by definition rs = r s. Hence φ(rs ) = φ(r)φ(s ) = φ(r s) = φ(r )φ(s). Dividing by φ(s)φ(s ) gives φ(r)/φ(s) = φ(r)φ(s )/φ(s)φ(s ) = φ(r )φ(s)/φ(s)φ(s ) = φ(r )/φ(s ). Hence φ(r/s) = φ(r)/φ(s) is independent of the choice of representative (r, s) [(r, s)]. It is then straightforward to check that φ is a (ring) isomorphism. To see that it is injective, suppose that φ(r/s) = φ(r /s ). Then φ(r)/φ(s) = φ(r )/φ(s ), and hence φ(rs ) = φ(r)φ(s ) = φ(r )φ(s) = φ(r s). Since φ is injective, rs = r s, and hence r/s = r /s. Thus φ is injective. Finally, if every element of F is of the form φ(r)/φ(s) for some r, s R with s 0, then φ is also surjective, hence an isomorphism. Here is a typical way we might apply the proposition: Lemma 3.4. Let R be an integral domain with field of quotients Q(R). Then Q(R[x]), the field of quotients of the integral domain R[x], is isomorphic to Q(R)(x), the field of rational functions with coefficients in Q(R). Proof. Since R is isomorphic to a subring of Q(R), there is a natural homomorphism from R[x] to Q(R)[x], and since Q(R)[x] is isomorphic to a subring of its field of quotients Q(R)(x), there is an injective homomorphism from R[x] to Q(R)(x), which amounts to viewing a polynomial with 7

8 coefficients in R as a particular example of a rational function with coefficients in Q(R). Hence, by the proposition, there is an injective homomorphism Q(R[x]) Q(R)(x). To see that it is surjective, it suffices to show that every rational function with coefficients in Q(R) is a quotient of two polynomials with coefficients in R. Given such a quotient f/g, suppose that f = n i=0 a ix i and g = m j=0 b jx j, with a i, b j Q(R). Then a i = r i /s i with r i, s i R and s i 0. Likewise, b j = t j /w j with t j, w j R and w j 0. We then proceed to clear denominators in the coefficients: Let N = s 0 s n w 0 w m = n i=0 s i m j=0 w j. Then N(r k /s k ) = r k i k s i m j=0 w j R, and similarly N(t j /w j ) R. Clearly Nf R[x] and Ng R[x]. Thus f g = f g N N = Nf Ng. It then follows that f/g = Nf/Ng is a quotient of two polynomials with coefficients in R. Hence Q(R[x]) = Q(R)(x). Another application of Proposition 3.3 is as follows: let F be a field of characteristic 0. As we have seen in the homework, the function f : Z F defined by f(n) = n 1 is a ring homomorphism. If char F = 0, the homomorphism f is injective. Hence by Proposition 3.3 there is an induced homomorphism f : Q F. Its image is the set of all quotients in F of the form n 1/m 1, with m 0. In particular, the image of f is a subfield of F isomorphic to Q. Thus every field of characteristic 0 contains a subfield isomorphic to Q, called the prime subfield. It is the smallest subfield of F, hence unique, and it can be described by starting with 1 and making sure that we can perform the operations of addition and subtraction and then automatically multiplication(to get the subring isomorphic to Z), and finally division to get the subfield isomorphic to Q. Here prime has nothing to do with prime numbers but simply means that the field Q is a basic, indivisible object. A similar statement holds if F is a field of positive characteristic, say char F = p where p is a prime number. In this case, the function f : Z F defined by f(n) = n 1 is still a ring homomorphism, but its kernel is p and hence its image, as an abelian group, is isomorphic to Z/pZ. The fact that f is a ring homomorphism implies that the image of f, as a ring, is isomorphic to Z/pZ = F p. Thus, every field of characteristic p contains a subfield isomorphic to F p, again called the prime subfield. The fields Q and F p are more generally called prime fields. They contain no proper 8

9 subfields, and every field F contains a unique subfield isomorphic either to Q, if char F = 0, or to F p, if char F = p. 9

### Algebra. Sample Solutions for Test 1

EPFL - Section de Mathématiques Algebra Fall semester 2008-2009 Sample Solutions for Test 1 Question 1 (english, 30 points) 1) Let n 11 13 17. Find the number of units of the ring Z/nZ. 2) Consider the

### Chapter 4, Arithmetic in F [x] Polynomial arithmetic and the division algorithm.

Chapter 4, Arithmetic in F [x] Polynomial arithmetic and the division algorithm. We begin by defining the ring of polynomials with coefficients in a ring R. After some preliminary results, we specialize

### Chapter 13: Basic ring theory

Chapter 3: Basic ring theory Matthew Macauley Department of Mathematical Sciences Clemson University http://www.math.clemson.edu/~macaule/ Math 42, Spring 24 M. Macauley (Clemson) Chapter 3: Basic ring

### ALGEBRA HANDOUT 2: IDEALS AND QUOTIENTS. 1. Ideals in Commutative Rings In this section all groups and rings will be commutative.

ALGEBRA HANDOUT 2: IDEALS AND QUOTIENTS PETE L. CLARK 1. Ideals in Commutative Rings In this section all groups and rings will be commutative. 1.1. Basic definitions and examples. Let R be a (commutative!)

### Appendix A. Appendix. A.1 Algebra. Fields and Rings

Appendix A Appendix A.1 Algebra Algebra is the foundation of algebraic geometry; here we collect some of the basic algebra on which we rely. We develop some algebraic background that is needed in the text.

### 3. Some commutative algebra Definition 3.1. Let R be a ring. We say that R is graded, if there is a direct sum decomposition, M n, R n,

3. Some commutative algebra Definition 3.1. Let be a ring. We say that is graded, if there is a direct sum decomposition, = n N n, where each n is an additive subgroup of, such that d e d+e. The elements

### Solutions to Homework Problems from Chapter 3

Solutions to Homework Problems from Chapter 3 31 311 The following subsets of Z (with ordinary addition and multiplication satisfy all but one of the axioms for a ring In each case, which axiom fails (a

### (a) Write each of p and q as a polynomial in x with coefficients in Z[y, z]. deg(p) = 7 deg(q) = 9

Homework #01, due 1/20/10 = 9.1.2, 9.1.4, 9.1.6, 9.1.8, 9.2.3 Additional problems for study: 9.1.1, 9.1.3, 9.1.5, 9.1.13, 9.2.1, 9.2.2, 9.2.4, 9.2.5, 9.2.6, 9.3.2, 9.3.3 9.1.1 (This problem was not assigned

### Math 210A: Algebra, Homework 1

Math 210A: Algebra, Homework 1 Ian Coley October 9, 2013 Problem 1. Let a 1, a 2,..., a n be elements of a group G. Define the product of the a i s by induction: a 1 a 2 a n = (a 1 a 2 a n 1 )a n. (a)

### Theorem Let G be a group and H a normal subgroup of G. Then the operation given by (Q) is well defined.

22. Quotient groups I 22.1. Definition of quotient groups. Let G be a group and H a subgroup of G. Denote by G/H the set of distinct (left) cosets with respect to H. In other words, we list all the cosets

### I. GROUPS: BASIC DEFINITIONS AND EXAMPLES

I GROUPS: BASIC DEFINITIONS AND EXAMPLES Definition 1: An operation on a set G is a function : G G G Definition 2: A group is a set G which is equipped with an operation and a special element e G, called

### 3 Congruence arithmetic

3 Congruence arithmetic 3.1 Congruence mod n As we said before, one of the most basic tasks in number theory is to factor a number a. How do we do this? We start with smaller numbers and see if they divide

### CHAPTER 6: RATIONAL NUMBERS AND ORDERED FIELDS

CHAPTER 6: RATIONAL NUMBERS AND ORDERED FIELDS LECTURE NOTES FOR MATH 378 (CSUSM, SPRING 2009). WAYNE AITKEN 1. Introduction In this chapter we construct the set of rational numbers Q using equivalence

### F1.3YE2/F1.3YK3 ALGEBRA AND ANALYSIS. Part 2: ALGEBRA. RINGS AND FIELDS

F1.3YE2/F1.3YK3 ALGEBRA AND ANALYSIS Part 2: ALGEBRA. RINGS AND FIELDS LECTURE NOTES AND EXERCISES Contents 1 Revision of Group Theory 3 1.1 Introduction................................. 3 1.2 Binary Operations.............................

### Cancelling an a on the left and a b on the right, we get b + a = a + b, which is what we want. Note the following identity.

15. Basic Properties of Rings We first prove some standard results about rings. Lemma 15.1. Let R be a ring and let a and b be elements of R. Then (1) a0 = 0a = 0. (2) a( b) = ( a)b = (ab). Proof. Let

### The Dirichlet Unit Theorem

Chapter 6 The Dirichlet Unit Theorem As usual, we will be working in the ring B of algebraic integers of a number field L. Two factorizations of an element of B are regarded as essentially the same if

### Revision of ring theory

CHAPTER 1 Revision of ring theory 1.1. Basic definitions and examples In this chapter we will revise and extend some of the results on rings that you have studied on previous courses. A ring is an algebraic

### Equivalence relations

Equivalence relations A motivating example for equivalence relations is the problem of constructing the rational numbers. A rational number is the same thing as a fraction a/b, a, b Z and b 0, and hence

### Algebra 2. Rings and fields. Finite fields. A.M. Cohen, H. Cuypers, H. Sterk. Algebra Interactive

2 Rings and fields A.M. Cohen, H. Cuypers, H. Sterk A.M. Cohen, H. Cuypers, H. Sterk 2 September 25, 2006 1 / 20 For p a prime number and f an irreducible polynomial of degree n in (Z/pZ)[X ], the quotient

### 3.1 The Definition and Some Basic Properties. We identify the natural class of integral domains in which unique factorization of ideals is possible.

Chapter 3 Dedekind Domains 3.1 The Definition and Some Basic Properties We identify the natural class of integral domains in which unique factorization of ideals is possible. 3.1.1 Definition A Dedekind

### Congruences. Robert Friedman

Congruences Robert Friedman Definition of congruence mod n Congruences are a very handy way to work with the information of divisibility and remainders, and their use permeates number theory. Definition

### Chapter 3, Rings. Definitions and examples.

Chapter 3, Rings Definitions and examples. We now have several examples of algebraic systems with addition and multiplication: Z, Z n, R, M n (R), 2Z = {2n n Z}. We will write down a system of axioms which

### Quotient Rings and Field Extensions

Chapter 5 Quotient Rings and Field Extensions In this chapter we describe a method for producing field extension of a given field. If F is a field, then a field extension is a field K that contains F.

### 1 Homework 1. [p 0 q i+j +... + p i 1 q j+1 ] + [p i q j ] + [p i+1 q j 1 +... + p i+j q 0 ]

1 Homework 1 (1) Prove the ideal (3,x) is a maximal ideal in Z[x]. SOLUTION: Suppose we expand this ideal by including another generator polynomial, P / (3, x). Write P = n + x Q with n an integer not

### it is easy to see that α = a

21. Polynomial rings Let us now turn out attention to determining the prime elements of a polynomial ring, where the coefficient ring is a field. We already know that such a polynomial ring is a UF. Therefore

### Solutions A ring A is called a Boolean ring if x 2 = x for all x A.

1. A ring A is called a Boolean ring if x 2 = x for all x A. (a) Let E be a set and 2 E its power set. Show that a Boolean ring structure is defined on 2 E by setting AB = A B, and A + B = (A B c ) (B

### Groups, Rings, and Fields. I. Sets Let S be a set. The Cartesian product S S is the set of ordered pairs of elements of S, S S = {(x, y) x, y S}.

Groups, Rings, and Fields I. Sets Let S be a set. The Cartesian product S S is the set of ordered pairs of elements of S, A binary operation φ is a function, S S = {(x, y) x, y S}. φ : S S S. A binary

### Factoring Polynomials

Factoring Polynomials Sue Geller June 19, 2006 Factoring polynomials over the rational numbers, real numbers, and complex numbers has long been a standard topic of high school algebra. With the advent

### Theorem (The division theorem) Suppose that a and b are integers with b > 0. There exist unique integers q and r so that. a = bq + r and 0 r < b.

Theorem (The division theorem) Suppose that a and b are integers with b > 0. There exist unique integers q and r so that a = bq + r and 0 r < b. We re dividing a by b: q is the quotient and r is the remainder,

### SOLUTIONS FOR PROBLEM SET 6

SOLUTIONS FOR PROBLEM SET 6 A. Suppose that G is a group and that H is a subgroup of G such that [G : H] = 2. Suppose that a,b G, but a H and b H. Prove that ab H. SOLUTION: Since [G : H] = 2, it follows

### Elementary Number Theory We begin with a bit of elementary number theory, which is concerned

CONSTRUCTION OF THE FINITE FIELDS Z p S. R. DOTY Elementary Number Theory We begin with a bit of elementary number theory, which is concerned solely with questions about the set of integers Z = {0, ±1,

### 5.1 Commutative rings; Integral Domains

5.1 J.A.Beachy 1 5.1 Commutative rings; Integral Domains from A Study Guide for Beginner s by J.A.Beachy, a supplement to Abstract Algebra by Beachy / Blair 23. Let R be a commutative ring. Prove the following

### 3. Prime and maximal ideals. 3.1. Definitions and Examples.

COMMUTATIVE ALGEBRA 5 3.1. Definitions and Examples. 3. Prime and maximal ideals Definition. An ideal P in a ring A is called prime if P A and if for every pair x, y of elements in A\P we have xy P. Equivalently,

### 4. CLASSES OF RINGS 4.1. Classes of Rings class operator A-closed Example 1: product Example 2:

4. CLASSES OF RINGS 4.1. Classes of Rings Normally we associate, with any property, a set of objects that satisfy that property. But problems can arise when we allow sets to be elements of larger sets

### Solutions to Assignment 4

Solutions to Assignment 4 Math 412, Winter 2003 3.1.18 Define a new addition and multiplication on Z y a a + 1 and a a + a, where the operations on the right-hand side off the equal signs are ordinary

### CHAPTER 5: MODULAR ARITHMETIC

CHAPTER 5: MODULAR ARITHMETIC LECTURE NOTES FOR MATH 378 (CSUSM, SPRING 2009). WAYNE AITKEN 1. Introduction In this chapter we will consider congruence modulo m, and explore the associated arithmetic called

### Modern Algebra Lecture Notes: Rings and fields set 4 (Revision 2)

Modern Algebra Lecture Notes: Rings and fields set 4 (Revision 2) Kevin Broughan University of Waikato, Hamilton, New Zealand May 13, 2010 Remainder and Factor Theorem 15 Definition of factor If f (x)

### 2. Let H and K be subgroups of a group G. Show that H K G if and only if H K or K H.

Math 307 Abstract Algebra Sample final examination questions with solutions 1. Suppose that H is a proper subgroup of Z under addition and H contains 18, 30 and 40, Determine H. Solution. Since gcd(18,

### CS 103X: Discrete Structures Homework Assignment 3 Solutions

CS 103X: Discrete Structures Homework Assignment 3 s Exercise 1 (20 points). On well-ordering and induction: (a) Prove the induction principle from the well-ordering principle. (b) Prove the well-ordering

### 26 Ideals and Quotient Rings

Arkansas Tech University MATH 4033: Elementary Modern Algebra Dr. Marcel B. Finan 26 Ideals and Quotient Rings In this section we develop some theory of rings that parallels the theory of groups discussed

### (x + a) n = x n + a Z n [x]. Proof. If n is prime then the map

22. A quick primality test Prime numbers are one of the most basic objects in mathematics and one of the most basic questions is to decide which numbers are prime (a clearly related problem is to find

### Further linear algebra. Chapter I. Integers.

Further linear algebra. Chapter I. Integers. Andrei Yafaev Number theory is the theory of Z = {0, ±1, ±2,...}. 1 Euclid s algorithm, Bézout s identity and the greatest common divisor. We say that a Z divides

### Introduction to Modern Algebra

Introduction to Modern Algebra David Joyce Clark University Version 0.0.6, 3 Oct 2008 1 1 Copyright (C) 2008. ii I dedicate this book to my friend and colleague Arthur Chou. Arthur encouraged me to write

### MATH 436 Notes: Examples of Rings.

MATH 436 Notes: Examples of Rings. Jonathan Pakianathan November 20, 2003 1 Formal power series and polynomials Let R be a ring. We will now define the ring of formal power series on a variable x with

### Factoring of Prime Ideals in Extensions

Chapter 4 Factoring of Prime Ideals in Extensions 4. Lifting of Prime Ideals Recall the basic AKLB setup: A is a Dedekind domain with fraction field K, L is a finite, separable extension of K of degree

### MATH 321 EQUIVALENCE RELATIONS, WELL-DEFINEDNESS, MODULAR ARITHMETIC, AND THE RATIONAL NUMBERS

MATH 321 EQUIVALENCE RELATIONS, WELL-DEFINEDNESS, MODULAR ARITHMETIC, AND THE RATIONAL NUMBERS ALLAN YASHINSKI Abstract. We explore the notion of well-definedness when defining functions whose domain is

### Spring As discussed at the end of the last section, we begin our construction of the rational

MATH 304: CONSTRUCTING THE REAL NUMBERS Peter Kahn Spring 2007 Contents 3 The Rational Numbers 1 3.1 The set Q................................. 1 3.2 Addition and multiplication of rational numbers............

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

COMMUTATIVE RINGS Definition: A commutative ring R is a set with two operations, addition and multiplication, such that: (i) R is an abelian group under addition; (ii) ab = ba for all a, b R (commutative

### Introduction to Algebraic Geometry. Igor V. Dolgachev

Introduction to Algebraic Geometry Igor V. Dolgachev August 19, 2013 ii Contents 1 Systems of algebraic equations 1 2 Affine algebraic sets 7 3 Morphisms of affine algebraic varieties 13 4 Irreducible

### Linear Maps. Isaiah Lankham, Bruno Nachtergaele, Anne Schilling (February 5, 2007)

MAT067 University of California, Davis Winter 2007 Linear Maps Isaiah Lankham, Bruno Nachtergaele, Anne Schilling (February 5, 2007) As we have discussed in the lecture on What is Linear Algebra? one of

### Pythagorean Triples, Complex Numbers, Abelian Groups and Prime Numbers

Pythagorean Triples, Complex Numbers, Abelian Groups and Prime Numbers Amnon Yekutieli Department of Mathematics Ben Gurion University email: amyekut@math.bgu.ac.il Notes available at http://www.math.bgu.ac.il/~amyekut/lectures

### POLYNOMIAL RINGS AND UNIQUE FACTORIZATION DOMAINS

POLYNOMIAL RINGS AND UNIQUE FACTORIZATION DOMAINS RUSS WOODROOFE 1. Unique Factorization Domains Throughout the following, we think of R as sitting inside R[x] as the constant polynomials (of degree 0).

### 11 Ideals. 11.1 Revisiting Z

11 Ideals The presentation here is somewhat different than the text. In particular, the sections do not match up. We have seen issues with the failure of unique factorization already, e.g., Z[ 5] = O Q(

### EXERCISES FOR THE COURSE MATH 570, FALL 2010

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

### Algebraic Structures II

MAS 305 Algebraic Structures II Notes 12 Autumn 2006 Factorization in integral domains Lemma If a, b, c are elements of an integral domain R and ab = ac then either a = 0 R or b = c. Proof ab = ac a(b

### S(A) X α for all α Λ. Consequently, S(A) X, by the definition of intersection. Therefore, X is inductive.

MA 274: Exam 2 Study Guide (1) Know the precise definitions of the terms requested for your journal. (2) Review proofs by induction. (3) Be able to prove that something is or isn t an equivalence relation.

### Math 4310 Handout - Quotient Vector Spaces

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

### Unique Factorization

Unique Factorization Waffle Mathcamp 2010 Throughout these notes, all rings will be assumed to be commutative. 1 Factorization in domains: definitions and examples In this class, we will study the phenomenon

### Free groups. Contents. 1 Free groups. 1.1 Definitions and notations

Free groups Contents 1 Free groups 1 1.1 Definitions and notations..................... 1 1.2 Construction of a free group with basis X........... 2 1.3 The universal property of free groups...............

### If I and J are graded ideals, then since Ĩ, J are ideal sheaves, there is a canonical homomorphism J OX. It is equal to the composite

Synopsis of material from EGA Chapter II, 2.5 2.9 2.5. Sheaf associated to a graded module. (2.5.1). If M is a graded S module, then M (f) is an S (f) module, giving a quasi-coherent sheaf M (f) on Spec(S

### Mathematics Course 111: Algebra I Part IV: Vector Spaces

Mathematics Course 111: Algebra I Part IV: Vector Spaces D. R. Wilkins Academic Year 1996-7 9 Vector Spaces A vector space over some field K is an algebraic structure consisting of a set V on which are

### Factorization in Polynomial Rings

Factorization in Polynomial Rings These notes are a summary of some of the important points on divisibility in polynomial rings from 17 and 18 of Gallian s Contemporary Abstract Algebra. Most of the important

### Proofs are short works of prose and need to be written in complete sentences, with mathematical symbols used where appropriate.

Advice for homework: Proofs are short works of prose and need to be written in complete sentences, with mathematical symbols used where appropriate. Even if a problem is a simple exercise that doesn t

### MATH 251 Homework 7 Solutions

MATH 51 Homework 7 Solutions 1. Let R be a ring and {S j j J} a collection of subrings resp. ideals of R. Prove that j J S j is a subring resp. ideal of R. Proof : Put S j J S j. Suppose that S j is a

### Quotient Rings of Polynomial Rings

Quotient Rings of Polynomial Rings 8-7-009 Let F be a field. is a field if and only if p(x) is irreducible. In this section, I ll look at quotient rings of polynomial rings. Let F be a field, and suppose

### Continued Fractions and the Euclidean Algorithm

Continued Fractions and the Euclidean Algorithm Lecture notes prepared for MATH 326, Spring 997 Department of Mathematics and Statistics University at Albany William F Hammond Table of Contents Introduction

### MATH 289 PROBLEM SET 1: INDUCTION. 1. The induction Principle The following property of the natural numbers is intuitively clear:

MATH 89 PROBLEM SET : INDUCTION The induction Principle The following property of the natural numbers is intuitively clear: Axiom Every nonempty subset of the set of nonnegative integers Z 0 = {0,,, 3,

### FACTORING IN QUADRATIC FIELDS. 1. Introduction. This is called a quadratic field and it has degree 2 over Q. Similarly, set

FACTORING IN QUADRATIC FIELDS KEITH CONRAD For a squarefree integer d other than 1, let 1. Introduction K = Q[ d] = {x + y d : x, y Q}. This is called a quadratic field and it has degree 2 over Q. Similarly,

### ADDITIVE GROUPS OF RINGS WITH IDENTITY

ADDITIVE GROUPS OF RINGS WITH IDENTITY SIMION BREAZ AND GRIGORE CĂLUGĂREANU Abstract. A ring with identity exists on a torsion Abelian group exactly when the group is bounded. The additive groups of torsion-free

### Subsets of Euclidean domains possessing a unique division algorithm

Subsets of Euclidean domains possessing a unique division algorithm Andrew D. Lewis 2009/03/16 Abstract Subsets of a Euclidean domain are characterised with the following objectives: (1) ensuring uniqueness

### 6. Fields I. 1. Adjoining things

6. Fields I 6.1 Adjoining things 6.2 Fields of fractions, fields of rational functions 6.3 Characteristics, finite fields 6.4 Algebraic field extensions 6.5 Algebraic closures 1. Adjoining things The general

### Topics in Number Theory

Chapter 8 Topics in Number Theory 8.1 The Greatest Common Divisor Preview Activity 1 (The Greatest Common Divisor) 1. Explain what it means to say that a nonzero integer m divides an integer n. Recall

### ABSTRACT ALGEBRA. Romyar Sharifi

ABSTRACT ALGEBRA Romyar Sharifi Contents Introduction 7 Part 1. A First Course 11 Chapter 1. Set theory 13 1.1. Sets and functions 13 1.2. Relations 15 1.3. Binary operations 19 Chapter 2. Group theory

### ALGEBRA HW 5 CLAY SHONKWILER

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

### Finite Sets. Theorem 5.1. Two non-empty finite sets have the same cardinality if and only if they are equivalent.

MATH 337 Cardinality Dr. Neal, WKU We now shall prove that the rational numbers are a countable set while R is uncountable. This result shows that there are two different magnitudes of infinity. But we

### Galois Theory III. 3.1. Splitting fields.

Galois Theory III. 3.1. Splitting fields. We know how to construct a field extension L of a given field K where a given irreducible polynomial P (X) K[X] has a root. We need a field extension of K where

### Computer Algebra for Computer Engineers

p.1/14 Computer Algebra for Computer Engineers Preliminaries Priyank Kalla Department of Electrical and Computer Engineering University of Utah, Salt Lake City p.2/14 Notation R: Real Numbers Q: Fractions

### 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

### The determinant of a skew-symmetric matrix is a square. This can be seen in small cases by direct calculation: 0 a. 12 a. a 13 a 24 a 14 a 23 a 14

4 Symplectic groups In this and the next two sections, we begin the study of the groups preserving reflexive sesquilinear forms or quadratic forms. We begin with the symplectic groups, associated with

### Sets and functions. {x R : x > 0}.

Sets and functions 1 Sets The language of sets and functions pervades mathematics, and most of the important operations in mathematics turn out to be functions or to be expressible in terms of functions.

### Introduction to Ring Theory. Math 228

Introduction to Ring Theory. Math 228 Unless otherwise stated, homework problems are taken from Hungerford, Abstract Algebra, Second Edition. Homework 5 - due February 23 3.1.5 d: Which of the following

### MATH32062 Notes. 1 Affine algebraic varieties. 1.1 Definition of affine algebraic varieties

MATH32062 Notes 1 Affine algebraic varieties 1.1 Definition of affine algebraic varieties We want to define an algebraic variety as the solution set of a collection of polynomial equations, or equivalently,

### Localization (Hungerford, Section 3.4)

Lemma. (1) If p is an ideal which is maximal with respect to the property that p is not finitely generated then p is prime. (2) If p is an ideal maximal with respect to the property that p is not principal,

### ORDERS OF ELEMENTS IN A GROUP

ORDERS OF ELEMENTS IN A GROUP KEITH CONRAD 1. Introduction Let G be a group and g G. We say g has finite order if g n = e for some positive integer n. For example, 1 and i have finite order in C, since

### 1. R In this and the next section we are going to study the properties of sequences of real numbers.

+a 1. R In this and the next section we are going to study the properties of sequences of real numbers. Definition 1.1. (Sequence) A sequence is a function with domain N. Example 1.2. A sequence of real

### Introduction to finite fields

Introduction to finite fields Topics in Finite Fields (Fall 2013) Rutgers University Swastik Kopparty Last modified: Monday 16 th September, 2013 Welcome to the course on finite fields! This is aimed at

### The Division Algorithm for Polynomials Handout Monday March 5, 2012

The Division Algorithm for Polynomials Handout Monday March 5, 0 Let F be a field (such as R, Q, C, or F p for some prime p. This will allow us to divide by any nonzero scalar. (For some of the following,

### Group Theory. Contents

Group Theory Contents Chapter 1: Review... 2 Chapter 2: Permutation Groups and Group Actions... 3 Orbits and Transitivity... 6 Specific Actions The Right regular and coset actions... 8 The Conjugation

### MODULAR ARITHMETIC. a smallest member. It is equivalent to the Principle of Mathematical Induction.

MODULAR ARITHMETIC 1 Working With Integers The usual arithmetic operations of addition, subtraction and multiplication can be performed on integers, and the result is always another integer Division, on

### A NOTE ON FINITE FIELDS

A NOTE ON FINITE FIELDS FATEMEH Y. MOKARI The main goal of this note is to study finite fields and their Galois groups. Since I define finite fields as subfields of algebraic closure of prime fields of

### Module MA3411: Abstract Algebra Galois Theory Appendix Michaelmas Term 2013

Module MA3411: Abstract Algebra Galois Theory Appendix Michaelmas Term 2013 D. R. Wilkins Copyright c David R. Wilkins 1997 2013 Contents A Cyclotomic Polynomials 79 A.1 Minimum Polynomials of Roots of

### Ideal Class Group and Units

Chapter 4 Ideal Class Group and Units We are now interested in understanding two aspects of ring of integers of number fields: how principal they are (that is, what is the proportion of principal ideals

### We now explore a third method of proof: proof by contradiction.

CHAPTER 6 Proof by Contradiction We now explore a third method of proof: proof by contradiction. This method is not limited to proving just conditional statements it can be used to prove any kind of statement

### Commutative Algebra seminar talk

Commutative Algebra seminar talk Reeve Garrett May 22, 2015 1 Ultrafilters Definition 1.1 Given an infinite set W, a collection U of subsets of W which does not contain the empty set and is closed under

### 5.1 Radical Notation and Rational Exponents

Section 5.1 Radical Notation and Rational Exponents 1 5.1 Radical Notation and Rational Exponents We now review how exponents can be used to describe not only powers (such as 5 2 and 2 3 ), but also roots

### 13 Solutions for Section 6

13 Solutions for Section 6 Exercise 6.2 Draw up the group table for S 3. List, giving each as a product of disjoint cycles, all the permutations in S 4. Determine the order of each element of S 4. Solution

### Define the set of rational numbers to be the set of equivalence classes under #.

Rational Numbers There are four standard arithmetic operations: addition, subtraction, multiplication, and division. Just as we took differences of natural numbers to represent integers, here the essence

### GROUPS SUBGROUPS. Definition 1: An operation on a set G is a function : G G G.

Definition 1: GROUPS An operation on a set G is a function : G G G. Definition 2: A group is a set G which is equipped with an operation and a special element e G, called the identity, such that (i) the