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

Size: px
Start display at page:

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

Transcription

1 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. Prove that ab c. Solution: By assumption, there exist r, s R such that ar = c = bs. In order to prove the statement, it is enough to show that a s. Since R is a UFD, there exist u, v, w, z R, n a, n b, n s, n r Z 0 and irreducible elements a 1,..., a na, r 1,..., r nr, b 1,..., b nb and s 1,..., s ns R such that a = ua 1 a na, r = vr 1 r nr, b = wb 1 b nb, s = zs 1 s ns. Then ar = bs reads a 1 a na (uvr 1 )r 2 r nr = b 1 b nb (wzs 1 )s 2 s ns (1) and by uniqueness of factorization each a i is associated to some b i or some s i. Suppose that, for some i, a i is associated to b j. Then a i gcd(a, b) R, a contradiction. Hence each a i divides one of the elements s j, and since in the decomposition (1) the associated elements on the two sides can be taken in a bijective correspondence, we can choose for each i a distinct j i such that s ji = λ i a i, so that so that a s and ab c as desired. λ 1 λ na a = λ 1 λ na a 1 a na = s j1 s jna s, 2. We use the notation 5 = i 5 C. Let Z[ 5] = {a + b 5 : a, b Z} C. Consider the map N : Z[ 5] Z sending x = a + b 5 a 2 + 5b 2 = xx. (a) Prove that N is a multiplicative map with image inside N. (b) Show that Z[ 5] = {x Z[ 5] : N(x) = 1} = {±1}. (c) Prove that the elements 2 and are irreducible in Z[ 5] and that they are coprime in the sense that the only common divisors of those are units. [Hint: If x y, then N(x) N(y)] (d) Using Exercise 1, deduce that Z[ 5] is not a UFD [Hint: 2 3 = 6 = (1 + 5)(1 5)] (e) Is 2 Z[ 5] a prime element? Solution: 1

2 (a) For each x, y Z[ 5], we see that N(xy) = xyxy = xxyy = N(x)N(y), meaning that N is a multiplicative map. As N(a + 5b) = a 2 + 5b 2 and a 2, 5b 2 are non-negative integers for all a, b Z, the norm has image inside N. (b) Let x Z[ 5] and write xy = 1 for some y Z[ 5]. Then N(x)N(y) = N(xy) = N(1) = = 1, so that N(x) 1 and since N(x) N by part (a), we conclude that N(x) = 1. Hence there is an inclusion of sets Z[ 5] {x Z[ 5] : N(x) = 1}. Now suppose that N(x) = 1 for x = a + 5b, i.e., a 2 + 5b 2 = 1. Then b = 0 (because else N(x) 5) from which it follows a 2 = 1 and the unique possibilities are x = ±1. Hence {x Z[ 5] : N(x) = 1} {±1}. Finally, the elements ±1 are clearly units, so that {±1} Z[ 5]. This allows us to conclude that the three sets are equal. (c) Suppose that 2 = xy for some x, y Z[ 5]. Then 4 = N(2) = N(xy) = N(x)N(y) so that N(x) {1, 2, 4}. Writing x = a + 5b, we see that the equality a 2 + 5b 2 = 2 cannot hold (because b must vanish, but then a 2 = 2 is a contradiction with a Z), so that N(x) = 1, in which case x is a unit, or N(x) = 4, in which case N(y) = 1 and y is a unit. Hence 2 is irreducible. Similarly, N(1 + 5) = 6 tells us that a proper divisor of could only have norm 2 or 3, which are both impossible situations (as the equality a 2 + 5b 2 = 3 cannot hold for a, b Z, too), so that is irreducible. Suppose by contradiction that 2 and 1+ 5 have a common divisor p R. As they are both irreducible, the only possibility is that p is associated to both 2 and so that 2 and are associated, a contradiction (because Z[ 5] = {±1}, but clearly 2 ±(1 + 5)). Hence 2 and are not coprime. (d) Suppose by contradiction that Z[ 5] is a UFD. Then 2 and are coprime in the sense that gcd(2, 1 + 5) R. Moreover, they both divide 6 = 2 3 = (1 + 5)(1 5) so that 2(1 + 5) 6 by Exercise 1. But N(2(1 + 5)) = = N(6), a contradiction. Hence Z[ 5] is not a UFD. (e) The element 2 is not prime, because it divides 6 = (1 + 5)(1 5) but it does not divide since N(2) = 4 6 = N(1 + 5). 3. Let K be a field and f K[X]. (a) Let x 1,..., x n K be distinct roots of f. Prove that n (X x i ) f. 2

3 (b) Let d = deg(f) and x 1,..., x d K be distinct roots of f. Prove that there exists c K such that d f = c (X x i ). Solution: (a) As seen in class and in previous assignments, if x K is a root of f, then X x f. By assumption, X x 1,..., X x n f. The elements X x i are pairwise coprime in K[X], so that for each k the elements k 1 (X x i) and X x k are coprime in K[X] and since K[X] is a UFD, we obtain by induction that n (X x i) f by Exercise 1. Alternative solution: The statement actually works also when K is just an integral domain. We can indeed prove that n (X x i) f by induction on n. The case n = 1 is clear by assumption. Now suppose that the statement is proven for n 1 and let x 1,..., x n K be distinct roots of f. Then n 1 (X x i) f by inductive hypothesis so that we can write n 1 f = g (X x i ). Since 0 = f(x n ) = g(x n ) n 1 (x n x i ) and all the terms x n x i are non-zero, the fact that K is an integral domain implies that g(x n ) = 0, so that X x n g and we are done. (b) By part (a), there exists g K[X] such that f = g d (X x i ) and by looking at the degrees we see that d = deg(f) = deg(g) + so that deg(g) = 0 and g K. d deg(x x i ) = deg(g) + d 4. Let K be a field. Prove that X and Y are coprime elements in K[X, Y ], but that the ideals (X) and (Y ) are not coprime. [Recall that two ideals I, J of R are called coprime if I + J = R] Solution: Let f K[X, Y ] and suppose that f X, Y. Clearly f 0. Moreover, deg Y (f) deg Y (X) = 0 and deg X (f) deg X (Y ) = 0, so that f is constant. Hence f K {0} = K. This implies that X and Y are coprime. On the other hand, the ideal (X, Y ) is proper, because its elements have all trivial evaluation at (0, 0), whereas 1 K[X, Y ] does not. 3

4 5. Let R be a UFD with K = Frac(R). Let f R[X] be a non-constant polynomial. Prove that f is irreducible in R[X] if and only if it is primitive and irreducible in K[X]. Solution: Recall that R[X] = R and K[X] = K, since R and K are integral domains. As f is assumed to be non-constant, it cannot be a unit neither trivial, so that the statement that saying that f is not irreducible is equivalent to saying that f is reducible, i.e., that it is the product of two not invertible elements. We prove the statement by contraposition, that is, we prove that f has a non-trivial decomposition in R[X] if and only if it is not primitive or it has a non-trivial decomposition in K[X]. Suppose that f is reducible in R[X], that is, f = f 1 f 2 for some f 1, f 2 R[X] R = R[X] R[X], without loss of generality with deg(f 1 ) deg(f 2 ). If deg(f 1 ) = 0, then f 1 R R divides the greatest common divisor of the coefficients of f, so that f is not primitive. Else, 0 < deg(f 1 ) deg(f 2 ) and f 1, f 2 K[X] K = K[X] K[X], so that f is reducible in K[X]. Hence, if f is reducible in R[X], then either it is not primitive or it is reducible in K[X]. Conversely, assume that if f is not primitive, then f = c(f)f 0 for some primitive f 0 R[X] and c(f) R R. This is a non-trivial factorization of f because f 0 R[X] as it is not constant (its degree coinciding with the one of f) and c(f) R = R[X] by assumption. Moreover, if f is reducible in K[X], i.e., f = f 1 f 2 with f i K[X] with deg(f i ) > 0, by Gauss lemma there exist α 1, α 2 K such that α 1 α 2 = 1 and α i f i R[X], so that f = (α 1 f 1 )(α 2 f 2 ) is a non-trivial decomposition of f in R[X]. Hence, if f is not primitive or it is reducible in K[X], then f is reducible in R[X]. 6. Let D := XW Y Z C[X, Y, Z, W ]. (a) Show that (D) is a prime ideal in C[X, Y, Z, W ]. [Hint: First, prove that D is an irreducible element] (b) Prove that C[X, Y, Z, W ]/(D) is not a UFD. [Hint: Let x = X + (D) C[X, Y, Z, W ]/(D). Is x prime? Is it irreducible?] Solution: We say that a polynomial f (in one or several variables) is homogeneous of degree d if all the monomials in f with non-zero coefficients are of degree d. Every polynomial can be uniquely written as a sum of homogeneous polynomials of different degrees. (a) Suppose by contradiction that D = fg for some f, g C[X, Y, Z, W ] C[X, Y, Z, W ] {0} = C[X, Y, Z, W ] C. Then deg(f)+deg(g) = deg(d) = 2 and since f and g cannot be constant the only possibility is that deg(f) = deg(g) = 1. Moreover, 0 = D(0, 0, 0, 0) = f(0, 0, 0, 0)g(0, 0, 0, 0) and without loss of generality we can say that f(0, 0, 0, 0) = 0 since C is a domain. This means that f is homogeneous of degree 1. Writing g = g 1 + g 0 with g 0 and g 1 4

5 homogeneous of degree 0 and 1 respectively, we see that XW Y Z = D = fg 0 +fg 1 and since fg 0 is homogeneous of degree 1, we conclude that fg 0 = 0 so that g 0 = 0 since f 0. Hence we can write f = f X X + f Y Y + f Z Z + f W W, g = g X X + g Y Y + g Z Z + g W W for some f X, f Y, f Z, f W, g X, g Y, g Z, g W C. Comparing the coefficients of X 2 in the equality D = fg, we see that f X g X = 0 and without loss of generality, we can assume that f X = 0. Then, comparing the coefficients of XW, we see that 1 = f X g W + f W g X = f W g X, so that f W 0 g X. Furthermore, a comparison of the coefficients of XY and XZ gives 0 = f X g Y + f Y g X = f Y g X = f Y = 0 0 = f X g Z + f Z g X = f Z g X = f Z = 0, so that f = f W W which means that W D, a contradiction (because it would imply that W XW D = Y Z which cannot hold because of additivity of the degree in W ). This implies that D is irreducible. Since C[X, Y, Z, W ] is a UFD (C is a UFD and for every UFD R, the polynomial ring R[T ] is a UFD as seen in class), then D is a prime element, i.e., the ideal (D) is prime. (b) The given quotient ring is an integral domain because (D) is a prime ideal by part (a). Hence we can talk about irreducible elements. Let x = X + (D) C[X, Y, Z, W ]/(D). The pre-image of the ideal (x) C[X, Y, Z, W ]/(D) under the canonical projection C[X, Y, Z, W ] C[X, Y, Z, W ]/(D) is J = (X, XW Y Z) = (X, Y Z) C[X, Y, Z, W ]. Since Y, Z J but Y Z J, the ideal J is not prime, so that (x) C[X, Y, Z, W ]/(D) is not prime by Exercise 5(b) from Assignment 5. This implies that x C[X, Y, Z, W ]/(D) is not a prime element. Suppose that x = fg for some f, g C[X, Y, Z, W ]/(D) and take representatives F, G C[X, Y, Z, W ]/(D) of f and g respectively. Write F = F F n and G = G G m where for each i the polynomials F i and G i are homogeneous of degree i. Up to adjusting F and G modulo (D) and reducing the numbers of summands n and m, we may assume that D F n and D G m. The condition x = fg then reads P C[X, Y, Z, W ] : X = F G + DP. (2) Writing P = P P q where P i is homogeneous of degree i, we notice that DP = DP DP q, where DP i is homogeneous of degree i + 2. In particular, comparing the homogeneous part of degree n + m in the equality (2), we see that if n + m 2 then 0 = F n G m + DP n+m 2 5

6 which implies that D F n G m, a contradiction since D is prime by part (a) and D F n, G m. Hence n + m < 2 and without loss of generality we can assume that n 1 and m 0, which implies that G C[X, Y, Z, W ] so that g = G + (D) (C[X, Y, Z, W ]/(D)). Hence x is irreducible. We proved that x is irreducible but not prime, which as seen in class can only happen if C[X, Y, Z, W ]/(D) is not a UDF. 6

### 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

### 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)

### D-MATH Algebra I HS18 Prof. Rahul Pandharipande. Solution 1. Arithmetic, Zorn s Lemma.

D-MATH Algebra I HS18 Prof. Rahul Pandharipande Solution 1 Arithmetic, Zorn s Lemma. 1. (a) Using the Euclidean division, determine gcd(160, 399). (b) Find m 0, n 0 Z such that gcd(160, 399) = 160m 0 +

### 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

### 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)

### 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

### D-MATH Algebra I HS17 Prof. Emmanuel Kowalski. Solution 12. Algebraic closure, splitting field

D-MATH Algebra I HS17 Prof. Emmanuel Kowalski Solution 1 Algebraic closure, splitting field 1. Let K be a field of characteristic and L/K a field extension of degree. Show that there exists α L such that

### 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)

### 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

### Exercise Sheet 3 - Solutions

Algebraic Geometry D-MATH, FS 2016 Prof. Pandharipande Exercise Sheet 3 - Solutions 1. Prove the following basic facts about algebraic maps. a) For f : X Y and g : Y Z algebraic morphisms of quasi-projective

### Example: This theorem is the easiest way to test an ideal (or an element) is prime. Z[x] (x)

Math 4010/5530 Factorization Theory January 2016 Let R be an integral domain. Recall that s, t R are called associates if they differ by a unit (i.e. there is some c R such that s = ct). Let R be a commutative

### 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

### Handout - Algebra Review

Algebraic Geometry Instructor: Mohamed Omar Handout - Algebra Review Sept 9 Math 176 Today will be a thorough review of the algebra prerequisites we will need throughout this course. Get through as much

### 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

### 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

### Part IX. Factorization

IX.45. Unique Factorization Domains 1 Part IX. Factorization Section IX.45. Unique Factorization Domains Note. In this section we return to integral domains and concern ourselves with factoring (with respect

### 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

### 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

### Quizzes for Math 401

Quizzes for Math 401 QUIZ 1. a) Let a,b be integers such that λa+µb = 1 for some inetegrs λ,µ. Prove that gcd(a,b) = 1. b) Use Euclid s algorithm to compute gcd(803, 154) and find integers λ,µ such that

### 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

### 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)

### 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

### (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

### 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

### 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

### 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

### 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

### Course 311: Michaelmas Term 2005 Part III: Topics in Commutative Algebra

Course 311: Michaelmas Term 2005 Part III: Topics in Commutative Algebra D. R. Wilkins Contents 3 Topics in Commutative Algebra 2 3.1 Rings and Fields......................... 2 3.2 Ideals...............................

### 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

### 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,

### TC10 / 3. Finite fields S. Xambó

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

### 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

### 8. Prime Factorization and Primary Decompositions

70 Andreas Gathmann 8. Prime Factorization and Primary Decompositions 13 When it comes to actual computations, Euclidean domains (or more generally principal ideal domains) are probably the nicest rings

### 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

### Recall, R is an integral domain provided: R is a commutative ring If ab = 0 in R, then either a = 0 or b = 0.

Recall, R is an integral domain provided: R is a commutative ring If ab = 0 in R, then either a = 0 or b = 0. Examples: Z Q, R Polynomials over Z, Q, R, C The Gaussian Integers: Z[i] := {a + bi : a, b

### 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

### D-MATH Algebraic Geometry FS 2018 Prof. Emmanuel Kowalski. Solutions Sheet 1. Classical Varieties

D-MATH Algebraic Geometry FS 2018 Prof. Emmanuel Kowalski Solutions Sheet 1 Classical Varieties Let K be an algebraically closed field. All algebraic sets below are defined over K, unless specified otherwise.

### x 3 2x = (x 2) (x 2 2x + 1) + (x 2) x 2 2x + 1 = (x 4) (x + 2) + 9 (x + 2) = ( 1 9 x ) (9) + 0

1. (a) i. State and prove Wilson's Theorem. ii. Show that, if p is a prime number congruent to 1 modulo 4, then there exists a solution to the congruence x 2 1 mod p. (b) i. Let p(x), q(x) be polynomials

### August 2015 Qualifying Examination Solutions

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

### 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

### Algebraic function fields

Algebraic function fields 1 Places Definition An algebraic function field F/K of one variable over K is an extension field F K such that F is a finite algebraic extension of K(x) for some element x F which

### MATH 326: RINGS AND MODULES STEFAN GILLE

MATH 326: RINGS AND MODULES STEFAN GILLE 1 2 STEFAN GILLE 1. Rings We recall first the definition of a group. 1.1. Definition. Let G be a non empty set. The set G is called a group if there is a map called

### 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

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,

### 2. THE EUCLIDEAN ALGORITHM More ring essentials

2. THE EUCLIDEAN ALGORITHM More ring essentials In this chapter: rings R commutative with 1. An element b R divides a R, or b is a divisor of a, or a is divisible by b, or a is a multiple of b, if there

### Solutions to Homework 1. All rings are commutative with identity!

Solutions to Homework 1. All rings are commutative with identity! (1) [4pts] Let R be a finite ring. Show that R = NZD(R). Proof. Let a NZD(R) and t a : R R the map defined by t a (r) = ar for all r R.

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

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

### 55 Separable Extensions

55 Separable Extensions In 54, we established the foundations of Galois theory, but we have no handy criterion for determining whether a given field extension is Galois or not. Even in the quite simple

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

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

### 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 4400 SOLUTIONS TO SOME EXERCISES. 1. Chapter 1

MATH 4400 SOLUTIONS TO SOME EXERCISES 1.1.3. If a b and b c show that a c. 1. Chapter 1 Solution: a b means that b = na and b c that c = mb. Substituting b = na gives c = (mn)a, that is, a c. 1.2.1. Find

### 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

### 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

### Homework due on Monday, October 22

Homework due on Monday, October 22 Read sections 2.3.1-2.3.3 in Cameron s book and sections 3.5-3.5.4 in Lauritzen s book. Solve the following problems: Problem 1. Consider the ring R = Z[ω] = {a+bω :

### 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

### CYCLOTOMIC POLYNOMIALS

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

### Polynomials over UFD s

Polynomials over UFD s Let R be a UFD and let K be the field of fractions of R. Our goal is to compare arithmetic in the rings R[x] and K[x]. We introduce the following notion. Definition 1. A non-constant

### MATH 2400 LECTURE NOTES: POLYNOMIAL AND RATIONAL FUNCTIONS. Contents 1. Polynomial Functions 1 2. Rational Functions 6

MATH 2400 LECTURE NOTES: POLYNOMIAL AND RATIONAL FUNCTIONS PETE L. CLARK Contents 1. Polynomial Functions 1 2. Rational Functions 6 1. Polynomial Functions Using the basic operations of addition, subtraction,

### 12. Hilbert Polynomials and Bézout s Theorem

12. Hilbert Polynomials and Bézout s Theorem 95 12. Hilbert Polynomials and Bézout s Theorem After our study of smooth cubic surfaces in the last chapter, let us now come back to the general theory of

### 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

### Institutionen för matematik, KTH.

Institutionen för matematik, KTH. Contents 7 Affine Varieties 1 7.1 The polynomial ring....................... 1 7.2 Hypersurfaces........................... 1 7.3 Ideals...............................

### 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.

### 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

### Factorization in Domains

Last Time Chain Conditions and s Uniqueness of s The Division Algorithm Revisited in Domains Ryan C. Trinity University Modern Algebra II Last Time Chain Conditions and s Uniqueness of s The Division Algorithm

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

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

### 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,

### 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

### Notes on Commutative Algebra 1

Notes on Commutative Algebra 1 Dario Portelli December 12, 2008 1 available at http://www.dmi.units.it/ portelli/ Contents 1 RINGS 3 1.1 Basic definitions............................... 3 1.2 Rings of

### 1. multiplication is commutative and associative;

Chapter 4 The Arithmetic of Z In this chapter, we start by introducing the concept of congruences; these are used in our proof (going back to Gauss 1 ) that every integer has a unique prime factorization.

### D-MATH Algebra I HS 2013 Prof. Brent Doran. Solution 3. Modular arithmetic, quotients, product groups

D-MATH Algebra I HS 2013 Prof. Brent Doran Solution 3 Modular arithmetic, quotients, product groups 1. Show that the functions f = 1/x, g = (x 1)/x generate a group of functions, the law of composition

### On Permutation Polynomials over Local Finite Commutative Rings

International Journal of Algebra, Vol. 12, 2018, no. 7, 285-295 HIKARI Ltd, www.m-hikari.com https://doi.org/10.12988/ija.2018.8935 On Permutation Polynomials over Local Finite Commutative Rings Javier

### 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

### Math Circle Beginners Group February 28, 2016 Euclid and Prime Numbers Solutions

Math Circle Beginners Group February 28, 2016 Euclid and Prime Numbers Solutions Warm-up Problems 1. What is a prime number? Give an example of an even prime number and an odd prime number. A prime number

### ALGEBRA HANDOUT 2.3: FACTORIZATION IN INTEGRAL DOMAINS. In this handout we wish to describe some aspects of the theory of factorization.

ALGEBRA HANDOUT 2.3: FACTORIZATION IN INTEGRAL DOMAINS PETE L. CLARK In this handout we wish to describe some aspects of the theory of factorization. The first goal is to state what it means for an arbitrary

### 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:

### 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

### D-MATH Algebra II FS18 Prof. Marc Burger. Solution 26. Cyclotomic extensions.

D-MAH Algebra II FS18 Prof. Marc Burger Solution 26 Cyclotomic extensions. In the following, ϕ : Z 1 Z 0 is the Euler function ϕ(n = card ((Z/nZ. For each integer n 1, we consider the n-th cyclotomic polynomial

### 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,

### Commutative Algebra and Algebraic Geometry. Robert Friedman

Commutative Algebra and Algebraic Geometry Robert Friedman August 1, 2006 2 Disclaimer: These are rough notes for a course on commutative algebra and algebraic geometry. I would appreciate all suggestions

### 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 =

### ALGEBRAIC GEOMETRY COURSE NOTES, LECTURE 2: HILBERT S NULLSTELLENSATZ.

ALGEBRAIC GEOMETRY COURSE NOTES, LECTURE 2: HILBERT S NULLSTELLENSATZ. ANDREW SALCH 1. Hilbert s Nullstellensatz. The last lecture left off with the claim that, if J k[x 1,..., x n ] is an ideal, then

### CHAPTER 1. AFFINE ALGEBRAIC VARIETIES

CHAPTER 1. AFFINE ALGEBRAIC VARIETIES During this first part of the course, we will establish a correspondence between various geometric notions and algebraic ones. Some references for this part of the

### IRREDUCIBILITY TESTS IN Q[T ]

IRREDUCIBILITY TESTS IN Q[T ] KEITH CONRAD 1. Introduction For a general field F there is no simple way to determine if an arbitrary polynomial in F [T ] is irreducible. Here we will focus on the case

### 10. Noether Normalization and Hilbert s Nullstellensatz

10. Noether Normalization and Hilbert s Nullstellensatz 91 10. Noether Normalization and Hilbert s Nullstellensatz In the last chapter we have gained much understanding for integral and finite ring extensions.

### Prof. Ila Varma HW 8 Solutions MATH 109. A B, h(i) := g(i n) if i > n. h : Z + f((i + 1)/2) if i is odd, g(i/2) if i is even.

1. Show that if A and B are countable, then A B is also countable. Hence, prove by contradiction, that if X is uncountable and a subset A is countable, then X A is uncountable. Solution: Suppose A and

### 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

### 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

### Modern Computer Algebra

Modern Computer Algebra Exercises to Chapter 25: Fundamental concepts 11 May 1999 JOACHIM VON ZUR GATHEN and JÜRGEN GERHARD Universität Paderborn 25.1 Show that any subgroup of a group G contains the neutral

### Problem 1.1. Classify all groups of order 385 up to isomorphism.

Math 504: Modern Algebra, Fall Quarter 2017 Jarod Alper Midterm Solutions Problem 1.1. Classify all groups of order 385 up to isomorphism. Solution: Let G be a group of order 385. Factor 385 as 385 = 5

### Modern Algebra Lecture Notes: Rings and fields set 6, revision 2

Modern Algebra Lecture Notes: Rings and fields set 6, revision 2 Kevin Broughan University of Waikato, Hamilton, New Zealand May 20, 2010 Solving quadratic equations: traditional The procedure Work in

### Math 40510, Algebraic Geometry

Math 40510, Algebraic Geometry Problem Set 1, due February 10, 2016 1. Let k = Z p, the field with p elements, where p is a prime. Find a polynomial f k[x, y] that vanishes at every point of k 2. [Hint:

### 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.

### Solutions of exercise sheet 2

D-MATH Algebra I HS 14 Prof. Emmanuel Kowalski Solutions of exercise sheet 2 1. Let k be a field with char(k) 2. 1. Let a, b k be such that a is a square in k(β), where β is an element algebraic over k

### 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

### Polynomials, Ideals, and Gröbner Bases

Polynomials, Ideals, and Gröbner Bases Notes by Bernd Sturmfels for the lecture on April 10, 2018, in the IMPRS Ringvorlesung Introduction to Nonlinear Algebra We fix a field K. Some examples of fields

### 1. Factorization Divisibility in Z.

8 J. E. CREMONA 1.1. Divisibility in Z. 1. Factorization Definition 1.1.1. Let a, b Z. Then we say that a divides b and write a b if b = ac for some c Z: a b c Z : b = ac. Alternatively, we may say that

### Homework 7 solutions M328K by Mark Lindberg/Marie-Amelie Lawn

Homework 7 solutions M328K by Mark Lindberg/Marie-Amelie Lawn Problem 1: 4.4 # 2:x 3 + 8x 2 x 1 0 (mod 1331). a) x 3 + 8x 2 x 1 0 (mod 11). This does not break down, so trial and error gives: x = 0 : f(0)

### Lecture 7.4: Divisibility and factorization

Lecture 7.4: Divisibility and factorization Matthew Macauley Department of Mathematical Sciences Clemson University http://www.math.clemson.edu/~macaule/ Math 4120, Modern Algebra M. Macauley (Clemson)