Algebra Interactive. Reader. Arjeh M. Cohen, Hans Cuypers, Hans Sterk. Eindhoven University of Technology

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1 Algebra Interactive Reader Arjeh M. Cohen, Hans Cuypers, Hans Sterk Eindhoven University of Technology

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3 iii Preface Algebra Interactive is designed as modern course material for undergraduate courses in Algebra. It can also be used by students for individual study. The many dynamic illustrations make it suitable for demonstrations in lectures as well. As such it is useful for instructors who wish to enliven their lectures with modern means. Algebra Interactive differs in content from classical material on algebra in that it focuses on algorithms (constructive aspects) and real life applications; the presentation deviates strongly from classical algebra books in that it brings to life the basic concepts of algebra through interactive examples and visualizations. With these modern techniques illustrations are presented in totally new ways that go beyond the traditional presentations. It provides students with a sophisticated learning environment with emphasis on computational and algorithmic aspects. New technology enriches the material in that many examples (in particular, the so called gapplets) allow for experimenting, dynamic illustrations (in particular, the applets under the Cool button) occur throughout the text, buttons enable focusing on specific aspects (proofs, examples, exercises, illustrations), on-line multiple choice tests are offered, on-line calculators pertaining to the subjects covered are available, various ways of cross referencing are supported. More information on the features of Algebra Interactive is given in the Appendix. Algebra Interactive offers a new challenge compared to traditional text books and we recommend that you grasp the material by experimenting with the examples, working through the exercises, testing yourself, etc. However, working with the interactive notes will not replace the full benefits of pad and pencil, it rather adds to them. So keep pad and pencil at hand while you work your way through. Also, for your convenience, we provide this reader, a paper excerpt of Algebra Interactive. It contains most of the text from the main window (in standard font size) and a fair portion of the proofs, examples, remarks, applications and explanations from the secondary window (in a smaller font size). Algebra Interactive covers topics ranging from arithmetic with integers and with polynomials (including modular arithmetic) to permutations in the first five chapters, and then unifies the many common

4 iv aspects of these arithmetic structures in the remaining chapters on groups and rings. Algebra Interactive grew out of algebra lectures given at the Eindhoven University of Technology, The Netherlands, over the past few years. It was developed as IDA: Interactive Document on Algebra. (We still refer to Algebra Interactive as IDA rather than AI.) Algebra Interactive is not the end of the story. It is part of the Open- Math project that ultimately aims at providing a flexible working environment for engineers and researchers using mathematics. In the not so distant future, we hope to produce a second version using the OpenMath standard for sharing mathematical expressions between the user interface and back engines (such as GAP). Arjeh M. Cohen, Hans Cuypers, Hans Sterk, Eindhoven, June 1999.

5 v Contents 1 Arithmetic

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7 1 1. Arithmetic In this chapter we study the properties of the set Z of integers. We mainly deal with its multiplicative structure and discuss notions such as the greatest common divisor (gcd) and the least common multiple (lcm) of two (or more) integers Divisors and multiples We know how to add integers, how to subtract them and how to multiply them. Division is a bit harder. Definition 1. Let a Z and b Z. We call b a divisor of a if there is an integer q Z such that a = qb. If b is a nonzero divisor of a, then the (unique) integer q with a = qb is called the quotient a over b and denoted by a/b. If b is a divisor of a, we also say that b divides a, or a is a multiple of b, or a is divisible by b. We write this as b a. Suppose that a is an integer. If a is nonzero, not every integer b is a divisor of a. If b (not equal to 0) does not divide a, then there is a remainder after division of a by b. Here is a precise statement about division with remainder. Theorem 2. If a Z and b N \ {0}, then there are exactly one q Z and one r Z such that a = qb + r and 0 r < b. This theorem states that there exist a quotient q and a remainder r, often denoted by a mod b, but it does not tell you how to find those two integers. However, the proof that we give here, is constructive: it provides an algorithm to find q and r. Proof. In the case where a, b > 0, the proof is roughly as follows: find the greatest multiple qb of b that is less than or equal to a; this can be accomplished by starting with q = 0 and increasing q by 1 until a (q + 1)b < 0. Then r = a qb. Existence We first consider the case where a 0 and use induction on a. If a < b, then we set q = 0 and r = a. If a b, then by the induction hypothesis, there exist integers q and r (with 0 r < b) such that a b = q b + r. Thus, a = (q + 1)b + r. Apparently, q = q + 1 and r = r satisfy the conditions of the theorem in this case. If a < 0, then by the above there are q and r with a = q b + r. But then a = (q b + r ) = (q + 1)b + ( r + b). So q = (q + 1) and r = ( r + b) satisfy the conditions of the theorem.

8 2 1 Arithmetic Uniqueness Suppose that a = qb + r and a = q b + r with 0 r, r < b. Suppose moreover that r r. (This is no essential restriction.) By subtracting these equalities we get r r = (q q)b. Now it follows that the multiple (q q)b of b satisfies 0 (q q)b = r r < b. This can only happen if q q = 0. In other words q = q. Now it also follows that r = r. Application. Suppose you need to fit rods of length b one after the other in a box of length a. Then the quotient tells you how many will fit and the remainder what the length of the space left open will be. For instance, in a box of length a = 23 we can fit two rods of length b = 11, and there will be an open space of length 1. We consider common divisors and multiples of two (or more) integers. Let a, b Z. Definition 3. An integer d is a common divisor of a and b if d a and d b. If a and b are not both 0, the largest common divisor of a and b (exists and) is called the greatest common divisor of a and b. If the greatest common divisor of a and b equals 1, then a and b are called relatively prime. Remark. If a and b are not both 0, then they have only a finite number of common divisors. Note that 1 is always a common divisor. Hence the greatest common divisor exists. We denote the greatest common divisor of a and b by gcd(a, b). Analogous to the greatest common divisor of two integers we can define the greatest common divisor of more than two numbers. We now discuss common multiples of two (or more) integers. Let a, b Z. Definition 4. The integer c is a common multiple of a and b if c is a multiple of a and of b (that is, a c and b c). The smallest nonnegative common multiple of a and b is called the least common multiple of a and b. We express the least common multiple of a and b with lcm(a, b) Euclid s algorithm The gcd of two integers a and b (which, for simplicity, we assume to be positive) can be determined with Euclid s algorithm.

9 1.2 Euclid s algorithm 3 Algorithm 5 (Euclid s algorithm). Input: two positive integers a and b. Output: the gcd of a and b. 1. Replace (simultaneously) a by b and b by the remainder of the division of a by b. 2. Repeat Step 1 until b is equal to Return a. Proof. We use three properties of the greatest common divisor: gcd(a, b) = gcd(b, a), gcd(a, b) = gcd(a, b ka), and gcd(a, 0) = a. These properties imply that in each step of the algorithm the gcd is left invariant. The variable b decreases with each step. Eventually we come to the point that b equals 0. Then the algorithm ends and it will give gcd(a, 0) = a as output. There is also an extended version of Euclid s algorithm, where integers x and y are determined such that ax + by = gcd(a, b). To do this, we record at each step of Euclid s algorithm how to express the intermediate results in the input integers. Algorithm 6 (The extended Euclidean algorithm). Input: positive integers a and b. Output: integers x and y such that gcd(a, b) = xa + yb. 1. Let x = v = 1 and y = u = Determine q and r such that a = qb + r and 0 r < b. Replace (simultaneously) a by b and b by r, x by u and y by v, u by x qu and v by y qv. 3. Repeat Step 2 until b equals Return x and y. Proof. Find the gcd of a and b using Euclid s algorithm. When after some steps using this algorithm a = xa + yb and b = ua + vb for certain integers x, y, u and v, then after the next step a = ua + vb and b = (ua + vb) q(xa + yb) = (x qu)a + (y qv)b, where q is the quotient of a and b. Since Euclid s algorithm will eventually return a as the gcd of a and b, the extended Euclidean algorithm will give x and y with gcd(a, b) = xa + yb. Remark. The integers x and y with xa + yb = gcd(a, b) are not unique: of course, for any integer t we have (x + tb)a + (y ta)b = gcd(a, b). The extended Euclidean algorithm leads us to the following characterization of the gcd. Let a and b be two positive integers.

10 4 1 Arithmetic Theorem 7 (Characterization of the gcd). The following three statements are equivalent. 1. gcd(a, b) = d. 2. d is a positive common divisor of a and b such that any common divisor c of a and b is a divisor of d. 3. d is the least positive integer that can be expressed as xa + yb with integers x and y. Proof. 1 2 Let d = gcd(a, b). Then d is a common divisor of a and b. By the extended Euclidean algorithm d = xa+yb for some integers x and y. If c is any common divisor of a and b, then it also divides d. This proves that the first assertion implies the second. As for the other way around, suppose that d is as in the second statement. Since gcd(a, b) is a common divisor of a and b it must divide d. On the other hand d cannot be larger than the greatest common divisor of a and b. Hence d and gcd(a, b) must be equal. This proves that the second statement implies the first. 1 3 Let d = gcd(a, b) and let e be the least positive integer that can be expressed as xa + yb with integers x and y. We show that d = e. Since d is a common divisor of a and b, the equality e = xa + yb implies that d divides e. So d e. Moreover, d itself can also be written as a linear combination of a and b. So d e by the defining property of e. Hence d must be equal to e. This proves the equivalence. This characterization of the gcd has the following trivial (but important) corollary. Corollary 8. Integers a and b are relatively prime if and only if there exist integers x and y such that xa + yb = 1. Proof. Apply the previous theorem with d = 1. Example. For all natural numbers m, n with m < n, the integers 2 m and 2 n 1 are relatively prime: 2 n m 2 m 1 (2 n 1) = 1. A consequence is the following useful result for deducing divisibility of one integer by another. Proposition 9. Let a, b, c be integers. If a and b are relatively prime, then a bc implies a c. Proof. There exist integers x and y such that xa + yb = 1. Multiply this relation by c to get xac + ybc = c. Since a xac and a bc (and hence also a ybc) we get that a xac + ybc = c.

11 1.3 Linear diophantine equations Linear diophantine equations We describe an algorithm to find all integer solutions to the equation xa + yb = c in the unknown x, y and given a, b, c Z. Such equations are known as linear diophantine equations. We first discuss a special case, the homogeneous equation. Let a, b Z. Lemma 10. If xa + yb = 0 and gcd(a, b) = 1, then there exists an integer n such that x = nb and y = na. Proof. Suppose that xa+yb = 0 and that gcd(a, b) = 1. From xa = by it follows that a by. Since gcd(a, b) = 1, we find a y, see Proposition 9 of Section 1.2. But that means that there exists an integer n such that an = y. Substitution of y in the original equation gives x = nb. This proves the lemma. Remark. If we interpret the equation over Q or R, then the solutions are, assuming b is not equal to 0, of the form (x, y) = (x, (c xa)/b). However, not all of these solutions are integral. From the lemma we conclude the following. Theorem 11. Suppose that a, b are not both 0. Then the integer solutions to the equation xa + yb = 0 are given by x = nb/d and y = na/d, where d = gcd(a, b) and n Z. Proof. Assume that xa + yb = 0 and gcd(a, b) = d. By dividing the equality xa + yb = 0 by d, we get the situation of the lemma. The theorem now follows. Example. The solutions to the equation 28x + 21y = 0 are x = 21n/7 = 3n and y = 28n/7 = 4n, with n Z, since gcd(28, 21) = 7. We are ready for the general case. Algorithm 12. Input: integers a, b, c, with a and b not both 0. Output: all integer solutions x, y to the equation xa + yb = c. 1. Find, using the extended Euclidean algorithm, integers x, y such that d := gcd(a, b) = x a + y b. 2. If d does not divide c, then return that there are no solutions to the equation. 3. If d c, then return x = (x c nb)/d and y = (y c + na)/d, with n Z. Proof. First note that if (x 0, y 0) and (x 1, y 1) are solutions to the equation ax + yb = c, then their difference (x 0 x 1, y 0 y 1) is a solution to the homogeneous equation ax + yb = 0. Hence all solutions of ax + yb = c, if there are any, are of the form (x 0,y 0) plus a solution to the homogeneous equation ax + yb = 0. Correctness of the algorithm is now a direct consequence of the previous lemma and theorem.

12 6 1 Arithmetic Example. Consider the equation 28x + 21y = 14. Using the extended Euclidean algorithm we find gcd(28, 21) = 7 = Moreover, it is easy to see that x = 2 and y = 2 is a solution of the equation. Hence all solutions are x = 2 21n/7 = 2 3n and y = n/7 = 2 + 4n, with n in Z. Note the structure of the solutions: x c/d, y c/d is one particular solution of the equation xa + yb = c, and all other solutions are obtained by adding all solutions of the homogeneous equation xa + yb = 0 to it Prime numbers In this section we discuss prime numbers, the building blocks for the multiplicative structure of the integers. Definition 13. A prime is an integer p larger than 1 that has no positive divisors other than 1 and p. The first five primes are 2, 3, 5, 7, and 11. But there are many more. Theorem 14. There are infinitely many primes. Proof. Suppose that there are only finitely many primes, say p 1,..., p r. Construct the integer m = p 1 p r + 1. Then m > 1. The integer m is not divisible by any of the p i (i = 1,..., r). The smallest divisor larger than 1 of m is a prime. This is a prime that is not in our list. Contradiction. Eratosthenes sieve is an algorithm to make a list of primes. Algorithm 15 (Eratosthenes sieve). Input: a positive integer n. Output: the list of primes less than or equal to n. 1. Construct the list L := [2,..., n] and the empty list M. 2. Let m be the smallest element in L. Add m to M. Remove all multiples of m from L. 3. Repeat Step 2 until L is empty. 4. Return M. Using this sieve we can find all the primes in the interval [1, n]. The number of such primes can be approximated as follows. Fact 16 (Prime number theorem). Let prime(n) be the number of primes in the interval [1, n]. Then we have prime(n) n/log(n) when n tends to infinity. (This means that lim n prime(n)/(n/ log(n)) = 1.)

13 1.5 Factorization 7 The next theorem gives a characterization of primes. Theorem 17. Let p > 1. Then p is a prime if and only if, for all integers b, c: p bc implies p b or p c. Proof. Suppose p is prime. Assume that p bc. If p b we are done. If p is not a divisor of b, then p and b have no common divisors > 1 and we can apply Proposition 9. Assume that p is not prime, but p = bc, for some b, c > 1. Then p divides the product bc, but it does not divide b or c. Example. Suppose a = bc, where a, b, c are natural numbers. If a is even, then so is at least one of b, c. It is one implication in the special case p = 2 of the theorem. This theorem has the following corollary. Corollary 18. If p is a prime and b 1,..., b s are integers such that p b 1 b 2 b s, then there is an index i {1,..., s} such that p b i. We leave the proof of the corollary to the reader Factorization We will show how integers are built out of primes. Theorem 19. Every positive integer a > 1 can be written as the product of finitely many primes: a = p 1 p s, where s is a positive integer and each p i is a prime. Up to the order of the factors, this factorization is unique. For a positive integer a we denote the number of times that the prime p occurs in its factorization by ord p (a). The factorization in primes of a can be written as a = p ordp(a). p prime Note that only a finite number of factors is distinct from 1. By definition a product that has the empty set as index set (the empty product) is 1. So, using this convention, we could have included this case a = 1 in the theorem. Proof. The proof is divided into two steps.

14 8 1 Arithmetic Existence We prove by induction on a that every positive integer can be written as a product of primes. The case a = 2 is trivial. So suppose that a is at least 3. If a is a prime then we are done. If a is not a prime then it has a divisor b such that 1 < b < a. According to the induction hypothesis b and a/b can both be written as a product of primes: b = p 1 p r, a/b = p r+1 p s. For a we have a = p 1 p r p r+1 p s. Uniqueness Here we also use induction. The case a = 2 is easy. Suppose that a > 2, and also suppose that uniqueness has been proven for the integers < a. If a = p 1 p r and a = q 1 q s are two ways of expressing a as a product of primes, then it follows that p 1 p 1 p r = q 1 q s. From Corollary 18 we may conclude that there exists an index k in the set {1,..., s} such that p 1 q k. But then we have p 1 = q k because q k is a prime. Now apply the induction hypothesis to the integer a/p 1 with the two expressions as products of primes a/p 1 = p 2 p r and a/p 1 = q 1 q k 1 q k+1 q s. These factorizations of a/p 1 are the same (up to the order of the factors) and therefore the two factorizations of a are the same. Application. If a is a square, then ord p(a) is even for each prime p. Using this observation it is not difficult to prove that 2 is not in Q. This means that there are no integers a, b such that (a/b) 2 = 2. For such a and b, we would have 2b 2 = a 2. Now ord 2(2b 2 ) is odd. But ord 2(a 2 ) is even. This is a contradiction. Therefore, the assumption that a and b with (a/b) 2 = 2 exist is false. Remark. There also exist arithmetic systems in which uniqueness of factorizations is not guaranteed. For example, in the system Z[ 5] of numbers of the form a + b 5 with a, b Z we can express 6 in two essentially different ways: 6 = 3 2 = (1 + 5) (1 5). The system Z[ 5] is an example of a ring, a structure we will encounter in Chapter 7. Here is an explicit description of the gcd and lcm in terms of prime factorizations. Theorem 20. If a and b are positive integers, then: gcd(a, b) = p prime pmin{ordp(a),ordp(b)}, lcm(a, b) = p prime pmax{ordp(a),ordp(b)}. In particular we have a b = gcd(a, b) lcm(a, b). Proof. We prove the first equality. The proof of the second is left to the reader. We certainly have min{ord p(a), ord p(b)} ord p(a) and min{ord p(a), ord p(b)} ord p(b). Hence the right-hand side of the equality is a common divisor of a and b. On the other hand, if ord p(ggd(a, b)) = r > 0 for some prime p, then p divides both a and b so that we can conclude that r ord p(a) and r ord p(b). Hence p prime pmin{ordp(a),ordp(b)} equals gcd(a, b). Example. Suppose 0 < p k m for some prime number p and integer k. Suppose k is maximal with this property. Then the binomial coefficient M = ( ) m p is not k divisible by p. Indeed, this binomial can be written as the quotient of m (m 1) (m p k + 1) by p k. Now for all 0 n p k we have ord p(m n) = min(k, ord p(n)) = ord p(p k n), and every factor p in the numerator is canceled by a factor p in the denominator.

15 1.6 Exercises Exercises 1. If c is a common multiple of a and b, then c is a multiple of lcm(a, b). Prove this. 2. For any positive integer n divide 10 3n by 10 n 1 and find the remainder. 3. Three cogwheels with 24, 15 and 16 cogs, respectively, touch as shown in the picture. Fig. 1. (a) What is the smallest positive number of times you have to turn the lefthand cogwheel before the right-hand cogwheel is back in its original position? (b) What is the smallest positive number of times you have to turn the lefthand cogwheel before all three wheels are back in their original position? 4. Prove that the square of an odd integer is again odd, where odd means not divisible by 2 or, equivently, leaving remainder 1 upon division by 2. Show that the remainder of division by 4 of the square of an odd integer is 1. Does the last statement hold if we replace 4 by 8? And by 16? 5. Prove each of the following statements. (a) If a b and b c then a c. (b) If a b and c d then ac bd. 6. Determine the gcd of each of the following pairs of numbers, and write the gcd as a linear combination of the given numbers: (a) 480, 175; (b) 5621, 219; (c) , Show that, for all positive integers x, y, z, gcd(zx, zy) = z gcd(x, y). 8. Prove: If a and b are integers, not both zero, and c = gcd(a, b), then c = min{xa + yb xa + yb > 0, x, y Z}. 9. Let a be a rational number such that both 18a and 25a are integers. Show that a itself is an integer. 10. Find all integers x and y such that 32x + 10y = Prove that there exist infinitely many primes of the form 4n For which positive integers n is n 2 1 prime? 13. Which integers of the form 13p + 1 (with p prime) are squares of integers? 14. Find all primes p and q such that 4p + 7q = pq. 15. Let a, b be distinct integers. (a) Show that a b a 2 b 2 and determine the quotient. (b) Show that a b a 3 b 3 and determine the quotient. (c) Prove that a b divides a n b n for all positive integers n. (d) Show that a + b a 2n+1 + b 2n+1 for each positive integer n. (e) Find the prime factorization of Prove that the cube root of 17 is not a rational number. 17. Find the factorization of

11 Ideals. 11.1 Revisiting Z

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