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

Notes for Math 345 Dan Singer Minnesota State University, Mankato August 23, 2006 Preliminaries 1. Read the To The Student section pp. xvi-xvii and the Thematic Table of Contents. 2. Read Appendix A: Logic and Proof. 3. Read Appendix C: Well Ordering and Induction. Exercises: 1, 2, 3, 4. 4. Homework guidelines: Write up all assigned problems. Feel free to work in groups of up to 3 people. I will sketch the solution to many of the problems in class. Chapter 1: Arithmetic in Z Revisited Section 1.1: The Division Algorithm. Well-Ordering Axiom: Every subset of non-negative integers contains a smallest element. Division Algorithm: For each pair of integers a and b, where b > 0, there exist integers q and r such that a = bq + r and 0 r < b. The choice of q and r is unique. Proof. First, we ll give an intuitive proof. Let b > 0 be given. Mark off multiples of b on the number line. The integer a will fall between two different multiples. qb is the smaller multiple, and r is the distance from qb to a. Clearly the choices of q and r are unique. Why we need the well-ordering principle: it s not clear that a will fall between two different multiples of b if the numbers a and b are very large. 1

Next we will prove that q and r exist, let alone are unique, assuming a 0 and using the Well-Ordering Axiom. Let S = {a, a b, a 2b,... }. Then S has at least one non-negative integer, namely a. Let S be the subset of non-negative integers in S. By the Well-Ordering Axiom, S contains a smallest element which we will call r. This number must have the form for some integer q. Therefore we have r = a qb a = qb + r. We would like to verify that 0 r < b. First observe that r b S because r b = a qb b = a (q + 1)b. Since r is the smallest non-negative integer in S, it must be the case that r b < 0. Therefore r < b. We already know that r is non-negative, therefore 0 r < b. Now consider a < 0. Let a = a ab. Then a = a(1 b) 0, therefore we have a = qb + r for some 0 r < b, therefore a = a + ab = (q + a)b + r. Finally, note that q and r are unique: if a = qb + r and a = q b + r, then we have qb + r = q b + r, r r = q q b. Since r r < b and r r is a multiple of b, it must be equal to 0. Therefore r = r and q = q. Exercises: 1, 2, 3, 6, 8 Section 1.2: Divisibility Definition: b divides a, written b a, if and only if a = qb for some integer q. In other words, the remainder r = 0. Lemma: Let a and b be two non-negative integers. a = b. If a b and b a, then 2

Proof. We have b = qa and a = q b. If a = 0 or b = 0 then a = b = 0. Otherwise, we can say b = qq b, therefore b(1 qq ) = 0, therefore 1 qq = 0, therefore qq = 1. This forces q = q = 1. Hence a = b. Corollary: If a and b are any two integers and a b and b a, then a = b. Proof. a and b divide each other. Use the Lemma. Divisor set of a: D(a) = {b 1 : b a}. Note that D(0) is the set of all positive integers. Note also that the divisors of a 0 are less than a, therefore D(a) is a finite set. Greatest common divisor of a and b, where a or b is non-zero: the largest number in D(a) D(b). Theorem: Let a and b be integers, at least one of which is non-zero. Then gcd(a, b) is the smallest positive integer in the set S = {ma + nb : m, n Z}. Proof. First note that a and b both belong to S, so S contains at least one positive integer. Let d be the smallest positive integer in S. We must show that d is the largest number in D(a) D(b). We will write d = m 0 a + n 0 b. We will first show that d D(a). Write a = qd + r, where 0 r < d. We must show that r = 0. We have r = a qd = a qm 0 a qn 0 b = (1 qm 0 )a + ( qn 0 )b, therefore r S. However, since r < d it cannot be positive, because d is the smallest positive integer in S. Therefore r = 0 and d D(a). An analogous argument shows that d D(b). We must now show that d is the largest number in D(a) D(b). We will actually show that every number in D(a) D(b) is a divisor of d, which does the job. Let x be any number in D(a) D(b). Then a = px and b = qx, therefore d = m 0 px + n 0 qx = (m 0 p + n 0 q)x. Therefore x d. 3

Corollary: D(a) D(b) D(gcd(a, b)). Proof. Read the last paragraph of the proof of the previous theorem carefully. Theorem: gcd(a, b) = 1 if and only if there there is a solution to xa+yb = 1. Proof. If gcd(a, b) = 1, then the previous theorem shows that 1 belongs to the set of linear combinations of a and b. Conversely, suppose 1 = xa + yb is possible. Then 1 is the smallest positive linear combination of a and b, therefore gcd(a, b) = 1. Theorem: if a bc and gcd(a, b) = 1, then a c. Proof. We have bc = qa and xa + yb = 1, therefore xac + ybc = c, therefore xac + yqa = c, therefore c = (xc + yq)a, therefore a c. Theorem: If a = qb + r, then gcd(a, b) = gcd(b, r). Proof. If x divides a and b, then it must divide r. Hence x divides b and r. Conversely, if x divides b and r, then it must divide a, hence it divides both a and b. Therefore D(a) D(b) = D(b) D(r), which implies gcd(a, b) = gcd(b, r). Euclidean Algorithm: Let a and b > 0 be integers. Assume b does not divide a. Use the division algorithm to obtain the statements a = q 0 b + r 0, b = q 1 r 0 + r 1, r 0 = q 2 r 1 + r 2,, r t 1 = q t+1 r t + r t+1, where b > r 0 > r 1 > > r t > r t+1 = 0. Then gcd(a, b) = r t. 4

Proof. We have gcd(a, b) = gcd(b, r 0 ) = gcd(r 0, r 1 ) = gcd(r 1, r 2 ) = = gcd(r t, r t+1 ) = gcd(r t, 0) = r t. Exercises: 1, 3, 7, 8, 10, 14, 19, 20, 32 Additional Problem: Prove that if x a and y a and gcd(x, y) = 1, then xy a. Find a counter-example if gcd(x, y) 1. Section 1.3: Primes and Unique Factorization Definition: a positive integer p > 1 is prime if and only if D(p) = {1, p}. Theorem: a positive integer p is prime if and only if, for all integers a and b, if p ab then p a or p b. Proof. First assume that p is a prime number. Now suppose that p ab. We will consider two cases. Case 1. p a. Then we have the desired conclusion. Case 2. p a. Therefore D(p) D(a) = {1}. Therefore gcd(p, a) = 1. A previous theorem says that gcd(p, a) = 1 and p ab implies p b. Conversely, assume that the condition on p holds. We will show that p is a prime number. Let x be a positive divisor of p. Then we have p = yx for some integer y. The number y is another divisor of p. Hence we have x p and y p. By the condition on p, we know that p x or p y. If p x then we have p and x dividing each other, hence x = p. If p y then we have p and y dividing each other, hence y = p, which forces x = 1. Therefore we have shown that D(p) {1, p}. Since {1, p} D(p) is also true, we know that D(p) = {1, p}. Therefore p is a prime number. Corollary: if p > 1 is a prime number and p a 1 a 2 a n, then p a i for some i n. 5

Proof. By induction on n 2. The base case is the theorem we have just proved. Assume the conclusion is true for any n-fold product of numbers. Now consider p a 1 a n+1. By the theorem, p a 1 a 2 a n or p a n+1. If the first condition holds, by the induction hypothesis we can say that p a i for some i n. If the second condition holds, then p a n+1. So either way p a i for some i n + 1. Theorem: Every integer n 2 is either a prime number or is the product of two or more prime numbers. Proof. Suppose there are integers n 2 which are neither prime nor the product of prime numbers. Let S be the set of such integers. By the Well- Ordering Axiom, there must be a smallest one, which we will call N. Since N is not a prime number, it has a divisor x, where 1 < x < N. Therefore we can write N = xy. Note that 1 < y < N also. Now x and y do not belong to S, therefore are either prime or the product of prime numbers. This implies that N itself is the product of prime numbers. Contradiction. Therefore there cannot exist any numbers which are neither prime nor the product of prime numbers. Fundamental Theorem of Arithmetic: Let n 2 be a given integer. Let n = p 1 p 2 p j and n = q 1 q 2 q k be two ways to factor n into prime numbers, where p 1 p 2 p j and q 1 q 2 q k. Then (p 1, p 2,..., p j ) = (q 1, q 2,..., q k ). Proof. Note that if n is prime then both lists consist of one prime each and the two lists are equal. So if the theorem is false, it must be false for a smallest number N, and we can assume that N is a composite number. If we write N = p 1 p 2 p j = q 1 q 2 q k, where (p 1, p 2,..., p j ) (q 1, q 2,..., q k ), 6

then we must have j > 1 and k > 1. Since p 1 p 2 p j = q 1 q 2 q k, we have p 1 q 1 q 2 q k and q 1 p 1 p 2 p j. Therefore p 1 q a for some a k and q 1 p b for some b j. Therefore p 1 = q a q 1 = p b p 1. Therefore p 1 = q 1. Let M = p 2 p j = q 2 q j. Since M < N, M cannot belong to S, hence it must be the case that Since p 1 = q 1, this implies (p 2,, p j ) = (q 2,, q k ). (p 1, p 2,, p j ) = (q 1, q 2,, q k ). Contradiction. So the theorem must be true. Exercises: 1, 6, 7, 12, 13, 17, 19, 20, 23, 28 (Outline of proof for problem 28: 8 p 2 q 2 is easy to show. 3 p 2 q 2 can be proved by considering p = 6a + r, q = 6b + s, 0 r, s < 6.) Additional Problem: Prove that gcd(x, y) = 1 if and only if x and y do not share a prime divisor. Chapter 2: Congruence in Z and Modular Arithmetic Section 2.1: Congruence and Congruence Classes Definition: Let n > 0 be given. a b (mod n) if and only if n (a b). Theorem: Let n > 0 be given. a b (mod n) if and only if a = pn + r and b = qn + r (same remainder r, 0 r < n). Proof. If a and b have this form, then clear n (a b). Conversely, assume n (a b). Write a = pn + r, b = qn + s. Then n r s. Since r s < n, this implies r s = 0, which implies r = s. Theorem: Congruence modulo n is an equivalence relation. Proof. Reflexive, symmetric, and transitive properties are easy to prove using the remainders modulo n. 7

Congruence class of a: The set of numbers which have the same remainder modulo n. Write a = qn + r, where 0 r < n. Then [a] = {pn + r : p Z}. The classes [0], [1],..., [n 1] are mutually disjoint. Every integer belongs to exactly one of these classes by the Division Algorithm. a and b are congruent if and only if they belong to the same class. Definition: Z n = {[0], [1], [2],..., [n 1]}. Properties of Congruence: (1) If [a] = [a ] and [b] = [b ] then [a + b] = [a + b ]. (2) If [a] = [a ] and [b] = [b ], then [ab] = [a b ]. Proof. To prove (1), we need to show that a + b a + b (mod n). Write a = pn + r, a = p n + r, b = qn + s, b = q n + s. Then therefore n ((a + b) (a + b )). (a + b) (a + b ) = (p p + q q )n, To prove (2), we need to show that ab a b (mod n). Write a = pn + r, a = p n + r, b = qn + s, b = q n + s. Then ab a b = (pn+r)(qn+s) (p n+r)(q n+s) = (pqn+ps+qr p q n p s q r)n, therefore n (ab a b ). Exercises: 1, 3, 4, 7, 10, 11, 12, 19, 30, 31, 33 Section 2.2: Modular Arithmetic We will construct addition and multiplication tables for Z n. Addition: Let 0 r, s < n be given. Then [r] [s] = [r + s]. Multiplication: Let 0 r, s < n be given. Then [r] [s] = [rs]. 8

For example, let n = 15. We wish to multiply [19] and [41]. Since neither 19 nor 41 is in range, we must first write [19] = [4] and [41] = [11]. Therefore [19] [41] = [4] [11] = [4 11] = [44] = [14]. This is a rather clumsy process. It would be nice to say that [19] [41] = [19 41] = [779]. However, there is some ambiguity here: we ve already computed [19] [41] = [14]. If [779] [14] then we re in trouble. We need to prove that [a] [b] produces the same class no matter how we perform the multiplication, [ab] versus [rs]. The same comment applies to [a + b] versus [r + s]. Theorem: Let a and b be arbitrary integers. Then [a] [b] = [a + b] and [a] [b] = [ab]. Proof. Write a = pn + r and b = qn + s. We must show that [r + s] = [a + b] and [rs] = [ab]. We know that [r] = [a] and [s] = [b], so the conclusion follows by properties of congruence (see above). Corollary: Let k be a positive integer. (1) k i=1 [a] = [ka]. (2) k i=1 [a] = [ak ]. Proof. Both statements can be proved by induction on k. Properties of arithmetic operations in Z n : See Theorem 2.7, page 34. We can prove some amazing things using modular arithmetic, which justifies its existence. For example, we can prove that 3 99 + 1 is a multiple of 7. Here s how: working in Z 7 we have [3 99 ] = [3] 99 = ([3] 3 ) 33 = [27] 33 = [ 1] 33 = [( 1) 33 ] = [ 1], 3 99 1 (mod 7), 9

7 3 99 + 1. Exercises: 1, 2, 3, 6, 7, 8, 10 Additional Exercise: Let a > 1 and b be two integers which satisfy gcd(a, b) = 1. Working in Z a, show that one can always find a positive integer k such that k i=1 [b] = [1]. Section 2.3: The Structure of Z p When p is Prime In algebra we learn that there is exactly one solution to the linear equation ax + b = c when a 0: x = c b. However, this is not necessarily the case a in Z n : For example, if n = 6 then there are two solutions to the equation [4][x] + [1] = [5], namely [x] = [1] and [x] = [4]. Moreover, there are no solutions to to the equation [4][x] + [2] = [5]. However, when n is a prime number the situation is a little more predictable. To demonstrate this, we will first establish the relevant properties of Z p. Theorem: Let p be a prime number. (1) For each [a] [0] there exists a [b] [0] such that [a][b] = [1]. (2) For each [a] [0] the equation [a][x] = [a][y] implies [x] = [y]. In particular the solution to [a][x] = [1] is unique. (3) [x][y] = [0] implies [x] = [0] or [y] = [0]. Proof. Let [a] [0] be given. Without loss of generality we can assume 1 a < p. Therefore gcd(a, p) = 1. Therefore there exist integers j and k such that ja + kp = 1. Therefore [1] = [ja + kp] = [ja] + [kp] = [ja] + [0] = [ja + 0] = [ja] = [j][a]. Now suppose [a] [0] and [a][x] = [a][y]. Choose [b] such that [a][b] = [1]. Then [b][a][x] = [b][a][y], therefore [1][x] = [1][y], therefore [x] = [y]. Now consider [x][y] = [0]. Then p xy. Therefore p x or p y. Therefore [x] = [0] or [y] = [0]. 10

We now return to the the problem of counting the number of solutions to the equation [a][x] + [b] = [c] over Z p. Theorem: Let p be a prime number. There is exactly one solution to the equation [a][x] + [b] = [c] when [a] [0]. Proof. We first show that at least one solution exists: Let [y] be the unique solution to [a][y] = [1]. Let [x] = [y(c b)]. Then [a][x] + [b] = [a][y(c b)] + [b] = [a][y][c b] + [b] = [1][c b] + [b] = [c b] + [b] = [c]. To show that [x] is unique, suppose that another solution is [x ]. Then we have [a][x] + [b] = [c] = [a][x ] + [b], Therefore therefore [a][x] = [c] + [ b] = [a][x ], [x] = [x ] by Property 2 of the previous theorem. Property 3 above is helpful for solving polynomial equations over Z p. For example, we will solve the quadratic equation over Z 11 : First note that [6][x] 2 + [4][x] + [1] = [0] [6][x] 2 + [4][x] + [1] = [6]([x] 2 + [8][x] + [2]) = [6]([x] + [9])([x] + [10]). Hence solving [6][x] 2 + [4][x] + [1] = [0] 11

is equivalent to solving ([x] + [9])([x] + [10]) = [0]. The only possibilities are [x] + [9] = [0] or [x] + [10] = [0], therefore [x] = [2] and [x] = [1] are the two solutions. Exercises: 1, 2, 4, 5, 6, 7, 8 10, 11, 13. Additional Exercise: Find all solutions to over Z 13. Chapter 3: Rings [3][x] 2 + [10][x] + [1] = [0] Read the introductory remarks beginning on page 41 carefully. They describe the essence of Abstract Algebra: beginning with a few well-understood examples of concrete mathematical objects exhibiting analogous properties, we pull back to an abstract object and prove theorems for this object. The theorems which hold for this abstract object will hold for the all the concrete objects it represents. In this chapter we pull back from algebraic operations in Z and Z n to what are called rings. We will also see many other examples of rings. Section 3.1: Definition and Examples of Rings. Rings are defined carefully on page 42. They are sets which are closed under addition and multiplication. (To say that a set is closed under an operation means that whenever you produce a new object by from existing ones using this operation, the new object also belongs to the set.) Addition has an additive identity called Zero, or 0 R if the name of the ring is R. Both addition and multiplication obey the associative property that all real numbers do. Addition and multiplication together obey the distribute law, which is the property we use to combine like terms when adding polynomials together. Note that the definition of ring on page 42 does not require that multiplication is commutative or that there exists a multiplicative identity. Hence there are commutative rings and rings with identity, but there exist rings without either property. Most of the rings we will be studying contain a 12

multiplicative identity, 1 R, but many rings of interest are not commutative. An example of a non-commutative ring is the set of all 2 2 matrices with real number entries, M 2 2 (R). If we set then we have and hence xy yx. xy = yx = ( ) 1 0 x =, y = 0 0 ( ) ( ) 1 0 1 2 0 0 3 4 ( ) ( ) 1 2 1 0 3 4 0 0 ( ) 1 2 3 4 = = ( ) 1 2 0 0 ( ) 1 0, 3 0 Note also that the definition of ring on page 42 allows for the possibility that xy = 0 R when neither x nor y is equal to the additive identity. For example, in the ring M 2 2 (R) we have x = ( ) 1 0 0 0 0 R, y = ( ) ( ) 1 0 0 0 xy = = 0 0 0 1 ( ) 0 0 0 0 1 R, ( ) 0 0 = 0 0 0 R. A ring which satisfies the property that xy = 0 R always implies x = 0 R or y = 0 R is called an integral domain. This book insists that an integral domain must also be commutative and have a multiplicative identity 1 R which is distinct from the additive identity 0 R. The set of integers Z is an example of a ring which is an integral domain. Note however that there is no solution to 2x = 1 where x Z. A field is an integral domain R for which the equation ax = 1 R has a solution x R for each a 0 R belonging to R. Two fields you are familiar with are the set of real numbers R and the set of complex numbers C. The hierarchy we have described so far is this: ring, integral domain, field. An integral domain is a ring meeting additional restrictions, and a field is an integral domain meeting additional restrictions. 13

There are many ways to construct new rings from existing ones. For example, if R is a ring then so is S = M 2 2 (R), the set of 2 2 matrices whose entries belong to R. The example above shows that if S can never be an integral domain when R is. Another way to construct a new ring from an existing ring R is to produce a proper subset S R which is closed with respect to addition, multiplication, and formation of additive inverses. For example, if R = Z, then S = {2x : x Z} is a ring. Note that S does not contain a multiplicative identity. Note also that the subset T = {2x + 1 : x Z} is not a ring. T is closed under multiplication and formation of additive inverses, but is not closed under addition. For example, 1 T and 3 T, but 1 + 3 T. Exercises: 1, 4, 5, 9, 13, 14, 15, 18, 22, 29, 31, 39 Section 3.2: Basic Properties of Rings This section is a bit tedious in my opinion. Suffice it to say that if you understand the axioms for a ring well, you can prove lots of other properties, some of which are interesting in their own right and some of which are useful for calculations. Read the statements of all the theorems, read the proofs of a few of them carefully to get the flavor of how these proofs are constructed, then try the exercises. The point of these exercises is to justify each step in a proof or a calculation by referring back to definitions. Exercises: 1, 2, 6, 7, 8, 11, 12, 16, 30, 31, 32, 35, 37 Section 3.3: Isomorphisms and Homomorphisms Definition: Let R and S be two rings. A homomorphism between R and S is a function f : R S which preserves addition and multiplication, in the sense that f(a + b) = f(a) + f(b) and f(ab) = f(a)f(b) for all pairs a, b R. Example of a homomorphism: f : C R defined by f(a + bi) = a. Note: Not all functions f : R S are homomorphisms. To take one example, let f : C R be defined by f(x + yi) = x 2 + y 2. 14

Then f preserves multiplication but does not preserve addition. To see this, let a = x 1 + y 1 i and b = x 2 + y 2 i. Then ab = (x 1 x 2 y 1 y 2 ) + (x 1 y 2 + x 2 y 1 )i. We have f(a) = x 2 1 + y1, 2 f(b) = x 2 2 + y2, 2 f(ab) = (x 1 x 2 y 1 y 2 ) 2 + (x 1 y 2 + x 2 y 1 ) 2 = x 2 1x 2 2 + y1y 2 2 2 + x 2 1y2 2 + x 2 2y1 2 = (x 2 1 + y1)(x 2 2 2 + y2) 2 = x 2 1 + y1 2 x 2 2 + y2 2 = f(a)f(b). This demonstrates that f preserves multiplication no matter what choices we make for a and b. Also, let a = 1 + 0i and b = 0 + 1i. Then a + b = 1 + i, f(a) = 1, f(b) = 1, f(a + b) = 2, therefore f(a + b) f(a) + f(b). This demonstrates that f fails to preserve addition in at least one instance, therefore f does not preserve addition for all pairs a, b C. Definition: Let R and S be two rings. An isomorphism between R and S is a homomorphism f : R S which is both one-to-one and onto. An isomorphism f : R S establishes a one-to-one correspondence between the elements of R and the elements of S, and preserves addition and multiplication. In some sense, R and S are the same ring. If we apply f to all the entries of the addition table for R, we will obtain the addition table for S. Similarly, the image of the multiplication table for R is the multiplication table for S. Example of an isomorphism: f : Z 15 Z 3 Z 5 defined by f([x] 15 ) = ([x] 3, [x] 5 ). We must be careful to check that f is well-defined. Namely, if [x] 15 = [x ] 15, is it also true that [x] 3 = [x ] 3 and [x] 5 = [x ] 5? Yes: 15 (x x ), therefore 3 (x x ) and 5 (x x ). We must also be careful to check that f is one-to-one and onto. One-to-one: Assume f([x] 15 ) = f([y] 15 ). We must show [x] 15 = [y] 15. We know that [x] 3 = [y] 3 and [x] 5 = [y] 5, therefore 3 (x y) and 5 (x y), therefore 15 (x y), therefore [x] 15 = [y] 15. Onto: Let ([a] 3, [b] 5 ) be given. We must find [x] 15 such that [x] 3 = [a] 3 and [x] 5 = [b] 5. This is equivalent to find a solution for p and q in the equation 15

3p + 5q = b a, for then we can simply set x = 3p + a and rest assured that we also have x = 5q + b. Here s how we can solve this problem: 3(7) + 5( 4) = 1, 3(7)(b a) + 5( 4)(b a) = b a, therefore p = 7(b a) and q = 4(b a). So for example if a = 2 and b = 4 then p = 14, q = 8, x = 44. Hence f([44] 15 ) = ([44] 3, [44] 5 ) = ([2] 3, [4] 5 ). Isomorphisms can be used to group rings into isomorphism classes. The two rings Z 15 and Z 3 Z 5 belong to the same isomorphism class. All rings in this class can be characterized as follows: They are all isomorphic images of Z 15. They all have 15 elements and are commutative. If R is in the isomorphism class of Z 15 and f : Z 15 R is an isomorphism, then the elements of R are multiplies of r = f([1] 15 ): every s R can be realized as k r for some 0 k < 15. Reason: if s = f([k] 15 ), where 0 k < 15, then s = f( k 15 [1] 15 ) = f([1] 15 ) = k r. i=1 i=1 The two rings Z 4 and Z 2 Z 2 are not isomorphic, hence belong to two different isomorphism classes. To prove this, we must show that the algebraic structure of these two rings are different. Observe that every element x Z 2 Z 2 satisfies x + x = 0, but this is not true of Z 4. Therefore no homomorphism f : Z 4 Z 2 Z 2 can be injective: we have and f([0] 4 ) = ([0] 2, [0] 2 ) f([2] 4 ) = f([1] 4 + [1] 4 ) = f([1] 4 ) + f([1] 4 ) = ([0] 2, [0] 2 ), hence f([0] 4 ) = f([2] 4 ) while [0] 4 [2] 4. (Note: this proof relies on the property that if f : R S is a homomorphism then it must be the case that f(0 R ) = 0 S. Verify this.) In general, to prove that the two rings R and S are not isomorphic, it is sufficient to show that no ring homomorphism 16

f : R S can be both one-to-one and onto. One or the other property must fail. Homomorphisms can be used to construct new rings from existing ones. To give just one example, consider the following theorem: Theorem: Let f : R S be a ring homomorphism. Then f(r) is a ring. Proof. We need to show that f(r) is closed under addition, multiplication, and formation of inverses. Let x, y f(r) be given. Then x = f(r) and y = f(s) for some r, s S. Therefore and x + y = f(r) + f(s) = f(r + s) f(r) xy = f(r)f(s) = f(rs) f(r). Hence f(r) is closed under addition and multiplication. Now let x = f(r) f(r) be given. We must show we can find x f(r) such that x + x = 0 S. We know we can find r R such that r + r = 0 R. Set x = f(r ). Then x + x = f(r) + f(r ) = f(r + r ) = f(0 R ) = 0 S. Note that if f : R S is a ring homomorphism, and f is one-to-one, then R is isomorphic to f(r) (verify this). Exercises: 1, 4, 6, 7, 10, 13, 16, 23, 28, 29, 33, 38, 40 Chapter 4: Arithmetic in F [x] Read the introductory remarks pp. 80-81 carefully. We will define polynomials precisely and develop their properties in an abstract setting. Section 4.1: Polynomial Arithmetic and the Division Algorithm Definition: Let R be a ring. Then we define the ring of polynomials R[x] to be the set of all expressions of the form p(x) = a i x i, i=0 17

where each a i belongs to R and only a finite number of these coefficients are not equal to 0 R. The degree of p(x) is the largest index n such that a n 0 R. (If all coefficients equal 0 R, then p(x) is the additive identity and by convention has degree.) Two polynomials are equal if and only if they have the same coefficients. It can be awkward to use the notation By convention we write p(x) = a i x i. k=0 p(x) = a 0 + a 1 x + + a n x n to indicate that the coefficient of x i is a i for i n and that a k = 0 R for all k > n. To illustrate the ins and outs of this notation, suppose R = M 2 2 [Z 3 ]. Then R[x] consists of polynomials whose coefficients are 2 2 matrices with entries in the ring Z 3. The polynomial ( ) [0] [1] p(x) = x + x 3 [2] [1] has degree 3, and its coefficients are ( ) ( ) [0] [0] [0] [1] a 0 =, a [0] [0] 1 =, a [2] [1] 2 = a 4 = a 5 = = ( ) [0] [0], a [0] [0] 3 = ( ) [0] [0]. [0] [0] ( ) [1] [0], [0] [1] If p(x) and q(x) are polynomials with coefficients a i and b i respectively, then p(x) + q(x) is the polynomial whose coefficients are of the form c i = a i + b i, and p(x)q(x) is the polynomial whose coefficients are of the form c i = a j b k = a 0 b i + a 1 b i 1 + + a i b 0. j+k=i 18

The additive identity is the polynomial 0 R[x] = 0 R x i. i=0 If R has a multiplicative identity 1 R, then the multiplicative identity in R[x] is 1 R[x] = 1 R x 0 + 0 R x i. If r R then typically we write r to denote the polynomial i=1 rx 0 + 0 R x i. i=1 Using this convention, the additive identity in R[x] is 0 R and the multiplicative identity is 1 R. Division Algorithm in F [x], where F is a field: Let a(x) and b(x) 0 F be polynomials. Then there exist a unique pair of polynomials q(x) and r(x) such that a(x) = q(x)b(x) + r(x), where r(x) = 0 F or r(x) 0 F and 0 deg r(x) < deg b(x). Proof. The book proves this by induction on deg a(x). For variety we will use the Well-Ordering Axiom. First, note that if a(x) = q(x)b(x) + r(x) is possible, then the choice of q(x) and r(x) is unique. For suppose a(x) = Q(x)b(x) + R(x). Then we have q(x)b(x) + r(x) = Q(x)b(x) + R(x), If q(x) Q(x), then (q(x) Q(x))b(x) = R(x) r(x). deg (q(x) Q(x))b(x) deg b(x) contradicts deg (R(x) r(x)) < deg b(x). 19

Therefore q(x) = Q(x), which forces r(x) = R(x). Secondly, note that the conclusion of the theorem is true if a(x) = 0 F. In this case we can write 0 F = 0 F b(x) + 0 F, hence we can choose q(x) = 0 F and r(x) = 0 F. Thirdly, note that the conclusion of the theorem is true if deg a(x) < deg b(x). In this case we can write a(x) = 0 F b(x) + a(x), hence we can choose q(x) = 0 F and r(x) = a(x). Suppose the theorem is not true. Then by the Well-Ordering Axiom there is a polynomial A(x) 0 F of smallest possible degree with the following property: there exists a polynomial B(x) 0 F such that A(x) cannot be expressed in the form A(x) = q(x)b(x) + r(x), where r(x) = 0 F or deg r(x) < deg B(x). We know from the argument above that we must have deg A(x) deg B(x). Write deg A(x) = n and deg B(x) = m. Let a(x) = A(x) a n b 1 m x n m B(x). Then a(x) = 0 F or deg a(x) < deg A(x). In either case, we know that a(x) can be expressed in the form a(x) = q(x)b(x) + r(x), where r(x) = 0 F or deg r(x) < deg B(x). But now we have that is where A(x) = a(x) + a n b 1 m x n m B(x) = q(x)b(x) + r(x) + a n b 1 m x n m B(x), A(x) = Q(x)B(x) + r(x) Q(x) = q(x) + a n b 1 m x n m. Contradiction. Therefore the theorem must be true after all. 20

Exercises: 4, 5, 6, 7, 11, 13, 15, 18, 19 Section 4.2: Divisibility in F [x]. In this section, F represents a field. coefficient equal to 1 F. A monic polynomial has leading Definition: Let f(x) and g(x) be polynomials in F [x]. Then f(x) g(x) if and only if g(x) = h(x)f(x) for some h(x) F [x]. Definition: Let f(x) F [x] be given. The divisor set of f(x) is D(f(x)) = {g(x) : g(x) divides f(x)}. Definition: Let f(x) and g(x) be given polynomials in F [x]. The greatest common divisor is the monic polynomial of largest degree in D(f(x)) D(g(x)). Note that this definition suggests that there is a unique monic polynomial in D(f(x)) D(g(x)) of largest degree. We will postpone the proof of this until we establish the following three theorems. Theorem 1: Let f(x) and g(x) be polynomials in F [x]. Let t(x) be a polynomial of least possible degree in the set S = {a(x)f(x) + b(x)g(x) : a(x), b(x) F [x]}. Then t(x) D(f(x)) D(g(x)). Proof. Write t(x) = a 0 (x)f(x) + b 0 (x)g(x). By the division algorithm, we can write f(x) = q(x)t(x) + r(x), where r(x) = 0 or 0 deg r(x) < deg t(x). We wish to show that r(x) = 0. If r(x) 0, we have r(x) = f(x) q(x)t(x) = (1 q(x)a 0 (x))f(x) + ( q(x)b 0 (x))g(x), hence r(x) S(x) and has degree smaller than t(x). Contradiction. Therefore r(x) = 0 and t(x) f(x). A similar argument shows that t(x) g(x). 21

Theorem 2: Let f(x) and g(x) be polynomials in F [x]. Let t(x) be a polynomial of least possible degree in the set S = {a(x)f(x) + b(x)g(x) : a(x), b(x) F [x]}. If h(x) D(f(x)) D(g(x)) then h(x) t(x). Proof. Write Then we have Therefore h(x) t(x). t(x) = a 0 (x)f(x) + b 0 (x)g(x), f(x) = p(x)h(x), g(x) = q(x)h(x). t(x) = (a 0 (x)p(x) + b 0 (x)q(x))h(x). Theorem 3: There is exactly one monic polynomial of smallest degree in the set S = {a(x)f(x) + b(x)g(x) : a(x), b(x) F [x]}. Proof. Let t(x) and T (x) be two such polynomials. We must show that t(x) = T (x). By Theorem 1, t(x) is a divisor of f(x) and g(x). Therefore, by Theorem 2, t(x) T (x). Similarly, we can say that T (x) t(x). Let us write and t(x) = q(x)t (x) T (x) = Q(x)t(x). Since t(x) and T (x) are both monic, q(x) and Q(x) must also be monic. We have t(x) = q(x)q(x)t(x). This forces q(x)q(x) = 1, which in turn forces q(x) = Q(x) = 1. Therefore t(x) = T (x). 22

We can now show that there is a unique monic polynomial of largest degree in D(f(x)) D(g(x)). Theorem 4: Any polynomial of largest degree in D(f(x)) D(g(x)) is a monic polynomial of smallest degree in Therefore it is unique. S = {a(x)f(x) + b(x)g(x) : a(x), b(x) F [x]}. Proof. Let d(x) be any monic polynomial of maximum degree in D(f(x)) D(g(x)). Let t(x) be the unique monic polynomial of smallest degree in S = {a(x)f(x) + b(x)g(x) : a(x), b(x) F [x]}. By Theorem 1, t(x) D(f(x)) D(g(x)). By Theorem 2, d(x) t(x). We will write t(x) = q(x)d(x). Since t(x) and d(x) are both monic, both belong to D(f(x)) D(g(x)), and deg t(x) deg d(x), it must be true that t(x) has the same degree as d(x). Therefore q(x) has degree 0. Since t(x) and d(x) are both monic, q(x) = 1. Therefore t(x) = d(x). Theorems 1 through 4 allow us to say that the greatest common divisor of f(x) and g(x) is the unique monic polynomial of smallest degree in S = {a(x)f(x) + b(x)g(x) : a(x), b(x) F [x]}. Definition: f(x) and g(x) are relatively prime if their greatest common divisor is equal to 1 F. Theorem: If f(x) and g(x) are relatively prime then f(x) g(x)h(x) f(x) h(x). 23

Proof. If f(x) h(x) then clearly f(x) g(x)h(x). Conversely, suppose f(x) g(x)h(x). Write a(x)f(x) + b(x)g(x) = 1. Then Write Substituting, we obtain a(x)f(x)h(x) + b(x)g(x)h(x) = h(x). g(x)h(x) = q(x)f(x). a(x)f(x)h(x) + b(x)q(x)f(x) = h(x). Therefore Therefore h(x) = (a(x)h(x) + b(x)q(x))f(x). f(x) h(x). The Euclidean Algorithm (pp. 4 5 of these notes) can be modified to find the greatest common divisor of two polynomials. See the example on page 93 of the textbook. Exercises: 2, 3, 5, 6ad, 9, 10, 14, 15. Section 4.3: Irreducibles and Unique Factorization In this section, F is a field. Definitions: The units of a ring R are those elements r R for which there exists an element s R such that r s = 1 R. Two elements r and s of a ring are associates if and only if r = u s for some unit u. It is easy to verify that in F [x], the only units are non-zero polynomials of degree 0: α + 0 F x i, i 1 where α 0. Hence two polynomials f(x) and g(x) are associates in F [x] if and only if f(x) = αg(x) for some nonzero α F. 24

Definition: A non-constant polynomial f(x) F [x] is irreducible if and only if its divisor set consists of units and associates. Note that if f(x) is irreducible, it cannot be factored into polynomials of degree d where 0 < d < deg f(x). In fact, in any factorization, its factors must all have degree 0 or degree deg f(x). An irreducible polynomial plays the same role in F [x] that a prime number plays in Z. Therefore irreducibles should have the same properties we encountered in Section 1.3. We will establish these properties in short order (go back and review pp. 5 7 of these notes, which I have slightly modified to meet the present need). Theorem: a polynomial p(x) is irreducible if and only if, for all polynomials a(x) and b(x), if p(x) a(x)b(x) then p(x) a(x) or p(x) b(x). Proof. First assume that p(x) is irreducible. Now suppose that p(x) a(x)b(x). We will consider two cases. Case 1. p(x) a(x). Then we have the desired conclusion. Case 2. p(x) a(x). Then none of the associates of p(x) divide a(x). Since D(p(x)) consists only of units and associates of p(x), D(p(x)) D(a(x)) consists only of units. Therefore gcd(p(x), a(x)) = 1 F. A previous theorem says that gcd(p(x), a(x)) = 1 F and p(x) a(x)b(x) implies p(x) b(x). Conversely, assume that the condition on p(x) holds. We will show that p(x) is irreducible. Let r(x) be a divisor of p(x). Then we have p(x) = s(x)r(x) for some polynomial s(x). The polynomial s(x) is another divisor of p(x). Hence we have r(x) p(x) and s(x) p(x). By the condition on p(x), we know that p(x) r(x) or p(x) s(x). If p(x) r(x) then we have p(x) and r(x) dividing each other, hence p(x) and r(x) are associates. If p(x) s(x) then we have p(x) and s(x) dividing each other, hence y(x) and s(x) are associates, which forces r(x) to be a unit. Therefore we have shown that all divisors of p(x) are either units or associates of p(x). Therefore p(x) is irreducible. Corollary: if p(x) is irreducible and p(x) a 1 (x)a 2 (x) a n (x), then p(x) a i (x) for some i n. 25

Proof. By induction on n 2. The base case is the theorem we have just proved. Assume the conclusion is true for any n-fold product of polynomials. Now consider p(x) a 1 (x) a n+1 (x). By the theorem, p(x) a 1 (x)a 2 (x) a n (x) or p(x) a n+1 (x). If the first condition holds, by the induction hypothesis we can say that p(x) a i (x) for some i n. If the second condition holds, then p(x) a n+1 (x). So either way p(x) a i (x) for some i n + 1. Theorem: Every non-constant polynomial is either irreducible or is the product of two or more irreducible polynomials. Proof. Suppose there is a non-constant polynomial which is neither irreducible nor the product of irreducible polynomials. Let S be the set of all such polynomials. By the Well-Ordering Axiom, there must be one of smallest possible degree, which we will call P (x). Since P (x) is not irreducible, it has a divisor r(x), where 0 < deg r(x) < deg p(x). Therefore we can write P (x) = r(x)s(x). Note that 0 < deg s(x) < deg p(x) also. Now r(x) and s(x) are non-constant polynomials which do not belong to S, therefore are either irreducible or the product of irreducible polynomials. This implies that P (x) itself is the product of irreducible polynomials. Contradiction. Therefore there cannot exist any non-constant polynomials which are neither irreducible nor the product of irreducible polynomials. The following theorem is the polynomial analogue of the Fundamental Theorem of Arithmetic. Theorem: Let p(x) be a non-constant polynomial. Let and p(x) = r 1 (x)r 2 (x) r j (x) p(x) = s 1 (x)s 2 (x) s k (x) be two ways to factor p(x) into irreducible polynomials. Then j = k and there is some permutation of s 1 (x) through s k (x) such that for each i, r i (x) and s σ(i) (x) are associates. 26

Proof. Note that if p(x) is irreducible then both lists consist of one irreducible each and the two lists are equal. So if the theorem is false, it must be false for a polynomial p(x) of smallest possible degree, and we can assume that p(x) is not irreducible. If we write p(x) = r 1 (x)r 2 (x) r j (x) = s 1 (x)s 2 (x) s k (x), where it is not possible to rearrange s 1 (x) through s k (x) so that they are associates of r 1 (x) through r j (x), then we must have j > 1 and k > 1. Since we have r 1 (x)r 2 (x) r j (x) = s 1 (x)s 2 (x) s k (x), r 1 (x) s 1 (x)s 2 (x) s k (x). Since r 1 (x) is irreducible, we must have r 1 (x) s a (x) for some a. Since s a (x) is irreducible, r 1 (x) must be an associate of s a (x) (it cannot be a unit because it has degree 1). We will write s a (x) = αr 1 (x). Set q(x) = p(x) αr 1 (x) = (α 1 r 2 (x)) r j (x). Then we also have q(x) = i a s i (x). Since q(x) has degree smaller than p(x), q(x) does not belong to S. Therefore j 1 = k 1 and there is a way to rearrange all the s i (x) polynomials for i a so that they are associates of α 1 r 2 (x) through r j (x), hence associates of r 2 (x) through r j (x). Since r 1 (x) and s a (x) are also associates, there is a way to rearrange s 1 (x) through s k (x) so that they are associates of r 1 (x) through r j (x). Contradiction. Therefore there are no counterexamples and the theorem is true. Exercises: 1, 3, 6, 9, 10, 12, 13, 15, 23, 26 Section 4.4: Polynomial Functions, Roots, and Reducibility Let F represent a field. following idea: All the results of this section flow from the 27

Let f(x) be a polynomial. Let a F be given. By the quotient-remainder theorem, we have f(x) = q(x)(x a) + r(x), where r(x) = 0 or 0 deg r(x) < 1. That is, r(x) is a constant, and we can write f(x) = q(x)(x a) + r. Substituting x = a we obtain To summarize: f(a) = r. Theorem: Let f(x) F [x] and a F be given. Then f(x) = q(x)(x a) + f(a) for some polynomial q(x). In fact we can compute q(x) as follows: q(x) = f(x) f(a). x a Corollary: Let f(x) F [x] and a F be given. Then x a divides f(x) if and only if f(a) = 0 F. Corollary: Let f(x) F [x] and a F be given. Then x a divides f(x) f(a). Corollary: Let f(x) F [x] be a polynomial of degree d 0. Then f(x) has at most d distinct roots in F. Proof. By induction on d 0. Base Case: f(x) has degree 0, therefore it is a constant polynomial: f(x) = α 0 for some α F. Therefore f(x) has no roots in F. Induction Hypothesis: all polynomials of degree d in F [x] have at most degree roots in F. Now let f(x) be a polynomial of degree d + 1. We will count the distinct roots of f(x). If f(x) has no roots, we are done. Otherwise, it has at least one root a. Therefore we can write f(x) = q(x)(x a) 28

for some polynomial q(x) of degree d. By the induction hypothesis, q(x) has at most d roots. All the roots of f(x) which are not equal to a must be roots of q(x). Therefore f(x) has at most d distinct roots which are not equal to a. Hence f(x) has at most d + 1 distinct roots. Corollary: Let f(x) and g(x) be two polynomials in F [x]. Suppose that f(a) = g(a) for n distinct values of a F and that deg(f(x) g(x)) < n. Then f(x) = g(x). Proof. If deg(f(x) g(x) = then f(x) g(x) = 0, therefore f(x) = g(x). But if deg(f(x) g(x) 0, then the hypothesis implies that f(a) g(a) has n distinct roots, and by the previous corollary we again have f(x) g(x) = 0. Applications: let f(x) = x 3 + 5x 18. Then f(1) = 10, therefore x 3 +5x 18 is not divisible by x 1. However, f(2) = 0, therefore x 3 +5x 18 is divisible by x 2. By long division we find that we can factor f(x) as follows: f(x) = (x 2)(x 2 + 2x + 9). Observe further that x 2 + 2x + 9 = (x + 1) 2 + 8. Since there is no real number a which satisfies (a + 1) 2 + 8 = 0, we know that we cannot find a linear (degree 1) factor of x 2 + 2x + 9. Therefore x 2 + 2x + 9 is irreducible in R[x]. Hence the prime factorization of x 3 + 5x 18 in R[x] is x 3 + 5x 18 = (x 2)(x 2 + 2x + 9). If we change our field from R to C, we get a different conclusion. Note that a = 1 + 8i is a root of (x + 1) 2 + 8 in C, therefore x + 1 8i is a divisor of x 2 + 2x + 9 in C[x]. By long division we find that we can factor x 2 + 2x + 9 as follows: x 2 + 2x + 9 = (x + 1 8i)(x + 1 + 8i). 29

Therefore the prime factorization of x 3 + 5x 18 in C[x] is x 3 + 5x 18 = (x 2)(x + 1 8i)(x + 1 + 8i). A method for proving that a degree 3 polynomial f(x) is irreducible in F [x]: If it is not irreducible, then it must factor as f(x) = p(x)q(x) where the degrees of p(x) and q(x) are equal to 1 or 2. Therefore one of these polynomials will be of the form x a. Therefore f(a) = 0. So if we can prove that f(x) does not have a root in F [x], then it is automatically irreducible. For example, x 3 +x+1 is irreducible in Z 5 [x]. It is easy to verify this by checking the five possible values of a Z 5 and showing that in each case f(a) 0. Note that if a polynomial has degree 4, then it is possible that it could be reducible even if it has no root in F. For example, the polynomial x 4 + 5x 2 + 6 has no root in R, yet we can factor it: x 4 + 5x 2 + 6 = (x 2 + 2)(x 2 + 3). Let n 1 be an integer. Let a = e i 2π n C. Let and let f(x) = (x a)(x a 2 ) (x a n ) g(x) = x n 1. Then f(x) and g(x) are both monic polynomials of degree n, which implies that deg(f(x) g(x)) < n. Observe that the n numbers a, a 2,..., a n are all roots of f(x) and roots of g(x). Therefore f(x) = g(x). We have proved n (x cos( 2π n ) sin(2π n )i) = xn 1. i=1 For example, setting n = 3 we have (x + 1 2 3 2 i)(x + 1 2 + 30 3 2 i)(x 1) = x3 1.

Exercises: 2, 3, 5, 6, 7, 8, 10, 15, 17, 19 Section 4.5: Irreducibility in Q[x] We have seen above that a polynomial can be irreducible over one ring and reducible over another ring. x 2 +1 is irreducible in R[x] yet reducible in C[x]. In this section we will prove that a polynomial with integer coefficients is irreducible in Z[x] if and only if it is irreducible in Q[x]. If p is a prime number which does not divide the leading coefficient of a polynomial f(x) with integer coefficients, then f(x) is irreducible in Q[x] provided it is irreducible in Z p [x]. We will also prove the Rational Root Test, which will help us to factor polynomials with integer coefficients, and Eisenstein s Criterion, which yields a large class of irreducible polynomials in Z[x]. We will start with a useful ring homomorphism: Let p be a prime number and define θ : Z[x] Z p [x] by θ(a 0 + a 1 x + + a n x n ) = [a 0 ] + [a 1 ]x + + [a n ]x n. θ can be used to prove the following result: Lemma: Let a(x) and b(x) be polynomials in Z[x] and assume that a prime number p divides all the coefficients of a(x)b(x). Then either p divides all the coefficients of a(x) or p divides all the coefficients of b(x). Proof. Suppose p divides all the coefficients of a(x)b(x). Then θ(a(x)b(x)) = 0. Since θ is a ring homomorphism, we have θ(a(x))θ(b(x)) = 0. Since Z p [x] does not have any zero-divisors, we must have θ(a(x)) = 0 or θ(b(x)) = 0. Therefore p divides all the coefficients of a(x) or it divides all the coefficients of b(x). Corollary: Let f(x) be a polynomial with integer coefficients. If f(x) is reducible in Q[x] then it is reducible in Z[x]. 31

Proof. Suppose f(x) = a(x)b(x), where a(x) and b(x) have rational coefficients. We will show that f(x) = A(x)B(x) is possible where A(x) and B(x) have integer coefficients and A(x) has the same degree as a(x) and B(x) has the same degree as b(x). Let D be any positive integer such that Da(x) and Db(x) have integer coefficients. Write a (x) = Da(x) and b (x) = Db(x). Then D 2 f(x) = a (x) b (x) is an equation involving polynomials with integer coefficients. We would like to divide through by D 2 without creating fractional coefficients on the right-hand side. Let D 2 = p 1 p 2 p k be the prime factorization of D 2. Since p 1 divides all the coefficients of D 2 f(x), it divides all the coefficients of a (x)b (x). By the Lemma above, p 1 either divides all the coefficients of a (x) or all the coefficients of b (x). Divide the appropriate polynomial by p 1, leave the other polynomial alone, call the new polynomials a (x) and b (x), and observe that they both have integer coefficients and satisfy the equation p 2 p 3 p k f(x) = a (x)b (x). Since p 2 divides all the coefficients of p 2 p 3 p k f(x), it either divides all the coefficients of a (x) or all the coefficients of b (x). So we can create a (x) and b (x) with integer coefficients such that p 3 p 4 p k f(x) = a (x)b (x). Keep on going until we have divided through by p 1 p 2 p k. We will be left with f(x) = A(x)B(x), where A(x) and B(x) have integer coefficients. To illustrate the use of this result, suppose we are attempting to factor the polynomial f(x) = 2x 4 +x 3 21x 2 14x+12 into a product of irreducible polynomials in Z[x]. One way to proceed is to look for a rational number r s such that f( r ) = 0. This tells us that s 2x 4 + x 3 21x 2 14x + 12 = q(x)(x r s ) 32

for some polynomial q(x) with rational coefficients. Having factored f(x) into two polynomials with rational coefficients, we know it is possible to factor it into two polynomials with integer coefficients. The Rational Root Test tells us how to guess the root r s. Rational Root Test: Let f(x) = a 0 + a 1 x + + a n x n be a polynomial with integer coefficients, and let r and s be non-zero relatively prime integers. Then f( r s ) = 0 implies r a 0 and s a n. Proof. We have therefore f( r s ) = a r 0 + a 1 s + + a r n 1 n 1 s + a r n n 1 n, s n s n f( r s ) = a 0s n + a 1 rs n 1 + + a n 1 r n 1 s + a n r n, therefore f( r s ) = 0, therefore a 0 s n + a 1 rs n 1 + + a n 1 r n 1 s + a n r n = 0, therefore r a 0 s n and s a n r n. Since gcd(r, s) = gcd(r, s n ) = gcd(r n, s) = 1, the latter condition implies r a 0 and s a n. To complete the example of how to factor 2x 4 + x 3 21x 2 14x + 12, see the bottom of page 108 of the textbook, in which it takes at most 16 guesses to find r, given that r 2 and s 12 and gcd(r, s) = 1. The result is r = 1 and s s 2 r s = 3, which yields 2x 4 + x 3 21x 2 14x + 12 = q(x)(x 1 )(x + 3) 2 for some polynomial q(x). We can use long division to compute q(x) = 2x 2 4x 8. 33

Since we are seeking to factor 2x 4 + x 3 21x 2 14x + 12 into polynomials with integer coefficients, we will re-write this as 2x 4 + x 3 21x 2 14x + 12 = (2x 1)(x + 3)(x 2 2x 4). Since x 2 2x 4 = (x 1) 2 5 has no rational roots, we know that x 2 2x 4 is irreducible in Q[x], hence in Z[x]. Hence we have found the prime factorization of 2x 4 + x 3 21x 2 14x + 12 in Z[x]. Another nice use of the ring homorphism θ : Z[x] Z p [x] is that it can be used to transfer questions of reducibility from Z[x] to Z p [x]. Let f(x) be a polynomial with integer coefficients. Then the degree of f(x) is equal to the degree of θ(f(x)) if and only if p does not divide the leading term of f(x). This property allows us to prove the following theorem: Theorem: Let f(x) be a polynomial with integer coefficients, and let p be any prime number which does not divide the leading term of f(x). If f(x) is irreducible in Z p [x] then it is irreducible in Q[x]. Proof. We will prove the contrapositive, namely that if f(x) is reducible in Q[x] then it is reducible in Z p [x]. Say that the degree of f(x) is n. Suppose f(x) is reducible in Q[x]. Then it can be factored as f(x) = a(x)b(x) for non-constant polynomials a(x) and b(x) of degree r and n r with integer coefficients and with 1 r n 1. Since f n = a r b n r and p f n, we know that p a r and p b n r. Therefore deg(θ(a(x)) = r and deg(θ(b(x)) = n r. Hence θ(f(x)) = θ(a(x))θ(b(x)) proves that θ(f(x)) is reducible in Z p [x]. Example: Let f(x) = x 5 + 8x 4 + 3x 2 + 4x + 7. Let p = 2. Then θ(f(x)) = x 5 + x 2 + [1]. Since this doesn t have any roots in Z 2, it has no linear factors in Z 2. If it factors as x 5 + x 2 + [1] = ([a] + [b]x + [c]x 2 )([d] + [e]x + [f]x 2 + [g]x 3 ) 34

then we must have [a] = [c] = [d] = [g] = [1]. This forces x 5 + x 2 + [1] = ([1] + [b]x + x 2 )([1] + [e]x + [f]x 2 + x 3 ). Again, because x 5 + x 2 + [1] has no roots in Z 2, we must have [b] = [1]. Now we have x 5 + x 2 + [1] = ([1] + x + x 2 )([1] + [e]x + [f]x 2 + x 3 ). The right-hand side has the same coefficient of x 2 and x 3 when you multiply the polynomials together, namely [1+e+f]. This is not true of the left-hand side. Therefore this factorization is not possible. Since θ(f(x)) is irreducible in Z 2 [x], it is irreducible in Q[x]. Our last task is to prove Eisenstein s Criterion. Eisenstein s Criterion: Let f(x) = a 0 + a 1 x + + a n x n be a non-constant polynomial with integer coefficients. If there is a prime number p such that p divides a 0 through p n 1, p a n, and p 2 a 0, then f(x) is irreducible in Q[x]. Proof. Suppose not. Then we can write f(x) = b(x)c(x) for some pair of polynomials b(x) and c(x) with integer coefficients, each of degree between 1 and n 1. Write and b(x) = b 0 + b 1 x + + b r x r c(x) = c 0 + c 1 x + + c n r x n r. Then a 0 = b 0 c 0, therefore p b 0 c 0, therefore p divides b 0 or c 0 but not both of them. Without loss of generality we will say that p b 0 and b c 0. We will now prove that p b k for all k by induction on k, 0 k r. This will yield a contradiction because a n = b r c n r and p b r implies p a n, contrary to hypothesis. 35

Base Case: p b 0. We already know this. Induction Hypothesis: p divides b 0 through b k for some k < r. We will now show that p b k+1. We have a k+1 = b k+1 c 0 + b k c 1 + b k 1 c 2 + + b 0 c k+1. Since k < r and r < n, k + 1 < n. Therefore we know that p a k+1. On the other hand, by the induction hypothesis we know that p divides b 0 through b k. Therefore p divides Since p b k+1 c 0 and p c 0, p b k+1. a k+1 (b k c 1 + + b 0 c k+1 ) = b k+1 c 0. This completes the induction proof. Hence p b r, which contradicts p a n. Therefore f(x) must be irreducible. Example: f(x) = 11 + 121x 5 + 5x 27 is irreducible in Q[x]. Exercises: 1, 2, 3, 4, 5, 7, 8, 9, 11, 18 Chapter 5: Congruence in F [x] and Congruence-Class Arithmetic In Chapter 2 we used the division algorithm for integers in order to classify all numbers according to their remainder after division by n. We placed all integers with remainder r into class [r] and said that two integers are congruent modulo n if and only if they fall into the same class. We then set Z n = {[0], [1],..., [n 1]} and defined addition and multiplication of classes. In Chapter 3 we learned that Z n has zero-divisors when n is a composite number and that Z p is a field when p is a prime number. We also learned that Z ab = Za Z b when a and b are relatively prime. In this chapter we will make all the analogous definitions for polynomials and obtain the same kind of results. In the next chapter we will repeat this process for an arbitrary ring. At this point you should review pp. 7 12 of these notes. I am going to copy here what I wrote there, editing it as necessary to make statements about polynomials. Section 5.1: Congruence in F [x] and Congruence Classes 36