Algebraic Systems, Fall 2012, Skeleton notes as of 11/16/12

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1 Algebraic Systems, Fall 2012, Skeleton notes as of 11/16/12 1. Algebraic Properties of the Integers Definition 1.1. A statement is a sentence that can be assigned a truth value. (In general there is a subject, verb and object in the statement). Ex. A: x 2 = 4, B: x = 2, C: x = ±2 If A and B are statements, A B means A implies B, that is, if A is true then B is true. A B means A is equivalent to B, that is, A is true iff B is true. Ex. Which are true? A C, A B, B A, A C. The symbols and are used between statements. The symbol = is used between objects (numbers, functions, sets, etc. ). Definition ) A binary operation on Z is a function that assigns to each ordered pair (a, b) of integers a unique integer denoted a b. 2) It is called commutative if a b = b a for all a, b Z. 3) It is called associative if a (b c) = (a b) c for all a, b, c Z. 4) An element e Z is called an identity element with respect to if a e = a and e a = a for all integers a. Example 1.1. Ordinary addition and multiplication are binary ops on Z; so is subtraction. Division fails. Addition and Multiplication are commutative and associative, and both have identities (what are they?). Definition 1.3. A subset S of Z is said to be closed with respect to if for any two a, b S we have a b S. Example 1.2. (1) Let a b = 2a + b. Is it binary op on Z? Is it commut? Is there an identity? Is O closed under? Is N closed under? (2) Let a b = ab. Is this a binary op on Z? 1.1. Deducing elementary properties of the integers from the axioms. In the following we will provide examples of two styles of proofs. The first is twocolumn style, where the right column provides the justification for each step. The second is text style, where the proof is written in paragraph form with complete sentences following all the rules of grammar. Example 1.3. Cancellation Law for Addition: Let a, x, y be integers such that a + x = a + y. Then x = y. Proof. a + x = a + y, assumption a + (a + x) = a + (a + y), addition is well defined ( a + a) + x = ( a + a) + y, associative law 0 + x = 0 + y, additive inverse property x = y, 0 is additive identity 1

2 2 Note 1.1. Look at the axioms required to prove the cancellation law. Any algebraic system satisfying those same axioms will also satisfy the cancellation law. Rings and Additive Groups are both examples of such systems that we will visit this semester. Example 1.4. Every integer has a unique additive inverse. Proof. (We ll do this one in text form.) By one of the axioms of Z, we know that every integer has an additive inverse, so our task here is to show that it is unique. Let a be a given integer. Suppose that b, c are additive inverses of a. Then a + b = 0 and a + c = 0. By the transitive law for equality, a + b = a + c. Thus by the cancellation law, b = c. Example 1.5. Subtraction-Equality principle: For any integers x, y, x y = 0 if and only if x = y. Proof. x y = 0, (x y) + y = 0 + y, (x + ( y)) + y = 0 + y, x + ( y + y) = 0 + y, x + 0 = 0 + y, x = y, assumption addition is well defined definition of subtraction associative law additive inverse property 0 is additive identity Note that because the statement was an if and only if statement we needed left-right arrows at each step. Example 1.6. For any integer n, n 0 = 0. Proof. The formal proof is homework but we ll give you a hint. Since 0 is linked with additive properties of Z and this theorem is a multiplicative statement, you will need to make use of the one axiom linking addition and multiplication (what is it?) Now start by writing 0 = (what property have I just used?) Example 1.7. Here is a proof written in text form. Property of Negatives: For any integer a, ( 1)a = a. Proof. (Here, we ll start with text form and then go to two-column form.) Our goal is to show that ( 1)a satisfies the property of an additive inverse, that is, ( 1)a + a = 0. Now, ( 1)a + a = ( 1)a + 1(a), 1 is the multiplicative identity = ( 1 + 1)a, distributive law = 0a, property of additive inverses = 0, by preceding example.

3 3 Why do we just use = here but in the earlier proofs. Comment on the General Associative/Commutative Law. What does mean? Note there are many different groupings one can give. The general associative law says all these values are equal. Comment on Discreteness Axioms for Z: 1) Well ordering property. Compare R. 2) Induction Principle. Proof by Induction: Let P (n) be a statement involving a natural number n. Suppose that (i) P (1) is true. (Base Case) (ii) If P (n) is true for a given n then P (n + 1) is true. (Note induction assumption.) Then P (n) is true for all n N. Example Sum of first n odd numbers. 2. Sum of first n numbers. Example 1.9. Prove that for any positive integer n, (1.1) n 3 = n2 (n + 1) 2. 4 Proof. Proof by induction. For n = 1 we have 1 3 = , a true statement. Suppose that statement (1.1) is true for a given n. Then for n + 1 we have n 3 + (n + 1) 3 = ( n 3 ) + (n + 1) 3 = n2 (n + 1) 2 + (n + 1) 3, by induction assumption (1.1), 4 (n + 1)2 = [n 2 + 4(n + 1)], 4 (n + 1)2 = [n 2 (n + 1)2 + 4n + 4] = [n + 2] 2 = (n + 1)2 ((n + 1) + 1) QED. Example n 3 n is a multiple of 3 for any integer n. Proof. Proof by induction. For n = 1 we note that = 0 = 0 3, a multiple of 3. Suppose that the statement is true for a given n, that is, n 3 n = 3k for some k Z. Then for n + 1 we have (n + 1) 3 (n + 1) = n 3 + 3n 2 + 3n + 1 n 1 = (n 3 n) + 3n 2 + 3n = 3k + 3n 2 + 3n, by induction assumption, = 3(k + n 2 + n) = 3 integer, since the integers are closed under addition and multiplication. QED. Example Let {F n } = 1, 1, 2, 3, 5, 8, 13,..., the Fibonacci sequence. Prove that (1.2) F 1 + F F 2k 1 = F 2k,

4 4 for any k N. Proof. Proof by induction on k. For k = 1 we have F 1 = 1 = F 2, so the statement is true. Suppose that the statement (1.2) is true for a given k. Then for k + 1 we have F 1 + F F 2k 1 + F 2k+1 = (F 1 + F F 2k 1 ) + F 2k+1 = F 2k + F 2k+1, by the induction hypothesis, = F 2k+2 = F 2(k+1), by the defining property of the Fibonacci sequence. QED. Definition 1.4. Let a, b Z, a 0. We say a divides b, written a b, if ax = b for some integer x. Ex since.., 5 12 since... Distinguish 3 12 from 3/12. Equivalent terms: a divides b. a is a divisor of b. a is a factor of b. b is divisible by a. b is a multiple of a. Example ) What are the divisors of 6? What are the divisors of 0? Goal: Fundamental Theorem of Arithmetic. Theorem 1.1. Basic divisibility properties. Let a, b, d be integers. (i) If d a and d b then d (a + b). (ii) If d a and d b then d (a b). (iii) If d a and d b then for any integers x, y, d (ax + by). Proof. (iii) Suppose that d a, d b and that x, y Z. Then a = dk and b = dl for some integers k, l. Thus, ax + by = (dk)x + (dl)y = d(kx) + d(ly) = d(kx + ly) = d(integer), since Z is closed under addition and multiplication. Thus d ax + by. Example Another way to think about them, is to use the word multiple. If a and b are multiples of d then so is a + b, etc. Let S be the set of all multiples of 5. Note S is closed under addition and subtraction. Theorem 1.2. Transitive law for divisibility. For any integers a, b, c, if a b and b c, then a c. Proof. Homework Definition 1.5. Let a, b be integers not both 0. The greatest common divisor of a, b, denoted gcd(a, b) is the largest integer that divides both a and b. ii) Two numbers are called relatively prime if gcd(a, b) = 1. Note gcd(0,0) is undefined. Why? 2. If a, b are not both zero, gcd(a,b) exists and is unique. (Why? Let S be the set of common divisors. It is a finite nonempty set, so it has a maximum element.) 3. gcd(0, n) = n. 4. gcd(a, b) = gcd(b, a)=gcd( a, b)= gcd( a, b). Example ) gcd(-16,-28)=4. 2) gcd(6,-16,-28) = 2.

5 5 Lemma 1.1. Subtraction Principle for GCDs. For any a, b Z, not both zero, and any integer q, gcd(a, b) = gcd(a qb, b). Proof. S, T be set of common divisors. Show S T and T S. Example Find gcd(1023, 1026). By subtraction principle this equals gcd(1023, 3). The latter equals 3 since Division of Integers with remainder. Ex = 7R2, that is, 38 = Quotient, remainder, divisor, dividend. Theorem 1.3. Division Algorithm. Let a, b be integers with b > 0. Then there exist integers q, r such that a = qb + r with 0 r < b. Moreover q, r are unique. q=quotient and r= remainder in dividing a by b. Proof. Existence: We let q be the greatest integer such that qb a, so that qb a < (q + 1)b. Then set r = a qb. Euclidean Algorithm. Example Find d = gcd(126, 49). (1) 126 = , d = gcd(28, 49) (2) 49 = , d = gcd(28, 21) (3) 28 = , d = gcd(7, 21) (4) 21 = 3 7, d = gcd(7, 0) = 7, ST OP Definition 1.6. A linear comb. of two integers a, b is an integer of the form ax+by where x, y Z. Claim: If d = gcd(a, b) then d can be expressed as a linear comb. of a and b. Example gcd(20,8)=4. By trial and error, 4 = ( 2)8. gcd(21,15)=3. By trial and error, 3 = Back Substitution: A method of solving the equation d = ax + by (with d = gcd(a, b)) by working backwards through the steps of the Euclidean algorithm. Example Use example above for gcd(126,49) to express 7 as a LC of 126 and 49. Use the method of back substitution. Start with equation (3): 7 = By (2) we have 21 = Substituting this into previous yields 7 = 28 (49 28) = By (1) we have 28 = Substituting this into previous yields 7 = 2 ( ) 49 = , QED. Array Method. Example Redo example using array method. Perform Euclidean Alg. on the numbers in top row, but do column operations on the array. Let C 1 be the column with top entry 126, C 2 the column with top entry 49, etc. Then C 3 = C 1 2C 2. C 4 = C 2 C 3, C 5 = C 3 C x + 49y x Thus, 7 = y

6 6 Example Find gcd(83, 17) and express it as a LC of 83 and x + 17y x Thus gcd = 1 and 1 = y Theorem 1.4. GCDLC. Let a, b be integers not both zero, d = gcd(a, b). Then d can be expressed as a LC of a and b. Note 1.3. (i) The set of all linear combinations of a, b is just the set of multiples of d. (ii) The gcd of a and b is the smallest positive LC of a and b. (iii) Every common divisor of a and b is a divisor of gcd(a, b). Solving Linear Equations in integers: Solve ax + by = c. GCDLC theorem tells us that this equation can be solved iff c is a multiple of d, that is d c. Theorem 1.5. Solvability of a Linear Equation. The linear equation ax + by = c has a solution in integers x, y iff d c where d = gcd(a, b). Example Solve the following equations or show that there is no solution. 120x 75y = 150,, 120x 75y = 11. By the array method we obtain 120(2) 75(3) = 15, the gcd of 120 and 75. Multiplying by 10 gives the solution (20, 30) to the first equation above. Since the second equation has no solution. Example A parcel costs $2 and we only have 13 cent and 17 cent stamps. How can we do it? 13x + 17y = 200. We know 200 is a lc since gcd=1. Use array to get (-50,50) then note that you can add (17,-13) to get another solution. Definition 1.7. Two integers a, b are called relatively prime if gcd(a, b) = 1. Lemma 1.2. Euclid s Lemma. If d ab and gcd(d, a) = 1 then d b. Note: This lemma fails if gcd(d, a) 1. For example 4 (2 2), but 4 2. Thus d ab does not imply that d a or d b. Note 1.4. Applications of Euclid s Lemma. (i) Every rational number can be uniquely expressed as a fraction in reduced form. Proof. Homework. (ii) If n is not a perfect square, then n is irrational. Proof. Homework. Definition 1.8. i) A positive integer p > 1 is called a prime if its only positive factors are 1 and itself. 2,3,5,7,... ii) A positive integer n > 1 is called a composite if it is not a prime, that is, n = ab for some positive integers a, b with a > 1 and b > 1. 4,6,8,9,... Note is not a prime or a composite. It is the multiplicative identity element. (Later, we will call it a unit.) Why? If 1 is a prime then we would violate unique factorization, eg 6 = 2 3 = Lemma 1.3. a) Let p be a prime such that p ab. Then p a or p b. b) Let p be a prime such that p a 1 a 2... a k where a i are integers. Then p a i for some i. Proof. Use Euclid s lemma for part (a) and induction for (b).

7 7 Theorem 1.6. FTA: Fundamental Theorem of Arithmetic. Any positive integer n > 1 can be expressed as a product of primes, and this expression is unique up to the order of the primes. Note 1.6. (i) 12 = = = 3 2 2, are all considered the same factorization. (ii) We say that a prime p has a trivial factorization as a product of primes. Strong Form of Induction Let P (n) be a statement involving the natural number n. Suppose (i) P (1) is true, and (ii) If P (k) is true for all k < n then P (n) is true. Then P (n) is true for all natural numbers n. Proof of FTA. Existence. Proof is by strong form of induction. Let P (n) be the statement that n has a factorization as a product of primes. P (2) is true. Suppose P (k) is true for all values k smaller than a given n. Consider P (n). If n is prime we are done. Otherwise n = ab for some integers a, b with 1 < a < n, 1 < b < n. By the induction assumption, a and b can be expressed as products of primes, say a = p 1 p k, b = q 1 q l. Then ab = p 1 p k q 1 q l, a product of primes. QED Uniqueness. Suppose that n is a positive integer with two representations as a product of primes, say, (1.3) n = p 1 p k = q 1 q r for some primes p i, q j, 1 i k, 1 j r. We may assume WLOG that k r. Then p 1 q 1... q r, so by lemma, p 1 q i1 for some i 1 {1, 2,..., r}. Since p 1 and q i1 are primes, we must have p 1 = q i1. Cancelling p 1 in (1.3) yields (1.4) p 2 p 3 p k = q 1 ˆq i1 q r, where ˆq i1 indicates that this factor has been removed. We can then repeat the argument with p 2 in place of p 1. After repeating this process k times we have that (1.5) p 1 = q i1, p 2 = q i2,..., p k = q ik for some distinct integers i 1, i 2,..., i k {1, 2,..., r}. Moreover, after cancelling each of the p i from (1.3) we are left with 1 on the LHS. If r > k then (1.3) would say that 1 is a product of primes, a contradiction. Therefore r = k, and so by (1.5), the primes p i are just a permutation of the primes q i. Theorem 1.7. There exist infinitely many primes. Proof. (Euclid) Proof by contradiction. Suppose that there are finitely many primes, say {p 1, p 2,..., p k }. Let N = p 1 p 2 p k +1. By FTA, N has a prime factor p i, for some i k. Thus, p i N and p i (p 1 p 2 p k ). Therefore p i (N p 1 p k ), that is, p i 1, a contradiction. Theorem 1.8. Basic primality test. Let a > 1 be a positive integer such that a is not divisible by any prime p with p a. Then a is a prime. Proof. Homework. Note 1.7. Sieve of Eratosthenes: This is the method of finding all of the primes in a given interval [a, b] by crossing out all multiples of primes p b.

8 8 2. Modular Arithmetic and the Ring of Integers (mod m) Example 2.1. What s the pattern? 3+5=8, 6+4=10, 7+6=1, 9+8=5, 9+2=11 Let m N. m =modulus. Definition 2.1. We say that two integers a, b are congruent modulo m, written a b (mod m), if a and b differ by a multiple of m, that is m (a b). Note: a b (mod m) is equivalent to a = b + mk for some integer k. Example 2.2. Let m = 12. Then 16 4 (mod 12) since 16 4 = (mod 12). In the example above we see = 17 5 (mod 12). How about 256 what is it (mod 12). 256 = , so (mod 12). Definition 2.2. The least residue of a (mod m) is the smallest nonnegative integer that a is congruent to (mod m). Note: The least residue of a (mod m) is the remainder in dividing a by m. Since 0 r < m l.r. is always in {0, 1, 2, 3,..., m 1}. Example 2.3. m = 5 Wrap the integers around a five hour clock. Theorem 2.1. Congruence is an equivalence relation. That is (i) Reflexive, (ii) Symmetric and (iii) Transitive. Theorem 2.2. Important properties of congruences. The substitution laws. Suppose a b (mod m), and c d (mod m). Then (i) a ± c b ± d (mod m). (ii) a c b d (mod m). (iii) a n b n (mod m) for any positive integer n. Example (mod 7) (mod 5). Proof. Two types for (i) and (ii). Induction for (iii). Example 2.5. Explore powers of 2 (mod 3), (mod 6), (mod 7), (mod 8), (mod 9). Note repeating pattern of length m. Use for finding (mod 6). Note 2.1. Trick for calculating a n (mod m) if gcd(a, m) = 1. First find a power k such that a k ±1 (mod m). Find (mod 5), (mod 7), (mod 7), (mod 17). A few applications of congruences: Example 2.6. Day of the week. What day of the week is it 10 years from today? What time will it be 486 hours from now? Divisibility tests: For numbers written in the base-10 (decimal) number system. What does 2715 mean? Is it divisible by 9? Theorem 2.3. Divisibility tests for 3,9 and 11. Let n be a positive integer with decimal rep. n = a k 10 k +... a 0, where the a i {0, 1, 2,..., 9}. (i) 3 n iff 3 (a k + + a 0 ). (ii) 9 n iff 9 (a k + + a 0 ). (iii) 11 n iff 11 a k a k 1 + a k 2 + ( 1) k a 0. Example 2.7. UPC symbols. A 12 digit code d 1, d 2,..., d 12. d 12 is the check digit. 3(d 1 + d d 11 ) + (d d 12 ) 0 (mod 10).

9 9 Definition 2.3. An integer x is called a multiplicative inverse of a (mod m) if ax 1 (mod m). We write x a 1 (mod m) in this case. Avoid fractions. Example 2.8. Find mult. inverse of 3 (mod 5), 4 (mod 6), by trial and error. Which numbers have mult. inverse (mod 10). Theorem 2.4. a has a mult inverse (mod m) iff gcd(a, m) = 1. Example 2.9. Find mult. inverse of 12 (mod 17). Then solve 12x 5 (mod 17). Example Solve 3x 5 (mod 6). Theorem 2.5. The congruence ax b (mod m) is solvable iff d b where d = gcd(a, m). Definition 2.4. The (residue class) congruence class of a (mod m), denoted [a] m is the set of all integers congruent to a (mod m). Thus [a] m = {a + km : k Z}. Example [2] 5 = {2, 7, 12,... } { 3, 8,... }. Note [7] 5, [12] 5 also represent the same class. Draw five hour clock. Note 2.2. [a] m = [b] m iff a b (mod m). Thus eg. [2] 5 = [12] 5. The values 2,7,12, etc. are called representatives for the class [2] 5. Definition 2.5. (i) Let m be a positive integer. The ring of integers (mod m) (residue class ring (mod m)) denoted Z m, is the set of all congruence classes (mod m). Z m = {[0] m,..., [m 1] m }. (ii) We define addition and multiplication on Z m as follows: For [a] m, [b] m Z m, [a] m + [b] m := [a + b] m, [a] m [b] m := [ab] m. Example [3] 5 + [4] 5 = [2] 5. [3] 5 [4] 5 = [2] 5. Note 2.3. Addition and multiplication are well defined on Z m, that is, if [a] m = [b] m and [c] m = [d] m then [a + c] m = [b + d] m and [ac] m = [bd] m. (That is, the sum and product do not depend on the choice of representatives for the congruence classes.) Proof. We ll do multiplication. The proof for addition is similar. First, the definition of multiplication in Z m is [x] n [y] m = [xy] m, for any [x] m, [y] m Z m. To show that the product is well defined we must show that the product does not depend on the choice of representatives for the congruence classes. Now lets begin the proof. Suppose that [a] m = [a ] m and [b] m = [b ] m. Our goal is to show that [ab] m = [a b ] m. By the definition of a congruence classes, we have a a (mod m) and b b (mod m). By the substitution property of congruences this implies that ab a b (mod m), that is, [ab] m = [a b ] m. QED. Note 2.4. The laws for Z hold for Z m as well: Commutative, Associative, Distributive, zero element, additive inverses (ii) Note one important property that Z has that Z m doesn t have. Integral domain property. Convention. If it is understood that we are working in Z m then the bracket notation can be dropped, and abbreviated Z m = {0, 1, 2,..., m 1}. We can say, in Z 6, 3 7 = 3. What is in Z 5? Find mult table for Z 4. Note 2 2 = 0 in Z 4. Definition 2.6. The group of units U m = {x Z m : gcd(x, m) = 1}.

10 10 Note 2.5. (i) U m is the set of elements of Z m that have multiplicative inverses. (ii) U m is closed under multiplication. Example U 9 and multiplication table. mult inverse. Note closed, each row and column, Definition 2.7. For any set S we define the cardinality of S, S, to be the number of elements in S. Write S = is S is infinite. Example Z 9 =, U 9 = 6, Z =. Definition 2.8. Euler phi-function. Note 2.6. By theorem above, φ(m) = U m. Find a formula for φ(m): Test p, p e, p e q f. Use Inclusion/Exclusion. Note φ(p e q f ) = U S T + S T = m m/p m/q + m/pq = m(1 1/p)(1 1/q). Theorem 2.6. Let m = p e pe k k. (i) φ(m) = φ(p e1 1 )φ(pe2 2 )... φ(pe k k ) = (pe1 1 pe1 1 1 )... (p e k k (ii) φ(m) = m(1 1 p 1 )... (1 1 p k ). Example φ(1500). pe k 1 k ). Euler s Theorem and Fermat s Little Theorem. Recall its useful for modular arith to find exponent k such that a k 1 (mod m). Theorem 2.7. Eulers Theorem. Let m N, and a Z with gcd(a, m) = 1. Then a φ(m) 1 (mod m). Example Find (mod 1500). Theorem 2.8. Fermats Little Theorem. Let p be a prime, and a Z, p a. Then a p 1 1 (mod p). Example Find (mod 37). Note 2.7. FLT is a special case of Euler s Theorem. If p a the theorem fails, but it can be restated a p a (mod p) for any a Z. Lemma 2.1. Permutation Lemma. Let m N and U m = {x 1, x 2,..., x r } where r = φ(m). Let a Z with gcd(a, m) = 1. Then U m = {ax 1, ax 2,..., ax r }, that is ax 1,..., ax r is just a permutation of the values x 1,..., x r. Example U 9 = {1, 2, 4, 5, 7, 8}. Test a = 2, a = 4. Note failure if a = 3. Proof. Note (i) for 1 i r, ax i U m. (ii) The values ax i are distinct, by cancellation law. Thus{ax 1,..., ax r } is a set of r distinct elements in U m, and so it must equal all of U m. Proof.. Proof of Eulers Theorem Standard. Public Key Cryptography. Idea is to send a secure message over a public medium such as radio, tv, cell phone, internet, etc. in such a way that only the intended recipient can decipher the message. First words are converted to numbers: A=01, B=02, etc. Hello = 805,121,215 Each person selects their own modulus m, encoding exponent e, and decoding exponent d. The first two are public and the latter top secret. e, d are chosen so that for any integer M with gcd(m, m) = 1, M de M (mod m).

11 Example Say John wishes to send the message M to Mary. He looks up Marys m and e in the phone book. Assume that M < m and gcd(m, m) = 1. John calculates M e M e (mod m) (encoded message). M e is then sent publicly to Mary. Mary then calculates Me d (mod m). Note Me d M de M (mod m). Thus Mary recovers the original message! Say M = 805,m = 1147 = 31 37, e = 23, d = 47. Note φ(m) = = If (M, m) = 1 by Euler s theorem M φ(m) 1 (mod m). Thus M de M 1081 M (mod m). M e 805 e 743 (mod 1147). M d 743 d 805 (mod m). In practice m is chosen to be a huge number (200 digits) that cannot be factored, and so φ(m) cannot be determined from the phone book information. Thus d remains secure. Security depends on the fact that we have no factoring algorithms for 200 digit numbers that can run in less time than the age of the universe. 11

12 12 3. Rings, Integral Domains and Fields Definition 3.1. A ring is a set R with two binary operations +, satisfying (1) Closed under + and (2) Associative law for both addition and multiplication. (3) Commutative law for addition. (4) Distributive laws hold. (5) R has a zero element 0. (6) Every element of R has an additive inverse. If R is a ring with commutative multiplication then R is called a commutative ring. If R is a ring with unity element 1 then R is called a ring with unity. (We require 1 0, so that R {0}.) Example 3.1. Z, R, Q, Z m are all rings. What type? Definition 3.2. Let R be a given ring. A subset S of R is called a subring if S is a ring under the same two binary operations. Example 3.2. Let E be the set of even numbers, O, the set of odd numbers. Is either of these a subring of Z? Example 3.3. Show that the set 3Z, of all multiples of 3 is a subring of Z. (1) Closed under addition: Let 3n, 3m 3Z, where m, n Z. Then 3n+3m = 3(n+m) Z. Also, 3n 3m = 3(3nm) 3Z. (2)-(4) The associative, commutative and distributive laws are inherited from Z. (5) 0 = 3 0 3Z. (6) If 3n 3Z then 3n = 3( n) 3Z. Thus all 6 properties hold, so 3Z is a subring of Z. Note 3.1. To show a subset S of a given ring R is a subring of R it suffices to verify (1) S is closed under + and, (5) 0 S, and (6) If x S then x S. All other properties are inherited from R. Example 3.4. E is a subring of Z. Z is a subring of Q. Q is a subring of R. Example 3.5. The subrings of Z are of the form nz := {nx : x Z}, with n a fixed integer. For instance E = 2Z, or 3Z = {0, ±3, ±6,... }. Example 3.6. If d m we say dz m = {0, d, 2d,..., ( m d 1)d}. Every subring of Z m is of the form dz m with d m. Consider Z 12. Find all subrings Polynomials. Definition 3.3. Let R be a given ring. a) A polynomial over R in the variable x is an expression of the form f(x) = a n x n + a n 1 x n a 0, where the a i are elements of R. b) The values a i are called coefficients of the polynomial. c) If a n 0 then a n is called the leading coefficient of the polynomial and the polynomial is said to be of degree n. d) A polynomial of the form f(x) = a with a R, is called a constant polynomial. If a 0 then it has degree 0. The zero polynomial, f(x) = 0, is not assigned a degree.

13 13 Definition 3.4. Let R be given ring. The polynomial ring in (the variable) x over R, denoted R[x], is the set of all polynomials in x with coefficients in R, R[x] = {a n x n + + a 0 : a i R, 0 i n, n 0}. Addition and multiplication are standard: Let f(x) = n i=0 a ix i, g(x) = n j=0 b jx j. Addition: f(x) + g(x) := n i=0 (a i + b i )x i. Multiplication: f(x) g(x) := n n i=0 j=0 a ib j x i+j = 2n k=0 ( i+j=k a ib j )x k. Note that since R is a ring, the coefficients of f(x) + g(x) and f(x)g(x) are again in R. We also have 0 R[x] and f(x) = n i=0 ( a i)x i R[x], so properties (5) and (6) are satisfied. It is routine, but tedious to verify that properties (2), (3) and (4) hold. Note 3.2. i) If R is ring with unity then so is R[x]. Indeed, if 1 R then 1 is a constant polynomial in R[x]. ii) If R is commutative then so is R[x]. This follows from the fact that a i b j = b j a i for all terms in the product definition above. Example 3.7. In Z 2 [x] find (1 + x) 2. In Z 3 [x] find (x + 1) 3. Definition 3.5. A nonzero element a R is called a zero divisor if ab = 0 or ba = 0 for some nonzero b R. Example is a zero divisor in Z 6 since 3 2 = 0 in Z 6. Example 3.9. Find all zero divisors in Z 9. Note that the remaining values are units. Note 3.3. (i) If p(x) = a n x n + a n 1 x n a 0, with a n 0, then the degree of p(x) is n, the leading term of p(x) is a n x n and the leading coefficient of p(x) is a n. ii) If p(x) = a n x n + + a 0, q(x) = b m x m + + b 0, with a 0 0, b 0 0, then p(x)q(x) = a n b m x m+n + + a 0 b 0. Note that if a n, b m are not zero divisors then a n b m 0 and so the degree of p(x)q(x) is m + n. Recall, the group of units for Z m, U m. Definition 3.6. Let R be a ring with unity. An element a R is called a unit if a has a multiplicative inverse in R, that is, ab = 1 = ba for some b R. Example Find all units in Z, Q, Z 6 Theorem 3.1. Let a Z m, a 0. Then a is a unit if (a, m) = 1 and a is a zero divisor if (a, m) > 1. Definition 3.7. An integral domain is a commutative ring with unity having no zero divisors, that is, if ab = 0 then either a = 0 or b = 0. Example Z is an integral domain. Theorem 3.2. Z m is an integral domain iff m is a prime. Note 3.4. The importance of integral domain is that we can solve equations in the same manner that you are used to: Solve x 2 3x + 2 = 0 in an integral domain R. Now, solve (x 1) 2 = 0 in Z 8, and note the difference because Z 8 is not an integral domain. Lemma 3.1. Let R be an integral domain and f(x), g(x) R[x] be nonzero polynomials of degrees n, m respectively. Then deg(f(x)g(x)) = n + m.

14 14 Proof. Homework. Theorem 3.3. If R is an integral domain, then R[x] is an integral domain. Proof. We already observed above that R[x] is a commutative ring with unity in this case, so we only need to show that R[x] has no zero divisors. This is a homework problem. Example More standard examples of integral domains: R[x] where R is a given integral domain, such as Z[x], R[x], etc. Note that the product of two nonzero polynomials with coefficients in R is always nonzero. Definition 3.8. A ring R is called a field if (i) R has a unity, (ii) R is commutative, (iii) Every nonzero element of R is a unit. Example Standard examples of fields: Q, R, C, Z p where p is a prime. Also, F (x) the set of all rational functions p(x)/q(x) with coefficients in a given field F. Theorem 3.4. If R is a field then R is an integral domain. Definition 3.9. A 2 by 2 matrix with entries in a given ring R is an array of the form [ ] a b, c d where a, b, c, d R. The entry position is given by specifying the row number first, column number second. Thus, a is the entry in the 1, 1 position, b the 1, 2 position, c the 2, 1 position and d the 2, 2 position. Definition Matrix Rings. Let R be a given ring. The ring of 2 by 2 matrices over R is given by {[ ] } a b M 2,2 (R) = : a, b, c, d R. c d Addition and[ multiplication ] [ are ] standard. [ ] a b e f a + e b + f Addition: + =. c d [ g ] [ h ] c + [ g d + h ] a b e f ae + bg af + bh Multiplication: = c d g h ce + dg cf + dh Note 3.5. Matrix multiplication is obtained by taking dot products of the rows of the left matrix with columns of the right matrix. Let A, B be the two matrices above. Let R 1, R 2 be the two rows of A and C 1, C 2 the two columns of B. Then the ij-th entry of AB is equal to R i C j. Note 3.6. M 2,2 (R) is in fact a ring. (1) Since R is closed under +, it follows that so is the matrix ring. Since R is closed under addition and mult, the product of any two matrices over R again has entries in R. (2) The associative law for addition follows immediately from the assoc. law for addition in R. The associative law for multiplication is not trivial. For people with more background with matrices: Let A = [a ij ], B = [b ij ], C = [c ij ]. The ij-th entry of (AB)C is given by k l (a ikb kl )c lj while the ij-th entry of A(BC) is given by k l a ik(b kl c lj ). Thus they are equal by the associative law for R. (3) The commutative law for addition is immediate.

15 15 (4) The distributive law: The ij-th entry of A(B + C) is given by 2 a ik (b kj + c kj ) = k=1 2 (a ik b kj + a ik c kj ) = k=1 2 a ik b kj + k=1 2 a ik c kj which is just the ij-th entry of AB + AC. [ ] 0 0 (5) The zero element in M 2,2 (R) is the matrix 0 =. 0 0 (6) The additive inverse of A = [a ij ] is the matrix A = [ a ij ], which is in M 2,2 (R). [ ] [ ] Note 3.7. (i) Matrix multiplication is not commutative. eg. compare, and its reverse. (ii) M 2,2 (R) has zero divisors. Indeed, for any a, b, c, d R, [ ] [ ] [ ] a =. b 0 c d 0 0 (iii) If R is a ring with unity 1, then M 2,2 (R) is a ring with unity I 2 given by [ ] 1 0 I 2 :=. 0 1 Example M 2,2 (Z 2 ), is a ring with 16 elements. [ ] a b Theorem 3.5. Let R be a commutative ring with unity, and A = c d M 2,2 (R). Put = ad bc, the determinant of A. Then A is a unit in M 2,2 (R) if and only if is a unit in R. k=1 Proof. In your homework you will show that if is a unit in R then [ ] A 1 = 1 d b. c a The converse is done in a matrix theory class. Definition The complex numbers C is the set of numbers, C := {a + bi : a, b R}, where i is the imaginary unit i = 1. (Draw complex plane with real and imaginary axes and indicate the point a + bi). ii) Let z = a + bi. Then a is called the real part of z and b is called the imaginary part. iii) Two complex numbers are equal iff they have the same real and imaginary parts. iv) The complex conjugate of z = a + bi, denoted z, is given by z = a bi. It is the reflection of z in the real axis. v) Addition in C is defined by (a + bi) + (c + di) = (a + c) + (b + d)i. vi) Multiplication in C is defined by (a + bi)(c + di) = (ac bd) + (bc + ad)i. vii) The modulus or absolute value of z = a + bi, denoted z, is given by z = a2 + b 2.

16 16 Note 3.8. i) One can verify that C is a commutative ring with unity 1. ii) Every nonzero complex number has a multiplicative inverse in C, indeed, if z = a + bi then Thus C is a field. iii) For any z C, zz = z 2. z 1 = a bi a 2 + b 2 = z z 2. Definition Polar coordinates r, θ, of a complex number z. i) The polar angle or argument of z, denoted θ, is the angle formed with respect to the positive real axis, (draw picture). It is not unique. One can add any multiple of 2π. ii) r = z, the modulus of z. It is unique and nonnegative, (unlike polar coordinates in R 2.) Definition i) The polar form of a complex number is given by z = r(cos(θ) + i sin(θ)), where r = z and θ is the polar angle of z. This identity follows from definition of the trig functions (cos θ is the x-coordinate on unit circle, sin θ is the y-coordinate. Illustrate). ii) The exponential polar form of a complex number is given by z = re iθ. To obtain the exponential polar form we need the following theorem. Theorem 3.6. For any real number t we have e it = cos(t) + i sin(t). Proof. Recall the Taylor expansions e z z k = k!, sin(t) = ( 1) k 1 t 2k 1 (2k 1)!, cos(t) = ( 1) k t2k (2k)!. k=0 k=1 Insert z = it, to get e it = cos(t) + i sin(t). Note 3.9. e iθ represents a complex number on the unit circle at polar angle θ. eg. e iπ/2 = i, e iπ/4 = i 2. Example A beautiful relationship. e iπ + 1 = 0. This equation has all the fundamental values, 0, 1, e, π and i in one equation. Theorem 3.7. A geometric interpretation of multiplication and division of complex numbers. a) If z, w C then zw is a complex number whose modulus is the product of the moduli of z, w, that is, zw = z w, and whose polar angle is the sum of the polar angles of z and w. b) If w 0, the quotient z/w is a complex number whose modulus is z / w and whose polar angle is the difference of the polar angles of z and w. Theorem 3.8. de Moivre s Formula for n-th powers. Let z be a complex number with exp. polar form z = re iθ. Then for any natural number n, z n = r n e inθ = r n (cos(nθ) + i sin(nθ)). k=0

17 17 Example (1 + i) 10. Start by writing 1 + i in exp. polar form 1 + i = 2e i π 4. Thus (1 + i) 10 = ( 2e i π 4 ) 10 = 2 5 e i 5 2 π = 2 5 e i π 2 = 32i. Definition Let n N, z C. The n-th roots of z denoted z 1/n are the set of complex numbers w satisfying w n = z. z 1/n = {w C : w n = z}. Recall convention that if x is a nonnegative real number then n x denotes the nonnegative n-th root of x. Example /2 = { 2, 2}. 1 1/4 = {1, 1, i, i}. 2 1/4 = /4 = {± 4 2, ± 4 2i}. Theorem 3.9. de Moivre s Formula for n-th roots: Let z be a complex number with exp. polar form z = re iθ. Then z 1/n = n re i( θ n + 2π n k), with k = 0, 1, 2..., n 1. (Technically, it is the set of these values, but the convention is to omit the set brackets.) Proof. Let w = ρe iα. Then w n = z is equivalent to ρ n e inα = re iθ, which means, ρ n = r and nα = θ + 2πk, for some k Z. Thus ρ = n r and α = θ n + 2π n k, for some k Z. Although k is allowed to be any integer, the polar angle for w repeats once k reaches n. Thus the distinct angles are obtained by letting k run from 0 to n 1. Note Every nonzero complex number has n distinct n-th roots. equally spaced around the circle of radius n r, centered at the origin. They are Example a) Find i 1/4. Start with the general exponential polar form of i, i = e i( π 2 +2πk), k Z. In the general form one allows all possible polar angles for i. Thus ( i 1/4 = e i( π +2πk)) 1/4 2 = e i( π 2 +2πk) 1 4 = e i( π 8 + π 2 k), with k = 0, 1, 2, 3. Plugging in these values of k, gives i 1/4 = {e i π 8, e i 5π 8, e i 9π 8, e i 13π 8 }. b) Find ( 3 + i) 1/5. By plotting the point z = 3 + i we see that its polar angle is 5 6 π. Also, z = = 2. Thus the general exp. polar form of z is 2e i( 5 6 π+2πk) and we obtain, z 1/5 = 5 2e i( 5 6 π+2πk) 1 5 = 5 2e i( 1 6 π+ 2 5 πk), with k = 0, 1, 2, 3, 4. c) Find all solutions of the equation x = 0, with x C. This is equivalent to solving the equation x 5 = 2, that is x = ( 2) 1/5. The general exp. polar form of 2 is 2 = 2e i(π+2πk), k Z. Thus with k = 0, 1, 2, 3, 4. ( 2) 1/5 = 5 2e (iπ+2πk) 1 5 = 5 2e i( π 5 + 2π 5 k),

18 18 4. Factoring Polynomials Definition 4.1. Let F be a field, and F [x] be the set of polynomials with coeff. in F. a) If f(x) F [x] we call f(x) a polynomial over F. b) The zero polynomial is the polynomial f(x) = 0 (with all coeff equal to zero). c) Say f(x) = a n x n + + a 0 with a n 0. Then a n is the leading coeff. of f(x), a n x n is the leading term, and n is the degree of f(x). d) f(x) is called monic if a n = 1. Definition 4.2. Let F be a field. a) A poly f(x) over F is called reducible over F if f(x) = g(x)h(x) for some nonconstant polys g(x), h(x). In particular 1 deg(g), deg(h) < deg(f). b) A poly f(x) over F is a called irreducible over F if deg(f) 1 and f(x) is not reducible. Note 4.1. Thus there are four types of polys in F [x]: 1) Zero, 2) Nonzero constant polys (these are the units), 3)Reducibles, 4) Irreducibles. Note analogy with Z. Example 4.1. Determine whether the following are irreducible over the given field, and if not, factor. a) 2x + 4 over Q, R b) x 2 2 over Q, R, C c) x over Q, R, C Definition 4.3. Let f(x), g(x) F [x]. We say that f(x) divides g(x) in F [x], written f(x) g(x) if f(x)h(x) = g(x) for some h(x) F [x]. f(x) is called a factor or divisor of g(x), etc. (same language as in Z.). Example 4.2. Factor x in R, C, Z 5. Theorem 4.1. Let F be a field and f(x), g(x) F [x] with g(x) 0. Then there exist polynomials q(x), r(x) such that f(x) = q(x)g(x) + r(x) with either r(x) = 0 or deg(r(x)) < deg(g(x)). q(x) is called the quotient and r(x) the remainder. Proof. Sketch. case i: Suppose deg(f) < deg(g). case ii: Suppose deg(f) deg(g). Say f = a n x n +..., g = b m x m +..., with b m 0. Then in the first step of long division we have a n (b 1 m )x n m. Subtract to get smaller degree etc. Example x 3 +3x 2 +1 x 2 1, (x 2 +2) (x i) in C[x]. (x 4 x+1) (x 2 +2) in Z 3 [x]. Note 4.2. f(x) g(x) iff the remainder in dividing f(x) by g(x) is zero. Example 4.4. x 3 1 = (x 1)(x 2 + x + 1) over any field F. Thus (x 1) and (x 2 + x + 1) are factors of x 3 1. Definition 4.4. Let f(x) F [x]. An element a F is called a zero or root of f if f(a) = 0. Theorem 4.2. Factor Theorem. Let F be a field, f(x) F [x], a F. a is a zero of f iff (x a) is a factor of f(x). Proof. Know this one. If (x a) is a factor then... Converse. Suppose a is a zero. Strategy, to show (x a) is a factor show remainder is zero.

19 19 Example 4.5. Given graph of 4-th degree polynomial with x-intercepts at -2,0,2, and tangent to x-axis at 0, determine the equation. Example 4.6. a) Given that x = 3 is a zero of f(x) = x 3 x 2 4x 6, factor f(x) completely over R, and over C. b) Factor x 3 + x + 1 completely over Z 3. Example 4.7. a) Factor x 2 + x + 1 over R, C. b) Factor x over C. c) Factor x 5 + x 2 + x + 1 over Z 2. Note analogy between Z and F [x]: 1)The four types, primes, composites, units, zero. 2)Definition of factor. 3) GCD. 4) Division algorithm. 5) Euclidean Algorithm. 6) GCDLC theorem. 7) Euclid s Lemma. 8) If p ab then p a or p b. 9) Unique factorization. Theorem 4.3. Unique Factorization Theorem for F [x]: Let F be a field and f(x) be a polynomial over F of degree 1. Then f(x) can be expressed as a product of irreducible polynomials over F and this factorization is unique up to the order of the factors and unit multiples. Proof. Sketch. Existence: By strong form induction on the degree of f. If f is irreducible done, otherwise f = gh with g, h of smaller degree. Uniqueness: Key Lemma. If p(x) f(x)g(x) and p(x) is irreducible, then p(x) f(x) or p(x) g(x). Then do exactly same proof as for Z. But how to get key Lemma: Need GCDLC: If f(x), g(x) F [x] and d(x) = gcd(f, g) then there exist polys a(x), b(x) such that f(x)a(x) + g(x)b(x) = d(x). Example 4.8. What do we mean by unique up to unit multiples. Factor x 2 3x + 2 over R. x 2 3x + 2 = (x 1)(x 2) = (x 2)(x 1) = (1 x)(2 x) =.. = (7x 7)( 1 7 x 2 7 ) Definition 4.5. Let F be a field and f(x) F [x]. A zero a of f(x) is said to have multiplicity m if (x a) m f(x), but (x a) m+1 f(x). Example 4.9. Suppose f(x) = (x + 1) 3 (x 2) 4 (x 2 + 1). Over R f(x) has a zero at -1 of mult 3 and zero at 2 of mult 4. Over C it has additional zeros at ±i each of mult. 1. Theorem 4.4. Number of zeros of a polynomial. Let F be any field, f(x) F [x] of degree n. Then the total number of zeros of f(x) in F counted with multiplicity is at most n. Proof. Let r 1,..., r k be the zeros of f(x) in F of mult. m 1, m 2,..., m k. Then f(x) = (x r 1 ) m1 (x r 2 ) m2... (x r k ) m k g(x) for some polynomial g(x) having no zero in F. Thus deg(f) = m 1 + m m k + deg(g) m 1 + m m k. Theorem 4.5. Some useful factoring formulas for any field F. a) For any n N, x n a n = (x a)(x n 1 + ax n a n 1 ). b) For any odd n N, x n + a n = (x + a)(x n 1 ax n 2 + a n 1 ). c) If F is a field in which 1 exists, then x 2 + a 2 = (x + a 1)(x a 1). (ex. F = C, or Z p, with p a prime, p 1 (mod 4). ex. In Z 5, 2 2 = 1. In Z 13, 5 2 = 1, etc.)

20 20 d) If F is a field in which 2 exists, then x 4 + a 4 = (x 2 2ax + a 2 )(x 2 + 2ax + a 2 ), provided that 2 F. (ex. F = R, Z p with p ±1 (mod 8). In Z 7, 3 2 = 2. In Z 17, 6 2 = 2. etc. Proof. a,b,c are basic. For d, suppose that a > 0. Then by de Moivre the zeros are aw, aw, aw, aw, where w = e 2πi/8 = i. Pair the conjugate factors to get the formula. Note 4.3. Here is a trick for sums of 4-th powers: x 4 +a 4 = x 4 +2a 2 x 2 +a 4 2a 2 x 2 = (x 2 + a 2 ) 2 ( 2ax) 2, which is a difference of two squares, and so can be factored easily. Example Factor x over R and Z 7. Note, there are no zeros, and yet the polynomial is not irreducible. This can t happen for cubic or quadratic polys. Theorem 4.6 (Conjugate Pair Theorem.). Let f(x) be a polynomial with real coefficients and z be a complex zero of f(x). Then z is also a zero of f(x). Note If z is a real number then z = z and so the conclusion of the theorem is trivial. 2. The theorem generalizes to other fields. For instance, F = Q. Suppose f(x) Q[x] and that a + b m is a zero of f(x), where m is not a perfect square. Then a b m is a zero of f(x). You ve seen this for quadratic equations. Theorem 4.7. Irreducibility of a Quadratic or Cubic polynomials: Let f(x) be a quadratic or cubic polynomial over a field F having no zero in F. Then f(x) is irreducible over F. Note: This does not generalize to higher degree polynomials. Factoring over Q. Theorem 4.8. Rational Root Test: (Descartes Criterion) Let f(x) = a n x n + + a 0 be a polynomial over Z and r s be a rational root of f(x) with r, s relatively prime integers. Then r a 0 and s a n. Example What are the possible rational zeros of 4x 3 + 7x 9. Example Let m Z such that m is not a perfect cube. Prove that 3 m is irrational. Example Test whether x 4 + 2x x + 1 is irreducible over Q. Note that the graph has two x-intercepts (using calculator). Describe the factorization over R and C. Use Gauss test to show it cannot factor as a product of two quadratics over the rationals. Theorem 4.9. Gauss Test for irreducibility. Let f(x) be a polynomial over Z such that f(x) is irreducible over Z that is f(x) g(x)h(x) for any polynomials of positive degree with coeff. in Z. Then f(x) is irreducible over Q. Factoring over C Theorem Fundamental Theorem of Algebra: Let f(x) be a nonconstant polynomial over C. Then f(x) has a zero in C.

21 21 Proof. Done in Complex Analysis. You first prove that if f(z) is differentiable on C and bounded f(z) C, then f(z) is a constant function. Apply this result to 1/f(z). If f(z) has no zero in C then it is differentiable everywhere. Furthermore 1/ f(z) 0 as z so it is bounded. Thus it would have to be constant, a contradiction. Theorem Linear Factorization Theorem for C[x] (Also called FTA) Any nonconstant polynomial over C can be expressed as a product of linear polynomials over C. More precisely, if f(x) is a polynomial over C of degree n 1 with leading coefficient a n, then there exist complex numbers r 1, r 2,..., r n such that f(x) = a n (x r 1 )(x r 2 )... (x r n ). Corollary 4.1. The only irreducible polynomials over C are linear polynomial. Factoring over R. Theorem Odd degree over R theorem. degree over R. Then f(x) has a zero in R. Let f(x) be a polynomial of odd This is easy to see by looking at the graph, since f(x) ± as x, and does just the opposite as x. Thus the graph must cross the x-axis. Theorem Factorization Theorem for R[x]: Let f(x) be a polynomial over R. Then i) f(x) is irreducible if and only if f(x) is linear, or quadratic with no zero in R. ii) In general, if f(x) is of degree n with leading coefficient a n and roots r 1,..., r j R (allowing repetition), then f(x) has factorization over R, f(x) = a n (x r 1 )(x r 2 )... (x r j )q 1 (x)q 2 (x)... q k (x), for some monic irreducible quadratic polynomials q 1 (x),..., q k (x) over R. Summary of irreducible factors: 1. Over C: only linear polynomials are irreducible. 2. Over R: linear or quadratics with no real zeros, that is, negative discriminants. 3. Over Q and Z p. There are irreducible polynomials of every degree. In general it is very difficult to tell whether the polynomial is irreducible. Cardano s Solution of the Cubic Equation in the year 1545 We wish to solve x 3 + ax 2 + bx + c = 0 over C. If we substitute x = y a/3 we obtain a cubic of the form y 3 + Ax + B = 0 where A = a2 3 2a2 a3 3 + b, B = 9 ab 3 + c a3 27. Thus we may assume there is no x 2 term. Note 4.5. Recall that every complex number z has three cube roots {α, αω, αω}, where α is a particular cube root of z and ω = e 2πi/3. Indeed, if z = re iθ then z 1 3 = 3 re i( θ 3 + 2kπ 3 ), k = 0, 1, 2, and so letting α = re iθ/3, we see that z 1 3 = {α, αω, αω 2 }. Note that ω 2 = ω.

22 22 Example Solve x 3 + x 1 = 0. Trick. Let x = u + v, to get u 3 + v 3 + (3uv + 1)(u + v) = 1. Set 3uv + 1 = 0, u 3 + v 3 = 0. The first becomes 27u 3 v 3 = 1. Set U = u 3, V = v 3, so that we have a system U + V = 1, 27UV = 1, which results in the quadratic equation 27U 2 27U 1 = 0. By symmetry, U, V are the distinct roots of this quadratic: U = , V = u, v are cube roots of U, V such that 3uv = 1, so that uv is real. Let ω = e 2πi/3 be a primitive cube root of unity, and α denote the real cube root of U, β the real cube root of V. Then, in order to make uv real, we need u = αω k, v = βω k, k = 0, 1, 2. Note that with this pairing of u and v we have (using UV = 1/27) 3uv = 3αω k βω k = 3αβ = 3 3 UV = 1. Finally, x = u + v = α + β, αω + βω, αω + βω. Cardano s Solution of the Quartic Equation in 1545 Cardano succeeded in solving the quartic equation ax 4 + bx 3 + cx 2 + dx + e = 0, by reducing it to a cubic equation and then using his formula for the solution of a cubic. For the next few hundred years, no further progress was made, that is, no formula could be obtained for the solution of a fifth degree or higher equation. It was finally proved by Abel and Ruffini in 1824, that there does not exist a formula for solving a fifth degree or higher polynomial. In order to succeed in proving this they needed to create a whole new branch of mathematics, called Group Theory.

23 23 5. Group Theory Definition 5.1. A group is a set G with binary operation such that i) G is closed under, that is for any x, y G, x y G. ii) is associative: For any x, y, z G, (x y) z = x (y z). iii) G has an identity element e satisfying x e = e x = x for all x G. iv) Inverses exist: For any element x G there is an element y G such that x y = y x = e. If in addition v) is commutative, then G is called an abelian group. Notation: 1. (G, ) denotes a group G with binary operation. 2. If + is used, generally 0 is used to denote the identity and a the inverse of a. 3. If is used, 1 is commonly used to denote the identity and a 1 the inverse. 4. Unless indicated otherwise, we shall use multiplicative notation for groups when stating theorems. Thus a product of two elements a, b G will simply be denoted ab, no matter what the binary operation is. Example 5.1. Examples of additive groups: For any ring R, (R, +) is an abelian group. For example, (Z m, +), (Z, +), or (M 2,2 (R), +). Example 5.2. Examples of multiplicative groups: 1) (U m, ), for any m N. U m is the multiplicative group of units (mod m). 2) (F, ) where F is any field. Definition 5.2. A subset H of a group (G, ) is called a subgroup of G if H is a group wrt. Note: 1. To show a subset is a subgroup it suffices to check properties (i), (iii) and (iv). Associativity is inherited. 2. If G is a finite set, then suffices to check just (i). One can prove that if (i) holds then so do (iii) and (iv). Example 5.3. Find all subgroups of (Z 6, +). 2Z 6 = {0, 2, 4}, 3Z 6 = {0, 3}, {0} and Z 6. Definition 5.3. If (G, ) is a group and a G then a) For any n N, a n = a a a, n-times and a n = (a n ) 1 = a 1 a 1. b) a 0 = e where e is the identity element in G. c) < a >= {a n : n Z}, called the subgroup of G generated by a. Note: 1. This set < a > is in fact a subgroup of G. 2. If + is the binary operation, then < a >= {na : n Z}. Example 5.4. a) In (Z 6, +), find < 1 >, < 2 >, etc. b) In (Z, +) find < 3 >. c) In (U 5, ), find < 1 >, < 2 >, < 3 >,.. Definition 5.4. Let G be a group with identity e. a) The order of a group G is the number of elements in G, denoted G ; it is also called the cardinality of G. b) The order of an element a of a group G, denoted ord(a) is the smallest positive integer n such that a n = e, (if such an n exists.). If no such n exists, a is said to have infinite order.