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Lecture 2 Basic Number Theory and Algebra In modern cryptographic systems,the messages are represented by numerical values prior to being encrypted and transmitted. The encryption processes are mathematical operations that turn the input numerical value into output numerical values. Building, analyzing, and attacking these cryptosystem requires mathematical tools. The most important of these is number theory, especially the theory of congruences. Outline Basic Notions Solving ax+by=d=gcd(a,b) Congruence The Chinese Remainder Theorem Fermat’s Little Theorem and Euler’s Theorem Primitive Root Inverting Matrices Mod n Square Roots Mod n Groups Rings Fields 1 Basic Notions 1.1 Divisibility Definition 1 Let a 0，and b be intergers . We say that a divides b, if there is an interger k such that b=ka. This is denoted by a|b. Another wa y to express this is that b is a multiple of a. If a does not divide b, we write a | b. 1.1 Divisibility (Continued) Propositio n 1 (1) For every a 0, a|0 and a|a. Also, 1|b for every b. (2) If a|b and b | c, then a | c. (3) If a|b and a|c , then a | ( sb tc) for all intergers s and t. Proof. (1) It is immediate from the Definition 1. (2) There exist k and l such that b ka and c lb. Therefore, c kla. (3) Write b k1a and c k 2 a. Then sb tc a ( sk1 tk2 ), so a|sb tc. 1.1 Divisibility (Continued) Theorem 1 (Division with remainder property). For two integers a, b with b>0, there exist unique integers q, r such that a=bq+r and 0 r<b. Proof. Consider t he integer sequence ,3b,2b,b,0, b,2b,3b, . The integer a must stay among two terms, i.e. there exists an integer q such that qb a (q 1)b. Set r a qb. It proves the existence of r and q. For uniqueness , suppose that a=bq+r and a=bq'+r ' , where 0 r<b and 0 r ' <b. Then subtractin g these two equations and rearrangin g terms, we obtain r' r=b(q q ' ). Now observe that by assumption , the left - hand side is less than b in absolute value. However, if q q' , then the right - hand side would be at least b in absolute value; therefore , we must have q q' . Furthermor e, we must have r=r'. 1.2 Prime Definition 2 A prime is a positive integer greater than 1 that is divisble by no positive integers other than 1 and itself. A positive integer greater th an 1 that is not prime is called composite. The primes less than 200: 2 3 5 7 11 13 17 19 23 29 31 37 41 43 47 53 59 61 67 71 73 79 83 89 97 101 103 107 109 113 127 131 137 139 149 151 157 163 167 173 179 181 191 193 197 199 1.2 Prime (Continued) Propositio n 2 There are infinitely many primes. Proof. By way of contradict ion, suppose that ther e were only finitely many primes; call them p1 2, p2 3, , pk . Then set n p1 p2 pk 1, and consider a prime p that divides n. There must be at least one such prime p, since n 2.Clearly, p cannot equal any of the p1 , p2 , , pk , since if it did, then p would divide n p1 p2 pk 1, which is impossible . Therefore, the prime p is not among p1 , p2 , , pk , which contradict s our assumption that these are the only primes. 1.2 Prime (Continued) Theorem 2 (Prime Number Theorem ) Let ( x) be the number of primes less than x. Then x ( x) , ln x in the sense that the ratio ( x) / ( x/ ln x) 1 as x . Proof . We omit the proof . In various applicatio ns, we' ll need large primes, say of around 100 digits. We can estimate the number of 100 - digit primes as follows : 100 99 10 10 97 (10100) (1099 ) 3 . 9 10 . 100 99 ln 10 ln 10 So there are certainly enough such primes. 1.2 Prime (Continued) Theorem 3 (Fundament al theorem of arithmetic ) Every positive integer is a product of primes. This factorizat ion into primes is unique, up to reordering . Proof. (Existence ) This amounts to showing that every positive integer n can be expressed as a product (possibly empty) of primes. We may prove this by induction on n. Let n > 1, and assume that every positive integer smaller th an n can be expressed as a product of primes. If n is a prime, then the statement is true, as n is the product of one prime; otherwise, n is composite, and so there exist 1<a<n,1<b<n, and n = ab; by the induction hypothesis , both a and b can be expressed as a product of primes, and so the same holds for n. 1.3 Greatest Common Divisor Definition 3 The great common divisor of a and b is the largest positive integer dividing both a and b and is denoted by either gcd( a, b)or by (a, b). We say that a and b are relatively prime if gcd( a, b) 1. There are two standard ways for finding the gcd : (1) If you can factor a and b into primes, i.e. a p11 p2 2 pn n and b p11 p2 2 pn n . Take the smaller of the two and get gcd( a, b) p1min(1 , 1 ) p2min( 2 , 2 ) pnmin( n , n ) . If a prime does not appear in a factorizat ion, then it cannot appear in the gcd. (2) The Euclidean algorithm. 1.3 Greatest Common Divisor (Continued) Example 1 Compute gcd(482, 1180). 1180 2 482 216 482 2 216 50 216 4 50 16 50 3 16 2 16 8 2 0. So, gcd(482, 1180) 2. Notice how the numbers are shift : remainder divisor dividend ignore. Using the example as a guideline, we can now give a more formal descriptio n of the Euclidean algorithm. Without loss of generality , suppose a b. We have a q1b r1 b q2 r1 r2 r1 q3r2 r3 rk 2 qk rk 1 rk rk 1 qk 1rk . Hence gcd( a, b) rk . (Without factorizat ion and fast speed) 1.3 Greatest Common Divisor (Continued) Theorem 4 Let a and b be two integers, with at least one of a, b nonzero, and let d gcd( a, b). Then there exist integers x, y such that ax by d . In particular , if a and b relatively prime, then ther e exist integers x, y with ax by 1. Proof. We can show that if r j is a remainder obtianed during the Euclidean algorithm, then ther e are integer x j , y j such that r j ax j by j .Taking x1 1 and y1 q1 , we have r1 ax1 by1. Similar, r2 a (q2 ) b(1 q1q2 ). Suppose we have ri axi byi for all i j. Then r j r j 2 q j r j 1 ax j 2 by j 2 q j (ax j 1 by j 1 ) a ( x j 2 q j x j 1 ) b( y j 2 q j y j 1 ). Continuing , we obtain the result for all j , in particular for r j k gcd( a, b). 1.3 Greatest Common Divisor (Continued) Corollary 1 If p is a prime and p | ab, then either p | a or p | b. More generally, if a prime p | ab z , then p must divide one of the factors a, b, , z. Proof. Consider t he case p|ab. If p divides a, we are done. Now assume p | a. Since p is prime, gcd( a, p ) 1 or p. Since p | a, gcd( a, p ) 1.Following Theorem 4, there exist integers x,y with ax py 1. Multiply by b to obtian abx pby b. Therefore, p | b. For the case p | ab z , if p | a, we' re done. Otherwise, p | b z. Either p | b or p divides the remianing factors. Continuing in this way, we can get the conclusion . 1.3 Greatest Common Divisor (Continued) Theorem 3 (Continued ) Proof. (Uniquenes s) Suppose that n p1 p2 ps q1q2 ql , where p1 , p2 , , ps and q1 , q2 , , ql are primes, and p1 p2 ps and q1 q2 ql . Since p1 | n, q1 | n, we know p1 | q j , q1 | pk . Since q j , pk are prime , we get p1 q j , q1 pk . Since p1 q1 , q1 p1. So p1 q1. Remove the same factor p1 to get p2 ps q2 ql . Continuing in this way, we can obtain s l , ps ql . 2 Solving ax+by=d=gcd(a,b) Based on the proof procedure of the Theorem 4, we get the following sequences : x1 1, x2 q2 , x j q j x j 1 x j 2 y1 q1 , y2 1 q1q2 , y j q j y j y j 2 Then axk byk gcd( a, b). In the Example 1, x1 1, x2 2, x3 2 x2 x1 5, x4 4 x3 x2 22, x5 3 x4 x3 71. Similarly, y5 29. An easy calculatio n shows that 482 71 1180 (29) 2 gcd( 482, 1180). The preceding mehtod is oftem called the extended Euclidean algorithm. 3 Congruences Definition 4 Let a, b, n be integers with n 0. We say that a b(mod n) (read : a is congruent to b mod n)if a b is a multiple (positive or negative ) of n. This can be rewritten as a b nk for some integer k (positive or negative). Example 2 32 （ 7 mod 5), 12 37(mod 7), 17 17(mod 13). Propostion 3 Let a, b, c, n be integers with n 0, (1) a 0(mod n) if and only if n | a. (2) a a (mod n). (3) b a (mod n) if and only if a b(mod n). (4) If a b(mod n), b c(mod n), then a c(mod n). Proof. We omit the proof. # Congruence behaves very much like equality. 3.1 Addition, Subtraction, Multiplication Propositio n 4 Let a, b, c, d , n be integers with n 0, and suppose a b(mod n), c d (mod n). Then a c b d (mod n), a c b d (mod n), ac bd (mod n). Proof. Write a b nk , c d nl , for integers k , l ,. Then, a c b d n(k l ), so a c b d (mod n). The proof that a c b d (mod n) is similar. For multiplica tion, we have ac bd n(dk bl nkl), so ac bd (mod n). # The propositio n says you can perform the usual arithmetic operations of addition, subtractio n, and multiplica tion with congruence s. 3.1 Addition, Subtraction, Multiplication (Continued) Example 3 Here is an example of how we can do algebra mod n. Consider t he following problem x 7 3(mod 17). Solution : x 3 7 4 13(mod 17). # There is nothing wrong with negative answers, but usually we write the final answer as an integer from 0 to n 1. 3.2 Division Propositio n 5 Let a, b, c, d , n be integers with n 0 and with gcd( a, n) 1. If ab ac(mod n), then b c(mod n), in other word s, if a, n are relatively prime , we can divide both sides of the congruence by a. Proof. Since gcd( a, n) 1, there exist x, y such that ax ny 1. Multiply by b c : (ab ac) x n(b c) y b c. Since n | ab ac, we can get n | b c. This means that b c(mod n). 3.2 Division (Continued) Example 4 Solve : 2 x 7 3(mod17) . Solution : 2 x 3 7 4. Since gcd(2,17) 1, x 2 15(mod 17). Example 5 Solve : 5 x 6 13(mod11) . 5 x 7, what does 7/5 mean (mod11)? Note that 5 x 7 18 29 40(mod11) . So x 8(mod 11). That is ,8 acts like 7/5. 3.2 Division (Continued) Propositio n 6 Suppose gcd( a, n) 1. Let s, t be integers such that as nt 1. Then as 1(mod n), so s is the multiplica tive inverse for a (mod n), witten as a 1 (mod n). s, t can be found using the extended Euclidean algorithm . Proof. Since as 1 nt , we see that as 1 is a multiple of n. Example 6 Solve 11111x 4(mod 12345). Solution : Using the extended Euclidean algorithm, we can gcd(11111,12345) 1, 11111 2471 12345 y 1. It means that 11111 2471 1(mod 12345). Hence, x 2471 4 9884(mod 12345). 3.3 Division (Continued) Solve congruence s of the form ax b(mod n) when gcd( a, n) d 1. The procdure is as follows : (1) Ifd | b, there is no solution. (2) Assume d | b. Consider t he new congruence (a / d ) x b / d (mod n / d ). Note that a / d , b / d , n / d are integers and gcd( a / d , n / d ) 1. Solve this congruence by the above procedure to obtian a solution x0 . (3) The solutions of the original congruence ax b(mod n) are x0 (mod n), x0 (n / d )(mod n), x0 2(n / d )(mod n), , x0 (d 1)( n / d )(mod n). 3.2 Division (Continued) Example 7 Solve 12 x 21(mod 39). Solution : gcd(12, 39) 3, which diviedes 21. Divide by 3 to obtian the new congruence 4 x 7(mod 39). A solution is x0 5. The solutions to the original congruence are x 5, 18, 31(mod 39). Working with fractions (1)In many situations , it will be convenient to work with fractions mod n. For example, 1 / 2(mod 12345) is easier to write than 6173(mod 12345). The general rule is that a fraction b/a can be used mod n if gcd( a, n) 1. b / a (mod n) really means a 1b(mod n). (2)The symbol 1 / 2 is simply a symbol with exactly one propery : if multiply 1 / 2 by 2, you get 1. So, 1 / 2(mod 12345) and 6713(mod 12345) can be interchang eable. (3) We can' t use fractions with arbitrary denominato rs. For example, 1 / 6(mod 6),1 / 2(mod 6).In general, if gcd( a, n) 1, it is not allowed. 4 The Chinese Remainder Theorem x 4(mod 7) x 25(mod 42) x 1(mod 6). The Chinese remainder theorem shows that this process can be reversed. Theorem 5 (Chinese Remainder Theorem) Suppose gcd( m, n) 1. Given a and b, there exists exactly one solution x(mod mn) to the simultaneo us congruence s x a (mod m), x b(mod n). Proof. There exist integers s, t such that ms nt 1. Let x bms ant. Then x ant a (mod m), x bms b(mod n). Suppose x1 is another solution. Then m | x x1 , n | x x1 , so x x1 mk nl , x x1 ( x x1 )( ms nt ) mn(ls kt), i.e. x x1 (mod mn). 4 The Chinese Remainder Theorem (Continued) Example 8 Solve x 3(mod 7), x 5(mod 15). Solution : Since 80(mod 7) 3(mod 7), 80(mod 15) 5(mod 15), x 80(mod 105). The theorem guarantees that such a solution exists and is uniquely determined by mod mn. Two methods to find the solution : (1)To list the numbers congruent to b(mod n) until you find one that is congruent to a (mod m). For example, the numbers congruent to 5(mod 15) are 5, 20, 35, 50, 65, 80. By mod 7, there are 5, 6, 0, 1, 2, 3. (2)The numbers congruent to b(mod n) are of the form b nk , so we need to solve b nk a (mod m). Obtain k (a b)n 1 (mod m), Substituti ng k back into b nk , then reducing mod nm, gives the answer. 4 The Chinese Remainder Theorem (Continued) Example 9 Solve x 7(mod 12345), x 3(mod 11111). Solution : 111111 (mod 12345) 2471.Therefore, k (7 3) 2471 9884(mod 12345).This yields x 3 11111 9884 109821127(mod 1111112345). # If you start with a congruence modulo a composite number n, you can break it into simultaneo us congruence s modulo each prime power factor of n, then recombine the resulting informatio n to obtian an answer mod n. The advatantag e is that often it is easier to analyze congruence modulo primes or modulo prime powers than to work modulo composite numbers. 4 The Chinese Remainder Theorem (Continued) Example 10 Solve x 2 1(mod 35) Solution : x 2 1(mod 35) x 2 1(mod 5) x 1(mod 5) 2 x 1(mod 7) x 1(mod 7). We can put togeth er in 4 ways : x 1(mod 5), x 1(mod 7) x 1(mod 35), x 1(mod 5), x 1(mod 7) x 6(mod 35), x 1(mod 5), x 1(mod 7) x 29(mod 35), x 1(mod 5), x 1(mod 7) x 34(mod 35). 4 The Chinese Remainder Theorem (Continued) Theorem 6 (CRT - General Form) Let m1 , m2 , , mk be integers with gcd( mi , m j ) 1 whenever 1 i j k . Given integers a1 , a2 , , ak , there exists exactly one solution x(mod m1m2 mk ) to the simultaneo us congruence s x a1 (mod m1 ), x a2 (mod m2 ), , xk ak (mod mk ). Proof. We can omit the proof. Therefore, in general, if n p1 p2 pr is the product of r distinct odd primes, then x 2 1(mod n) has 2 r solutions. 5 Fermat’s Little Theorem and Euler’s Theorem Theorem 7 (Fermat' s Little Theorem ) If p is a prime and p | a, then a p 1 1(mod p ). Proof. Let S {1,2,3, , p 1}. Consider t he map S S : ( x) ax(mod p ).Clearly, ( x) 0(mod p). Now, suppose x y S . We have ax ay (mod p ). Therefore , (1), (2), , ( p 1) are distinct elements of S . It follows that 1 2 3 ( p 1) (1) (2) ( p 1) (a 1) (a 2) (a 3) (a ( p 1)) a p 1 (1 2 3 ( p 1))(mod p ). Since gcd( j , p ) 1 for j S , we can divide this congruence by 1,2,3, , p 1. What remains is 1 a p 1 (mod p ). 5 Fermat’s Little Theorem and Euler’s Theorem (Continued) Example 11 210 (mod 11), 253 (mod 11). 210 1024 1(mod 11） . From this, we can evaluate 253 (210 )5 23 23 8(mod 11). In other word s, from 53 3(mod 10), we deduce 253 23 (mod 11). Search for prime numbers using the Fermat' s little Theorem Choose a starting point n0 and successive ly test each odd number n n0 to see whether 2 n 1 ? 1(mod n). If n fails the test, discard it and proceed to the next n. When passes the test, use more sophistica ted techniques . # The advantage is that this procedure is much faster and eliminate many numbers quickly. However, there exist the exceptions such as 561 3 11 17, 2560 1(mod 561). 5 Fermat’s Little Theorem and Euler’s Theorem (Continued) Definition 5 Let (n) be the number of integers 1 a n such that gcd( a, n) 1. Often is called Euler' s - function . Propositio n 7 If n p1a1 p2a2 pkak is the prime power factorizat ion , 1 a 1 a then (n) n 1 , in particular , ( p ) 1 p . pi p i 1 Proof. We omit the proof. k Example 12 (10) (2 5) 10(1 1 / 2)(1 1 / 5) 4, (120) (23 3 5) 120(1 1 / 2)(1 1 / 3)(1 1 / 5) 32. 5 Fermat’s Little Theorem and Euler’s Theorem (Continued) Theorem 7 (Euler ' s Theorem) If gcd( a, n) 1, then a ( n ) 1(mod n). Proof. The proof of this theorem is almost the same as the one given for Fermat ' s theorem . Let S be the set of integers 1 x n with gcd( x, n) 1. Let S S be defined by ( x) ax(mod n). Clearly, the numbers ( x) are the numbers in written in S some order. Therefore, x ( x) a ( n ) x(mod n), xS xS xS Dividing out the factors, we obtain a ( n ) 1(mod n). 5 Fermat’s Little Theorem and Euler’s Theorem (Continued) Example 13 What are the last three digits of 7803 ? Solution : Knowing the last three digits is the same as working modulo 1000. Since (1000) 1000(1 1 / 2) (1 1 / 5) 400, we have 7803 (7 400) 2 7 3 73 343(mod 1000). Example 14 Compute 2 43210 (mod 101). Solution : From Fermat' s theorem, we know that 2100 1(mod 101). Therefore, 2 43210 (2100) 432 210 1024 14(mod 101). 5 Fermat’s Little Theorem and Euler’s Theorem (Continued) Basic Principle 1 Let a, n, x, y be integers with n 1, gcd( a, n) 1. If x y (mod (n)), then a x a y (mod n). In other word s, if you want t o work modulo n, you should work modulo (n) in the exponent. Proof. Write x y (n)k . Then a x a y ( n ) k a y (a ( n ) ) k a y (1) k a y (mod n). # Work with the exponent using modulo (n) not n. 6 Primitive Root Consider t he powers of 3(mod 7) : 31 3, 32 2, 33 6, 34 4, 35 5, 36 1. Note that we obtain all the nonzero congruence classes modulo 7 as powers of 3. This means that 3 is a primitive root modulo 7. But, 33 1(mod 13), so only 1, 3, 9 are powers of 3. Therefore, 3 is not a primitive root mod 13. In gereral, when p is a prime, a primitive root modulo p is a number who se powers yield every nonzero class modulo p. # There are ( p 1) primitive root modulo p. 6 Primitive Root (Continued) Propositio n 8 Let g be a primitive root for the prime p. (1) If n is an integer , then g n 1(mod p) if and only if n 0(mod p 1). (2) If j and k are integers , then g j g k (mod p) if and only if j k (mod p 1). Proof. (1) If n 0(mod p 1), then n ( p 1)m for some m. Therefore, g n ( g m ) p 1 1(mod p) by Fermat ' s theorem . Suppose g n 1(mod p). Write n ( p 1)q r , with 0 r p 1. We have 1 g n ( g q ) p 1 g r g r g r (mod p). Suppose r 0. The powers of g (mod p) yield g (mod p), g 2 (mod p), , g r 1 (mod p). Since r p 1, this contradict s the assumption that g is a primitive root. So r 0. (2) Assume that j k . Suppose that g j g k (mod p). Dividing both sides by g k yields g j k 1(mod p). By (1), j k 0(mod p 1), so j k (mod p 1). If j k (mod p 1), then j k 0(mod p 1), so g j k 1(mod p) by (1), i.e. g j g k (mod p). 7 Inverting Matrices Mod n Finding the inverse of a matrix modulo n can be accomplish ed by the usual methods for inverting a matrix. The basic fact we need is that a square matrix is invertible modulo n if and only if its determinan t and n are relatively prime. For example, a b c d 1 b 1 d b 1 d (ad bc) (mod n) ad bc c a c a 7 Inverting Matrices Mod n (Continued) 1 2 Example 15 Invert (mod 11). 4 3 1 2 2, 2 5 1(mod 11), we obtain Solution : Since 3 4 1 1 2 1 4 2 4 2 9 1 (mod 11). 5 3 4 5 7 1 3 1 3 2 A quick calculatio n shows that 1 2 9 1 23 11 1 0 3 4 7 5 55 23 0 1 (mod 11). 1 1 1 1 Example 16 Invert 1 2 3 (mod 11). 1 4 9 1 1 1 1 6 5 1 3 3 6 1 1 1 Solution : 1 2 3 2,2 6 1(mod 11), 1 2 3 6 6 8 2 8 4 10 (mod 11). 2 3 1 1 4 6 1 4 9 1 4 9 7 Inverting Matrices Mod n (Continued) Why do we need the determinan t and n to be relatively ？ Suppose MN I (mod n), where I is the identity matrix. Then , | M || N || MN || I | 1(mod n). Therefore, | M | has an inverse modulo n, which means that | M | and n must be relatively prime . 8 Square Roots Mod n Consider x 2 71(mod 77). How do we find one solution and all solutions ? Let' s start with the case of sqare roots modulo a prime p. The easiest case is when p 3(mod 4). Propositio n 9 Let p 3(mod 4) be prime and let y be an integer . Let x y ( p 1) / 4 (mod p ). (1) If y has a square root modulo p, then the square roots of y mod p are x. (2) If y has no square root mod p, then y has the square roots modulo p, and the square roots of y are x. 8 Square Roots Mod n (Continued) Proof. If y 0(mod p), all the statements are trivial. So assume y 0(mod p). By the Fermat ' s theorem, we have x 4 y p 1 y 2 y p 1 y 2 (mod p), This implies that ( x 2 y )( x 2 y ) 0(mod p), so x 2 y (mod p). Therefore, at least one of y and y is a square modulo p. Suppose both y and y are squares modulo p, say a 2 y (mod p) and b 2 y (mod p). Then 1 （a / b）2 (mod p), which means 1 is a square mod p. This is impossible , because 1 (1) ( p 1) / 2 (a / b) p 1 (mod p). It contradict s Fermat' s little theorem . 8 Square Roots Mod n (Continued) Example 17 Find the square root of 5(mod11) . Solution : Since ( p 1)/4 12/4 3, x 53 4(mod 11). We can compute 4 2 5(mod 11). So the square roots of 5(mod11) are 4. Example 18 Find the square root of 2(mod11) . Solution : Since ( p 1)/4 12/4 3, x 23 8(mod 11). But 82 9 2(mod 11), so 2 has no square root mod11, the square roots of 2(mod11) are 8. Example 19 Solve x 2 71(mod 77). Solution : It means that x 2 71 1(mod 7), x 2 71 5(mod 11). Therefore , x 1(mod 7), x 4(mod 11). We can combine in four ways, i.e. x 1(mod 7) x 1(mod 7) x 1(mod 7) x 1(mod 7) , , , , x 4(mod 11) x 4(mod 11) x 4(mod 11) x 4(mod 11) Using the Chinese remainder theorem, we can compute x 15,29,29,15 (mod 77). 8 Square Roots Mod n (Continued) Square Root Oracle Suppose n pq is the product of two primes and we know the four solutions x a,b(mod n) of x 2 y (mod n). From Example 19, we know that a b(mod p )( a b(mod q )) a b(mod q )( a b(mod p )).Therefore, p | a b(q | a b) but q | a b ( p | a b), i.e. gcd( a b, n) p (q ). In Example 19, gcd(15 29,77) 7 gives a nontrivial factor of 77. Basic Principle 2 Suppose n pq is the product of two primes congruent to 3(mod 4) and y is a number relatively prime to n which has a squere root mod n. Then finding the four solutions x a,b to x 2 y (mod n) is computatio nally equivalent to factoring n. 9 Groups, Rings, Fields 9.1 Groups Definition 6 A group (G, * ) consists of a set G with a binary operation * on G satisfying the following three axioms. (1) The group operation is associativ e. That is, a* (b* c) (a*b) * c for all a, b, c G. (2) There is an element 1 G, called the identity element, such that a *1 1*a a for all a G. (3) For each a G there exists an element a 1 G, called the inverse of a, such that a * a 1 a 1 *a 1. A group G is abelian (or commutativ e) if, furthermor e, (4) a*b b*a for all a, b G. 9.1 Groups (Continued) Example 20 (1) The set of integers Z with the operation of addition forms a group. The identity element is 0 and the inverse of an integer a is the integer a. (2) The set Z n , with the operation of addition modulo n, forms a group. The set Z n with the operation of multiplica tion modulo n is not a group, since not all elements have multiplica tive inverses. However, the set Z n* is a group under the operation of multiplica tion modulo n, with identity element 1. (3) The set {T , F }, with the operation of XOR, form a group, with identity element F , T 1 T . 9.2 Rings Definition 7 A ring ( R,,) consists of a set R with two binary operations arbitraril y denoted (addition) and (multiplic ation) on R, satisfying the following axioms. (1) (R, ) is an abelian group with identity denoted 0. (2) The operation is associativ e. That is, a (b c) (a b) c for all a, b, c R. (3) There is a multiplica tive identity denoted 1, with 1 0, such that 1 a a 1 a for all a R. (4) The operation is distributi ve over . That is, a (b c) (a b) (a c) and (b c) a (b a ) (c a ) for all a, b, c R. The ring is a commutativ e ring if a b b a for all a, b R. 9.2 Rings (Continued) Example 21 (1) The set of integers Z with the usual operations of addition and multiplica tion is a commutativ e ring. (2) The set Z n with addition and multiplica tion performed modulo n is a commutativ e ring. 9.3 Fields Definition 8 A field is a commutativ e ring in which all non - zero elements have multiplica tive inverses. Example 22 (1) The set of integers under the usual operations of addition and multiplica tion is not a field, since the only non - zero integers with multiplica tive inverses are 1 and 1. However, the rational numbers Q, the real numbers R, and the complex numbers C form fields under the usual operations . (2) Z n is a field (under the usual operations of addition and multiplica tion modulo n) if and only if n is a prime number. If n is prime. # A algebra structure is finite if the number of elements is finite. The number of elements is called its order. Thank you!

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