Solutions I.N. Herstein- Second Edition

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1 Solutions I.N. Herstein- Second Edition Sadiah Zahoor Please me if any corrections at R is a ring in all problems. Problem 0.1. If a, b, c, d R, evaluate (a + b)(c + d). Proof. We use the distributive law in R to get (a+b)(c+d) = a(c+d)+b(c+d) as (c + d) R by closure in R. Then we again use distributive law a(c + d) + b(c + d) = ac + ad + bc + bd. Problem 0.2. Prove that if a, b R, then (a + b) 2 = a 2 + ab + ba + b 2. Proof. We use the distributive law in R to get (a+b)(a+b) = a(a+b)+b(a+b) as (a + b) R by closure in R. Then we again use distributive law a(a + b) + b(a + b) = a 2 + ab + ba + b 2. Problem 0.3. Find the form of the binomial theorem in a ring; in other words, find an expression for (a + b) n, where n is a positive integer. Proof. From above (a + b) 2 = a 2 + ab + ba + b 2 = x 1 x 2. Similarly, (a + b) 3 = (a + b)(a + b) 2 = (a + b)(a 2 + ab + ba + b 2 ) = a 3 + a 2 b + aba + ab 2 + ba 2 + bab + b 2 a + b 3 = x 1 x 2 x 3. For (a+b) 4 = (a+b)(a+b) 3 = (a+b)(a 3 +a 2 b+aba+ab 2 +ba 2 +bab+b 2 a+b 3 ) = a 4 + a 3 b + a 2 ba + a 2 b 2 + aba 2 + abab + ab 2 a + ab 3 + ba 3 + ba 2 b + baba + bab 2 + b 2 a 2 + b 2 ab + b 3 a + b 4 = x 1 x 2 x 3 x 4 Thus, (a + b) n = x 1 x 2 x 3...x n = n x i. Problem 0.4. If every x R satisfies x 2 = x, prove that R must be commutative. (A ring in which x 2 = x for all elements is called a Boolean Ring). Proof. Let a, b R. Then we have, a + b = (a + b) 2 = a 2 + b 2 + ab + ba = a + b + ab + ba ab + ba = 0 ab = ba. Also, we have, a b = (a b) 2 = a 2 ab ba + b 2 = a + b ab ba b + ab + ba = b b = b. Thus, ab = ba = ba. Since, a, b R were arbitrary, every Boolean Ring is commutative. 1

2 Solutions by Sadiah Zahoor 2 Problem 0.5. If R is a ring, merely considering it as an abelian group under its addition, we have defined, in Chapter 2, what is meant by na, where a R and n is an integer. Prove that if a, b R and n, m are integers, then (na)(ma) = (nm)(ab). Proof. We know na = n a. We use the distributive law in R to get, (na)(mb) = ( n a)( m b) j=1 = ( n a( m b)) j=1 = ( n ( m (ab))) j=1 = ( n (m(ab))) = (nm)(ab) Problem 0.6. If D is an integral domain and D is of finite characteristic, prove that the characteristic of D is a prime number. Proof. We are given D has finite characteristic, that is, there exists m Z such that ma = 0 for all a D. Then we can define the characteristic of D to be the smallest positive integer n such that na = 0 for all a D. Suppose n is not prime. Then n is composite and we can write n = n 1 n 2 where 1 < n 1, n 2 < n. Then (n 1 n 2 )a = 0. In particular for a = 1, we have (n 1 n 2 )1 = 0 (n 1.1)(n 2.1) = 0 n 1.1 = 0 or n 2.1 = 0 because D is an integral domain. This implies n 1.a = 0 or n 2.a = 0 for any a D. Thus, there exists a positive integer, say n 1, 1 < n 1 < n such that n 1 a = 0. Since a was arbitrary, we get n 1 is a positive integer less than the characteristic of D for which n 1 a = 0 for all a D. This is a contradiction to the definition of the characteristic of an integral domain. Thus, the characteristic of D must be a prime. Problem 0.7. Give an example of an integral domain which has infinite number of elements, yet of finite characteristic. Proof. Consider the example Z p [x], the set of polynomials over Z p where p is a prime. Then the characteristic of the polynomials in Z p [x] is p and we can have infinitely many elements because we can have polynomial of any degree in Z p [x]. Also, Z p [x] is an integral domain because Z p is a field and we know F [x] is an integral domain for a field F. Problem 0.8. If D is an integral domain and if na = 0 for some a 0 in D and some integer n 0, prove that D is of finite characteristic. Proof. We have na = 0 for some a 0 in D and some integer n 0. Now, let b D, then n(ab) = 0, which further implies (1.n)(ab) = 0 (a)(nb) = 0 a = 0 or nb = 0 because D is an integral domain. Now we are given a 0. Then nb = 0. Since b was arbitrary, we get D is of finite characteristic.

3 Solutions by Sadiah Zahoor 3 Problem 0.9. If R is a system satisfying all the conditions for a ring with the unit element with the possible exception of a + b = b + a, prove that the axiom a + b = b + a must hold in R and that R is thus a ring. (Hint: Expand (a + b)(1 + 1) in two ways.) Proof. Let a, b R, a ring with unity in which the axiom of commutativity under addition is not stated. Then and Thus, we get: (a + b)(1 + 1) = a(1 + 1) + b(1 + 1) = a + a + b + b (a + b)(1 + 1) = (a + b)1 + (a + b)1 = a + b + a + b a + a + b + b = a + b + a + b a + b = b + a Since, a, b R were arbitrary, we get (R, +) is abelian. Problem Show that the commutative ring D is an integral domain if and only if a, b, c D with a 0, the relation ab = ac implies that b = c. Proof. Let D be an integral domain. Let a 0 be an element of D. Also, let b, c D such that ab = ac. Then ab ac = 0 a(b c) = 0. We have a, b c D by closure and existence of additive inverses in D. Since D is an integral domain, it has no zero divisors and it follows, a = 0 or b c = 0. Since a 0, we get b = c. Conversely, let the cancellation law hold for all non-zero elements in a commutative ring D. We need to show that D is an integral domain. Let a, b be two elements in D such that a 0 and ab = 0. Then ab + aa = 0 + aa which implies a(b + a) = aa and by the cancellation law, b + a = a b = 0. Thus, we cannot have the product of two non-zero elements equal to zero in D. This shows absence of zero divisors in D and D is thus, an integral domain. Problem Prove that Lemma is false if we drop the assumption that the integral domain is finite. Proof. We present the infinite integral domain Z under usual multiplication as a counter example. The set of integers under usual multiplication of integers is not a field because there are elements in Z for which the multiplicative inverse does not lie in Z. Problem Prove that any field is an integral domain. Proof. We know that in a field, every non-zero element is a unit. We show that a unit cannot be a zero divisor and hence it will follow that our field has no zero divisor and is therefore an integral domain. Let F be a field and a F be a unit. Suppose there exists a non-zero element b F such that ab = 0. Then ua = 1 for some u F. This implies b = 1b = (ua)b = u(ab) = u0 = 0 which is a contradiction. Thus a unit cannot be a zero divisor. Problem Using the Pigeon-hole Principle, prove that if m and n are relatively prime integers and a and b are any integers, there exists an integer x such that x a mod m and x b mod m. (Hint: Consider the remainders of a, a + m, a + 2m,..., a + (n 1)m on division by n.)

4 Solutions by Sadiah Zahoor 4 Proof. Pigeon hole principle If n objects are to be distributed over m places, and if n > m, then some place receives at least two objects. We will be using the pigeon hole principle to show that the a set of elements which contains n elements, all distinct remainders on division by n is equivalent to the set of remainders on division by n, {0, 1, 2,..., n 1}. Let m, n be relatively prime integers, (m, n) = 1. Let a be any integer. Consider the set {a, a + m, a + 2m,..., a + (n 1)m}. Suppose a + km a + k m mod n where k, k {0, 1, 2,..., n 1}. Then (k k )m 0 mod n. This implies n (k k )m and we know (m, n) = 1. Thus, it follows that n k k and we get k k mod n. But since k, k {0, 1, 2,..., n 1}, we get k = k. This means that a + km = a + k m. This shows that any two elements in the set under consideration are distinct modulo n. Now for any integer b, b mod n can have only one of the n distinct values between 0 and n 1, both included. By the pigeon hole principle, if we have to place a, a + m, a + 2m,..., a + (n 1)m, b mod n objects in holes which represent remainders 0, 1,..., n 1 on division by n, since the objects are n + 1, atleast one hole must get two objects. It follows that one of the elements in our set is congruent to b mod n. Let x = a + km, where k {0, 1, 2,..., n 1} be that element. Then, x b mod n. and also x a mod n as required. Problem Using the pigeon hole principle, prove that the decimal expansion of a rational number must, after some point, become repeating. Proof. Let p/q be our rational number where p, q Z, q > 0. Then we use the division algorithm and get: p = k 0 q + r 0 where 0 r 0 < q p q = k 0 + r 0 q where 0 r 0 q < 1 Here k 0 is the non-decimal, integer part of the rational number. We want to find the first decimal place now. This will be given by the division algorithm on evaluating 10r 0 /q. 10r 0 = k 1 q + r 1 where 0 r 1 < q r 0 q = k r 1 10q where 0 r 1 10q < r 1 10q where 0 r 1 10q < 1 10 Here, k 1 represents the first decimal place. We want to find the second decimal place now. This will be given by the division algorithm on evaluating 10r 1 /q. We also note that it s possible that k 1 = 0. In that case r 1 = 10r 0 and we will be evaluating 100r 0 /q. 10r 1 = k 2 q + r 2 where 0 r 2 < q r 1 10q = k r 2 100q where 0 r 2 100q < k r 2 100q where 0 r 2 100q < 1 100

5 Solutions by Sadiah Zahoor 5 Here, k 2 represents the second decimal place. Similarly we can show that the n th decimal place can be given by evaluating 10r n 1 /q. 10r n 1 = k n q + r n where 0 r n < q r n 1 10 n 1 q = k n 10 n + r n 10 n where 0 r n 10 n < 1 10 n 10 + k k n 10 n + r n 10 n where 0 r n 10 n < 1 10 n Now our rational number p/q = k 0.k 1 k 2...k n... We consider the following set: {r 0, r 1,..., r n } where n > q. For each 0 i n, we have 0 r i < q. If we want to distribute them among q remainders using pigeon hole principle, we get that atleast for some 0 i < n, we have r n = r i. This implies that 10r n = 10r i which implies k i+1 q + r i+1 = k n+1 q + r n+1 (k i+1 k n+1 )q = r n+1 r i+1 r n+1 r i+1 mod q and 0 r n+1, r i+1 < q r n+1 = r i+1 Thus, we get k i+1 = k n+1. Also, r i+1 = r n+1 10r i+1 = 10r n+1 which will further using the same argument as above imply k i+2 = k n+2. Thus, we get, k i+j = k n+j where j N Thus, we get p/q = k 0.k 1 k 2...k i k i+1 k i+2...k n k i+1 k i+2...k n... = k 0.k 1 k 2...k i k i+1 k i+2...k n which means that the decimal expansion of a rational number after some point becomes repeating.