103b_06w_1s

103b_06w_1s - . (b) Just compute 4 ( x ) for x Z 6 and list...

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Math 103B First Hour Exam Solutions 1 February 2006 1. (a) Since x 2 = 1 in R , there are just four elements: A , 1 + A , x + A and x + 1 + A , where A = a x 2 + 1 A > . (b) We can exhibit zero divisors: ( x + 1 + A ) 2 = x 2 + 2 x + 1 + A = x 2 + 1 + A = A . 2. Since multiplication is componentwise ( r, s ) is a zero divisor if and only if r is a zero divisor of Z 3 or s is a zero divisor of Z 4 and, in addition, ( r, s ) n = (0 , 0) Hence the zero divisors are (0 , s ), ( r, 0) and ( r , 2), where s is any nonzero element of Z 4 , r is any nonzero element of Z 3 and r is any element of Z 3 . That’s enough for full credit. If you want to count them, there are 3 + 2 + 3 1 = 7, where the 1 arises because (0 , 2) is counted twice, once with s = 2 and once with r = 0. 3. (a) Regardless of the value of k , ϕ k ( a + b ) = k ( a + b ) = ka + kb = ϕ k ( a ) + ϕ k ( b ). Thus we only need to check multiplication. If k 2 = k , ϕ k ( ab ) = kab = kakb = ϕ k ( a ) ϕ k ( b ). Thus k 2 = k is su±cient. To see that it is necessary, use the hint and the fact that 1 · 1 = 1: We must have ϕ k (1) 2 = ϕ k (1). In other words, k 2 = k
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Unformatted text preview: . (b) Just compute 4 ( x ) for x Z 6 and list those x that give 0: { , 3 } . 4. (a) Suppose b is a nonzero nilpotent element. Let n be the smallest power of b that is zero. Since b n = 0, n > 1. Thus 0 = b n = b n 1 b = bb n 1 . Since n is as small as possible, b n 1 n = 0 and so b is a zero divisor. (b) Call the set of nilpotent elements N . Suppose a, b N and r R . We must show that a b , ar and ra all lie in N . The rst follows from the remark preceding this part of the problem. If a n = 0, commutativity gives us ( ra ) n = ( ar ) n = a n r n = r n = 0 and so we are done. (c) Using the hint and letting c = a + b , note that a 2 = b 2 = 0 and c 2 = I , the 2 2 identity matrix. Thus a and b are nilpotent but c is not since c 2 n = I and c 2 n +1 = c ....
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