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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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 Spring '08
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 Math, Algebra

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