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Transcript
Chapter 13 solutions 14. Show that the nilpotent elements of a commutative ring form a subring. Solution: If a, b A, a m 0 b m , for A a commutative ring, then (ab) min( m,n ) 0 , and, when (a b)m + n is expanded, each term either has a raised to at least the n or b to the m power, it is clear that the set of nilpotent elements is a subring. Commutativity is used heavily. 46. Let R be a commutative ring with unity 1 and prime characteristic. If a R is nilpotent, prove that there is a positive integer k such that (1 a) k = 1. Solution: You need to use the previous problem (45), a really beautiful one, to show that, in a commutative ring R of prime characteristic p, for every positive integer n, n n n ( x y) p x p y p , x, y R. The first part of that problem is proved by noting that the binomial coefficients appearing in the sum, except for the 0 and p terms, have p in the numerator and not in the denominator, and hence are divisible by p. Part b is proved by induction on the power of p, giving you a proof of another fact about the binomial coefficients that is hard to prove directly, by the way. The proof in the current problem is n accomplished by taking n from 45 large enough so that a p 0. You could, for example, take k = n. 50. Let R be a ring with m elements. Show that the characteristic of R divides m. Solution: Since the additive group is abelian, the result follows from the basis theorem for finite abelian groups, or one of its preliminary results. We have, however, remarked previously that, in a non-abelian group, the exponent of the group need not be the order of an element. So you must invoke abelian if you are going to use the characteristic being the order of an element. The exponent still divides the order of a non-abelian group, though. It’s a supplementary problem on groups. 58. Let F be a finite field with n elements. Prove that xn-1 = 1 for all nonzero x in F. Solution: The set of non-zero elements of F forms a group of order n-1 under multiplication, so the result follows by a corollary to Lagrange.