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Elements Of Modern Algebra
Elements Of Modern Algebra
8th Edition
ISBN:9781285463230
Author:Gilbert, Linda, Jimmie
Publisher:Gilbert, Linda, Jimmie
Chapter6: More On Rings
Section6.1: Ideals And Quotient Rings
Problem 30E: a. For a fixed element a of a commutative ring R, prove that the set I={ar|rR} is an ideal of R....
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Show that the set N of all nilpotent elements in a commutative ring R forms an ideal. Also show that R/N has no nonzero nilpotent element. Let us show that the set N of nilpotent elements in a commutative ring R forms an ideal: a) Check the under addition: Let a and b be nilpotent elements, iea" = 0 and bn = 0. Then (a + b)+m can be expanded in Newton's binomial: (a + b)n+m = n+m Σ(" + m) a² + 1 k=0 ak* bn+m-k Since R is commutative, a and b commute, and each term contains either a" or bm and then each term is zero. Therefore,(a + b)n+m = 0, and a + b is also a nilpotent element. b) Check the under multiplication by elements of the ring: Let a EN and r ER. Then (ra)" = r² * a"=rn* 0 = 0, hence ra is also a nilpotent element. Thus N forms an ideal in R. Let us show that there is no nonzero nilpotent element in R/N: Let x + NE R/N be a nilpotent element, that is, (x + N) = x² + N = N for some k > 0. This means that x* E N and x* -is a nilpotent element. But since x* * 0 (because x + N + N), x is itself a nilpotent element, and x E N. Hence x + N = N, which contradicts the assumption that x+N‡N. Therefore, there is no non-zero nilpotent element in R/N.
Transcribed Image Text:Show that the set N of all nilpotent elements in a commutative ring R forms an ideal. Also show that R/N has no nonzero nilpotent element. Let us show that the set N of nilpotent elements in a commutative ring R forms an ideal: a) Check the under addition: Let a and b be nilpotent elements, iea" = 0 and bn = 0. Then (a + b)+m can be expanded in Newton's binomial: (a + b)n+m = n+m Σ(" + m) a² + 1 k=0 ak* bn+m-k Since R is commutative, a and b commute, and each term contains either a" or bm and then each term is zero. Therefore,(a + b)n+m = 0, and a + b is also a nilpotent element. b) Check the under multiplication by elements of the ring: Let a EN and r ER. Then (ra)" = r² * a"=rn* 0 = 0, hence ra is also a nilpotent element. Thus N forms an ideal in R. Let us show that there is no nonzero nilpotent element in R/N: Let x + NE R/N be a nilpotent element, that is, (x + N) = x² + N = N for some k > 0. This means that x* E N and x* -is a nilpotent element. But since x* * 0 (because x + N + N), x is itself a nilpotent element, and x E N. Hence x + N = N, which contradicts the assumption that x+N‡N. Therefore, there is no non-zero nilpotent element in R/N.
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