Check if the solution is correct. Correct the mistakes

Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
icon
Related questions
Question
100%

Check if the solution is correct. Correct the mistakes

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.
Expert Solution
steps

Step by step

Solved in 4 steps with 5 images

Blurred answer
Recommended textbooks for you
Advanced Engineering Mathematics
Advanced Engineering Mathematics
Advanced Math
ISBN:
9780470458365
Author:
Erwin Kreyszig
Publisher:
Wiley, John & Sons, Incorporated
Numerical Methods for Engineers
Numerical Methods for Engineers
Advanced Math
ISBN:
9780073397924
Author:
Steven C. Chapra Dr., Raymond P. Canale
Publisher:
McGraw-Hill Education
Introductory Mathematics for Engineering Applicat…
Introductory Mathematics for Engineering Applicat…
Advanced Math
ISBN:
9781118141809
Author:
Nathan Klingbeil
Publisher:
WILEY
Mathematics For Machine Technology
Mathematics For Machine Technology
Advanced Math
ISBN:
9781337798310
Author:
Peterson, John.
Publisher:
Cengage Learning,
Basic Technical Mathematics
Basic Technical Mathematics
Advanced Math
ISBN:
9780134437705
Author:
Washington
Publisher:
PEARSON
Topology
Topology
Advanced Math
ISBN:
9780134689517
Author:
Munkres, James R.
Publisher:
Pearson,