If you find any mistakes, please make a comment! Thank you.

In a noncommutative ring, the set of nilpotent elements need not be an ideal


Solution to Abstract Algebra by Dummit & Foote 3rd edition Chapter 7.3 Exercise 7.3.31

Solution: We begin with a lemma.

Lemma: Let $R$ be a ring with $1 \neq 0$. If $x \in R$ is a unit, then $x$ is not nilpotent.

Proof: Suppose to the contrary that $xy = yx = 1$ and $x^n = 0$. Then we have $$1 = 1^n = (xy)^n = x^ny^n = 0,$$ a contradiction. $\square$

Evidently, $A^2 = B^2 = 0$, while $(A+B)^2 = 1$. By the lemma, $A+B$ is not nilpotent. Since $\mathfrak{N}(\mathsf{Mat}_2(\mathbb{Z}))$ is not closed under addition, it is not an ideal of any ring.


Linearity

This website is supposed to help you study Linear Algebras. Please only read these solutions after thinking about the problems carefully. Do not just copy these solutions.

Leave a Reply

Close Menu