# Is the Set of Nilpotent Element an Ideal?

## Problem 620

Is it true that a set of nilpotent elements in a ring $R$ is an ideal of $R$?

If so, prove it. Otherwise give a counterexample.

Contents

## Proof.

We give a counterexample.
Let $R$ be the noncommutative ring of $2\times 2$ matrices with real coefficients.
Consider the following matrices $A, B$ in $R$.
$A=\begin{bmatrix} 0 & 1\\ 0& 0 \end{bmatrix} \text{ and } B=\begin{bmatrix} 0 & 0\\ 1& 0 \end{bmatrix}.$ Direct computation shows that $A^2$ and $B^2$ are the zero matrix, hence $A, B$ are nilpotent elements.

However, the sum $A+B=\begin{bmatrix} 0 & 1\\ 1& 0 \end{bmatrix}$ is not nilpotent as we have
\begin{align*}
\begin{bmatrix}
0 & 1\\
1& 0
\end{bmatrix}^n
=\begin{cases} \begin{bmatrix}
0 & 1\\
1& 0
\end{bmatrix} & \text{if $n$ is odd}\10pt] \begin{bmatrix} 1 & 0\\ 0& 1 \end{bmatrix} & \text{ if n is even}. \end{cases} \end{align*} Hence the set of nilpotent elements in R is not an ideal as it is not even an additive abelian group. ## Comment. If a ring R is commutative, then it is true that the set of nilpotent elements form an ideal, which is called the nilradical of R. Sponsored Links ### More from my site • Nilpotent Element a in a Ring and Unit Element 1-ab Let R be a commutative ring with 1 \neq 0. An element a\in R is called nilpotent if a^n=0 for some positive integer n. Then prove that if a is a nilpotent element of R, then 1-ab is a unit for all b \in R. We give two proofs. Proof 1. Since a […] • A Ring Has Infinitely Many Nilpotent Elements if ab=1 and ba \neq 1 Let R be a ring with 1. Suppose that a, b are elements in R such that \[ab=1 \text{ and } ba\neq 1. (a) Prove that $1-ba$ is idempotent. (b) Prove that $b^n(1-ba)$ is nilpotent for each positive integer $n$. (c) Prove that the ring $R$ has infinitely many […]
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##### Boolean Rings Do Not Have Nonzero Nilpotent Elements

Let $R$ be a commutative ring with $1$ such that every element $x$ in $R$ is idempotent, that is, $x^2=x$....

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