Is the Set of Nilpotent Element an Ideal?

Problems and solutions of ring theory in abstract algebra

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.

 
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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 $R$.


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