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.

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 $R$.

The post Is the Set of Nilpotent Element an Ideal? first appeared on Problems in Mathematics.

Let $R$ be a ring with unit $1\neq 0$.

Prove that if $M$ is an ideal of $R$ such that $R/M$ is a field, then $M$ is a maximal ideal of $R$.
(Do not assume that the ring $R$ is commutative.)

Proof.

Let $I$ be an ideal of $R$ such that
\[M \subset I \subset R.\] Then $I/M$ is an ideal of $R/M$.

Since $R/M$ is a field by assumption, the only ideals of the field $R/M$ is $\bar{0}=M/M$ or $R/M$ itself.
So the ideal $I/M$ is either $\bar{0}$ or $R/M$.

By the fourth (or lattice) isomorphism theorem for rings, there is a one-to-one correspondence between the set of ideals of $R$ containing $M$ and the set of ideals of $R/M$. Hence $I$ must be either $M$ or $R$.

(Since the fourth isomorphism theorem gives the correspondence $M \leftrightarrow M/M=\bar{0}$ and $R \leftrightarrow R/M$, and there is no ideal of $R/M$ between $\bar{0}$ and $R/M$.)

Therefore the ideal $M$ is maximal.

The post If the Quotient Ring is a Field, then the Ideal is Maximal first appeared on Problems in Mathematics.