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 commutative ring with $1$ such that every element $x$ in $R$ is idempotent, that is, $x^2=x$. (Such a ring is called a Boolean ring.)

(a) Prove that $x^n=x$ for any positive integer $n$.

(b) Prove that $R$ does not have a nonzero nilpotent element.

Proof.

(a) Prove that $x^n=x$ for any positive integer $n$.

By assumption, $x^n=x$ is true for $n=1, 2$.

Suppose that $x^k=x$ for some $k \geq 2$ (induction hypothesis).
Then we have
\begin{align*}
x^{k+1}&=xx^k\\
&=xx &&\text{by induction hypothesis}\\
&=x^2=x &&\text{by assumption.}
\end{align*}

Thus, we conclude that $x^n=x$ for any positive integer $n$ by induction.

(b) Prove that $R$ does not have a nonzero nilpotent element.

Let $x$ be a nilpotent element in $R$. That is, there is a positive integer $n$ such that $x^n=0$.

It follows from part (a) that $x=x^n=0$.
Thus every nilpotent element in $R$ is $0$.

The post Boolean Rings Do Not Have Nonzero Nilpotent Elements first appeared on Problems in Mathematics.

A square matrix $A$ is called nilpotent if some power of $A$ is the zero matrix.
Namely, $A$ is nilpotent if there exists a positive integer $k$ such that $A^k=O$, where $O$ is the zero matrix.

Suppose that $A$ is a nilpotent matrix and let $B$ be an invertible matrix of the same size as $A$.
Is the matrix $B-A$ invertible? If so prove it. Otherwise, give a counterexample.

Solution.

We claim that the matrix $B-A$ is not necessarily invertible.
Consider the matrix
\[A=\begin{bmatrix}
0 & -1 \\0& 0
\end{bmatrix}.\] This matrix is nilpotent as we have
\[A^2=\begin{bmatrix}
0 & -1 \\0& 0
\end{bmatrix}
\begin{bmatrix}
0 & -1 \\0& 0
\end{bmatrix}
=
\begin{bmatrix}
0 & 0 \\0& 0
\end{bmatrix}.\]

Also consider the matrix
\[B=\begin{bmatrix}
1 & 0 \\1& 1
\end{bmatrix}.\] Since the determinant of the matrix $B$ is $1$, it is invertible.

So the matrix $A$ and $B$ satisfy the assumption of the problem.
However the matrix
\[B-A=\begin{bmatrix}
1 & 1 \\1& 1
\end{bmatrix}\] is not invertible as its determinant is $0$.
Hence we found a counterexample.

Related Question.

Here is another problem about a nilpotent matrix.

The solution is given in the post ↴
Nilpotent Matrices and Non-Singularity of Such Matrices

The post Is the Sum of a Nilpotent Matrix and an Invertible Matrix Invertible? first appeared on Problems in Mathematics.

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 nilpotent elements.

Proof.

(a) Prove that $1-ba$ is idempotent.

We compute
\begin{align*}
(1-ba)^2&=(1-ba)(1-ba)=1-ba-ba+b\underbrace{ab}_{=1}a\\
&=1-ba-ba+ba=1-ba.
\end{align*}
Thus, we have $(1-ba)^2=1-ba$, and hence $1-ba$ is idempotent.

(b) Prove that $b^n(1-ba)$ is nilpotent for each positive integer $n$.

As a lemma, we show that $(1-ba)b=0$.
To see this, we calculate
\begin{align*}
(1-ba)b=b-b\underbrace{ab}_{=1}=b-b=0.
\end{align*}

Now we compute
\begin{align*}
b^n(1-ba)\cdot b^n(1-ba)&=b^n\underbrace{(1-ba)b}_{=0 \text{ by lemma}}b^{n-1}(1-ba)=0.
\end{align*}
This proves that $b^n(1-ba)$ is nilpotent.

(c) Prove that the ring $R$ has infinitely many nilpotent elements.

In part (a), we showed that the element $b^n(1-ba)$ is a nilpotent element of $R$ for each positive integer $n$.
We claim that $b^n(1-ba)\neq b^m(1-ba)$ for each pair of distinct integers $m, n$.
Without loss of generality, we may assume that $m > n$.

We state simple facts which are needed below.
We have
\begin{align*}
a^nb^n&=1\\
a^nb^m&=b^{m-n}.
\end{align*}
Note that $a^nb^n$ and $a^nb^m$ look like
\[aa\cdots a\cdot bb\cdots b.\] Then we use the relation $ab=1$ from the middle successively, and we obtain the right-hand sides.

Now we prove that $b^n(1-ba)\neq b^m(1-ba)$ for each pair of distinct integers $m, n$.
Assume on the contrary $b^n(1-ba)= b^m(1-ba)$ for $m > n$.
Then we multiply by $a^n$ on the left and get
\begin{align*}
a^n b^n(1-ba)= a^n b^m(1-ba).
\end{align*}

Using the facts stated above, we obtain
\[1-ba=b^{m-n}(1-ba).\] Note that the left-hand side is a nonzero idempotent element by part (a).
On the other hand, the right-hand side is nilpotent by part (b).
Since a nonzero idempotent element can never be nilpotent, this is a contradiction.

Therefore, $b^n(1-ba)\neq b^m(1-ba)$ for each pair of distinct integers $m, n$.
Hence there are infinitely many nilpotent elements in $R$.

The post A Ring Has Infinitely Many Nilpotent Elements if $ab=1$ and $ba \neq 1$ first appeared on Problems in Mathematics.

Prove that if $A$ is a diagonalizable nilpotent matrix, then $A$ is the zero matrix $O$.

Definition (Nilpotent Matrix)

A square matrix $A$ is called nilpotent if there exists a positive integer $k$ such that $A^k=O$.

Proof.

Main Part

Since $A$ is diagonalizable, there is a nonsingular matrix $S$ such that $S^{-1}AS$ is a diagonal matrix whose diagonal entries are eigenvalues of $A$.

As we show below, the only eigenvalue of any nilpotent matrix is $0$.
Thus, $S^{-1}AS$ is the zero matrix.
Hence $A=SOS^{-1}=O$.

The only eigenvalue of each nilpotent matrix is $0$

It remains to show that the fact we used above: the only eigenvalue of the nilpotent matrix $A$ is $0$.

Let $\lambda$ be an eigenvalue of $A$ and let $\mathbf{v}$ be an eigenvector corresponding to $\lambda$.
That is,
\[A\mathbf{v}=\lambda \mathbf{v}, \tag{*}\]

Since $A$ is nilpotent, there exists a positive integer $k$ such that $A^k=O$.

Then we use the relation (*) inductively and obtain
\begin{align*}
A^k\mathbf{v}&=A^{k-1}A\mathbf{v}\\
&=\lambda A^{k-1}\mathbf{v} && \text{by (*)}\\
&=\lambda A^{k-2}A\mathbf{v}\\
&=\lambda^2 A^{k-2}\mathbf{v} && \text{by (*)}\\
&=\dots =\lambda^k \mathbf{v}.
\end{align*}

Hence we have
\[\mathbf{0}=O\mathbf{v}=A^k\mathbf{v}=\lambda^k \mathbf{v}.\]

Note that the eigenvector $\mathbf{v}$ is a nonzero vector by definition.
Thus, we must have $\lambda^k=0$, hence $\lambda=0$.
This proves that the only eigenvalue of the nilpotent matrix $A$ is $0$, and this completes the proof.

Another Proof of the Fact

Even though the fact proved above is true regardless of diagonalizability of $A$, we can make use that $A$ is diagonalizable to prove the fact as follows.

Let $\lambda_1, \dots, \lambda_n$ be the eigenvalues of the $n\times n$ nilpotent matrix $A$.
Then we have
\[S^{-1}AS=D,\] where
\[D:=\begin{bmatrix}
\lambda_1 & 0 & \dots & 0 \\
0 &\lambda_2 & \dots & 0 \\
\vdots & \vdots & \ddots & \vdots \\
0 & 0 & \dots & \lambda_n
\end{bmatrix}.\]

Then we have
\(\require{cancel}\)
\begin{align*}
D^k&=(S^{-1}AS)^k\\
&=(S^{-1}A\cancel{S})(\cancel{S}^{-1}A\cancel{S})\cdots (\cancel{S}^{-1}AS)\\
&=S^{-1}A^kS\\
&=S^{-1}OS=O.
\end{align*}

Since
\[D^k=\begin{bmatrix}
\lambda_1^k & 0 & \dots & 0 \\
0 &\lambda_2^k & \dots & 0 \\
\vdots & \vdots & \ddots & \vdots \\
0 & 0 & \dots & \lambda_n^k
\end{bmatrix},\] it yields that $\lambda_i^=0$, and hence $\lambda_i=0$ for $i=1, \dots, n$.

Related Question.

The converse of the above fact is also true: if the only eigenvalue of $A$ is $0$, then $A$ is a nilpotent matrix.

See the post ↴
Nilpotent Matrix and Eigenvalues of the Matrix
for a proof of this fact.

The post Every Diagonalizable Nilpotent Matrix is the Zero Matrix first appeared on Problems in Mathematics.

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 $a$ is nilpotent, we have $a^n=0$ for some positive integer $n$.
Then for any $b \in R$, we have $(ab)^n=a^nb^n=0$ since $R$ is commutative.

Then we have the following equality:
\[(1-ab)(1+(ab)+(ab)^2+\cdots+(ab)^{n-1})=1.\] Therefore $1-ab$ is a unit in $R$.

Proof 2.

There exists $n \in \N$ such that $a^n=0$ since $a$ is nilpotent.
Assume that $1-ab$ is not a unit for some $b \in R$.
Then there exists a prime ideal $\frakp$ of $R$ such that $\frakp \ni 1-ab$.

Since $a^n=0\in \frakp$, we have $a\in \frakp$ since $\frakp$ is an prime ideal.
Then $ab \in \frakp$ and we have
\[1=(1-ab)+ab \in \frakp.\]

However, this implies that $\frakp=R$ and this is a contradiction. Thus $1-ab$ is a unit of the ring $R$ for all $b \in R$.

The post Nilpotent Element a in a Ring and Unit Element -ab$ first appeared on Problems in Mathematics.