# A Ring Has Infinitely Many Nilpotent Elements if $ab=1$ and $ba \neq 1$

## Problem 543

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.

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Contents

## 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 $R$.

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