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 $R$.

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, $x^2=x$. (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 […]

Is the Set of Nilpotent Element an Ideal?
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 […]

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 $a$ […]

If the Localization is Noetherian for All Prime Ideals, Is the Ring Noetherian?
Let $R$ be a commutative ring with $1$.
Suppose that the localization $R_{\mathfrak{p}}$ is a Noetherian ring for every prime ideal $\mathfrak{p}$ of $R$.
Is it true that $A$ is also a Noetherian ring?
Proof.
The answer is no. We give a counterexample.
Let […]

Ring of Gaussian Integers and Determine its Unit Elements
Denote by $i$ the square root of $-1$.
Let
\[R=\Z[i]=\{a+ib \mid a, b \in \Z \}\]
be the ring of Gaussian integers.
We define the norm $N:\Z[i] \to \Z$ by sending $\alpha=a+ib$ to
\[N(\alpha)=\alpha \bar{\alpha}=a^2+b^2.\]
Here $\bar{\alpha}$ is the complex conjugate of […]

Equivalent Conditions For a Prime Ideal in a Commutative Ring
Let $R$ be a commutative ring and let $P$ be an ideal of $R$. Prove that the following statements are equivalent:
(a) The ideal $P$ is a prime ideal.
(b) For any two ideals $I$ and $J$, if $IJ \subset P$ then we have either $I \subset P$ or $J \subset P$.
Proof. […]

Is the Given Subset of The Ring of Integer Matrices an Ideal?
Let $R$ be the ring of all $2\times 2$ matrices with integer coefficients:
\[R=\left\{\, \begin{bmatrix}
a & b\\
c& d
\end{bmatrix} \quad \middle| \quad a, b, c, d\in \Z \,\right\}.\]
Let $S$ be the subset of $R$ given by
\[S=\left\{\, \begin{bmatrix}
s & […]

The Ring $\Z[\sqrt{2}]$ is a Euclidean Domain
Prove that the ring of integers
\[\Z[\sqrt{2}]=\{a+b\sqrt{2} \mid a, b \in \Z\}\]
of the field $\Q(\sqrt{2})$ is a Euclidean Domain.
Proof.
First of all, it is clear that $\Z[\sqrt{2}]$ is an integral domain since it is contained in $\R$.
We use the […]