# Idempotent Elements and Zero Divisors in a Ring and in an Integral Domain ## Problem 516

Prove the following statements.

(a) If $a\neq 1$ is an idempotent element of $R$, then $a$ is a zero divisor.

(b) Suppose that $R$ is an integral domain. Determine all the idempotent elements of $R$. Add to solve later

## Definitions (Idempotent, Zero Divisor, Integral Domain)

Let $R$ be a ring with $1$.

• An element $a$ of $R$ is called idempotent if $a^2=a$.
• An element $a$ of $R$ is called zero divisor if there exists a nonzero element $x$ of $R$ such that $ax=0$ or $xa=0$.
• A commutative ring that does not have a nonzero zero divisor is called an integral domain

## Proof.

### (a) If $a\neq 1$ is an idempotent element of $R$, then $a$ is a zero divisor.

By definition of an idempotent element, we have $a^2=a$.
It yields that
\begin{align*}
a(a-1)=a^2-a=0.
\end{align*}
Since $a\neq 1$, the element $a-1$ is a nonzero element in the ring $R$.
Thus $a$ is a zero divisor.

### (b) Suppose that $R$ is an integral domain. Determine all the idempotent elements of $R$.

Suppose that $a$ is an idempotent element in the integral domain $R$.
Thus, we have $a^2=a$.
It follows that we have
\begin{align*}
a(a-1)=a^2-a=0. \tag{*}
\end{align*}

Since $R$ is an integral domain, there is no nonzero zero divisor.
Hence (*) yields that $a=0$ or $a-1=0$.

Clearly, the elements $0$ and $1$ are idempotent.
Thus, the idempotent elements in the integral domain $R$ must be $0$ and $1$. Add to solve later ##### 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....