Idempotent Elements and Zero Divisors in a Ring and in an Integral Domain

Problems and solutions of ring theory in abstract algebra

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 $R$.

 
LoadingAdd to solve later

Sponsored Links


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 $1$.


LoadingAdd to solve later

Sponsored Links

More from my site

Please Login to Comment.

This site uses Akismet to reduce spam. Learn how your comment data is processed.

More in Ring theory
Problems and solutions of ring theory in abstract algebra
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....

Close