Boolean Rings Do Not Have Nonzero Nilpotent Elements

Problems and solutions of ring theory in abstract algebra

Problem 618

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.

 
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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 $0$.


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  1. 11/28/2017

    […] (Remark: every Boolean ring has no nonzero nilpotent elements.) […]

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