Question

An element a of a ring is nilpotent if ${\mathit{a}}^{\mathbf{n}}\mathbf{=}\mathit{R}$for some positive integer n. Prove that R has no non-zero nilpotent elements if and only if ${\mathbf{0}}_{\mathbf{R}}$is the unique solution of the equation${\mathit{x}}^{\mathbf{2}}\mathbf{=}{\mathbf{0}}_{\mathbf{R}}$ .

Short answer

It is proved that R has no non-zero nilpotent element.

Step-by-step solution

Show that R has no non-zero nilpotent elements if and only if 0R is the unique solution of the equation x2=0R 

Assume that R contains no non-zero nilpotent elements.

When $a\in R$is a solution to $x^2=0_R$, then$a^2=0_R$ from the definition of nilpotent.

Assume that $a=0_R$.

Assume that $0_R$would be the unique solution to the equation, $x^2=0_R$.

Consider that $a\in R$as a nilpotent element and $n\in Z$as the smallest positive integer in which $a^n=0_R$.

When. $n=1,a=0$

Now, assume that $n>1$. This signifies that$a\ne 0_R$in particular.

Consider that $m=\frac{n}{2}$, $m$would be the smallest integer which is larger or equal to$\frac{n}{2}$ . Then, m<n according to the minimality of n, it has been that $a^m\ne 0_R$.

It is observed that $2m\ge n$and as a result, ${\left(a^m\right)}^2=a^{2m}=0_R$. This means that $a^m=0_R$which is a contradiction.

Hence, it is proved that R has no non-zero nilpotent elements.