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## Homework Statement

An element a of a ring is nilpotent if a^n = 0 for some positive integer n.

Prove that R has no nonzero nilpotent elements if and only if 0 is the unique

solution of the equation x^2 = 0

## Homework Equations

I think nilpotent means that not only that a^n = 0 for some positive integer

n but that n is the least positive integer such that a^n = 0. ?

## The Attempt at a Solution

Proving ->

if R has no nonzero nilpotent elements then there is no nonzero element a such that a^n = 0, and specifically in the case of n = 2, there is no such a.

Hence 0 is the unique solution of x^2 = 0

Proving <-(via contradiction)

if 0 is the unique solution of x^2 = 0 then

if R does have some nonzero nilpotent element a, then a^n = 0 for

some positive inetger n.

then either

n is even where a^n = a^(n/2) * a^(n/2) = 0

but since 0 is the unique solution of x^2 = 0, then

a^(n/2) = 0. Which contradicts the fact that n is the least

positive integer for which a^n = 0.

Or n is odd, where

a^n = 0 ---> (a^n)*(a) = 0 * (a) (multiplying both sides by a)

---> a^(n+1) = 0, now n+1 is an even integer,

which can then be factored into a square

so a^(n+1/2) * a^(n+1/2) = 0, but since 0 is the unique

solution of x^2 = 0, then a^(n+1/2) = 0, which contradicts the

fact that n is the least positive integer for which a^n = 0.

I would very much appreciate any evaluation of my attempt. Especially because I feel unsure of the n is odd case, since I just multiplied both sides by a so i could get the square.