"Proving Nilpotent Element in R has Unique Solution

[rookandpawn]

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.

 

[Dick]

What could be wrong with that? I think you are just lacking confidence. You may want to say explicitly that a^n=0 and a^m!=0 for m<n. As you say, make sure n is the least nilpotent power of a.