Let $x$ be a nilpotent element of a commutative ring $R$.
Prove that $x$ is either zero or a zero divisor.
My attempt : I got the answer here but i didn't understand the answer
My proof :since $x$ is nilpotent , so there exist some index $ m$ with $x^m=0.$
if $m=1 $ then $x=0$
if $m\neq 1$ ,then $x\neq 0$ i,e $xx^{m-1}=0$ where $x^{m-1}=0$ but $x\neq 0$
for example take $x= \begin{bmatrix} 0&1\\0&0\end{bmatrix} \neq0$ but $x^{m-1}= \begin{bmatrix} 0&0\\0&0\end{bmatrix}$ where $m-1 >0$
Is my proof is correct or not ?
Best Answer
Suppose as you said that $x$ is nilpotent where the index $m$ is the lowest integer for which we have $x^m = 0$.
If $m = 1$ we have as you said $x=0$.
If $m > 1$ then $x \ne 0$ and $x\cdot x^{m-1} = 0$
But at the beginning we supposed $m$ as the lowest integer for which we have $x^m = 0$, so $x^{m-1} \ne 0$.
Then you can see that the case $m > 1$ lead us to the conclusion that $x$ is a zero-divisor.