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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 ?

user26857
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jasmine
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1 Answers1

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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.

Gabrielek
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    Re your comment below Q - no reason to apologise. Putting up correct answers is always a good idea. People who put answers in comments usually do so because they are in a hurry, don't want to take the time to check carefully, and don't want to risk a downvote. We need correct answers so that the Q gets cleared! – almagest Dec 21 '20 at 18:11
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    If you first handle the case $m=1$, it’s still possible that $m=0$. In that case, $R$ has only one element $1_R=0_R$. So the two cases $m=1$ and $m>1$ are not exhaustive. – Steve Kass Dec 21 '20 at 18:11
  • Thank you for the approval and the clafirification @almagest, I agree on everything. – Gabrielek Dec 21 '20 at 19:27
  • Answering to @SteveKass your sentence is a border phenomena.. Considering the limit case of a commutative ring $R$ with only one element is like consider a single point in $\mathbb{R}$ an interval – Gabrielek Dec 21 '20 at 19:27
  • @SteveKass I think some authors define nilpotent elements to be such that $x^m = 0$ for some $m \in \mathbb{Z}^{+}$ – no lemon no melon Feb 16 '21 at 21:02
  • @Gabrielek does the fact that we can choose the lowest positive integer for which $x^m = 0$ follow from the Well Ordering Principle (every nonempty set of natural numbers contains a smallest element)? – no lemon no melon Feb 16 '21 at 21:05
  • Yes, of course. – Gabrielek Feb 17 '21 at 11:20