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Let $R$ be a ring. An element $r \in R$ is called nilpotent if $r^n=0$ for some integer $n \ge 1$. Show that if $r$ is nilpotent in a ring with identity, then $1-r$ is a unit in $R$.

Proof. Recall that in any ring we have $(−a)(−b) = (ab)$ Thus in any ring with 1 (commutative or not) we have the following identities:

$1 −(−1)^na^n = (1 + a)(1 − a + a^2 − · · · + (−1)^{−1}a^{n−1})$

If $a^n = 0$ then we obtain an inverse $1 − a$

Now, I'm not so familiar with series and am not sure if this suffices. I also assume this can be shown with matrices, integers....as long as it is a ring

cele
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  • This is the third question you've asked recently closed because it duplicates many previous questions. Please take advantage of the search feature before you ask a question. – rschwieb Apr 22 '15 at 23:16
  • @rschwieb, I find that the search is very specific and that you have to be very general to get decent results. But then there are too many questions to sift through. I have not found a good way to filter down the results, but thanks. I have gotten away from searching recently because of that. I will make a better effort to do so – cele Apr 22 '15 at 23:18

1 Answers1

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$$1=1-r^n=(1-r)(1+r+\ldots+r^{n-1})$$

Timbuc
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