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Here is the question I wanna solve:

Let $x$ be a nilpotent element of the commutative ring $R$ (an element $x$ in $R$ is called nilpotent if $x^m = 0$ for some $m \in \mathbb Z^{+}$).

Prove that $1 + x$ is a unit in $R.$

I found the following solution on the internet:

enter image description here

But I do not know how should I have guessed that this is the inverse of 1+x? Could someone clarify to me the intuition behind this please? How can I find this inverse before showing that it is actually an inverse as the solution is doing.

Brain
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  • See the linked dupes for how to invert a unit + nilpotent. As explained here we can view it as an analog of rationalizing the denominator. – Bill Dubuque Aug 30 '23 at 00:59
  • All the links you gave does not have a clean answer for my question @BillDubuque, could you please allow my question to receive answers? – Brain Aug 30 '23 at 01:54
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    As explained here in the first dupe, similar to rationalizing denominators, we invert $1-x$ it by looking for some multiple that it simpler to invert. An obvious choice is the multiple $,1-x^m = 1.,$ What is not "clean" about that? As explained in the Remark there we can view the method as arising from a constructive proof of divisors of units are units. See the second linked dupe on the method of simpler multiples for many analogs. – Bill Dubuque Aug 30 '23 at 02:03
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    Note also, as explained in the first dupe, we can generalize the idea to compute inverses for any factor of $,x^n-1,$ for all $,n\ge m,,$ i.e. for polynomials $\Phi_n(x)$ of cyclotomic type. – Bill Dubuque Aug 30 '23 at 02:51
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    I added some remarks and links here and more general conceptual ways to view this. – Bill Dubuque Aug 31 '23 at 02:38

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