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Let $R$ be a commutative local ring. I want to show that then $x\in R$ is not a unit implies $1-x$ is a unit.

My idea was the following:

Since $R$ is a local ring we can chose $\mathfrak{m}$ to be it's unique maximal ideal. Now let $x\in R$ be a nonunit and assume that $1-x$ is also a nonunit. Then $I:=(1-x)\subsetneq R$. Since $\mathfrak{m}$ is a maximal ideal $I\subseteq \mathfrak{m}$. Now I wanted to write $1=1-x+x$ and I can deduce that $1-x+x$ is a unit. But somehow I don't see how to conclude from there. Can someone help me?

Martin Brandenburg
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1 Answers1

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  1. If $R$ is local with maximal ideal $\mathfrak{m}$, then $R^{\times} = R \setminus \mathfrak{m}$. See for example SE/3913530.
  2. If $R$ is any ring and $I \subseteq R$ is any proper ideal, then $x \in I$ implies $1-x \notin I$. In fact, if $1-x \in I$, then $1 = x + (1-x) \in I$, so $I = R$.

The claim follows from 1. and 2.

Martin Brandenburg
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  • Please strive not to post more (dupe) answers to dupes of FAQs, cf. recent site policy announcement here. – Bill Dubuque Oct 22 '23 at 17:01
  • I searched for like 5 mins and couldn't find it. PS: You dont have to write these comments anymore to me. I know your policy. – Martin Brandenburg Oct 22 '23 at 17:13
  • It's not my policy - it is site policy that is now being enforced (users who ignore it have been suspended - including some high rep users). If it seems likely that it is a dupe please delay answering so that it can be properly processed by others. – Bill Dubuque Oct 22 '23 at 17:18