# Determining certain units in a local ring. [closed]

I've been stuck on this problem for a while:

Let R be a commutative ring with $1 \neq 0$. If R has a unique maximal ideal (i.e. R is local), then either $x$ or $1-x$ (or both) are units in R.

-

## closed as off-topic by user26857, Eric Stucky, Silvia Ghinassi, Michael Albanese, Brandon CarterFeb 17 at 5:11

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – user26857, Eric Stucky, Silvia Ghinassi, Michael Albanese, Brandon Carter
If this question can be reworded to fit the rules in the help center, please edit the question.

Let M be the unique maximal ideal in R. Suppose for contradiction that $x \in R$ such that $x \notin M$ is not a unit. Consider $(x)$. Since every proper ideal in a ring with identity is contained in some maximal ideal, it follows that $(x) \subset I$ for some maximal ideal $I \subset R$. Moreover, it follows that $I = M$ since M is the unique maximal ideal in $R$. Thus, $(x) \subset M$, which is a contradiction because $x \notin M$. Hence, any $x \in R$ such that $x \notin M$ is a unit.

Now, let $x \in M$. Suppose for contradiction that $1-x \in M$. Then, it follows that $1-x = m$ for some $m \in M$ and moreover, $1 = x + m$. Since M is an ideal and therefore closed under addition, $1 \in M$. Thus, $M=R$ contradicting the construction of M. Hence, $1-x \notin M$ and by my previous argument $1-x$ is a unit.

-

Suppose $I=(x)$ and $J=(1-x)$ are proper ideals.

Every proper ideal is contained in a maximal ideal, but there is only one, say $M$. Then $I \subset M$ and $J \subset M$ and so $x, (1-x) \in M \implies x+(1-x)=1 \in M$ which is absurd.

Then $I$ or $J$ or both are not proper, e.g. $I=R \implies \exists a \in R$ such that $ax=1 \implies x$ is a unit.

-