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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.

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

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up vote 4 down vote accepted

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.

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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.

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