So I'm trying to prove that if $M$ is a maximal ideal in $R$ such that for all $x\in M$, $x+1$ is a unit, then $R$ is a local ring with maximal ideal $M$, that is to say $R$ has a unique maximal ideal.
I've been at this one for a day now and I just can't figure it out. I have that $R$ being a local ring is equivalent to there being a proper ideal $I$ of $R$ which contains all non-units of $R$, and also equivalent to the set of non-units of $R$ being an ideal.
The set of $x+1$ for $x\in M$ is itself multiplicative, but I'm not sure where to go with that since inverting that set just gives back $M$ (since they're all units). I haven't been successful at proving anything about elements of $R$ which are not either in $M$ nor of the form $x+1$ for $x\in M$.
I also tried just assuming there was some other maximal ideal $N$ and then trying to draw out a contradiction by looking at the ideal $M+N$, clearly if $M+N$ doesn't contain $1$ then I've got my contradiction, but I don't seem to have enough information to pursue that path.
Can anyone give me some guidance? Thanks.
Best Answer
Actually upon writing this up I believe I've solved it:
Assume there existed another maximal ideal $N$. Then if $1\in M+N$ then there exists $1 = m+n\in M+N$ and thus $n = -m+1$ and since $m\in M$ implies $-m\in M$, this means that $n = -m+1$ is a unit and thus $N=R$. Therefore $M+N$ must not contain $1$ and thus we have found a proper ideal of $R$ which contains $M$, contradicting the fact that $M$ is maximal.