Let $M$ be a maximal ideal in $R$ such that for all $x\in M$, $x+1$ is a unit. Show that $R$ is a local ring with maximal ideal $M$

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.

Solution 1:

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.

Solution 2:

You can find the answer in Atiyah-Macdonald. I'll repeat it here:

Two steps:

1. If every $y \in R-M$ is a unit, then $R$ is a local ring. In fact every ideal $\neq (1)$ consists of non-units, hence is contained in $M$.

2. In this case, let $y\in R-M$, we show that $y$ is a unit. Since $M$ is maximal, $(y) + M = (1)$, so there exists $r\in R$ and $x\in M$, such that $ry-x=1$, so $ry=1+x$ is a unit, then $y$ itself is a unit.

Solution 3:

Given $x\in M$, the hypothesis implies that $1+xr $ is invertible for every $r\in R$.
Hence $x\in Jac(R)$, the Jacobson radical of $R$ (Atiyah-Macdonald, Proposition 1.9), so $M\subset Jac(R)$
But the Jacobson radical is by definition the intersection $Jac(R)=\bigcap M_i$ of all the maximal ideals $M_i$ of $R$.
Thus we have $M\subset\bigcap M_i $, which immediately implies that $M$ is the sole maximal ideal of $R$ .

Solution 4:

Hint $\ $ Put $\rm\,J = \it M\ $ in $(2\Rightarrow 1)$ below, yielding $\rm\:1+{\it M}\, \subset U\:$ $\Rightarrow$ $\, M\,$ lies in every maximal ideal, therefore $\,M\,$ is the only maximal ideal.

Theorem $\ $ TFAE in ring $\rm\:R\:$ with units $\rm\:U,\:$ ideal $\rm\:J,\:$ and Jacobson radical $\rm\:Jac(R)\:.$

$\rm(1)\quad J \subseteq Jac(R),\quad $ i.e. $\rm\:J\:$ lies in every max ideal $\rm\:M\:$ of $\rm\:R\:.$

$\rm(2)\quad 1+J \subseteq U,\quad\ \ $ i.e. $\rm\: 1 + j\:$ is a unit for every $\rm\: j \in J\:.$

$\rm(3)\quad I\neq 1\ \Rightarrow\ I+J \neq 1,\qquad\ $ i.e. proper ideals survive in $\rm\:R/J\:.$

$\rm(4)\quad M\:$ max $\rm\:\Rightarrow M+J \ne 1,\quad $ i.e. max ideals survive in $\rm\:R/J\:.$

Proof $\: $ (sketch) $\ $ With $\rm\:i \in I,\ j \in J,\:$ and max ideal $\rm\:M,$

$\rm(1\Rightarrow 2)\quad j \in all\ M\ \Rightarrow\ 1+j \in no\ M\ \Rightarrow\ 1+j\:$ unit.

$\rm(2\Rightarrow 3)\quad i+j = 1\ \Rightarrow\ 1-j = i\:$ unit $\rm\:\Rightarrow I = 1\:.$

$\rm(3\Rightarrow 4)\ \ \ $ Let $\rm\:I = M\:$ max.

$\rm(4\Rightarrow 1)\quad M+J \ne 1 \Rightarrow\ J \subseteq M\:$ by $\rm\:M\:$ max.