Prove that if $M$ is a maximal ideal and $x \notin M$, then $x+M$ is a unit in $R/M$ I want to prove that if $M$ is a maximal ideal and $R$ is a commutative ring and $x \notin M$, then $x+M$ is a unit in $R/M$.
I am thinking of letting $(x,M) = \{rx+b \mid r \in R \, \text{and} \, b \in M\} \subset R$ and showing that $(x,M)$ is an ideal such that $M \subset (x,M) \subset R$, but I not sure how to proceed with this proof.
 A: Just follow your nose from where you started.
$(M, x)$ cannot be equal to $M$ (since $x$ isn't in it) and therefore $(M,x)=R$. Then there has to be $m+xr=1$ for some $r\in R$.
Mod $M$, that reads $xr\equiv 1\pmod{M}$: in other words, $x$ is a unit.
A: If $R$ is not a commutative unital ring, then the statement in the question is incorrect.
If it's not commutative then we can see that $M_2 (\mathbb{R})$ - the ring of $2\times 2$ matrices with real entries - is a simple ring, and therefore ${0}$ is a maximal ideal, however there are non-zero $2\times 2$ matrices which are not units, for example $\pmatrix{1&0\\0&0}$.
If $R$ is not unital then we can take $R=2\mathbb{Z}$, then $M=4\mathbb{Z}$ is a maximal ideal, however $2+M$ is non-zero in $R/M$ but $(2+M)^2=0$.
Finally if $R$ is a commutative unital ring then let $x+M\in R/M$ such that $x+M\ne M$.
Now the set $I=(x+M)R/M = \{(x+M)(y+M)\mid y\in R\}$ is an ideal of $R/M$, and $x+M\in I$ so $I$ is a non-trivial ideal.
Since $M$ is maximal, $R/M$ has only two ideals: ${0}$ and $R/M$ (by the correspondence theorem for the natural homomorphism $\phi : R\to R/M$ defined by $x\mapsto x+M$).
We've shown that $I\ne 0$ and therefore $I=R/M$, so $1+M\in I$ and therefore $x+M$ is a unit.
