We have that a commutative ring $R$ with $1$ is called local if $R − R^×$ is an ideal of $R$. I have to proof the following:
$R$ is local $\iff$ $R$ has exactly one maximal ideal.
We have that every ideal $I$ $\subseteq$ $R$ is contained in a unique maximal ideal. $R$ is local so that the only maximal ideal is $M$. So, $I$ $\subseteq$ $M$. The homomorphism $\phi$: $R$ $\rightarrow$ $R/I$ tells me that the ideals of $R/I$ are all contained in $\phi$(M). $\phi$(M) is a maximal ideal in $R/I$ so that it is also the unique maximal ideal in $R/I$ since any other maximal ideal is contained in it.
So, $R/I$ has a unique maximal ideal $\phi$(M) and hence it is local.
I am not sure if this is a complete proof of the above statement. Any help would be grateful.