# Prove: $R$ is local $\iff$ $R$ has exactly one maximal ideal.

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.

Assume $$R$$ is local and let $$M=R\setminus R^\times$$. By assumption $$M$$ is an ideal. It is also maximal because any ideal which properly contains it must contain invertible elements and hence it is all $$R$$. Now let $$I\subseteq R$$ be any proper ideal. Then all the elements of $$I$$ must be non invertible and hence $$I\subseteq M$$. So this shows any proper ideal is contained in $$M$$, hence it must be the only maximal ideal of $$R$$.
Second direction: suppose there is exactly one maximal ideal and let's call it $$M$$. We will show that $$M=R\setminus R^x$$. Obviously $$M\subseteq R\setminus R^x$$. Now suppose there is a non invertible element $$y\in R\setminus R^x$$ such that $$y\notin M$$. Then define:
$$I=\{ry: r\in R\}$$
It is easy to see that $$I$$ is an ideal of $$R$$. It is also a proper ideal because $$1\notin I$$. (because $$y$$ is not invertible). Then $$I$$ is contained in some maximal ideal. Since by assumption $$M$$ is the only maximal ideal we must have $$I\subseteq M$$. But this means $$y\in M$$, a contradiction. Hence we must have $$M=R\setminus R^x$$, so $$R$$ is indeed local.
• We don't know yet that $M=R\setminus R^x$. In this direction of the proof I defined $M$ to be the only maximal ideal of $R$. The equality $M=R\setminus R^x$ is what we are trying to prove. – Mark May 26 '20 at 9:37