Non Units of Commutative Ring All Being Contained in Some Ideal M which is not R

Question

Let R be a commutative ring with $1_R$.

Then I'd like to prove the equivalence between those statements, particularly the point from $(ii) \to (iii)$ :

(i) all non units of R are contained in some ideal M $\ne R$

(ii) the set of all non units of R forms an ideal of R

(iii) for any r,s $\in$ R, if $r+s = 1$ then one of r or s in a unit in R

Following is what I've tried:

To use contradiction, suppose $r$ and $s$ both are non-unit in $R$.

Since $r, s$ both are non-unit they are in an ideal which is set of all non-units in $R$ (Let's call it $M$).

then by definition of ideal, $r\cdot s = s\cdot r\in M$.

I can't figure out where to go further from this point.. any hint to proceed?

Answer 1

How to prove $(ii) \Rightarrow (iii)$:

Let $r,s\in R$ be non units such that $r+s=1$. Then (as the non units form an ideal) we have that $r+s=1$ is also a non unit. This means, $1$ is a non unit, which is a contradiction.

I think the easiest way to prove the equivalence of the three statements is to prove $(i) \Rightarrow (iii) \Rightarrow (ii) \Rightarrow (i)$.

$(i) \Rightarrow (iii)$: Assume by contradicton that there exists $r,s\in R$ non units such that $r+s=1$. We know that $r,s\in M$ and $M$ is an ideal, hence, $1=r+s\in M$, which contradicts $M\neq R$.

$(iii) \Rightarrow (ii)$: Take $r,s\in R$ non units and assume by contradiciton that $r+s$ is a unit. Then there exists a unit $t\in R$ such that $t(r+s)=1$. However, then $tr+ts=1$. By (iii) you have that either $tr$ or $ts$ is a unit. Multiply with $t^{-1}$ and get that either $r$ or $s$ is a unit, which is a contradiction.

$(ii) \Rightarrow (i)$: Take $M$ equal the non units, by $(ii)$ this is an ideal and as $1\notin M$, we also have $M\neq R$.

Answer 2

Hint: assume (ii) and let $N$ be the set of non-units. Assume $r + s = 1$. If $r$ and $s$ are both non-units, then $r, s \in N$, hence $1 = r + s \in N$, because by (ii) $N$ is an ideal. Is $1$ a non-unit?

Answer 3

If you have (ii), then the sum of two nonunits is a nonunit, so it can't be $1$.

For (iii)$\implies$(i), suppose there is no ideal $M\ne R$ containing all nonunits. Note that the set $N$ of nonunits is closed under products by arbitrary elements: if $x\in N$ and $r\in R$, then $rx\in N$. Thus, the fact $N$ fails to be contained in a proper ideal means there are $r,s\in N$ with $r+s\notin N$. Then $u=r+s$ is a unit and therefore $1=u^{-1}r+u^{-1}s$, with $u^{-1}r$ and $u^{-1}s$ nonunits.