If $1+a$ is a unit for all non-units $a$ in a nonzero ring $R$, then $R-R^*$ is an ideal Suppose that $R$ is a nonzero commutative ring with $1$. Suppose $1+a$ is a unit for all non-units $a$ in a nonzero ring $R$. I'd like to understand why this implies that $R-R^*$ is an ideal.
In particular, I'd like to know why the sum of two non-units is a non-unit.
So suppose $a,b$ are non-units. We'd like to show that $a+b\in R-R^*$ as well. From the assumption, we have that $1+a,1+b\in R^*$, and so $(1+a)(1+b)=1+a+b+ab\in R^*$.
But now I'm stuck. Any suggestions?

These are equivalent definitions of local rings.
For a nonzero commutative ring $R$, we show the equivalence of the following statements:

*

*For every $a \in R \backslash R^\times$, we have $1 + a \in R^\times$.

*$R$ has a unique maximal ideal.

*If $a, b \in R \backslash R^\times$, then $a + b \in R\backslash R^\times$.

We may prove in the following order: $1 \implies 2 \implies 3 \implies 1$.
$1 \implies 2$: let $I$ be a maximal ideal of $R$. We want to show that any proper ideal $J \subseteq R$ is contained in $I$.
Suppose there is a proper ideal $J$ not contained in $I$. Then the ideal $I + J$ must be $R$, as $I$ is maximal. Thus there exist $i \in I$ and $j \in J$ such that $i + j = 1$. But both $-i$ and $1 + (-i) = j$ are non-units, contradiction.
$2 \implies 3$: both $aR$ and $bR$ are proper ideals of $R$, hence are contained in the unique maximal ideal $M$. Therefore their sum is also in $M$ and is not a unit.
$3 \implies 1$: suppose there exists $a \in R$ such that both $a$ and $1 + a$ are non-units. Then $1 = (1 + a) + (-a)$ is a non-unit, contradiction.
