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?

Consider a (nonzero) commutative (unital) ring $R$ with multiplicative identity $1_R$ and multiplicative group of units $R^\times$ with the property that $1_R + a$ is a unit whenever $a$ is a non-unit (i.e., whenever $a$ is in $R \setminus R^\times$). We claim that $R \setminus R^\times$ is an ideal.
Observe that the product $ra$ of any element $r$ of $R$ and any non-unit $a$ must be a non-unit. (Why?) Consequently, it remains to show that $R \setminus R^\times$ is closed under addition.
Given non-units $a$ and $b,$ as you have written, we have that $1_R + a$ and $1_R + b$ are units (by hypothesis). Our aim is to show that $a + b$ is a non-unit. On the contrary, assume that $a + b$ is a unit so that there exists a $u$ such that $1_R = (a + b)u = au + bu.$ By rearranging, we have that $bu = 1_R - au.$ But this implies that $bu$ is a unit so that $b$ is a unit --- a contradiction. (Why?)

Closely related to this post is the Jacobson radical of the ring $R,$ i.e., the ideal $$\operatorname{Jac}(R) = \bigcap \{\mathfrak m \,|\, \mathfrak m \text{ is a maximal ideal of } R\}.$$ One can show that $a$ is in $\operatorname{Jac}(R)$ if and only if $1_R + ra$ is a unit for all $r \in R.$ For your case, the non-units of $R$ belong to a single maximal ideal $\mathfrak m,$ hence we have that $\operatorname{Jac}(R) = \mathfrak m = R \setminus R^\times.$
