If $1+a$ is a unit for all non-units $a$ in a nonzero ring $R$, then $R-R^*$ is an ideal

Question

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?

Answer 1

These are equivalent definitions of local rings.

For a nonzero commutative ring $R$, we show the equivalence of the following statements:

1. For every $a \in R \backslash R^\times$, we have $1 + a \in R^\times$.
2. $R$ has a unique maximal ideal.
3. 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.

Answer 2

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?)

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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 $\operatorname{Jac}(R) = \mathfrak m = R \setminus R^\times.$