Proposition 1

Let A be a nonzero ring, and let $\mathfrak{r}$ be the set of noninvertible elements of A. The following properties are equivalent:

(i) The set $\mathfrak{r}$ is a two-sided ideal of A.

(ii) The set $\mathfrak{r}$ is stable under addition.

(iii) The ring A has a unique maximal left ideal.

(iv) For every $a\in A$, one of the elements $a$ and $1-a$ is invertible.

(v) For every $a\in A$, one of the elements $a$ and $1-a$ is left invertible.

The implication (i) $\Rightarrow$ (ii) follows from the definition of an ideal. Since 1 does not belong to $\mathfrak{r}$, we have (ii) $\Rightarrow$ (iv).

We have $\mathfrak{r}\not= A$, and the set $\mathfrak{r}$ contains every left ideal of A not equal to A. If $\mathfrak{r}$ is a left ideal of A, it is therefore the unique maximal left ideal of A. This proves that (i) implies (iii).

Suppose that A has a unique maximal left ideal $\mathfrak{m}$. Take $b\in A-\mathfrak{m}$. The left ideal $Ab$ is not contained in any maximal left ideal of A, hence is equal to A (I, §8, No. 6, p. 104, Theorem 1), and $b$ is left invertible. For every $a\in A$, one of the elements $a$ and $1-a$ belongs to A $-\mathfrak{m}$ because $\mathfrak{m}$ is an ideal that does not contain 1. Thus, (iii) implies (v).

Suppose that property (v) holds. Let $b$ be a left invertible element of A. Let $c\in A$ be such that $cb= 1$. We have $(1-bc)b= 0$ and $b\not= 0$; hence, $1-bc$ is not left invertible. By property (v), $bc$ is left invertible and, a fortiori, $c$ is left invertible. But then $c$ is invertible; $b$ is its inverse, so $b$ is invertible. It follows that (v) implies (iv).

It remains to prove that (iv) implies (i). Suppose that (iv) holds. Then $\mathfrak{r}$ is a two-sided ideal of A by the following assertions a) through d):

a) We have $0\in \mathfrak{r}$ because the ring A is nonzero.

b) The product of two elements of A, one belonging to $\mathfrak{r}$ and the other to A $-\mathfrak{r}$, belongs to $\mathfrak{r}$.

c) The set $\mathfrak{r}$ is stable under addition.

Indeed, let $a$ and $b$ be elements of $\mathfrak{r}$ such that $s=a+b$ is invertible. By b), the elements $s^{-1}a$ and $s^{-1}b$ of A belong to $\mathfrak{r}$; since $s^{-1}b= 1-s^{-1}a$, this contradicts the assumption that (iv) holds.

d) The set $\mathfrak{r}$ is stable under multiplication.

Indeed, let $a$ and $b$ be two elements of $\mathfrak{r}$. The element $a'=-a(1-b)$ belongs to $\mathfrak{r}$ by b), so that the element $ab$, which is equal to $a+a'$, belongs to $\mathfrak{r}$ by c); assertion d) follows.