Proposition 1

Let $\mathbf{A}$ be a ring and $I$ the set of non-invertible elements of $\mathbf{A}$. The set $I$ is the union of the ideals of $\mathbf{A}$ which are distinct from $\mathbf{A}$. Moreover, the following conditions are equivalent: (a) $I$ is an ideal. (b) The set of ideals of $\mathbf{A}$ distinct from $\mathbf{A}$ has a greatest element. (c) $\mathbf{A}$ has a unique maximal ideal.

The relation $x \in I$ is equivalent to $1 \notin xA$ and hence $xA \neq \mathbf{A}$. If $a$ is an ideal of $\mathbf{A}$ distinct from $\mathbf{A}$ and $x \in a$, then $xA \subset a$, hence $xA \neq \mathbf{A}$ and $x \in I$. Hence every ideal of $\mathbf{A}$ distinct from $\mathbf{A}$ is contained in $I$ and every element $x \in I$ belongs to a principal ideal $xA \neq \mathbf{A}$. This proves the first assertion, which immediately implies the equivalence of (a), (b) and (c).

Remark (1). Note that, if (c) holds, $I$ is the Jacobson radical of the ring $\mathbf{A}$ (Algebra, Chapter VIII, §6, no. 3, Definition 3).