Theorem 1

Let $\mathbf{A}$ be a ring. The following conditions are equivalent:

(a) $\mathbf{A}$ is a field; (b) $\mathbf{A}$ is not reduced to 0 and the only left ideals of $\mathbf{A}$ are 0 and $\mathbf{A}$.

Suppose that $\mathbf{A}$ is a field. Then $\mathbf{A}$ is not reduced to 0. Let $a$ be a left ideal of $\mathbf{A}$ distinct from 0. There exists a non-zero $a$ belonging to $a$. For all $x \in \mathbf{A}$, $x = (xa^{-1})a \in a$; hence $a = \mathbf{A}$.

Suppose that $\mathbf{A}$ satisfies condition (b). Let $x \neq 0$ be in $\mathbf{A}$. We need to prove that $x$ is invertible. The left ideal $Ax$ contains $x$ and hence is not zero, whence $Ax = \mathbf{A}$. There thus exists $x' \in \mathbf{A}$ such that $x'x = 1$. Then $x' \neq 0$ since $1 \neq 0$. Applying the above result to $x'$, it is seen that there exists $x'' \in \mathbf{A}$ such that $x''x' = 1$. Then $x'' = x'' . 1 = x''x'x = 1 . x = x$, hence $xx' = 1$. Hence $x'$ is the inverse of $x$.