Theorem 1 (Jacobson)

The radical of a ring A consists of the elements $x\in A$ such that $1 +ax$ is left invertible (I, §2, No. 3, p. 15) for every $a\in A$. It is also the largest two-sided ideal $\mathfrak{a}$ such that $1 +x$ is invertible for every $x\in \mathfrak{a}$. The radical of A contains every left nil ideal of A.

The element 1 generates the A-module $A_s$, and $1 +ax$ is left invertible if and only if it generates the A-module $A_s$. The first assertion of Theorem 1 is therefore a specific case of the corollary of Proposition 2 (VIII, p. 153).

Let $x\in \mathfrak{R}(A)$. By the above, $1 +x$ is left invertible; let $y$ be an element of A such that $y(1 +x) = 1$. We then have $1-y=yx$, so $1-y$ belongs to $\mathfrak{R}(A)$; consequently, $y$ is left invertible. Since $y$ is also right invertible, it is invertible (I, §2, No. 3, p. 16, Proposition 3), and so is its right inverse $1 +x$.

Let $\mathfrak{a}$ be a left ideal of A such that $1 +x$ is invertible for every $x\in \mathfrak{a}$. This holds, for example, when $\mathfrak{a}$ is a nil ideal because the relation $x^n= 0$ implies that $1-x+\cdots + (-x)^{n-1}$ is the inverse of $1 +x$. Let $x\in \mathfrak{a}$. For every $a\in A$, we have $ax\in \mathfrak{a}$, so $1 +ax$ is invertible; hence, we have $x\in \mathfrak{R}(A)$. It follows that $\mathfrak{a}$ is contained in $\mathfrak{R}(A)$.