Characterisation of Jacobson Radical

Theorem
Let $A$ be a commutative ring.

Let $A^\times$ be the group of units of $A$

Let $\operatorname{Jac}(A)$ be the Jacobson radical of $A$.

Then


 * $\operatorname{Jac}(A)=\left\{a\in A:1-ax\in A^\times\text{ for all }x\in A\right\}$

Proof
First suppose that $1-xy\notin A^\times$.

Then it is contained in some maximal ideal $\mathfrak m\subseteq A$.

Then $x\in \operatorname{Jac}(A)\subseteq\mathfrak m$ implies that $y\in\mathfrak m$ and therefore $1\in \mathfrak m$, which is impossible.

This shows that $\operatorname{Jac}(A)\subseteq\left\{a\in A:1-ax\in A^\times\text{ for all }x\in A\right\}$.

Now suppose that $x\notin \mathfrak m$ for some maximal ideal $\mathfrak m$ of $A$.

Since $\mathfrak m$ is maximal, $x$ and $\mathfrak m$ generate $A$.

Therefore there exist $w\in\mathfrak m$ and $y\in A$ such that $w+xy=1$.

Thus $1-xy\in\mathfrak m$, and $1-xy\notin A^\times$.

This completes the proof.