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 $1 - x y \notin A^\times$.

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 $1 \in \mathfrak m$.

But if $1 \in \mathfrak m$ then from Ideal of Unit is Whole Ring $\mathfrak m = R$ and so by definition $\mathfrak m$ would not be maximal.

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 $w + x y = 1$.

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

This completes the proof.